Chapter 4: Thermodynamics
In the early 1700s, the deep coal mines of England had a serious problem: water kept flooding them. Thomas Newcomen built a machine to solve it. His engine burned coal to make steam, pushed the steam into a large cylinder, then sprayed cold water inside to condense that steam. The cylinder collapsed under atmospheric pressure, the piston moved, and water was pumped out of the mine. It worked — but it drank enormous quantities of coal for very little useful pumping, as suggested in Fig. 4.0.
Diagram
Fig. 4.0 – A Newcomen-type beam engine at a coal mine: boiler, cylinder with piston, and the beam driving a water pump.
In the 1760s, a young instrument maker at the University of Glasgow named James Watt was asked to repair a small working model of this engine. He noticed something the builders had missed. The same cylinder was being heated by steam and then chilled by water in every single stroke. Most of the fuel was not moving the piston at all — it was being spent simply reheating metal that had just been cooled.
Watt’s answer was to condense the steam in a separate vessel and keep the working cylinder hot. The engine suddenly needed far less coal for the same work. A machine that converts heat into motion had been made better, not by burning more, but by understanding where the heat was going.
That question — where does heat go, and how much of it can be turned into useful work — is the whole of this chapter.
This chapter will give you the physics needed to answer every one of these questions.
4.1 Introduction
In Std. XI you studied the thermal properties of matter — how substances expand on heating, how they conduct heat, and how much heat is needed to raise their temperature. Those ideas described what happens to a material when heat is given to it. In this chapter we go one step further and study the laws that govern the behaviour of thermal energy itself.
In particular, we shall study processes in which work is converted into heat, and processes in which heat is converted into work. By work we mean the same quantity you met in mechanics: force acting through a displacement, measured in joule (J). By thermal energy we mean the energy a body possesses because its particles are in constant motion.
Work Produces Heat: A Familiar Observation
When we drive a vehicle, its engine becomes warmer after some time. Similarly, when we exercise, we feel warmth in our body. The physics involved in both cases is the same.
The engine of a vehicle and our muscles both do some work, and both produce some heat while doing it. The work is the useful output; the warmth appears along with it, whether we want it or not.
It is therefore natural to turn the question around. If the work done by an engine or by our muscles produces some heat, then heat should also be able to ‘do’ some work. Thermodynamics is mostly the study of the conversion of work (or any form of energy) into heat, and of the reverse conversion as well.
Something Is Exchanged Between Hot and Cold Bodies
When a hot object is placed in contact with a cold object, we notice that both objects reach the same temperature after some time. The hot object gets cooler and the cold object becomes warmer.
Neither object gains or loses any matter in this process, yet both change. That means something is exchanged between the two objects. This ‘something’ is heat. According to the modern theory, heat is a form of energy.
This is worth pausing on, because it is a comparatively recent idea. For a long time heat was thought to be an invisible, weightless fluid stored inside hot bodies, which flowed out of them when they cooled. On that picture, heat was a substance — and a substance can only be squeezed out of a body, never created without limit.
The Experiment That Settled the Question
In the year 1798 it was observed by Benjamin Thomson, a British scientist (later known as Count Rumford), that tremendous heat is produced when brass cannons are bored, as shown in Fig. 4.29. The heat thus produced was large enough to boil water.
Diagram
Fig. 4.29 – A brass cannon barrel being bored by a horse-driven drill, with the barrel immersed in a trough of water that is boiling.
A very important observation was that the amount of heat produced was related to the work done in turning the drill that was used to bore the cannon. It was also noticed that more heat was produced when the drill bored for a longer time.
Crucially, the heat produced did not depend on the sharpness of the drills used. This detail is the heart of the argument. A sharper drill would have removed more metal, and according to the older theory of heat — which assumed heat to be some form of a fluid — removing more metal should have released more of the stored heat fluid. That is not what was seen.
This observation could be explained only if heat was a form of energy and not any fluid. As long as the drill was turned, work was being done, and heat kept appearing; the supply never ran out.
It is natural to conclude from these observations that energy can be converted from one form to another form. In this particular case, a very important law of physics can be proposed, that ‘the work done by a system is converted into heat’. (The drills used to bore the cannons ‘do’ the work, and the cannons get heated up.)
What Thermodynamics Studies
We can now state what this branch of physics deals with.
It is the field of study that allows us to understand the nature of many of the fundamental interactions in the universe. It can explain phenomena as simple as water boiling in a vessel, and also something as complex as the creation of a new star. Thermodynamics is an important branch of physics having many practical applications.
Notice how wide that range is. The same small set of laws applies to a kettle, to a car engine, to a living cell and to a collapsing cloud of interstellar gas, because those laws say nothing about what the system is made of — only about how energy may be moved and transformed.
The Road Ahead in This Chapter
In this chapter we will try to understand a thermodynamic system, thermodynamic variables, thermodynamic processes, and the laws that govern these processes.
We will also study the most important and useful applications of thermodynamics, the heat engines and their efficiency.
In examination questions, this introductory material is rarely tested by itself. It is tested indirectly: you are given a situation and asked whether energy has entered a system as heat or as work, and in which direction. Getting that identification right at the start is what makes the numerical work later in the chapter straightforward.
4.2 Thermal Equilibrium
The thermal properties of materials discussed in Std. XI are useful to understand the behaviour of a material when it is heated or cooled. We now use them to look at what happens when two bodies at different temperatures are brought together.
When you put a piece of ice in water at room temperature, the ice melts. This is because the water at room temperature is at a higher temperature than the ice, so it transfers its heat to the ice and helps the ice melt. The water, having given away energy, becomes colder.
Similarly, when hot water is mixed with cold water, the hot water transfers its heat to the cold water. The hot water cools down and the cold water warms up, as shown in Fig. 4.30.
Diagram
Fig. 4.30 – Two beakers, one of hot water and one of cold water, poured together into a third beaker whose thermometer reads a single intermediate temperature.
When the Transfer Stops
In both these examples, we notice that the two components reach a stage where there is no more transfer of heat. The exchange does not go on forever, and it does not reverse itself either.
In such cases, we assume that heat is something that is transferred from a substance at a higher temperature to that at a lower temperature. This transfer continues till the level of heat content in both the substances is the same — that is, until neither substance is any longer hotter than the other.
Then we say that a thermal equilibrium is reached between the two substances. We can say that when two objects are at the same temperature, they are in thermal equilibrium.
Note carefully what equilibrium does not mean. It does not mean that the molecules have stopped moving, or that energy has stopped crossing the boundary between the two bodies. It means that whatever energy passes one way is exactly matched by energy passing the other way, so there is no net transfer, and no observable change in either body.
Why This Idea Matters
This concept of thermal equilibrium is used in the Zeroth Law of thermodynamics, which we take up in the next section.
It is called the Zeroth Law because it was proposed after the First and the Second laws of thermodynamics were already formulated. Once physicists realised that it was the assumption on which the very idea of temperature rests, it had to be placed before the other two — and the only number left was zero.
4.3 Zeroth Law of Thermodynamics
The Zeroth law is very important as it helps us to define the concept of a temperature scale. The formal statement of the Zeroth law of thermodynamics is as follows.
Reading the Statement from the Diagram
Figure 4.1 shows a schematic representation of the Zeroth law of thermodynamics, in which the double arrow represents thermal equilibrium between systems.
Diagram
Fig. 4.1 – Schematic representation of the Zeroth law: A with C, A with B, therefore B with C, each pair joined by a double arrow.
Follow the logic in Fig. 4.1 one step at a time. If system A and system C are in thermal equilibrium, and systems A and B are in thermal equilibrium, then systems B and C must be in thermal equilibrium. Then systems A, B and C are at the same temperature.
The important point is that B and C reach this conclusion without ever being brought into contact with each other. System A acts as a go-between, and its verdict on B and on C is enough to settle the relation between B and C.
Why a Thermometer Works
For example, when we use a thermometer to measure the temperature of an object, we use the same principle. When the thermometer and the object are in thermal equilibrium, the thermometer indicates the temperature of the object.
Here the thermometer is playing the part of system A in Fig. 4.1. The zeroth law, therefore, enables us to use a thermometer to compare the temperatures of different objects. If the same thermometer balances against two different objects at the same reading, those two objects are at the same temperature, even though they were never touched to each other. This is schematically shown in Fig. 4.2.
Diagram
Fig. 4.2 – Two separate beakers of liquid, each with a thermometer dipped in it, both thermometers showing the same reading.
It also implies that temperature is a measurable quantity. The science of measuring temperatures is called Thermometry, which involves different temperature scales and methods of measuring temperature. This has already been discussed in Std. XI.
4.4 Heat, Internal Energy and Work
Earlier in this chapter, we saw that when two substances, initially at different temperatures, are brought in contact with each other, the substance at higher temperature loses its heat and the substance at lower temperature gains it.
That description tells us which way heat moves, and when the movement stops. But it leaves a deeper question untouched.
The Question We Have Not Yet Asked
We did not discuss the reasons why any substance can ‘have’ that heat in the first place, and what exactly is the nature of the heat content of that substance. We simply took it for granted that a hot body has “more” of something and a cold body has less.
The examples we discussed in the previous section, together with the study of transfer of heat in Chapter 7 of Std. XI — where you learnt how energy travels by conduction, convection and radiation — help us understand the transfer of heat from one body to the other. They describe the journey of heat between bodies very well.
But they do not help us in explaining two other very ordinary observations.
The first is why the action of rubbing our palms together generates warmth. Here nothing hot has been brought near the palms at all; no body at a higher temperature is supplying them with heat. Yet they become warmer.
The second is why an engine gets warmer when it is running. Again, the engine is not being heated from outside — it is doing work, and heat appears as a by-product.
What Is Needed to Explain Them
These and similar phenomena can be explained on the basis of three ideas taken together:
- the concept of the internal energy of a system,
- the conversion of work and heat into each other, and
- the laws governing these inter-conversions.
The subsections that follow build these ideas one at a time. We first ask what a system actually stores inside itself, then define carefully what we mean by a system and by a process, then sharpen the meaning of heat, and finally see the two distinct routes by which the energy stored inside a system can be changed.
4.4.1 Internal Energy
We know that every system, large or small, consists of a large number of molecules. Even a spoonful of water contains molecules in numbers too large to picture, and every one of them is in ceaseless, random motion.
That motion carries energy, and it is this energy that we now name.
Ordered Energy and Disordered Energy
Internal energy is different from the macroscopic ordered energy of a moving object. The word ordered is the key: in a moving cricket ball, every molecule travels in the same direction with the same average velocity, and that shared motion is what we call the kinetic energy of the ball.
For example, consider a glass of water kept on a table, as shown in Fig. 4.31. It has no kinetic energy because it is not moving. Its potential energy can also be taken as zero, since we may choose the table top as our reference level.
Diagram
Fig. 4.31 – A glass of water at rest on a table, with an enlarged circular inset showing water molecules moving in random directions with arrows of different lengths.
But we know, from the kinetic theory, that the water molecules in the glass at the given temperature move at a random speed. Notice in Fig. 4.31 how the arrows in the inset point every which way — that is exactly what “disordered” means. The velocities do not add up to any motion of the glass as a whole, because for every molecule moving one way there is another moving the opposite way.
Thus, we can say that the internal energy of a substance is the total energy of all its atoms/molecules.
What Goes into Internal Energy for Different Substances
The contents of the internal energy are not the same for every kind of substance.
For an ideal monatomic gas such as argon, the internal energy is just the translational kinetic energy of the atoms having a linear motion. This was discussed in Chapter 3. A single atom can only fly from place to place; it has nothing else it can do with energy.
For a polyatomic gas such as carbon dioxide, we consider the rotational and vibrational kinetic energy of the molecules in addition to their translational kinetic energy. A molecule made of several atoms can also spin about its axes, and its atoms can vibrate against the bonds joining them, so there are more ways for it to hold energy.
In case of liquids and solids, we need to consider the potential energy of the molecules due to the intermolecular attractive forces amongst them. In these states the molecules are close enough to pull on one another, and the energy stored in that mutual attraction is part of the total.
Remember that all of this is again at the molecular level, that is, on the microscopic scale only. This internal energy of a system is denoted by \(U\), and like every form of energy it is measured in joule (J).
The reason the two answers differ is the one given above. An argon atom stores energy only in its motion from place to place, while an oxygen molecule can also rotate, and so it holds a larger share of energy at the same temperature.
In application-based questions this idea is usually tested indirectly. You are told that the temperature of an ideal gas did not change during a process, and you are expected to conclude at once that its internal energy did not change either — no matter how much its pressure and volume were altered along the way.
4.4.2 Thermodynamic System and Thermodynamic Process
Let us understand what is meant by a thermodynamic system and a thermodynamic process first. Almost every statement made later in this chapter begins by naming a system, so it is worth being precise about what the word means.
System, Boundary and Surrounding
A thermodynamic system is a collection or a group of objects that can form a unit which may have the ability to exchange energy with its surroundings.
Anything that is not a part of the system is its surrounding, or its environment. The choice is ours to make, but once made it must be kept fixed throughout the problem.
For example, water kept in a vessel is a system, the vessel is its boundary, and the atmosphere around it is its surrounding. Figure 4.3 (a) shows this schematically.
Diagram
Fig. 4.3 (a) – An irregular shaded blob labelled “System”, enclosed by a line labelled “Boundary”, with the region outside labelled “Surrounding (Environment)”.
Three Kinds of Systems
Thermodynamic systems can be classified on the basis of the possible transfer of heat and matter to the environment. Based on this, they are classified as open, closed or isolated systems.
An open system is a system that freely allows exchange of energy and matter with its environment.
For example, water boiling in a kettle is an open system. Heat escapes into the air — this is the exchange of energy with the surroundings. At the same time, steam also escapes into the air — this is exchange of matter with the surroundings.
A closed system, on the other hand, does not allow the exchange of matter but allows energy to be transferred.
For example, water boiling in a boiler is a closed system. It allows heat, that is energy, to be transferred from the source of heat (a burner) to the water, which is the system, inside. Similarly, heat is also transferred to the surroundings. Steam, which is matter, is not allowed to escape as long as the valve is kept closed.
An isolated system is completely sealed, that is, isolated from its environment. Matter as well as heat cannot be exchanged with its environment. A thermos flask is a very familiar example of an isolated system.
Figure 4.3 (b) shows an open system, a closed system, and an isolated system schematically. Notice in Fig. 4.3 (b) how the arrows disappear one by one as we move from left to right: the open system has arrows for both heat and matter, the closed system has an arrow for heat alone, and the isolated system has none.
Diagram
Fig. 4.3 (b) – Three boxes side by side: open system with arrows for heat and matter leaving, closed system with only a heat arrow, isolated system with no arrows.
What a Thermodynamic Process Is
A thermodynamic process is a process in which the thermodynamic state of a system is changed.
For example, water contained in a vessel with a lid on it is an open system. When the pot is heated externally, water starts boiling after some time, and steam is produced which exerts pressure on the walls of the vessel.
In this case, the state of the water in the container is changed. This is because the temperature (\(T\)), the volume (\(V\)) and the pressure (\(P\)) of the water inside the vessel change when it starts boiling.
Thus, we can describe the state of a system by using temperature, pressure and volume as its variables. Here \(T\) is measured in kelvin (K), \(V\) in cubic metre (m\(^3\)) and \(P\) in pascal (Pa) in SI units.
This is a considerable simplification, and it is worth appreciating why it is remarkable. A litre of water contains an unimaginable number of molecules, yet three measured numbers are enough to fix its condition completely. We will discuss these variables in some detail at a later stage, in Section 4.6.
4.4.3 Heat
Let us now try to understand heat and its relation with the internal energy of a system. We have used the word “heat” freely so far; it is now time to fix its meaning exactly.
A System and Its Environment at Different Temperatures
Consider a glass filled with water on a table. The glass, along with the water in it, forms a system. Let the temperature of this system be \(T_S\).
The table on which the glass is kept, and the other relevant parts of the room, will then be its surrounding or the environment. Let the temperature of the environment be \(T_E\). Both temperatures are measured in kelvin (K).
We notice that if \(T_S\) and \(T_E\) are not the same, then \(T_S\) will change until both the temperatures are equal and a thermal equilibrium will be reached between the ‘system’ and the ‘environment’.
\(T_E\) will also change to some extent, but the end result is that the ‘system’ and the ‘environment’ reach thermal equilibrium. If the environment is very large, the change in \(T_E\) may not be measurable, but it is certainly not zero.
Such a change in temperature is caused by the transfer of internal energy between the system and its environment. In this case, the transfer of energy is between the glass of water and its surrounding.
The Sign Convention for Energy Transfer
When transfer of energy takes place between a system and its environment, we observe the following conventions. These signs are not decoration; they carry the direction of the transfer, and every numerical problem later in this chapter depends on getting them right.
- When the energy is transferred to a system from its environment, it is positive. We say that the system gains (or absorbs) energy.
- When the energy is transferred from the system to its environment, it is negative. We say that the system loses (or releases) energy.
Notice that the convention is written from the system’s point of view. A quantity is positive when the system is the receiver. If you ever find yourself unsure of a sign, ask the single question: is the system gaining or losing?
The Three Possible Cases
Consider Fig. 4.4, which shows energy transfer between a system and its environment. Let \(T_S\) and \(T_E\) be the temperatures of the system and its environment respectively, and let \(Q\) be the energy transferred between the system and its environment, measured in joule (J).
As shown in Fig. 4.4 (a), when \(T_S < T_E\), the system gains energy, and \(Q\) is positive.
Diagram
Fig. 4.4 (a) – A system blob with an arrow pointing inward from the environment, labelled energy flows into the system, \(T_S < T_E\), \(Q > 0\).
In Fig. 4.4 (b), \(T_S > T_E\), the system loses energy, and \(Q\) is negative.
Diagram
Fig. 4.4 (b) – A system blob with an arrow pointing outward to the environment, labelled energy flows from the system, \(T_S > T_E\), \(Q < 0\).
In Fig. 4.4 (c), \(T_S = T_E\), the system and the environment are in thermal equilibrium and there is no transfer of energy, so \(Q = 0\).
Diagram
Fig. 4.4 (c) – A system blob with no arrows, labelled no transfer of energy, \(T_S = T_E\), \(Q = 0\).
In all three cases the deciding factor is the same: the temperature difference. Remove the difference, as in Fig. 4.4 (c), and the transfer stops, however much internal energy each side may hold.
The Definition of Heat
Using these observations, we can now define heat.
Read that definition carefully, because it says something stronger than it first appears. Heat is not a quantity stored inside a body. It is energy caught in the act of crossing a boundary, and it exists only while the transfer is going on.
4.4.4 Change in Internal Energy of a System
In the previous discussion we have seen that the internal energy of a system can be changed — it can be gained or released — due to exchange with its environment. Now we will try to understand how this transfer of energy between a system and its environment is possible.
Consider the following experiment.
The Arrangement
Figure 4.5 (a) shows a cylinder filled with some gas in it. This cylinder is provided with a movable, massless and frictionless piston at one end, as shown.
The gas inside the cylinder is our system and the rest is its environment. Let the temperature of the gas be \(T_S\) and that of the environment be \(T_E\).
The piston is taken to be massless and frictionless so that no energy is wasted in moving the piston itself or in overcoming friction at the walls. Every joule that enters or leaves can then be accounted for cleanly. This idealised cylinder-and-piston will be our standard system for the rest of the chapter.
Internal energy of the system, that is of the gas, can be changed in two different ways, or by both together.
Method One: Heating the System
The cylinder can be brought in contact with a source of heat such as a burner, as shown in Fig. 4.5 (a).
Diagram
Fig. 4.5 (a) – A gas cylinder with a movable piston standing over a lit burner, with an arrow showing heat entering the gas.
As discussed previously, the temperature difference between the source of heat, which is the environment here, and the system will cause a flow of energy — heat — towards the gas in the cylinder. This is because \(T_E > T_S\). Thus, there will be an increase in the internal energy of the gas.
Such exchange of energy is possible in another way also. If the surrounding is at a temperature lower than the gas, that is \(T_S > T_E\), the gas will lose energy to its environment and cool down.
So heating is really a two-way route: it raises the internal energy when the surrounding is hotter, and lowers it when the surrounding is colder. The direction is decided entirely by which side is at the higher temperature.
Method Two: Doing Work on the System
The other way to increase the internal energy of the gas is to quickly push the piston inside the cylinder, so that the gas is compressed, as shown in Fig. 4.5 (b).
Diagram
Fig. 4.5 (b) – The same gas cylinder with an external force pushing the piston inward, compressing the gas, no burner present.
In this case, we know that the piston does some work on the gas in moving it through some distance. The gas gains energy and its temperature is increased.
Note carefully what is missing from Fig. 4.5 (b). There is no burner, no hot body, no temperature difference of any kind. The energy has entered through the moving boundary rather than through a temperature difference — and yet the gas becomes hotter, exactly as if it had been heated.
On the other hand, if the gas pushes the piston out, so that the gas is expanded, some work is done by the gas. It loses some of its energy and the gas cools down.
The Two Routes Are Equivalent
Thus, we see that the internal energy of a system can be changed in two different ways: 1) by heating it, or 2) by doing work on it.
The experiment we discussed just now can be carried out in a very meticulous way so that we achieve the same change in temperature of the gas by both the methods.
This last sentence deserves attention. It means that once the gas has reached a certain temperature, it carries no memory of how it got there. Heat and work are two different routes, but the destination — the internal energy of the system — is one and the same quantity.
Where This Leads
The conclusion of this experiment leads us to a very important principle of thermodynamics.
It is related to the work done on the system, or by the system, and the change in the internal energy of the system. Both are related through the energy that is transferred to, or by, the system and the heat that is involved in the process.
This leads us to the First Law of Thermodynamics, which we take up in the next section.
4.6 Thermodynamic State Variables
Earlier, we have discussed thermal equilibrium and understood the concept of temperature and the Zeroth law of thermodynamics. Thermodynamics, however, is not the study of changes in temperature of a system only.
As we have seen earlier, when the temperature of a system changes — that is, when it gains or releases energy — its other properties can also change. Let us understand these properties. We will define the term property of a thermodynamic system first.
Property of a System or a System Variable
A property of a system is any measurable or observable characteristic or property of a system when the system remains in equilibrium. A property is also called a state variable of the system.
Note the condition attached: the system must be in equilibrium. A quantity such as “the pressure of the gas” has a single definite value only when the pressure is the same everywhere in the system. During a violent, uncontrolled change there is no one number to quote, and the property is simply not defined.
We will use the term variable to describe a characteristic of a system. For example, pressure, volume, temperature, density and mass of a system are some of the variables that are used to describe a system.
These are measurable properties and are called macroscopic variables of a system. The word macroscopic is a reminder of what was said in Section 4.4.1: these variables describe the system as a whole, not the individual molecules inside it.
Intensive and Extensive Variables
Intensive variables do not depend on the size of the system. Extensive variables depend on the size of the system.
There is a simple test that separates the two, and it is worth learning because it works every time. Consider a system in equilibrium. Let this system be divided into two equal compartments, each with half the original volume, as shown in Fig. 4.32.
Diagram
Fig. 4.32 – A container of gas in equilibrium, and beside it the same container split by a partition into two equal halves, with \(p\), \(T\) and \(\rho\) marked unchanged in each half and \(M\) and \(U\) marked as halved.
We notice in Fig. 4.32 that the pressure \(p\), the temperature \(T\), and the density \(\rho\) are the same in both compartments. Halving the amount of gas has not made it hotter, denser or more compressed. These are intensive variables.
The total mass \(M\), and the internal energy \(U\) of the system are equally divided in the two compartments and are extensive variables of the system. Each half holds half the mass and half the stored energy.
4.6.1 Thermodynamic Equilibrium
A system is in thermodynamic equilibrium if the following three conditions of equilibrium are satisfied simultaneously. These are, 1) Mechanical equilibrium, 2) Chemical equilibrium, and 3) Thermal equilibrium.
The word simultaneously is the important one. Meeting two of the three conditions is not enough; a system that is uniform in temperature but still has a pressure difference inside it will keep changing, and so it is not yet in thermodynamic equilibrium. The three conditions are illustrated together in Fig. 4.33.
Diagram
Fig. 4.33 – Three panels of the same gas container: one showing balanced pressure arrows (mechanical), one showing uniform composition with no reaction arrows (chemical), and one showing a uniform temperature (thermal).
Mechanical Equilibrium
When there are no unbalanced forces within the system and between the system and its surrounding, the system is said to be in mechanical equilibrium.
The system is also said to be in mechanical equilibrium when the pressure throughout the system and between the system and its surrounding is the same. These two ways of putting it agree with each other, because a difference in pressure between two regions is precisely what produces an unbalanced force on the boundary between them.
Whenever some unbalanced forces exist within the system, they will get neutralized with time to attain the condition of equilibrium. A denser, higher-pressure region pushes outward into its neighbours until the imbalance is evened out.
Chemical Equilibrium
A system is said to be in chemical equilibrium when there are no chemical reactions going on within the system, or there is no transfer of matter from one part of the system to the other due to diffusion.
Both possibilities have to be ruled out, because either one would keep altering the make-up of the system from place to place. A reaction changes which substances are present; diffusion changes where they are.
Thermal Equilibrium
When the temperature of a system is uniform throughout and does not change with time, the system is said to be in thermal equilibrium. We have discussed thermal equilibrium at length earlier, in Section 4.2.
4.6.2 Thermodynamic State Variables and Equation of State
Every equilibrium state of a thermodynamic system is completely described by specific values of some macroscopic variables, also called state variables.
For example, an equilibrium state of a gas is completely described by the values of its pressure \(p\), volume \(V\), temperature \(T\), and mass \(m\). Quote these four numbers and you have said everything there is to say about the gas; no further measurement can reveal anything new about its state.
Consider a mixture of gases or vapours, as in the case of the fuel in an automobile engine. Its state can be described by the state variables, but we also need its composition to describe its state — how much of the mixture is fuel vapour and how much is air makes a real difference to its behaviour.
When a System Has No State to Describe
A thermodynamic system is not always in equilibrium. Figure 4.9 shows such a case.
Diagram
Fig. 4.9 – An inflated ball beside the same ball just after being punctured, with air rushing out and swirling, marked “non-equilibrium state”.
For example, when an inflated ball is punctured, the air inside it suddenly expands to the atmosphere. This is not an equilibrium state. During the rapid expansion, the pressure of the air may not be uniform throughout — it is already low near the hole while it is still high deep inside.
Similarly, the fuel, a mixture of petrol vapour and air, in the cylinder of an automobile engine undergoing an explosive chemical reaction when ignited by a spark is not an equilibrium state. This is because its temperature and pressure are not uniform. The flame front has passed through part of the cylinder and not yet through the rest.
Such a system which is not in equilibrium cannot be described in terms of the state variables. There is no single number to call “the pressure” when the pressure differs from point to point, and no single number to call “the temperature” when the temperature does.
Eventually, the air in the first case, and the fuel in the second case, reach a uniform temperature and pressure and attain thermal and mechanical equilibrium with their surroundings. Thus each of them attains thermodynamic equilibrium.
In simple words, thermodynamic state variables describe the equilibrium states of a system.
The Equation of State
The various state variables are not always independent. They can be mathematically related.
For example, for an ideal gas, the equation of state is the ideal gas equation, \[pV = nRT \qquad \text{--- (4.5)}\]
Here, \(p\), \(V\) and \(T\) are the pressure, the volume and the temperature of the gas, measured in pascal (Pa), cubic metre (m\(^3\)) and kelvin (K) respectively; \(n\) is the number of moles of the gas, measured in mol; and \(R\) is the gas constant, measured in J mol\(^{-1}\) K\(^{-1}\).
Equation (4.5) has a consequence worth stating separately. For a fixed amount of the gas, that is, for a given \(n\), there are thus only two independent variables. It could be \(p\) and \(V\), or \(p\) and \(T\), or \(V\) and \(T\).
The p-V Diagram
The graphical representation of the equation of state of a system, that is of a gas, is called the \(p\)-\(V\) diagram, or the \(p\)-\(V\) curve (the pressure–volume curve), or the indicator diagram of the system.
Figure 4.10 shows a typical \(p\)-\(V\) diagram for an ideal gas at some constant temperature. Notice in Fig. 4.10 that as the gas moves from state A at \((V_1, p_1)\) to state B at \((V_2, p_2)\), the pressure falls as the volume rises, tracing a smooth curve.
Diagram
Fig. 4.10 – A typical \(p\)-\(V\) diagram: a falling curve from point A\((V_1, p_1)\) to point B\((V_2, p_2)\), with dashed lines to both axes.
Every point on this curve is one equilibrium state of the gas, with its own definite pressure and volume — and, through Eq. (4.5), its own definite temperature.
The pressure-volume curve for a constant temperature is called an isotherm.
Real gases may have more complicated equations of state and therefore, a complicated \(p\)-\(V\) diagram. The Van der Waals equation, for example, with its various corrections, is complicated for a real gas and is equally interesting.
The equation of state of a system — usually a gas confined to a cylinder with a movable, frictionless and massless piston — and its \(p\)-\(V\) diagram are very useful in studying its behaviour. In the following sections, we will discuss some systems and their behaviour using \(p\)-\(V\) diagrams.
4.6.3 The p - V Diagram
Consider Eq. (4.2), that is, \[W = \int_{V_i}^{V_f} dW = \int_{V_i}^{V_f} P\,dV\]
The integral in this equation can be evaluated if we know the relation between the pressure \(p\) and the volume \(V\), or the path between the limits of integration.
This is the crucial point. Knowing only the starting volume \(V_i\) and the finishing volume \(V_f\) is not enough to work out the integral, because the pressure may take different values along the way in different processes. We must know how \(p\) varies as \(V\) changes.
Equation (4.2) can be represented graphically, and this is where the \(p\)-\(V\) diagram earns its usefulness.
Three Ways of Changing the Volume
A gas confined to a cylinder with a movable, frictionless and massless piston can be changed in three different ways that we shall now compare:
- it can be expanded with varying pressure, as in Fig. 4.11 (a),
- it can be compressed with varying pressure, as in Fig. 4.11 (b), or
- it can expand at constant pressure, as in Fig. 4.11 (c).
Area Under the Curve Is the Work Done
The area under the curve in the \(p\)-\(V\) diagram is the graphical representation of the value of the integral in Eq. (4.2).
The reason is the meaning of an integral. Each thin vertical strip under the curve has width \(dV\) and height \(p\), so its area is \(p\,dV\) — which is exactly the small amount of work \(dW\) from Eq. (4.1). Adding all the strips from \(V_i\) to \(V_f\) adds up all the small contributions of work.
Since this integral represents the work done in changing the volume of the gas, the area under the \(p\)-\(V\) curve also represents the work done in this process.
To get you started, recall that pressure is force per unit area and volume is a length cubed; multiplying them should leave you with force multiplied by length, which is the dimension of work and of energy. In SI units, Pa \(\times\) m\(^3\) = (N m\(^{-2}\)) \(\times\) m\(^3\) = N m = J.
Expansion with Varying Pressure
Figure 4.11 (a) shows expansion of the gas. Its volume changes due to outward displacement of the piston and the pressure of the gas decreases.
Diagram
Fig. 4.11 (a) – A \(p\)-\(V\) curve falling from point 1 at \((V_1, P_1)\) to point 2 at \((V_2, P_2)\) with the area under it shaded, arrow along the curve pointing right; captioned positive work with varying pressure.
The work done by the gas in this case is positive because the volume of the gas has increased. Notice in Fig. 4.11 (a) that the process runs from left to right along the curve, in the direction of increasing volume, and the shaded area beneath it is the magnitude of the work done.
Compression with Varying Pressure
Similarly, Fig. 4.11 (b) shows compression due to inward displacement of the piston. The pressure of the gas is increased and the work done by the gas is now negative.
Diagram
Fig. 4.11 (b) – A \(p\)-\(V\) curve traced from point 1 at \((V_1, P_1)\) to point 2 at \((V_2, P_2)\) in the direction of decreasing volume, with the area under it shaded; captioned negative work with varying pressure.
The shaded area in Fig. 4.11 (b) has the same meaning as before, but the sign is reversed. The direction of travel along the curve, and not the shape of the curve, decides the sign: moving towards smaller volume means work is being done on the gas, so the work done by the gas is negative.
Expansion at Constant Pressure
Figure 4.11 (c) shows the \(p\)-\(V\) diagram when the volume of the gas changes from \(V_1\) to \(V_2\) at a constant pressure. The curve is actually a line parallel to the volume axis, because the pressure keeps the same value throughout the change.
Diagram
Fig. 4.11 (c) – A horizontal line at pressure \(P\) from point 1 at \(V_1\) to point 2 at \(V_2\), with the rectangular area beneath shaded and labelled Work = Area = \(P(V_2 - V_1) > 0\).
The work done during volume change at constant pressure is \[W = P\,(V_2 - V_1)\]
Only in this case is the integration simply \(P(dV)\), with the pressure taken outside the integral. The area under the curve in Fig. 4.11 (c) is a plain rectangle of height \(P\) and width \((V_2 - V_1)\), so no integration is really needed — the area can be read off at a glance.
When the Volume Does Not Change
When the volume is constant in any thermodynamic process, the work done is zero because there is no displacement.
This follows directly from Eq. (4.1): if \(dV = 0\) at every stage, then \(dW = p\,dV = 0\) however large the pressure may be. On the \(p\)-\(V\) diagram such a process is a vertical line, and a vertical line encloses no area beneath it.
These changes are very slow. We will discuss such processes in some detail in a later section.
4.7 Thermodynamic Process
A thermodynamic process is a procedure by which the initial state of a system changes to its final state.
During such a change, there may be a transfer of heat into the system from its environment, which is positive heat — for example, when water boils, heat is transferred to the water. Heat may also be released from the system to its environment, which is negative heat.
Similarly, some work can be done by the system, which is positive work, or some work can be done on the system, which is negative work. When the piston in a cylinder is pushed in, some work is done on the system.
So in any process there are exactly four possibilities to keep track of, and the sign convention of Section 4.4.3 covers all of them. Heat may go in or out; work may be done by the system or on it. Both signs must be settled before the first law can be applied.
Changes That Are Slow Enough to Follow
We know that these changes should occur infinitesimally slowly so that the system is always in thermodynamic equilibrium.
The reason goes back to Section 4.6.2. A system that is not in equilibrium has no definite pressure or temperature, so it cannot be marked as a point on a \(p\)-\(V\) diagram at all. If we want to draw the whole change as a curve, every intermediate stage must itself be an equilibrium state — and that is possible only if the change is made slowly enough for the system to settle at every step.
Such processes, in which changes in the state variables of a system occur infinitesimally slowly, are called quasi static systems.
Paths on the p-V Diagram
When a thermodynamic system changes from its initial state to its final state, it passes through a series of intermediate states.
This series of intermediate states, when plotted on a \(p\)-\(V\) diagram, is called a path.
The \(p\)-\(V\) curve, or the \(p\)-\(V\) diagram, shown in Fig. 4.11 is such a path. It tells us the way a system has gone through a change — not merely where it started and where it finished, but the route taken between them.
That distinction will turn out to matter a great deal. In the next two subsections we shall find that two quantities we have been treating casually, the work done and the heat transferred, both depend on which path was taken.
4.7.1 Work Done During a Thermodynamic Process
Let us understand the relation between a path and the work done along a path.
Consider Fig. 4.12 (a), which describes different ways in which we can change the state of a system.
Diagram
Fig. 4.12 (a) – A \(p\)-\(V\) diagram showing initial point A at \((V_i, p_i)\) and final point B at \((V_f, p_f)\), joined by three differently shaped paths labelled 1, 2 and 3.
The system is initially at state A on the \(p\)-\(V\) diagram. Its pressure is \(p_i\) and its volume is \(V_i\). We say that the state is indicated by the coordinates \((V_i, p_i)\).
The final state of the system is shown by the point B, with its coordinates given by \((V_f, p_f)\).
The whole point of Fig. 4.12 (a) is that the two end points are fixed while the routes joining them are not. Both A and B are the same for every path drawn; only the journey between them differs.
The First Path: Both Pressure and Volume Change Together
The curve 1, path 1, shown in Fig. 4.12 (a) is one of the many ways, or paths, in which we can change the system from state A to the state B.
When the system changes itself from A to B along path 1, both its pressure and volume change. The pressure decreases while the volume increases.
The work done by the system is positive, because the volume increases. It is given by the area under the curve 1, as shown in Fig. 4.12 (b).
Diagram
Fig. 4.12 (b) – The same \(p\)-\(V\) diagram with only path 1 drawn from A to B and the entire area beneath it shaded, marked as the work done.
The Second Path: Constant Pressure, Then Constant Volume
The second way to change the state from A to state B is path 2, as shown in Fig. 4.12 (c).
In this case, the volume increases to \(V_f\) from the point A up to the point C at the constant pressure \(p_i\). This first leg is a horizontal line on the diagram, of the kind we met in Fig. 4.11 (c).
The pressure then decreases to \(p_f\) as shown, and the volume remains constant during this change. This second leg is a vertical line, and along it no work is done at all, since there is no change in volume.
The system is now in the state B with its coordinates given by \((V_f, p_f)\).
The work done in this process is represented by the shaded area under the curve 2 as in Fig. 4.12 (c).
Diagram
Fig. 4.12 (c) – A \(p\)-\(V\) diagram with path A to C horizontal at pressure \(p_i\), then C to B vertical down to \(p_f\), with the rectangular area under the horizontal leg shaded.
Notice in Fig. 4.12 (c) that the shaded region is a rectangle of height \(p_i\) and width \((V_f - V_i)\). The expansion has taken place entirely at the higher pressure.
The Third Path: Constant Volume, Then Constant Pressure
The third way to change the state from A to state B is path 3, as shown in Fig. 4.12 (d).
In this case, the pressure decreases from \(p_i\) to \(p_f\) but the volume remains the same. Next, the volume changes to \(V_f\) at constant pressure \(p_f\).
The work done in this process is represented by the shaded area under the curve 3 as in Fig. 4.12 (d).
Diagram
Fig. 4.12 (d) – A \(p\)-\(V\) diagram with path A to D vertical down from \(p_i\) to \(p_f\), then D to B horizontal at pressure \(p_f\), with the rectangular area under the horizontal leg shaded.
Here the expansion has taken place entirely at the lower pressure, so the shaded rectangle in Fig. 4.12 (d) is visibly shorter than the one in Fig. 4.12 (c), although its width is the same.
It is easily noticed that in the three cases we discussed, the amount of work done is not the same.
The Conclusion
Remember that these are only three paths amongst many along which the system can change its state.
It is interesting to note that in all these cases, though the work done during the change of state is different, the initial and the final state of the system is the same. The gas ends up with exactly the same pressure, volume and temperature whichever route it took — yet the work delivered along the way differs.
We conclude that the work done by a system depends not only on the initial and the final states, but also on the intermediate states, that is, on the paths along which the change takes place.
This is why it is meaningless to speak of “the work contained in a system.” Work is not a property of a state; it is a quantity associated with a process, and asking how much work a gas “has” is as confused as asking how much heat a cup of tea “has”.
In numerical questions, this idea is tested by giving you two states and a specific path, and expecting you to compute the area under that path. A common mistake is to use the constant-pressure formula \(W = p(V_f - V_i)\) on a path where the pressure is not constant. Always check the shape of the path before choosing the formula.
4.7.2 Heat Added During a Thermodynamic Process
The thermodynamic state of a system can be changed by adding heat also. Having found that the work done depends on the path, we now ask the same question about the heat transferred.
Consider a thermodynamic system consisting of an ideal gas confined to a cylinder with a movable, frictionless and massless piston. Suppose we want to change the initial volume \(V_i\) of the gas to the final volume \(V_f\) at a constant temperature.
There are two different ways in which this change in volume can be made, and comparing them is the whole of this section.
The First Method: A Slow, Controlled Expansion
Figure 4.13 (a) shows the first method.
Diagram
Fig. 4.13 (a) – A gas cylinder with a movable piston standing on a burner, heat entering from below and the piston being pushed outward; labelled isothermal expansion, \(W > 0\), \(Q > 0\).
In this case, the gas is heated slowly, in a controlled manner, so that it expands at a constant temperature. It reaches the final volume \(V_f\) isothermally.
The system absorbs a finite amount of heat during this process. Notice in Fig. 4.13 (a) that the gas also does work, because the piston is pushed outward through a real distance against the pressure outside. Both \(Q\) and \(W\) are positive here.
The Second Method: A Sudden, Uncontrolled Expansion
In the second case, shown in Fig. 4.13 (b), the gas cylinder is now surrounded by an insulating material, and it is divided into two compartments by a thin, breakable partition.
Diagram
Fig. 4.13 (b) – An insulated cylinder divided by a thin partition into compartment X containing gas and empty compartment Y, shown before and after the partition is broken; labelled \(W = 0\), \(Q = 0\).
The compartment X has a volume \(V_i\) and the compartment Y has a volume \(V'_i\), so that \[V_i + V'_i = V_f\]
The compartment X of the cylinder is filled with the same amount of gas at the same temperature as that in the first case shown in Fig. 4.13 (a). The compartment Y is empty; it contains no gas particles or any other form of matter.
The initial state of the system is therefore the same in both cases. This is essential to the argument — we are comparing two journeys that begin from the same place.
The partition is now suddenly broken. This causes a sudden, uncontrolled expansion of the volume of the gas. The gas occupies the volume that was empty before the partition was broken.
There is no exchange of heat between the gas and its environment, because the cylinder is now surrounded by an insulating material, so \(Q = 0\).
The final volume of the system after the partition is broken is \(V_f\).
In this case, the gas has not done any work during its expansion, because it has not pushed any piston or any other surface for its expansion, so \(W = 0\).
Such expansion is called free expansion.
A common example of free expansion is the abrupt puncturing of an inflated balloon or a tyre.
Same End Points, Different Paths
It is experimentally observed that when an ideal gas undergoes a free expansion, there is no change of temperature.
This fits what we learnt in Section 4.4.1. Since both \(Q\) and \(W\) are zero, the first law gives \(\Delta U = 0\); and since the internal energy of an ideal gas depends only on its temperature, the temperature cannot have changed either.
Therefore, the final state of the gas in this case also is the same as in the first case — same amount of gas, same final volume \(V_f\), same temperature, and hence the same pressure.
The intermediate states, or the paths, during the change of state in the first and the second case are different. But the initial and the final states are the same in both cases.
Figures 4.13 (a) and 4.13 (b) represent two different ways of taking a system from the initial state to the final state. This means we have two different paths connecting the same initial and the final states of a system.
Now compare the energy accounting along the two paths.
In the case of the method shown in Fig. 4.13 (a), there is an exchange of heat — the burner supplies a definite quantity of it.
In the case of the method shown in Fig. 4.13 (b), there is no exchange of heat at all, and also the system does not do any work whatsoever, because there is no displacement of any piston or any other surface.
To conclude, heat transferred to a system also depends on the path.
This is why \(U\) is called a state variable while \(Q\) and \(W\) are not. Speaking of “the heat of a system” is as meaningless as speaking of “the work of a system” — both name transfers, not contents. We shall meet free expansion once more in the next subsection, where it takes its place among the standard thermodynamic processes.
4.7.3 Classification of Thermodynamic Processes
As we have seen earlier, a thermodynamic state can be described by its pressure \(p\), volume \(V\), and temperature \(T\). These are the state variables of a system. At present, we will restrict our description of a thermodynamic system only to its pressure, volume and temperature.
A process by which two or more of these variables can be changed is called a thermodynamic process or a thermodynamic change.
As we have discussed earlier, there can be a number of different ways to change these parameters, that is, there are different thermodynamic processes. But in practice, for the sake of measurement, any one of the state variables is held constant and the other two are varied.
Holding one variable fixed is not merely a convenience. It removes one unknown from the equation of state, so that the remaining two variables are tied together by a simple relation which can be drawn as a definite curve on the \(p\)-\(V\) diagram. This leads us to a very useful way of classifying thermodynamic processes.
Reversible and Irreversible Processes
We know that when two objects at different temperatures are brought in thermal contact, they reach a thermal equilibrium. In this process, the object at higher temperature loses its heat and the object at lower temperature gains heat.
But we never observe that after some time, the two objects are back to their initial temperatures. The object that was previously hot never becomes hot again, and the previously cold object never becomes cold again, once they have reached thermal equilibrium.
That means two objects at different temperatures reaching thermal equilibrium is an irreversible process. Such processes do not restore the initial state of the system.
Puncturing an inflated balloon or a tyre, rubbing our palms together, and burning a candle are some familiar examples of irreversible thermodynamic processes.
Some processes, such as melting of ice, freezing of water, boiling of water and condensation of steam, can be reversed. That means the initial states of the system can be restored. These are some familiar thermodynamic processes that are reversible.
A thermodynamic process, or change, can therefore be a reversible process or it can be an irreversible process.
Reversible and Irreversible Paths on the p-V Diagram
Earlier, we have seen that a thermodynamic process can be represented by a \(p\)-\(V\) diagram. This gives us a way of seeing the difference between the two kinds of process.
A reversible process is a change that can be retraced in the reverse, that is opposite, direction. The path of a reversible thermodynamic process is the same in the forward and the reverse direction.
Figure 4.14 (a) shows the path of a reversible thermodynamic process.
Diagram
Fig. 4.14 (a) – A \(p\)-\(V\) diagram with a single smooth curve from A to B carrying arrowheads in both directions, showing the same path used for expansion and compression.
Notice in Fig. 4.14 (a) that there is only one curve. This path shows a reversible expansion of a gas followed by its reversible compression, and the return journey lies exactly on top of the outward one.
Such changes are very slow and there is no loss of any energy in the process, and the system is back to its initial state after it is taken along the reverse path.
Reversible processes are ideal processes. A real thermodynamic process will always encounter some loss due to friction or some other dissipative forces.
An irreversible process is a change that cannot be retraced in the reverse, that is opposite, direction. The path of an irreversible thermodynamic process is not the same in the forward and the reverse direction.
Figure 4.14 (b) shows the path of an irreversible thermodynamic process.
Diagram
Fig. 4.14 (b) – A \(p\)-\(V\) diagram from A at \((V_i, p_i)\) to B at \((V_f, p_f)\) with two separate curves, the outward and return paths not coinciding, enclosing a loop.
Observe in Fig. 4.14 (b) that the two journeys trace different curves and enclose an area between them. There is a permanent loss of energy from the system due to friction or other dissipative forces in an irreversible process.
The change of state depends on the path taken to change the state during an irreversible process. An irreversible process shows a hysteresis — the system’s behaviour on the way back is not a mirror image of its behaviour on the way out, and the gap between the two curves is the record of the energy lost.
Most real life thermodynamic processes that we deal with are irreversible.
Cause of Irreversibility
There are two main reasons for the irreversibility of a thermodynamic process.
The first is that many processes, such as a free expansion or an explosive chemical reaction, take the system to non-equilibrium states. As we saw in Section 4.6.2, a system passing through such states cannot even be described by state variables, let alone made to retrace them in order.
The second is that most processes involve friction, viscosity or some other dissipative forces. For example, an object sliding on a surface stops after moving through some distance due to friction, and loses its mechanical energy in the form of heat to the surface, and it gets heated itself.
The dissipative forces are always present everywhere, and can be minimized at best, but cannot be fully eliminated.
This idea — that nature has a preferred direction — will return as the central theme of the Second law of thermodynamics later in this chapter.
Assumptions for Discussion of Thermodynamic Processes
We will be discussing various types of thermodynamic systems in the following sections. Here are the assumptions we make for this discussion.
The majority of the thermodynamic processes we will be discussing in the following sections are reversible. That is, they are quasistatic in nature. They are extremely slow and the system undergoes an infinitesimal change at every stage, except the adiabatic processes. The system is, therefore, in thermodynamic equilibrium during all the change.
The ‘system’ involved in all the processes is an ideal gas enclosed in a cylinder having a movable, frictionless and massless piston. Depending on the requirements of the process, the walls of the cylinder can be good thermal conductors, for an isothermal process, or can be thermally insulating, for an adiabatic process.
The ideal gas equation is applicable to the system.
Keep the second assumption in mind, because it explains why the same picture — a cylinder and a piston — reappears in every process that follows. Only the walls change their character from one process to the next.
Isothermal Process
A process in which change in pressure and volume takes place at a constant temperature is called an isothermal process or isothermal change. For such a system \(\Delta T = 0\).
Isothermal process is a constant temperature process. This is possible when a system is in good thermal contact with its environment, and the transfer of heat from, or to, the system is extremely slow so that thermal equilibrium is maintained throughout the change.
Both conditions are necessary. Good thermal contact allows heat to move in or out freely so that any tendency of the gas to warm up or cool down is immediately corrected by the surroundings. Slowness gives that correction time to happen.
For example, melting of ice, which takes place at constant temperature, is an isothermal process.
The second half of that statement is worth reading twice. “Isothermal” means constant temperature, not constant heat. Heat flows freely in and out of the system throughout the process — that is precisely how the temperature is kept from changing.
Thermodynamics of Isothermal Process
The temperature of a system remains constant in an isothermal change, and Boyle’s law can be applied to study these changes. Therefore, the equation of state for an isothermal change is given by, \[pV = \text{constant} \qquad \text{--- (4.6)}\]
If \(p_i\), \(V_i\) and \(p_f\), \(V_f\) are the variables of a system in its initial and the final states respectively, then for an isothermal change, \[p_i V_i = p_f V_f = \text{constant}\]
This follows from the equation of state \(pV = nRT\) of Section 4.6.2: with \(n\) fixed and \(T\) fixed, the whole right-hand side is a constant, so the product \(pV\) must be a constant too.
Consider the isothermal expansion of an ideal gas. Let its initial volume be \(V_i\) and the final volume be \(V_f\).
The work done in an infinitesimally small isothermal expansion is given by Eq. (4.1), \(dW = p\,dV\). The total work done in bringing out the expansion from the initial volume \(V_i\) to the final volume \(V_f\) is given by, \[W = \int_{V_i}^{V_f} p\,dV\]
We cannot take \(p\) outside this integral, because the pressure falls continuously as the gas expands. Instead we must substitute for \(p\) in terms of \(V\).
But we know that for an ideal gas, \(pV = nRT\), so that \(p = nRT/V\). Using this in the previous equation we get, \[W = nRT \int_{V_i}^{V_f} \frac{dV}{V}\]
Here \(n\), \(R\) and \(T\) are all constants during an isothermal change, so they come outside the integral together. The remaining integral of \(dV/V\) is a standard one and gives a natural logarithm. \[\therefore W = nRT \ln \frac{V_f}{V_i} \qquad \text{--- (4.7)}\]
In this equation \(W\) is in joule (J), \(n\) is in mole (mol), \(R = 8.314\) J mol\(^{-1}\) K\(^{-1}\), \(T\) is in kelvin (K), and \(V_f\) and \(V_i\) are in cubic metre (m\(^3\)). Since the ratio \(V_f/V_i\) is a pure number, the logarithm is a pure number as well.
For an ideal gas, its internal energy depends on its temperature. Therefore, during an isothermal process, the internal energy of an ideal gas remains constant, \(\Delta U = 0\), because its temperature is constant, \(\Delta T = 0\).
The first law of thermodynamics, Eq. (4.4), when applied to an isothermal process would now read as, \[Q = W \qquad \text{--- (4.8)}\] \[\therefore Q = W = nRT \ln \frac{V_f}{V_i} \qquad \text{--- (4.9)}\]
Thus, the heat transferred to the gas is completely converted into the work done, that is, into the expansion of the gas. Nothing is retained inside as internal energy, because there is nowhere for it to go while the temperature is held fixed.
From Eq. (4.8) it is obvious that when the gas absorbs heat, it does positive work and its volume expands. When the gas is compressed, it releases heat and it does negative work.
Any change of phase occurs at a constant temperature, and therefore, it is an isothermal process.
Figure 4.15 shows the \(p\)-\(V\) diagram of an isothermal process. It is called an isotherm.
Diagram
Fig. 4.15 – A \(p\)-\(V\) diagram showing a smooth hyperbola-like isotherm falling from A at \((V_i, p_i)\) to B at \((V_f, p_f)\).
Isobaric Process
It is a constant pressure process. Boiling water at constant pressure, normally at atmospheric pressure, is an isobaric process.
Figure 4.16 shows the \(p\)-\(V\) diagram of an isobaric process. It is called an isobar.
Diagram
Fig. 4.16 – A \(p\)-\(V\) diagram showing a horizontal line from A at \(V_i\) to B at \(V_f\) at constant pressure, marked as an isobar.
The different curves shown on the maps provided by the meteorology department are also called isobars. They indicate the locations having the same pressure in a region.
For an isobaric process, none of the quantities \(\Delta U\), \(Q\) and \(W\) is zero. This makes it the most “complete” of the processes we study: heat enters, work is done, and the internal energy changes, all at the same time.
Thermodynamics of Isobaric Process
The pressure of a system remains constant in this process, that is \(\Delta p = 0\).
Consider an ideal gas undergoing volume expansion at constant pressure. If \(V_i\) and \(T_i\) are its volume and temperature in the initial state of a system, and \(V_f\) and \(T_f\) are its final volume and temperature respectively, the work done in the expansion is given by \[W = P\,dV = P\left(V_f - V_i\right) = nR\left(T_f - T_i\right) \qquad \text{--- (4.10)}\]
The last step uses the equation of state twice: \(PV_f = nRT_f\) and \(PV_i = nRT_i\) at the same pressure \(P\), so subtracting one from the other gives \(P(V_f - V_i) = nR(T_f - T_i)\). This form is useful when a problem gives temperatures rather than volumes.
Also, the change in the internal energy of a system is given by, \[\Delta U = nC_V \Delta T = nC_V \left(T_f - T_i\right) \qquad \text{--- (4.11)}\]
Where \(C_V\) is the specific heat at constant volume, measured in J mol\(^{-1}\) K\(^{-1}\), and \(\Delta T = (T_f - T_i)\) is the change in its temperature during the isobaric process, in kelvin (K).
According to the first law of thermodynamics, the heat exchanged is given by, \[Q = \Delta U + W\]
Using the previous two equations we get, \[Q = nC_V\left(T_f - T_i\right) + nR\left(T_f - T_i\right)\] \[Q = \left(nC_V + nR\right)\left(T_f - T_i\right)\] \[Q = nC_P\left(T_f - T_i\right) \qquad \text{--- (4.12)}\]
Where \(C_P\) is the specific heat at constant pressure, and \[\therefore C_P = C_V + R\]
This last relation has a clear physical meaning. To raise the temperature of a gas by one kelvin at constant pressure, we must supply not only the energy needed to raise its internal energy, which is \(C_V\), but also the extra energy the gas spends in pushing back its surroundings as it expands, which is \(R\). That is why \(C_P\) is always greater than \(C_V\).
Equation (4.12) tells us that the temperature of a system changes in an isobaric process, and therefore its internal energy also changes, as given by Eq. (4.11).
The heat exchanged, Eq. (4.12), is partly used for increasing the temperature and partly to do some work. The change in the temperature of the system depends on the specific heat at constant pressure \(C_P\).
Isochoric Process
It is a constant volume process. A system does no work on its environment during an isochoric change.
Figure 4.17 shows the \(p\)-\(V\) diagram of an isochoric process.
Diagram
Fig. 4.17 – A \(p\)-\(V\) diagram showing a vertical line from A at \((V, P_i)\) down to B at \((V, P_f)\) at constant volume.
Notice in Fig. 4.17 that the path is a vertical line, and a vertical line has no area beneath it — which is the graphical statement of the fact that no work is done.
For an isochoric process, \(\Delta V = 0\), and we have, from the first law of thermodynamics, \[\Delta U = Q\]
This means that for an isochoric change, all the energy added in the form of heat remains in the system itself and causes an increase in its internal energy. Nothing is spent on pushing back the surroundings.
Heating a gas in a constant volume container, or diffusion of a gas in a closed chamber, are some examples of an isochoric process.
Thermodynamics of Isochoric Process
For an isochoric process, we have \(\Delta V = 0\). The system does not do any work and all the energy supplied to the system is converted into its internal energy.
The first law of thermodynamics for an isochoric process is \[Q = \Delta U \qquad \text{--- (4.13)}\]
The change in internal energy is given by \[\Delta U = nC_V \Delta T\]
The work done is given by \[W = p\Delta V = 0 \quad (\text{because } \Delta V = 0)\]
The heat exchanged is given by the first law of thermodynamics, \[Q = \Delta U + W = \Delta U = nC_V \Delta T \qquad \text{--- (4.14)}\]
Adiabatic Process
It is a process during which there is no transfer of heat from or to the system.
Figure 4.18 shows the \(p\)-\(V\) diagram of an adiabatic process.
Diagram
Fig. 4.18 – A \(p\)-\(V\) diagram showing an adiabatic curve falling from an initial point to a final point, drawn steeper than an isotherm through the same initial point.
For an adiabatic change, \(Q = 0\). Heat transfer to or from the system is prevented by either perfectly insulating the system from its environment, or by carrying out the change rapidly so that there is no time for any exchange of heat.
The second route is the important one in practice. Heat takes time to flow; if the change is completed in a fraction of a second, hardly any heat can cross the boundary even if the walls are excellent conductors.
Puncturing an inflated balloon or a tyre are some familiar examples of adiabatic changes.
For an adiabatic change, putting \(Q = 0\) into Eq. (4.3) gives \[\Delta U = -W \qquad \text{--- (4.15)}\]
When a system expands adiabatically, \(W\) is positive — work is done by the system — and \(\Delta U\) is negative, so the internal energy of the system decreases.
When a system is compressed adiabatically, \(W\) is negative — work is done on the system — and \(\Delta U\) is positive. The internal energy of the system increases in this adiabatic process.
It is observed that for many systems, temperature increases when internal energy increases, and decreases when the internal energy is decreased.
Thermodynamics of Adiabatic Process
For an adiabatic process we have, \[PV^{\gamma} = \text{constant} = C \qquad \text{--- (4.16)}\]
where \(\gamma\) is the ratio of the specific heat at constant pressure to the specific heat at constant volume, that is, \[\gamma = \frac{C_P}{C_V}\]
\(\gamma\) is also called the adiabatic ratio. Since \(C_P\) is always greater than \(C_V\), as we saw in the isobaric process, \(\gamma\) is always greater than 1. It is a pure number with no units.
For moderate temperature changes, the value of \(\gamma\) is \(\dfrac{5}{3}\) for monoatomic gases, \(\dfrac{7}{5}\) for diatomic gases, and \(\dfrac{8}{6} = \dfrac{4}{3}\) for triatomic gases. The values have been calculated in Chapter 3; Equations (3.30), (3.33) and (3.39) can be extended to obtain these values.
An adiabatic system is thermally isolated from its environment, and therefore it cannot exchange heat with it. Therefore, when a system undergoes an adiabatic change, its temperature and internal energy both change.
The change in internal energy is, \[\Delta U = C_V (\Delta T) \qquad \text{--- (4.17)}\]
The work done is, \[W = \int_{V_i}^{V_f} P\,dV\]
Using Eq. (4.16), which gives \(P = C/V^{\gamma}\), we have, \[W = C \int_{V_i}^{V_f} \frac{dV}{V^{\gamma}}\]
Integrating the power of \(V\) in the usual way, \[W = C\left[\frac{V^{1-\gamma}}{1-\gamma}\right]_{V_i}^{V_f}\]
where \(V\) changes from \(V_i\) to \(V_f\). Substituting the limits, \[W = \frac{C}{1-\gamma}\left[V_f^{\,1-\gamma} - V_i^{\,1-\gamma}\right] \qquad \text{--- (4.18)}\]
From Eq. (4.16) we have, \[pV^{\gamma} = C\] \[\text{or,} \quad p_i V_i^{\gamma} = p_f V_f^{\gamma}\]
We now use this to replace the constant \(C\) separately in each of the two terms, writing \(C = p_f V_f^{\gamma}\) in the first and \(C = p_i V_i^{\gamma}\) in the second. Therefore, we can write Eq. (4.18) as, \[W = \frac{1}{1-\gamma}\left[p_f V_f^{\gamma} V_f^{\,1-\gamma} - p_i V_i^{\gamma} V_i^{\,1-\gamma}\right] \qquad \text{--- (4.19)}\]
Adding the powers of \(V\) in each bracket, \(\gamma + (1-\gamma) = 1\), so each product collapses to a simple \(pV\) term, \[W = \frac{1}{1-\gamma}\left[p_f V_f - p_i V_i\right]\] \[W = \frac{p_f V_f - p_i V_i}{1-\gamma} \qquad \text{--- (4.20)}\]
Equation (4.20) implies that when work is done by the gas, that is when the gas expands, \(W > 0\), and \(T_i > T_f\). This means that the gas will cool down.
Similarly, if the work is done on the gas, that is if the gas is compressed, \(W < 0\), and \(T_i < T_f\). This means that the gas will warm up.
Cyclic Process
A thermodynamic process that returns a system to its initial state is a cyclic process. In this process, the initial and the final state is the same.
Figure 4.19 shows the \(p\)-\(V\) diagram of a cyclic process.
Diagram
Fig. 4.19 – A \(p\)-\(V\) diagram showing a closed loop from A at \((V_i, P_i)\) to B at \((V_f, P_f)\) along one path and back along another, with arrows marking the direction of travel.
For a cyclic process, the total change in the internal energy of a system is zero, \(\Delta U = 0\). This must be so, because internal energy is a state variable and the final state is identical to the initial one.
According to the first law of thermodynamics, we have, for a cyclic process, \[Q = W \qquad \text{--- (4.21)}\]
This single sentence explains why the cyclic process matters more than any other in the rest of this chapter. An engine that is to run continuously must return to its starting state at the end of every cycle, ready to begin again.
Free Expansion
These expansions are adiabatic expansions, and there is no exchange of heat between a system and its environment. Also, there is no work done on the system or by the system.
Therefore \(Q = W = 0\), and according to the first law of thermodynamics, \[\Delta U = 0\]
For example, when a balloon is ruptured suddenly, or a tyre is suddenly punctured, the air inside rushes out rapidly but there is no displacement of a piston or any other surface.
Free expansion is different from the other thermodynamic processes we have discussed so far, because it is an uncontrolled change. It is an instantaneous change and the system is not in thermodynamic equilibrium.
A free expansion cannot be plotted on a \(p\)-\(V\) diagram. Only its initial and the final state can be plotted, as isolated points with nothing joining them — there is no path to draw, because the states in between simply do not exist as equilibrium states.
Working from a V-T diagram to a \(p\)-\(V\) diagram in this way is a common examination task. The method is always the same: identify which variable is held constant along each leg by using \(pV = nRT\), name the process, and then draw the leg with the shape that belongs to that process.
4.8 Heat Engines
As mentioned earlier, thermodynamics is related to the study of different processes involving conversion of heat and work into each other.
In this section, we will study the practical machines that convert some heat into work. These are heat engines.
4.8.1 Heat Engine
Heat engines are devices that transform heat partly into work or mechanical energy. Heat engines work by using cyclic processes and involve thermodynamic changes. Automobile engines are familiar examples of heat engines.
A heat engine receives heat from a source called a reservoir and converts some of it into work.
Remember that all the heat absorbed is not converted into work by a heat engine. Some heat is lost in the form of exhaust. The word partly in the definition is not an admission of poor engineering; as we shall see in Section 4.10, it is a requirement of nature.
The Elements of a Heat Engine
A typical heat engine has the following elements.
A working substance. It is called the system. It can be an ideal gas for an ideal heat engine, to be discussed later. For a practical heat engine, the working substance can be a mixture of fuel vapour and air in a gasoline (petrol) or diesel engine, or steam in a steam engine. It is the working substance that absorbs heat and does work.
Hot and cold reservoirs. The working substance interacts with the reservoirs.
The hot reservoir is the source of heat. It is at a relatively high temperature and is capable of providing a large amount of heat at a constant higher temperature, \(T_H\). It is also called the source.
The cold reservoir absorbs a large amount of heat from the working substance at a constant lower temperature, \(T_C\). It is also called the sink.
Both reservoirs are taken to be large enough that their temperatures do not change even though heat is drawn from one and dumped into the other, cycle after cycle. This is the idea we met in Section 4.4.3: a sufficiently large environment holds its temperature fixed.
A cylinder. Generally, the working substance is enclosed in a cylinder with a moving, frictionless and massless piston.
The working substance does some work by displacing the piston in the cylinder. This displacement is transferred to the environment using some arrangement such as a crank shaft, which transfers mechanical energy to the wheels of a vehicle.
Two Basic Types
Heat engines are of two basic types. They differ in the way the working substance absorbs heat.
In an external combustion engine, the working substance is heated externally, as in the case of a steam engine. The fuel burns outside the cylinder, and its heat is passed to the working substance through a wall.
In the case of the internal combustion engine, the working substance is heated internally, similar to an automobile engine using gasoline or diesel. Here the fuel burns within the working substance itself, so no wall stands between the flame and the gas that does the work.
The Three Basic Steps
Any heat engine works in the following three basic steps.
- The working substance absorbs heat from a hot reservoir at a higher temperature.
- Part of the heat absorbed by the working substance is converted into work.
- The remaining heat is transferred to a cold reservoir at a lower temperature.
Heat engines are classified according to the working substance used and the way these steps are actually implemented during their operation.
Heat engines are diagrammatically represented by an energy flow diagram, schematically shown in Fig. 4.20. Energy exchange takes place during the various stages of working of a heat engine.
Diagram
Fig. 4.20 – Energy flow diagram of a heat engine: a hot reservoir at \(T_H\) at the top feeding a wide input pipeline \(Q_H\) into a circle marked “Engine”, a side branch of work \(W\) leaving, and a narrower output pipeline \(Q_C\) going down into a cold reservoir at \(T_C\).
Let \(Q_H\) be the heat absorbed by the working substance at the source, and \(Q_C\) be the heat rejected by it at the sink. In a heat engine, \(Q_H\) is positive and \(Q_C\) is negative. Also, let \(W\) be the work done by the working substance.
The signs follow the convention of Section 4.4.3 without any change: heat entering the system is positive, heat leaving it is negative.
Reading the Energy Flow Diagram
In Fig. 4.20, the circle represents the engine. The ‘heat pipelines’ shown in the diagram represent the heat absorbed, rejected, and converted into work.
The width of the heat ‘pipeline’ indicated by \(Q_H\) is proportional to the amount of heat absorbed at the source.
The width of the branch indicated by \(Q_C\) is proportional to the magnitude \(|Q_C|\) of the amount of heat rejected at the sink.
The width of the branch of the pipeline indicated by \(W\) is proportional to the part of the heat converted into mechanical work.
This is a picture worth studying, because it makes the energy accounting visible. The incoming pipeline splits into two, and nothing is lost at the junction — the widths of the two outgoing branches must add up to the width of the one coming in.
One single execution of the steps mentioned above is one operating ‘cycle’ of the engine. Several such cycles are repeated when a heat engine operates.
The quantities \(Q_H\) and \(Q_C\) represent the amount of heat absorbed (positive) and rejected (negative) respectively during one cycle of operation.
Heat and Work in One Cycle
The net heat \(Q\) absorbed per operating cycle is, \[Q = Q_H + Q_C = |Q_H| - |Q_C| \qquad \text{--- (4.22)}\]
The two forms agree because \(Q_C\) is itself a negative quantity: adding a negative number is the same as subtracting its magnitude.
The net work done in one operating cycle, by the working substance, is given by using the first law of thermodynamics, \[W = Q = |Q_H| - |Q_C| \qquad \text{--- (4.23)}\]
Equation (4.23) is simply Eq. (4.21) applied to an engine. Over one complete cycle the working substance returns to its starting state, so \(\Delta U = 0\), and the whole of the net heat appears as work.
Efficiency of a Heat Engine
Ideally, we would expect a heat engine to convert all the heat absorbed, \(Q_H\), into work. Practically, this is not possible. There is always some heat lost, that is, \(Q_C \neq 0\).
The thermal efficiency \(\eta\) of the heat engine is defined as, \[\eta = \frac{W}{Q_H} \qquad \text{--- (4.24)}\]
Thus, the thermal efficiency, or simply the efficiency, of a heat engine is the ratio of the work done by the working substance and the amount of heat absorbed by it. It is the ratio of the output, in the form of the work done \(W\) by the engine, to the input, in the form of the heat supplied \(Q_H\).
In simple words, the efficiency of a heat engine is the fraction of the heat absorbed that is converted into work, as given by Eq. (4.24).
In terms of the energy flow diagram of Fig. 4.20, the ‘pipeline’ representing the work is as wide as possible, and the pipeline representing the exhaust is as narrow as possible, for the most efficient heat engine.
There is a fundamental limit on the efficiency of a heat engine set by the second law of thermodynamics, which we will discuss later.
Using Eq. (4.22) and Eq. (4.23), we can write the efficiency of a heat engine as, \[\eta = \frac{W}{Q_H} = 1 - \frac{|Q_C|}{|Q_H|} \qquad \text{--- (4.25)}\]
This form is often the more useful one in problems, because it needs only the two heats and not the work. It also states the position plainly: the efficiency falls short of 1 by exactly the fraction of the input heat that is dumped into the sink.
Equation (4.25) gives the thermal efficiency of a heat engine. It is a ratio of the quantities which represent energy. Therefore, it has no units, but we must express \(W\), \(Q_H\) and \(Q_C\) in the same units.
4.8.2 The Heat Engine Cycle and the p-V Diagram
As discussed previously, the working of a heat engine is a well defined sequence of operations. It is a cyclic thermodynamic process.
We know that a thermodynamic process can be represented by a \(p\)-\(V\) diagram. We will now discuss the \(p\)-\(V\) diagram of a heat engine.
Keep in mind that this is a \(p\)-\(V\) diagram of a general heat engine. There are different ways of operating a heat engine, and we will discuss some such heat engines in the following sections.
The Idea Behind the Cycle
A heat engine uses energy absorbed in the form of heat to do work, and then rejects the heat which cannot be used to do work.
The three tasks are not carried out all at once. Heat is absorbed in one part of the cycle, work is done in another part, and the unused heat is rejected in yet another part of the cycle.
The \(p\)-\(V\) diagram of a typical heat engine is shown in Fig. 4.21.
Diagram
Fig. 4.21 – A closed \(p\)-\(V\) loop of a typical heat engine with four corners marked A, B, C and D: A to B vertical (pressure rises at constant volume), B to C expansion, C to D vertical (pressure falls at constant volume), D to A compression back to the start.
Walking Around the Cycle
The operating cycle begins at the point A in the cycle.
The working substance, the gas in this case, absorbs heat at constant volume, and no work is done by the gas or on the gas. This is the isochoric process of Section 4.7.3, in which \(W = 0\) and \(Q = \Delta U\).
The pressure is increased till the point B is reached. The temperature of the gas also increases and its internal energy increases. On the diagram, this stage is the vertical line A to B.
The gas then starts expanding by pushing the piston away, and its volume changes from the point B to the point C. Because the gas expands, its pressure is reduced.
The gas does work in this part of the cycle. This is the stage from which the engine delivers its useful output, and it takes place while the gas is at its highest pressures — which, as we saw in Section 4.7.1, is exactly the condition for the work delivered to be large.
When the point C is reached, the excess heat — the heat that is not utilized in doing work by the gas — is rejected. The gas cools down and its internal energy decreases. This process is again at constant volume.
The pressure of the gas is reduced, and point D on the \(p\)-\(V\) diagram is reached.
The gas is now compressed. Its volume decreases and its pressure increases. The change continues till the point A is reached.
The cycle is complete, and the system is ready for the next cycle.
Why the Diagram Is So Useful
Thus, the \(p\)-\(V\) diagram is a visual tool for the study of heat engines. The working substance of a heat engine is usually a gaseous mixture.
Study of the \(p\)-\(V\) diagram helps us understand the behaviour of the three state variables of a gas throughout the operational cycle.
The operation of a heat engine is a cyclic process, and therefore its \(p\)-\(V\) diagram is a closed loop. The area of the loop represents the work done during one complete cycle.
Since work is done by the gas, or on the gas, only when its volume changes, the \(p\)-\(V\) diagram provides a visual interpretation of the work done during one complete cycle.
The reasoning is the one established in Section 4.6.3. The area under the upper part of the loop is the work done by the gas as it expands; the area under the lower part is the work done on the gas as it is compressed; and what remains between them — the enclosed area — is the difference, which is the net work.
Similarly, the internal energy of the gas depends upon its temperature. Hence, the \(p\)-\(V\) diagram, along with the temperatures calculated from the ideal gas law, determines the changes in the internal energy of the gas.
Once the work and the change in internal energy are known for each stage, we can calculate the amount of heat added or rejected from the first law of thermodynamics.
Thus, a \(p\)-\(V\) diagram helps us analyse the performance of any heat engine which uses a gas as its working substance.
4.9 Refrigerators and Heat Pumps
So far, we have discussed a heat engine which takes heat from a source at higher temperature and rejects it to a sink at lower temperature.
The input provided to the working substance in a heat engine — a gas, or a mixture of gasoline and air — is in the form of heat, which is converted into mechanical work as output. Figure 4.20 shows this in the form of an energy flow diagram.
Refrigerators and heat pumps are heat engines that work in the backward direction. They convert mechanical work into heat.
In other words, every arrow in the energy flow diagram of Fig. 4.20 is reversed. Work is now supplied rather than delivered, heat is drawn from the cold side rather than the hot side, and heat is delivered to the hot side rather than taken from it.
4.9.1 Heat Flow from a Colder Region to a Hotter Region
According to the second law of thermodynamics, to be discussed in the next article, heat cannot flow from a region of lower temperature to a region of higher temperature on its own.
The words “on its own” carry the whole meaning. They do not say that such a transfer is impossible; they say that it never happens spontaneously, without something else being paid for it.
We can force heat to flow from a region of lower temperature to a region of higher temperature by doing work on the system, that is, on the working substance of a heat engine.
Refrigerators or air-conditioners and heat pumps are examples of heat engines which cause heat to be transferred from a cold region to a hot region.
How the Transfer Is Actually Achieved
Usually, this is achieved with the aid of a phase change of a fluid, called the refrigerant.
The refrigerant is forced to evaporate and then condense by successively decreasing and increasing its pressure. It can, therefore, ‘pump’ energy from a region at lower temperature to a region of higher temperature.
The reason a phase change is used goes back to the latent heat you studied in Std. XI. When a liquid evaporates, it must absorb a large quantity of energy — its latent heat of vaporization — without any rise in temperature, and it takes that energy from whatever surrounds it. When the vapour condenses again, exactly that quantity of energy is released. A substance that can be made to evaporate in one place and condense in another is therefore an excellent carrier of heat.
It extracts the heat of vaporization of the refrigerant from the cold region and rejects it to the hotter region outside the refrigerator. This results in cooling down the cold region further.
The Mechanism in Outline
Figure 4.22 shows the concept of transferring heat from a cold region to a hot region in a schematic way.
Diagram
Fig. 4.22 – Schematic diagram of heat being transferred from a cold region on the left to a hot region on the right by a circulating refrigerant, with work supplied to the loop in between.
Heat from the cold region is carried to the hot region by the refrigerant. It extracts heat from the cold region due to forced evaporation.
The heat of evaporation of the refrigerant thus absorbed is rejected by compressing and condensing it into liquid at a higher temperature.
All this process is carried out in a mechanism involving a compressor and closed tubing, such as is seen at the back of a household refrigerator. The tubing is closed because the same refrigerant is used over and over again — it evaporates, is compressed, condenses, expands, and returns to evaporate once more, cycle after cycle.
4.9.2 Refrigerator
Refrigeration is a process of cooling a space or substance of a system, and of maintaining its temperature below its ambient temperature. In simple words, refrigeration is artificial cooling.
Note that a refrigerator has two jobs, not one. It must first bring the contents down to a low temperature, and then keep them there in spite of the heat that leaks in continuously from the warmer room outside. This is why a refrigerator switches on and off repeatedly instead of running once and stopping.
A refrigerator extracts heat from a cold region — the inside of the chamber, or the compartments — and delivers it to the surrounding, which is the atmosphere, thus further cooling the cold region.
That is the reason why, if you place your hand behind a working refrigerator, you can feel the warm air. But the interior of the refrigerator is cold. An air conditioner also works on similar principles.
Inside the Machine
Figure 4.23 (a) shows the schematics of the mechanism used in a typical refrigerator.
Diagram
Fig. 4.23 (a) – Schematic of a refrigerator: a closed loop of tubing carrying refrigerant, with the cooling coil inside the chamber, an expansion valve, and the condenser coil outside, with a compressor in the loop.
It consists of a compressor, an expansion valve, and a closed tube which carries the refrigerant.
Part of the tube, called the cooling coil, is in the region which is to be cooled, at lower temperature and lower pressure. The other part, which is exposed to the surrounding — generally, the atmosphere — is at a higher temperature and higher pressure.
Notice in Fig. 4.23 (a) that the compressor and the expansion valve sit at the two junctions between these regions. They are the two devices that create and maintain the pressure difference, and everything else follows from it.
A fluid such as isobutane is used as refrigerant. Normally, the cold and the hot part of the coil contain the refrigerant as a mixture of liquid and vapour phase in equilibrium.
This mixed state is deliberate. A liquid in equilibrium with its own vapour stays at a fixed temperature while it boils or condenses, which is exactly what is wanted: the cooling coil holds a steady low temperature, and the outside coil a steady high one.
The Energy Flow Diagram
Figure 4.23 (b) shows the energy flow diagram of a refrigerator.
Diagram
Fig. 4.23 (b) – Energy flow diagram of a refrigerator: a narrow pipeline \(Q_C\) rising from the cold reservoir into a circle, a work input \(W\) joining it from the side, and a wider pipeline \(Q_H\) leaving into the hot reservoir.
As you can see, the heat extracted from the cold reservoir is supplemented by the mechanical work done on the refrigerant by the compressor, and the total energy is rejected at the hot reservoir.
Compare this with Fig. 4.20. There, the wide pipeline was at the top and the work branched away from it; here, the wide pipeline is at the bottom, and the work joins the flow instead of leaving it. This is the meaning of “an engine running backwards”, and it also tells us that the heat delivered to the room must be greater than the heat removed from the food — by exactly the amount of work supplied.
The Four Steps of One Cycle
The refrigerant goes through the following steps in one complete cycle of refrigeration.
Step 1. The fluid passes through a nozzle and expands into a low-pressure area. Similar to the way carbon dioxide comes out of a fire extinguisher and cools down, the fluid turns into a gas and cools down. This is essentially an adiabatic expansion.
The cooling here is the adiabatic cooling of Section 4.7.3: the fluid expands too quickly for heat to enter, so it does work at the expense of its own internal energy and its temperature drops. This step is what makes the refrigerant colder than the inside of the fridge.
Step 2. The cool gas is in thermal contact with the inner compartment of the fridge. It heats up as heat is transferred to it from the contents of the fridge. This takes place at constant pressure, so it is an isobaric expansion.
This is the step that actually does the cooling. Heat flows in the ordinary direction — from the warmer contents of the fridge to the colder refrigerant — because Step 1 has made the refrigerant the colder of the two.
Step 3. The gas is transferred to a compressor, which does most of the work in this process. The gas is compressed adiabatically, heating it and turning it back to a liquid.
Here the work supplied from outside enters the cycle. Adiabatic compression raises the temperature of the refrigerant, and it must be raised above the temperature of the room, or the next step could not happen.
Step 4. The hot liquid passes through coils on the outside of the fridge, and heat is transferred to the atmosphere. This is an isobaric compression process.
Again the heat flows in its natural direction, from the hot refrigerant to the cooler room air. These are the coils that feel warm to the hand. The refrigerant, having given up its load, returns to the nozzle and the cycle begins once more.
The compressor is driven by an external energy source, and it does the work \(|W|\) on the working substance during each cycle.
4.9.3 Performance of a Refrigerator
Consider the energy flow diagram of a refrigerator, Fig. 4.23 (b). It shows the relation between the work and heat involved in transferring heat from a low temperature region to a high temperature region.
This is a cyclic process, in which the working substance — the refrigerant in this case — is taken back to the initial state.
Setting Up the Signs
For a refrigerator, the heat absorbed by the working substance is \(Q_C\) and the heat rejected by it is \(Q_H\).
A refrigerator absorbs heat at lower temperature and rejects it at higher temperature. Therefore, we have \(Q_C > 0\), \(Q_H < 0\), and \(W < 0\).
Hence, we write \(|W|\) for the magnitude of the work supplied, and \(|Q_H| = -\,Q_H\) for the magnitude of the heat rejected.
Compare these signs with those of the heat engine in Section 4.8.1, where \(Q_H\) was positive and \(Q_C\) negative. Every sign has been reversed, which is precisely what “running the cycle backwards” means.
In this case, we apply the first law of thermodynamics to the cyclic process. For a cyclic process, the internal energy of the system in the initial state and the final state is the same. Therefore, from Eq. (4.21), we have, \[Q_H + Q_C = W\] or \[Q_H + Q_C - W = 0\] \[\therefore\ Q_H = W - Q_C\]
For a refrigerator, \(Q_H < 0\) and \(W < 0\). Therefore, writing the same relation in terms of magnitudes, \[|Q_H| = |Q_C| + |W| \qquad \text{--- (4.26)}\]
From Fig. 4.23 (b) we realize that the heat \(|Q_H|\) rejected by the working substance at the hot reservoir is always greater than the heat \(Q_C\) received by it at the cold reservoir.
Equation (4.26) says this in the plainest possible way. Everything that is taken from the cold side, together with everything supplied by the compressor, must be delivered to the hot side, since nothing can accumulate inside a system that returns to its initial state each cycle.
Note that Eq. (4.26), derived for a refrigerator, and Eq. (4.23), derived for a heat engine, are the same. They are valid for a heat engine and also for a refrigerator.
The Coefficient of Performance
The ratio \(\dfrac{|Q_C|}{|W|}\) indicates the performance of a refrigerator, and is called the coefficient of performance (CoP), \(K\), or the quality factor, or the Q-value of a refrigerator.
The larger the ratio, the better the refrigerator. That means a refrigerator has the best performance when the heat extracted by the refrigerant at the cold reservoir is maximum by doing minimum work in one operating cycle.
Notice how the definition is built. The numerator is what we want from the machine — heat removed from the food. The denominator is what we pay for it — electrical work at the compressor. This is the same “output over input” idea used for the efficiency of a heat engine in Eq. (4.24), applied to a machine with a different purpose.
From Eq. (4.26), \[|W| = |Q_H| - |Q_C|\]
\[\therefore\ K = \frac{|Q_C|}{|W|} = \frac{|Q_C|}{|Q_H| - |Q_C|} \qquad \text{--- (4.27)}\]
All the quantities on the right side of Eq. (4.27) represent energy and are measured in the same energy units. The coefficient of performance, \(K\), of a refrigerator is, therefore, a dimensionless number.
For a typical household refrigerator, \(K \approx 5\).
In numerical questions on this topic, the commonest error is to divide the wrong pair of quantities. Keep the roles clear: for an engine, the useful output is the work and the input is \(|Q_H|\); for a refrigerator, the useful output is \(|Q_C|\) and the input is the work.
4.9.4 Air Conditioner
The working of an air conditioner and a refrigerator is exactly similar. It differs from a refrigerator only in the volume of the chamber or room it cools down.
The physics is therefore nothing new. What changes is the scale of the job and the placement of the two coils, and the second of these is worth attention.
For an air conditioner, the evaporator coils are inside the room that is to be cooled, and the condenser is outside the room, as shown in Fig. 4.36.
Diagram
Fig. 4.36 – Schematic of a room air conditioner: evaporator coil and fan inside the room drawing heat from the room air, compressor and condenser coil outside the wall rejecting heat to the atmosphere.
The air cooled by the evaporator coils inside the room is circulated by a fan placed inside the air conditioning unit.
Notice in Fig. 4.36 that the cold coil and the hot coil are separated by the wall of the room. This is the essential difference from a refrigerator, whose cold coil is inside its cabinet while its hot coil sits in the very room the cabinet stands in. An air conditioner delivers its rejected heat to the outdoors, and that is why it can cool a room while a refrigerator with its door open cannot.
Measuring the Performance
The performance of an air conditioner is defined by \[K = \frac{|Q_C|}{|W|}\]
This is the same coefficient of performance defined for a refrigerator in Eq. (4.27), with the same meaning: heat removed divided by work supplied.
For an air conditioner, however, it is important to consider the rate at which heat is removed, \(H\), and the power \(P\) required for removing the heat. A machine that removes a great deal of heat very slowly is of no use in a warm room, where heat is leaking in continuously through the walls and windows.
We define the rate of heat removed as the heat current \[H = \frac{|Q_C|}{t}\] where \(t\) is the time in which the heat \(|Q_C|\) is removed. Here \(H\) is measured in joule per second, that is in watt (W).
Therefore, the coefficient of performance of an air conditioner can be calculated as, \[K = \frac{|Q_C|}{|W|} = \frac{Ht}{Pt} = \frac{H}{P} \qquad \text{--- (4.28)}\]
Here \(P\) is the electrical power supplied to the machine, also measured in watt (W). The time \(t\) cancels between the numerator and the denominator, which is why the same number \(K\) describes the machine whether we compare energies over one cycle or rates over any period we please.
Typical values of \(K\) are 2.5 to 3.0 for room air conditioners.
4.9.5 Heat Pump
A heat pump is a device which works similar to a refrigerator. It is used to heat a building or a similar larger structure by cooling the air outside it.
A heat pump works like a refrigerator operating inside out. In this case, the evaporator coils are outside and absorb heat from the cold air from outside. The condenser coils are inside the building. They release the absorbed heat to the air inside the room, thus warming the building.
Diagram
Fig. 4.37 – Schematic of a heat pump: evaporator coil outdoors absorbing heat from cold outside air, compressor, and condenser coil indoors releasing heat into the room.
Compare Fig. 4.37 with the air conditioner of Fig. 4.36 and you will see that the two coils have simply exchanged places. Many modern units do exactly this by means of a valve that reverses the direction of the refrigerant, so that the same machine cools a room in summer and warms it in winter.
4.10 Second Law of Thermodynamics
We now come to the law that has been anticipated at several points in this chapter — in the one-way flow of heat from hot to cold, in the impossibility of a perfect engine, and in the work that must always be paid to run a refrigerator.
The subsections that follow first examine what the first law leaves undecided, and then state the second law itself in its two standard forms.
4.10.1 Limitations of the First Law of Thermodynamics
The First law of thermodynamics tells us that heat can be converted into work, and work can also be converted into heat.
It is merely a quantitative statement of the equivalence of heat and work. The word merely is important here: the law tells us how much, but it never tells us whether. It has the following limitations.
It Does Not Tell Us Which Direction Nature Takes
- It does not tell us whether any particular process can actually occur.
According to the first law of thermodynamics, heat may, on its own, flow from an object at higher temperature to one at lower temperature, and it can also, on its own, flow from an object at lower temperature to one at higher temperature.
Both possibilities balance the books equally well. If a hot body loses 100 J and a cold body gains 100 J, energy is conserved; and if a cold body loses 100 J and a hot body gains 100 J, energy is conserved just as exactly.
We know that practically, heat cannot flow from an object at lower temperature to another at higher temperature. The First law of thermodynamics does not predict this practical observation.
It Does Not Limit How Much Heat Can Become Work
- According to the First law, we could convert all — that is 100% — of the heat available to us into work. Similarly, all the work could be converted into heat.
Again, we know that practically this is not possible.
Notice that the two halves of this statement are not on the same footing in practice. Converting work entirely into heat is easy; rubbing your palms together, as in Section 4.4.4, does it perfectly. It is the other direction that nature resists.
Thus, the First law of thermodynamics does not prevent us from converting heat entirely into work, or work entirely into heat.
These limitations lead to the formulation of another law of thermodynamics, called the Second law of thermodynamics.
What Experience Actually Shows
We have seen earlier in Section 4.7.3 that an irreversible process defines the preferred direction of a thermodynamic process. Heat flows one way on its own; friction turns motion into warmth and never the reverse.
It is also found that it is impossible to build a heat engine that has 100% efficiency, as defined in Eq. (4.25). That is, it is not possible to build a heat engine that can completely convert heat into work. In the language of Section 4.8.1, \(Q_C\) can be made small but never zero.
Similarly, for a refrigerator, it is impossible to remove heat without doing any work on a system. That is, the coefficient of performance, Eq. (4.28), of a refrigerator can never be infinite.
These practical observations form the basis of a very important principle of thermodynamics, the Second law of thermodynamics.
An Analogy: Height and Temperature
Consider an object A at a certain height \(h\) above the ground, and another object B of the same mass at a height \(h/2\), as shown in Fig. 4.24 (a).
Diagram
Fig. 4.24 (a) – Two identical objects, A at height \(h\) above the ground and B at height \(h/2\), with the ground marked as the common reference level.
We know that the potential energy of the object B is half that of the object A. That means we can extract only half the energy from the object B.
The reason is that what can be extracted depends not on the height alone but on the drop available — on how far the object can fall before it reaches the ground and can fall no further.
Similarly, if a heat engine as shown in Fig. 4.24 (b) operates between the temperatures of 800 K and 400 K — that is, if it receives heat at 800 K and rejects it at 400 K — its maximum efficiency can be 50%.
Diagram
Fig. 4.24 (b) – A heat engine drawn between a hot reservoir at 800 K and a cold reservoir at 400 K, with its maximum efficiency marked as 50%.
Compare the two parts of Fig. 4.24. Heat “falls” from a high temperature to a low one much as an object falls from a height, and only the portion of the fall that actually takes place inside the engine can be harnessed. Heat arriving at 800 K and leaving at 400 K has descended only half the way to absolute zero, and at most half of it can be converted into work.
We will discuss the Second law of thermodynamics in the next subsection.
4.10.2 The Second Law of Thermodynamics, Statement
We now know that heat can be converted into work by using a heat engine. However, our practical experience says that the entire heat supplied to the working substance can never be converted into mechanical work.
The Second law of thermodynamics helps us to understand this. It exists in two standard forms, one framed around engines and the other around refrigerators, and we shall take them in turn.
The Engine Statement
According to the second law of thermodynamics:
Read the statement carefully and notice how modest it is. It does not say that engines are inefficient because of friction, or poor materials, or careless design. It says that a certain quantity of heat must be thrown away, no matter how perfect the machine.
Figure 4.25 is a diagrammatic representation of the application of the Second law of thermodynamics to a heat engine.
Diagram
Fig. 4.25 (a) – A “perfect” heat engine taking \(Q_H\) from the hot reservoir and delivering all of it as work \(W\), with a large cross drawn through the diagram to show that it is impossible.
As you can see from the diagram, Fig. 4.25 (a), all heat engines lose some heat to the environment, unlike the perfect heat engine shown there. All the heat \(Q_H\) extracted cannot be used to do work.
Figure 4.25 (b) shows the energy flow diagram for such a situation.
Diagram
Fig. 4.25 (b) – Energy flow diagram of the engine statement: \(Q_H\) entering the engine from the hot reservoir, work \(W\) leaving, and an unavoidable \(Q_C\) passing on to the cold reservoir.
Compare Fig. 4.25 (a) with Fig. 4.25 (b). The forbidden machine has only one outgoing branch; every real engine must have two. In the language of Eq. (4.25), \(|Q_C|\) can never be zero, and so \(\eta\) can never reach 1.
This form of statement of the Second law of thermodynamics is called the Kelvin-Planck statement, or the ‘First form’ of the Second law of thermodynamics. Sometimes it is also called the ‘Engine Law’ or the ‘Engine Statement’ of the Second law of thermodynamics.
The Refrigerator Statement
We have seen how the efficiency of a heat engine is restricted by the second law of thermodynamics. A heat engine is one form of ‘heat – work conversion’. Let us see what happens in the case of a refrigerator, the other form of ‘work – heat conversion’.
This means that energy will not flow spontaneously from an object at low temperature to an object at a higher temperature. This rules out the possibility of a perfect refrigerator.
The statements about refrigerators are also applicable to air conditioners and heat pumps, which work on the same principles.
This is the ‘Second form’, or Clausius statement, of the Second law of thermodynamics. Sometimes it is also called the ‘Refrigerator Law’ or the ‘Refrigerator Statement’ of the Second law of thermodynamics.
Figure 4.26 is a diagrammatic representation of the application of the Second law of thermodynamics to a refrigerator.
Diagram
Fig. 4.26 (a) – Energy flow diagram of a practical refrigerator: heat drawn from the cold reservoir, work \(W\) supplied from outside, and the total rejected to the hot reservoir.
As you can see from the energy flow diagram, Fig. 4.26 (a), a practical refrigerator requires work \(W\) to be done in order to extract heat from a cold reservoir and reject it to a hot reservoir.
Diagram
Fig. 4.26 (b) – A “perfect” refrigerator moving heat from the cold reservoir to the hot reservoir with no work input, with a large cross through it to show that it is impossible.
The statement means that the spontaneous flow of heat from an object at a cold temperature to a hotter one is not possible, as indicated in Fig. 4.26 (b).
This qualification matters. A cold body does radiate energy towards a hot one, and some fast-moving molecules do carry energy the “wrong” way across a boundary. The Second law is a statement about the balance of these exchanges, not about every individual one — which is exactly the point made about thermal equilibrium in Section 4.2.
4.11 Carnot Cycle and Carnot Engine
In Section 4.7.3, we have discussed the concept of a reversible and an irreversible process at length.
Now we will discuss why reversibility is such a basic and important concept in thermodynamics.
4.11.1 Significance of Reversibility in Thermodynamics
We know that a reversible process is a ‘bidirectional’ process, that is, it follows exactly the same steps in either direction.
This requires the process to take place in infinitesimally small steps. Also, the difference between the state variables in the two infinitesimally close states should be very small.
The two requirements are really one. If a step were large — a sudden shove of the piston, or contact with a much hotter body — the system would be thrown out of equilibrium during that step, and the return journey would not retrace it. Only a change small enough to leave the system always balanced can be undone exactly.
This would be possible if the system is in thermodynamic equilibrium with its environment throughout the change. Reversibility is therefore not an extra property that some processes happen to possess; it is what remains when every source of imbalance has been removed.
An irreversible process, on the contrary, is a unidirectional process. It can take place in only one direction. Any irreversible process is not in thermal equilibrium with its environment.
4.11.2 Maximum Efficiency of a Heat Engine and Carnot’s Cycle
We know that conversion of work to heat, as in the refrigerator of Section 4.9.2, is an irreversible process.
A heat engine would convert maximum heat into work if all irreversible processes could be avoided. In that case, the efficiency of the heat engine can be maximum.
Sadi Carnot, the French engineer and scientist, proposed a hypothetical ideal engine in 1824 which has the maximum efficiency.
Building the Ideal Cycle
In a Carnot engine, there are basically two kinds of process.
- Exchange of heat, which corresponds to the steps A to B and C to D in Fig. 4.21. For this to be reversible, the heat exchange must be isothermal.
This is possible if the working substance is at the temperature \(T_H\) of the source while absorbing heat. The working substance should be at the temperature of the cold reservoir, \(T_C\), while rejecting the heat.
The reason is the one identified in Section 4.11.1. If heat were allowed to flow across a real temperature gap — from a source at 800 K into a gas at 500 K, say — the flow could never be run backwards, and the step would be irreversible. Heat may cross a boundary reversibly only when the two sides are at the same temperature, which is precisely the isothermal condition.
- Work done, which corresponds to the steps B to C and D to A. For the work done to be reversible, the process should be adiabatic.
Here the working substance has to be moved from the temperature of the source to that of the sink, and back again. It cannot be put in contact with anything at an intermediate temperature without an irreversible heat flow, so the only reversible route is to change its temperature by doing work on it, with no heat crossing the boundary at all.
Thus, the cycle includes two isothermal and two adiabatic processes for maximum efficiency. The corresponding \(p\)-\(V\) diagram will then be as shown in Fig. 4.27.
Diagram
Fig. 4.27 – Carnot cycle on a \(p\)-\(V\) diagram: A\((V_1, p_1)\) to B\((V_2, p_2)\) isothermal expansion at \(T_H\) absorbing \(Q_H\), B to C\((V_3, p_3)\) adiabatic expansion, C to D\((V_4, p_4)\) isothermal compression at \(T_C\) rejecting \(Q_C\), D to A adiabatic compression.
Follow the loop in Fig. 4.27 once around. Along AB the gas is in contact with the source and expands while heat \(Q_H\) flows in. Along BC it is isolated and expands further, cooling from \(T_H\) to \(T_C\). Along CD it is in contact with the sink and is compressed while heat \(Q_C\) is given up. Along DA it is isolated again and is compressed back to the start, warming from \(T_C\) to \(T_H\). Every step is either isothermal or adiabatic, and every step is reversible.
The Efficiency of a Carnot Engine
By using the expressions for the work done during an adiabatic and an isothermal process, Eq. (4.7) and Eq. (4.20), we can derive an expression for the efficiency of a Carnot cycle, or Carnot engine, as \[\eta = \frac{W}{Q_H} = 1 - \frac{|Q_C|}{|Q_H|} = 1 - \frac{T_C}{T_H} \qquad \text{--- (4.29)}\]
Here \(T_C\) and \(T_H\) are the absolute temperatures of the sink and the source respectively, and both must be in kelvin (K). Using degrees Celsius in Eq. (4.29) gives a wrong answer, and this is one of the commonest mistakes made in examinations.
The remarkable feature of Eq. (4.29) is what has vanished from it. The heats have gone, and so has every detail of the machine — its size, its working substance, its fuel. Only the two temperatures remain.
Thus, while designing a heat engine for maximum efficiency, the source temperature \(T_H\) should be as high as possible, and the sink temperature \(T_C\) should be as low as possible.
Point 4 is the meeting place of this section with the Second law. Equation (4.29) gives \(\eta = 1\) only if \(T_C = 0\) K, and absolute zero is unattainable. The Kelvin-Planck statement is thus not merely an observation about real machines; it survives even in the ideal case.
4.11.3 Carnot Refrigerator
We know that a refrigerator is nothing but a heat engine operated in the reverse direction.
Because each step in the Carnot cycle is reversible, the entire Carnot cycle is reversible. This is exactly what makes the reversal possible: a cycle built only of reversible steps can be traced backwards step for step, whereas a cycle containing even one irreversible step cannot.
If we operate the Carnot engine in the reverse direction, we get the Carnot refrigerator. On the \(p\)-\(V\) diagram of Fig. 4.27, this simply means going round the same loop in the opposite sense — anticlockwise instead of clockwise — which, as Solved Example 4.8 showed, turns a machine that delivers work into one that consumes it.
Using the definition of the coefficient of performance, Eq. (4.28), we can write the coefficient of performance of a Carnot refrigerator as, \[K = \frac{|Q_C|}{|Q_H| - |Q_C|} = \frac{\dfrac{|Q_C|}{|Q_H|}}{1 - \dfrac{|Q_C|}{|Q_H|}} \qquad \text{--- (4.30)}\]
The second form is obtained from the first by dividing both the numerator and the denominator by \(|Q_H|\). This is done so that the two heats appear only as a ratio, and that ratio is something we already know for a Carnot cycle.
Using \[\frac{|Q_C|}{|Q_H|} = \frac{T_C}{T_H}\] in Eq. (4.30), we have the coefficient of performance of a Carnot refrigerator as, \[K = \frac{T_C}{T_H - T_C} \qquad \text{--- (4.31)}\]
Here \(T_C\) and \(T_H\) are the absolute temperatures of the cold and the hot reservoir respectively, both in kelvin (K), and \(K\) is a pure number with no unit.
Equation (4.31) gives the coefficient of performance of an ideal refrigerator, or the Carnot refrigerator. Just as with Eq. (4.29) for the engine, every detail of the machine has dropped out and only the temperatures remain.
What the Formula Tells Us
Equation (4.31) says that the coefficient of performance of a Carnot refrigerator also depends only on the temperatures of the hot and the cold reservoir.
When the temperature difference is very small, the coefficient is very large. In this case, a large quantity of heat can be removed from the lower temperature to the higher temperature by doing a very small amount of work.
The coefficient of performance is very small when the temperature difference is large. That means a small quantity of heat will be removed even when a large amount of work is done.
4.11.4 The Second Law of Thermodynamics and the Carnot Cycle
We have made two very important statements here.
- The Carnot engine is the most efficient heat engine, and
- The efficiency of a Carnot engine is independent of its working substance.
The first statement turns Eq. (4.29) into a universal ceiling. Whatever engine anyone builds between a source at \(T_H\) and a sink at \(T_C\), its efficiency cannot exceed \(1 - T_C/T_H\). An engine claimed to do better is not an engineering triumph but an arithmetical error.
The second statement explains why the working substance never appeared in Eq. (4.29). Air, steam, helium or petrol vapour — between the same two temperatures, a reversible engine using any of them performs identically.
We can also show that a Carnot refrigerator has the greatest coefficient of performance among all the refrigerators working between the same two temperatures. This is the same argument applied to the reversed cycle, and Eq. (4.31) is likewise an upper limit rather than a typical value.
4.12 Sterling Cycle
This is a closed thermodynamic cycle. The Sterling engine is based on this cycle, shown in Fig. 4.28.
Diagram
Fig. 4.28 – \(p\)-\(V\) diagram of the Sterling cycle: isothermal expansion A to B at \(T_H\) absorbing \(Q_H\), the step B to C at constant volume \(V_b\) releasing \(Q\), isothermal compression C to D at \(T_C\) rejecting \(Q_C\), and the step D to A at constant volume \(V_a\) absorbing \(Q\).
The working substance used in a Sterling engine is air, helium, hydrogen, nitrogen, and so on.
All the processes in the Sterling cycle are reversible processes. This places it, like the Carnot cycle, among the ideal cycles — but with an important practical difference, since a Sterling engine can actually be built and run.
When the gas is heated, the Sterling engine produces useful work. When work is done on the gas, it works as a refrigerator. This is the reverse working of a Sterling cycle.
The Four Processes
The ideal Sterling cycle has two isothermal processes, AB and CD. Two constant-volume processes, BC and DA, connect the two isothermal processes, as the vertical lines at \(V_b\) and \(V_a\) in Fig. 4.28 show.
Heat is absorbed at constant temperature \(T_H\) and rejected at constant temperature \(T_C\). The four processes in a Sterling cycle are described briefly in the following.
Isothermal expansion (AB). The gas is heated by supplying heat \(Q_H\) at constant temperature \(T_H\). Useful work is done by the gas in this part of the cycle.
Isochoric process (BC). Part of the heat absorbed, \(Q_H\), by the gas in the previous part of the cycle is released by the gas to the regenerator. This heat, \(Q\), is used in the next part of the cycle. The gas cools down to temperature \(T_C\).
Isothermal compression (CD). The heat generated in this part of the cycle, \(Q_C\), is rejected to the coolant, which is the sink. The temperature of the gas is maintained at \(T_C\) during this process.
Isochoric heat absorption (DA). The compressed gas absorbs the heat \(Q\) during this process. Its temperature is increased to \(T_H\).
The cycle repeats when the process reaches the point A.
Why the Stored Heat Matters
Notice what happens to the heat \(Q\) in the two constant-volume steps. It is given up by the gas along BC, held, and returned to the gas along DA. It is neither taken from the source nor thrown away to the sink.
4.13 Summary
Basic Ideas
- Thermodynamics is the branch of physics that deals with heat and temperature, and with the inter-conversion of heat and other forms of energy.
- Two objects at the same temperature are in thermal equilibrium: placed in thermal contact, they transfer no net heat between each other.
- Zeroth law of thermodynamics: if two systems are each in thermal equilibrium with a third system, they are also in thermal equilibrium with each other. Systems in thermal equilibrium are at the same temperature. This law is what allows a thermometer to be used to compare temperatures.
Internal Energy, Heat and Work
- Internal energy \(U\) is the energy associated with the random, disordered motion of the molecules of a system. It is measured in joule (J), and is different from the ordered macroscopic energy of a moving body.
- For an ideal monatomic gas the internal energy per molecule is \(\dfrac{3}{2}kT\); for a diatomic gas it is \(\dfrac{5}{2}kT\), where \(k\) is the Boltzmann constant and \(T\) the absolute temperature in kelvin (K). For a polyatomic gas, rotational and vibrational energies must also be counted; for liquids and solids, the potential energy of intermolecular attraction must be included.
- Heat \(Q\) is the energy transferred between a system and its environment because of a temperature difference between the two. Heat is energy in transit, not a quantity stored in a body.
- Sign convention: energy transferred into the system is positive; energy transferred out of the system is negative. Work \(W\) is positive when done by the system.
- A thermodynamic system may be open (exchanges energy and matter), closed (exchanges energy only) or isolated (exchanges neither).
- The internal energy of a system can be changed in exactly two ways — by transferring heat, or by doing work.
Work Done by a Gas
- For an infinitesimal change of volume, \[dW = p\,dV \qquad \text{--- (4.1)}\]
- For a finite change, \[W = \int_{V_i}^{V_f} p\,dV = p\left(V_f - V_i\right) \qquad \text{--- (4.2)}\] the last step being valid only when the pressure is constant.
- The area under the \(p\)-\(V\) curve represents the work done. The area enclosed by a closed loop is the net work done in one complete cycle.
- If the volume does not change, the work done is zero, however large the pressure may be.
First Law of Thermodynamics
\[\Delta U = Q - W \qquad \text{--- (4.3)}\] \[Q = \Delta U + W \qquad \text{--- (4.4)}\]
- The change in internal energy of a system is the difference between the heat supplied to it and the work done by it.
- The first law is a generalisation of the law of conservation of energy to systems that are not isolated.
- \(\Delta U\) may be positive, negative or zero, since \(Q\) and \(W\) may each be positive, negative or zero.
State Variables and Equations of State
- A state variable is any measurable characteristic of a system in equilibrium. Intensive variables (\(p\), \(T\), \(\rho\)) do not depend on the size of the system; extensive variables (\(M\), \(U\)) do.
- A system is in thermodynamic equilibrium only when it is in mechanical, chemical and thermal equilibrium simultaneously. A system not in equilibrium cannot be described by state variables.
- For an ideal gas, the equation of state is \[pV = nRT \qquad \text{--- (4.5)}\] where \(n\) is the number of moles and \(R\) the gas constant in J mol\(^{-1}\) K\(^{-1}\). For a fixed \(n\), only two of \(p\), \(V\), \(T\) are independent.
- A graph of \(p\) against \(V\) is the \(p\)-\(V\) diagram, or indicator diagram. A constant-temperature curve on it is an isotherm.
Processes and Paths
- A quasi static process changes the state variables infinitesimally slowly, so the system stays in equilibrium throughout; only such processes can be drawn as a path.
- A reversible process retraces exactly the same path in the reverse direction; an irreversible process does not. Irreversibility arises from non-equilibrium states and from dissipative forces such as friction and viscosity. All spontaneous natural processes are irreversible.
- Both the work done and the heat transferred depend on the path, not only on the initial and final states. Their difference, \(\Delta U\), does not.
The Standard Processes
Isothermal (\(\Delta T = 0\)): $pV = $ constant, so \(p_iV_i = p_fV_f\); \(\Delta U = 0\) for an ideal gas. \[W = nRT \ln\frac{V_f}{V_i} \qquad \text{--- (4.7)}\] \[Q = W \qquad \text{--- (4.8), (4.9)}\]
Isobaric (\(\Delta p = 0\)): none of \(\Delta U\), \(Q\), \(W\) is zero. \[W = P\left(V_f - V_i\right) = nR\left(T_f - T_i\right) \qquad \text{--- (4.10)}\] \[\Delta U = nC_V\left(T_f - T_i\right) \qquad \text{--- (4.11)}\] \[Q = nC_p\left(T_f - T_i\right), \qquad C_p = C_V + R \qquad \text{--- (4.12)}\]
Isochoric (\(\Delta V = 0\)): \(W = 0\), and all the heat supplied changes the internal energy. \[Q = \Delta U = nC_V\,\Delta T \qquad \text{--- (4.13), (4.14)}\]
Adiabatic (\(Q = 0\)): the system is insulated, or the change is too rapid for heat exchange. \[PV^{\gamma} = \text{constant}, \qquad \gamma = \frac{C_p}{C_V}\] \[\Delta U = -W \qquad \text{--- (4.15)}\] \[W = \frac{p_f V_f - p_i V_i}{1-\gamma} \qquad \text{--- (4.20)}\] \(\gamma\) is \(\dfrac{5}{3}\) for monoatomic, \(\dfrac{7}{5}\) for diatomic and \(\dfrac{8}{6} = \dfrac{4}{3}\) for triatomic gases. Adiabatic expansion cools a gas; adiabatic compression heats it. The adiabatic curve is steeper than the isotherm.
Cyclic: the system returns to its initial state, so \(\Delta U = 0\) and \[Q = W \qquad \text{--- (4.21)}\] The net work is the area of the loop: positive for a clockwise cycle, negative for an anticlockwise one.
Free expansion: a sudden, uncontrolled expansion into vacuum, with \(Q = W = 0\) and hence \(\Delta U = 0\). It is not quasi static and cannot be plotted on a \(p\)-\(V\) diagram.
Heat Engines
- A heat engine converts part of the heat supplied to it into work, working in a cycle between a source at \(T_H\) and a sink at \(T_C\). \[Q = Q_H + Q_C = |Q_H| - |Q_C| \qquad \text{--- (4.22)}\] \[W = Q = |Q_H| - |Q_C| \qquad \text{--- (4.23)}\]
- Thermal efficiency: \[\eta = \frac{W}{Q_H} = 1 - \frac{|Q_C|}{|Q_H|} \qquad \text{--- (4.24), (4.25)}\] \(\eta\) has no unit; \(W\), \(Q_H\) and \(Q_C\) must be in the same units.
- Engines are external combustion (steam engine) or internal combustion (petrol, diesel) according to how the working substance is heated.
Refrigerators, Air Conditioners and Heat Pumps
- These are heat engines run backwards: work is supplied to move heat from a cold region to a hot one, usually by repeatedly evaporating and condensing a refrigerant. \[|Q_H| = |Q_C| + |W| \qquad \text{--- (4.26)}\]
- Coefficient of performance: \[K = \frac{|Q_C|}{|W|} = \frac{|Q_C|}{|Q_H| - |Q_C|} \qquad \text{--- (4.27)}\] \(K\) is dimensionless; \(K \approx 5\) for a household refrigerator.
- For an air conditioner, with heat current \(H = |Q_C|/t\) and power \(P\), \[K = \frac{|Q_C|}{|W|} = \frac{Ht}{Pt} = \frac{H}{P} \qquad \text{--- (4.28)}\] Typical values of \(K\) are 2.5 to 3.0 for room air conditioners.
- A heat pump warms a building by taking heat from the colder air outside it.
- Heat flow from hot to cold is spontaneous; work is always needed to move heat from cold to hot.
Second Law of Thermodynamics
- The first law does not tell us which direction a process will take, nor does it forbid the complete conversion of heat into work. The second law supplies both restrictions.
- Kelvin-Planck (engine) statement: it is impossible to extract heat \(Q_H\) from a hot reservoir and use all of it to do work \(W\); some heat \(Q_C\) must be exhausted to a cold reservoir. A perfect heat engine is impossible.
- Clausius (refrigerator) statement: heat cannot flow from a colder body to a warmer body without work being done to accomplish the flow. A perfect refrigerator is impossible.
- These statements refer to the net transfer of energy.
Carnot Cycle and Carnot Engine
- The Carnot engine, proposed by Sadi Carnot in 1824, is a hypothetical ideal engine made entirely of reversible steps: two isothermal processes (during which heat is exchanged) and two adiabatic processes (during which work is done). \[\eta = \frac{W}{Q_H} = 1 - \frac{|Q_C|}{|Q_H|} = 1 - \frac{T_C}{T_H} \qquad \text{--- (4.29)}\] with both temperatures in kelvin. For maximum efficiency, \(T_H\) should be as high as possible and \(T_C\) as low as possible.
- Efficiency reaches 100% only if \(T_C = 0\) K, which is not attainable; even an ideal engine cannot be perfectly efficient.
- Reversing the cycle gives the Carnot refrigerator: \[K = \frac{|Q_C|}{|Q_H| - |Q_C|} \qquad \text{--- (4.30)}\] \[K = \frac{T_C}{T_H - T_C} \qquad \text{--- (4.31)}\] \(K\) is large when the temperature difference is small, and small when it is large.
- The Carnot engine is the most efficient heat engine, and all Carnot engines working between the same two temperatures have the same efficiency, whatever the working substance. Likewise, a Carnot refrigerator has the greatest coefficient of performance among all refrigerators working between the same two temperatures.
Sterling Cycle
- A closed, fully reversible cycle with two isothermal processes (AB and CD) connected by two constant-volume processes (BC and DA). Heat is absorbed at \(T_H\) and rejected at \(T_C\), while the heat exchanged along the connecting legs is stored and returned within the machine.
- Run forward it produces work; run in reverse it acts as a refrigerator, and the reversed Sterling cycle is widely used in cryogenics to reach extremely low temperatures and to liquefy gases.
4.14 Objective Questions
- A gas in a closed container is heated with 10 J of energy, causing the lid of the container to rise 2 m with 3 N of force. What is the total change in the internal energy of the system?
- 10 J (B) 4 J (C) −10 J (D) −4 J
- Which of the following is an example of the first law of thermodynamics?
- The specific heat of an object explains how easily it changes temperature.
- While melting, an ice cube remains at the same temperature.
- When a refrigerator is unplugged, everything inside of it returns to room temperature after some time.
- After falling down the hill, a ball’s kinetic energy plus heat energy equals the initial potential energy.
- Efficiency of a Carnot engine is large when
- \(T_H\) is large (B) \(T_C\) is low (C) \(T_H - T_C\) is large (D) \(T_H - T_C\) is small
- The second law of thermodynamics deals with transfer of:
- work done (B) energy (C) momentum (D) mass
- During a refrigeration cycle, heat is rejected by the refrigerant in the:
- condenser (B) cold chamber (C) evaporator (D) hot chamber
4.15 Check Your Concepts
The Zeroth law was formulated after the First and the Second laws, yet it is numbered zero. Explain what makes it more fundamental than the two laws that were stated before it.
A student writes, “A cup of hot tea contains a large amount of heat.” Identify what is wrong with this statement and rewrite it correctly.
A sealed bottle of water is carried in a fast-moving train. Its kinetic energy has increased, but its internal energy has not. Explain the difference between the two kinds of energy involved.
Why does the work done in Eq. (4.2) require a knowledge of the path, while the change in internal energy does not?
In the equation \(\Delta U = Q - W\), the heat carries a positive sign and the work a negative sign. Explain why the two quantities enter the same equation with opposite signs.
A gas expands freely into an evacuated chamber. It has clearly increased its volume, yet the work done is zero. Explain how both statements can be true.
Why is \(C_p\) always greater than \(C_V\) for a gas? Show how the relation \(C_p = C_V + R\) expresses this difference physically.
An isothermal process has \(\Delta U = 0\), yet heat flows into the gas continuously throughout the process. Explain why there is no contradiction.
Two curves are drawn through the same point on a \(p\)-\(V\) diagram, one an isotherm and the other an adiabatic. Which is steeper, and what is the physical reason for the difference?
The barrel of a bicycle pump becomes warm when a tyre is inflated quickly. Identify the thermodynamic process involved and explain the temperature rise using the first law.
A student proposes to cool a closed room on a summer afternoon by leaving the refrigerator door open. What will actually happen to the temperature of the room, and why?
Why must every heat engine reject some heat to a cold reservoir? Answer with reference to the requirement that an engine must operate in a cycle.
The coefficient of performance of a refrigerator can be greater than 1, while the efficiency of a heat engine can never be greater than 1. Explain why this is not a violation of energy conservation.
A deep freezer consumes far more electricity per litre of storage than an ordinary refrigerator compartment of the same volume. Explain this using Eq. (4.31).
No real process is exactly quasi static, and no real engine is exactly reversible. Why, then, are these idealisations useful in thermodynamics?
Explain why the efficiency of a Carnot engine does not depend on the working substance used, while the efficiency of a real petrol engine certainly depends on its design.
A free expansion cannot be drawn as a curve on a \(p\)-\(V\) diagram, although its initial and final states can both be marked on it. Explain why.
An inventor claims to have built an engine that takes in heat at 400 K, rejects heat at 300 K, and has an efficiency of 40%. Without doing a detailed calculation of the machinery, explain what is wrong with the claim.
When a cyclic process is traced clockwise on a \(p\)-\(V\) diagram the net work is positive, and when it is traced anticlockwise the net work is negative. Explain this difference in terms of the pressures at which expansion and compression occur.
A metal rod has one end held in steam and the other in melting ice. After some time the temperature at every point along it stops changing. Is the rod in thermal equilibrium? Justify your answer.
A gas contained in a cylinder surrounded by a thick layer of insulating material is quickly compressed. (a) Has there been a transfer of heat? (b) Has work been done?
Give an example of some familiar process in which no heat is added to or removed from a system, but the temperature of the system changes.
Give an example of some familiar process in which heat is added to an object, without changing its temperature.
What sets the limits on the efficiency of a heat engine?
Why should a Carnot cycle have two isothermal and two adiabatic processes?
A mixture of hydrogen and oxygen is enclosed in a rigid insulating cylinder. It is ignited by a spark. The temperature and the pressure both increase considerably. Assuming that the energy supplied by the spark is negligible, what conclusions may be drawn by application of the first law of thermodynamics?
A resistor held in running water carries electric current. Treat the resistor as the system. (a) Does heat flow into the resistor? (b) Is there a flow of heat into the water? (c) Is any work done? (d) Assuming the state of the resistance to remain unchanged, apply the first law of thermodynamics to this process.
A mixture of fuel and oxygen is burned in a constant-volume chamber surrounded by a water bath. It was noticed that the temperature of the water increased during the process. Treating the mixture of fuel and oxygen as the system, (a) Has heat been transferred? (b) Has work been done? (c) What is the sign of \(\Delta U\)?
Draw a \(p\)-\(V\) diagram and explain the concept of positive and negative work. Give one example of each.
A solar cooker and a pressure cooker are both used to cook food. Treating them as thermodynamic systems, discuss the similarities and differences between them.
4.16 Practice with Numericals
A system absorbs 760 J of heat from its surroundings and does 290 J of work on them. Calculate the change in its internal energy and state whether it increases or decreases.
During a process, 450 J of work is done on a gas while it gives out 180 J of heat to its surroundings. Write \(Q\) and \(W\) with their correct signs and calculate \(\Delta U\).
A gas expands at a constant pressure of \(1.5 \times 10^{5}\) Pa from a volume of \(2.0 \times 10^{-3}\) m\(^3\) to \(6.0 \times 10^{-3}\) m\(^3\). Calculate the work done by the gas. If 900 J of heat is supplied during the expansion, find the change in the internal energy of the gas.
5 mol of an ideal gas at 350 K expands isothermally until its volume becomes four times the initial volume. Calculate the work done by the gas and the heat absorbed by it. (Take \(R = 8.314\) J mol\(^{-1}\) K\(^{-1}\).)
0 mol of an ideal gas is compressed isothermally at 300 K from \(8.0 \times 10^{-3}\) m\(^3\) to \(2.0 \times 10^{-3}\) m\(^3\). Calculate the work done on the gas and the heat rejected by it. (Take \(R = 8.314\) J mol\(^{-1}\) K\(^{-1}\).)
A rigid sealed vessel contains 4.0 mol of an ideal gas with \(C_V = 12.5\) J mol\(^{-1}\) K\(^{-1}\). The gas is heated until its temperature rises from 300 K to 375 K. Calculate the work done by the gas, the change in its internal energy, and the heat supplied.
0 mol of a diatomic ideal gas is heated at constant pressure so that its temperature rises from 290 K to 390 K. Taking \(C_V = 20.8\) J mol\(^{-1}\) K\(^{-1}\) and \(R = 8.314\) J mol\(^{-1}\) K\(^{-1}\), calculate the work done by the gas, the change in its internal energy, and the heat supplied to it.
A monatomic ideal gas at a pressure of \(1.0 \times 10^{5}\) Pa and a temperature of 300 K is compressed adiabatically to one-eighth of its initial volume. Taking \(\gamma = 5/3\), calculate its final pressure and its final temperature.
A diatomic ideal gas occupying \(3.0 \times 10^{-3}\) m\(^3\) at a pressure of \(2.0 \times 10^{5}\) Pa expands adiabatically to twice its volume. Taking \(\gamma = 1.4\), calculate the final pressure and the work done by the gas.
A gas is carried around a rectangular cycle on a \(p\)-\(V\) diagram in the clockwise sense, the pressure varying between \(0.8 \times 10^{5}\) Pa and \(2.4 \times 10^{5}\) Pa and the volume between \(1.0 \times 10^{-3}\) m\(^3\) and \(4.0 \times 10^{-3}\) m\(^3\). Calculate the net work done by the gas per cycle, and the net heat absorbed per cycle. If the cycle is repeated 40 times, find the total work done.
A gas is taken from a state at \(4.0 \times 10^{5}\) Pa and \(1.0 \times 10^{-3}\) m\(^3\) to a state at \(1.0 \times 10^{5}\) Pa and \(4.0 \times 10^{-3}\) m\(^3\), first by expanding at the initial pressure and then dropping the pressure at constant volume, and afterwards, in a second trial, by dropping the pressure first at the initial volume and then expanding at the final pressure. Calculate the work done by the gas along each route.
A heat engine absorbs 1250 J of heat per cycle from its source and rejects 875 J per cycle to its sink. Calculate the work done per cycle and the thermal efficiency of the engine as a percentage.
A heat engine of efficiency 25% delivers 500 J of work in each cycle. Calculate the heat absorbed from the source and the heat rejected to the sink in each cycle.
An engine completes 1800 cycles per minute and does 240 J of work in each cycle. Calculate the power output of the engine in kilowatt.
A Carnot engine works between a source at 750 K and a sink at 300 K. Calculate its efficiency. If it absorbs 1800 J of heat per cycle from the source, find the work done and the heat rejected per cycle.
A Carnot engine has an efficiency of 45% when its sink is at 300 K. Calculate the temperature of its source. To what temperature must the source be raised if the efficiency is to become 60%, the sink remaining at 300 K?
A refrigerator extracts 2400 J of heat from its cold chamber in each cycle while its compressor does 480 J of work per cycle. Calculate the coefficient of performance of the refrigerator and the heat delivered to the room in each cycle.
A Carnot refrigerator maintains its cold chamber at 255 K in a room at 300 K. Calculate its coefficient of performance and the work needed to remove 1800 J of heat from the chamber.
A room air conditioner removes heat at the rate of \(3.6 \times 10^{3}\) W while consuming electrical power of 1.2 kW. Calculate its coefficient of performance, and the heat delivered to the outside air in 30 minutes.
A gas contained in a cylinder fitted with a frictionless piston expands against a constant external pressure of 1 atm from a volume of 5 litres to a volume of 10 litres. In doing so it absorbs 400 J of thermal energy from its surroundings. Determine the change in internal energy of the system.
A system releases 130 kJ of heat while 109 kJ of work is done on the system. Calculate the change in internal energy.
The efficiency of a Carnot cycle is 75%. If the temperature of the hot reservoir is 727 °C, calculate the temperature of the cold reservoir.
A Carnot refrigerator operates between 250 K and 300 K. Calculate its coefficient of performance.
An ideal gas is taken through an isothermal process. If it does 2000 J of work on its environment, how much heat is added to it?
An ideal monatomic gas is adiabatically compressed so that its final temperature is twice its initial temperature. What is the ratio of the final pressure to its initial pressure?
A hypothetical thermodynamic cycle is shown in Fig. 4.38. Calculate the work done in 25 cycles.
Diagram
Fig. 4.38 – The \(p\)-\(V\) diagram of the hypothetical thermodynamic cycle referred to in question 26, as given in the original exercise.
- Figure 4.39 shows the V-T diagram for one cycle of a hypothetical heat engine which uses an ideal gas. Draw the \(p\)-\(V\) diagram and the \(p\)-\(T\) diagram of the system.
Diagram
Fig. 4.39 – The V-T diagram for one cycle of the hypothetical heat engine referred to in question 27, as given in the original exercise.
- A system is taken to its final state from its initial state along the hypothetical paths shown in Fig. 4.40. Calculate the work done in each case.
Diagram
Fig. 4.40 – The \(p\)-\(V\) diagram showing the hypothetical paths AB, BC, CD and DA referred to in question 28, as given in the original exercise.