11.1 Introduction
Chapter 11 — Thermodynamics
In the year 1798, an American-born military engineer named Benjamin Thompson — better known by his title, Count Rumford — was supervising the boring of brass cannons in a workshop in Munich. Cannons were made by drilling a solid brass cylinder with a blunt metal borer, and Rumford noticed something that puzzled him deeply. The drilling produced enormous amounts of heat — so much that the water used to cool the metal actually began to boil, seemingly without limit.
At that time, most scientists believed heat was an invisible fluid called “caloric” that was stored inside objects and squeezed out during processes like drilling. But Rumford saw a problem. If heat were a fluid trapped in the metal, then a sharper drill should scoop out more of it, and eventually the supply would run dry. Instead, he found the opposite: a blunt drill produced just as much heat, and the heat kept coming as long as the horses kept turning the drill. The amount of heat depended only on the mechanical work being done — not on the drill’s sharpness.
Rumford concluded that heat was not a fluid at all. It was a form of energy, produced here by converting the work of the horses. This single observation helped overturn a centuries-old idea and opened the door to the science of thermodynamics — the study of heat, work, and how energy flows between them.
Figure to come
Fig. 11.0 – A horse-driven drill boring a brass cannon submerged in water, with the water boiling and steam rising, illustrating mechanical work being converted into heat.
By the end of this chapter, you will be able to answer each of these questions.
In the previous chapter we studied the thermal properties of matter — how substances expand on heating, how heat flows, and how temperature is measured. In this chapter we go one step further and study the laws that govern thermal energy itself: the rules that decide how heat and work turn into each other, and how far that conversion can go.
The central theme of this chapter is the two-way traffic between work and heat. Work can be converted into heat, and heat can be converted into work. Both of these happen around us all the time.
Think of a cold winter morning. When you rub your palms together briskly, they grow warm. Here you are doing mechanical work by rubbing, and that work reappears as heat in your skin. This is the conversion of work into heat.
Now think of the reverse. In a steam engine, the “heat” carried by hot, high-pressure steam is used to push a piston back and forth. That moving piston turns the wheels of a locomotive. Here heat is being converted into useful work. So heat and work are two sides of the same coin, and thermodynamics is the study of how they trade places.
Before we can state the laws precisely, we have to be careful about words. In everyday speech we use “heat,” “temperature,” and “work” loosely, but in physics each has a sharp meaning. Getting these meanings right took scientists a surprisingly long time.
From “caloric fluid” to heat as energy
For a long time, heat was misunderstood. Before the modern picture, scientists imagined heat to be a kind of invisible, weightless fluid that soaked into the tiny pores of a substance. They gave this imagined fluid a name — caloric.
In this old picture, when a hot body touched a cold body, caloric was thought to flow from the hotter body into the colder one, until both settled at the same temperature. This was pictured as being just like water flowing through a horizontal pipe that connects two tanks filled to different heights, as shown in Fig. 11.0a. Water flows from the higher tank to the lower one until the two water levels become equal. In the caloric picture, heat was supposed to flow in the same way until the “caloric levels” — that is, the temperatures — of the two bodies became equal.
Figure to come
Fig. 11.0a – Two water tanks at different heights joined by a horizontal pipe, water flowing from the higher level to the lower until levels equalise, shown side-by-side with a hot body and a cold body exchanging “caloric” until their temperatures equalise.
The caloric idea seemed to explain a lot, but it was wrong. As we saw in the chapter opener, Count Rumford’s observation while boring cannons could not be explained by a stored fluid of heat: the heat produced kept flowing endlessly as long as work was being done, and it depended on the work done, not on how much “fluid” a sharper or blunter drill could scrape out of the pores. The only sensible conclusion was that heat is not a substance stored inside a body at all — heat is a form of energy, and it can be produced by converting work into it.
Rumford’s cannon experiment was a qualitative clue. A little later, the idea was put on a firm quantitative footing.
Once heat was recognised as energy, the whole subject of thermodynamics could be built on the solid foundation of energy conservation.
Thermodynamics is a macroscopic science
Let us now describe more carefully what thermodynamics actually is. Thermodynamics is the branch of physics that deals with the concepts of heat and temperature and with the inter-conversion of heat and other forms of energy.
The key word here is macroscopic. Thermodynamics deals with matter in bulk — a full cylinder of gas, a block of metal, a beaker of water — and does not concern itself with the individual atoms and molecules inside. In fact, its laws were worked out in the nineteenth century, before the molecular picture of matter was firmly established. That is a remarkable point: the science works perfectly well without ever needing to talk about molecules.
Because it is macroscopic, thermodynamics describes a system using only a small number of everyday, common-sense quantities — ones we can actually measure with instruments. These are called macroscopic variables, and they include pressure, volume, temperature, mass, and composition (the mixture of gases present, if any).
Compare this with two other ways of describing the same gas. A truly microscopic description would mean listing the position and velocity of every single molecule in the gas — an impossibly huge amount of information. The kinetic theory of gases, which you will meet in the next chapter, is less demanding than that: it does not track every molecule, but it does deal with how molecular velocities are distributed. Thermodynamics is simpler still — it avoids the molecular description altogether and works only with the handful of bulk variables listed above.
How thermodynamics differs from mechanics
It is worth pausing to see clearly how thermodynamics differs from the mechanics you have already learnt.
In mechanics, our interest is in the motion of a particle or a body as a whole — how it moves under the action of forces and torques. We ask about its position, velocity, acceleration, and kinetic energy.
Thermodynamics, on the other hand, is not concerned with the motion of the system as a whole. It cares about the internal macroscopic state of the body — its temperature, its internal energy, and how heat and work change these.
A simple example makes the difference sharp. When a bullet is fired from a gun, what changes is its mechanical state — chiefly its kinetic energy, because it is now moving fast. Its temperature, however, is not raised just because it is flying quickly, as shown in Fig. 11.0b. But when that bullet strikes a block of wood and comes to a stop, its kinetic energy does not simply vanish — it is converted into heat. This heating raises the temperature of both the bullet and the surrounding layers of wood.
Figure to come
Fig. 11.0b – A bullet in flight (fast but not hot) on the left; the same bullet embedded in a wooden block after impact, with heat/warmth radiating from the point of impact, on the right.
The lesson is this: temperature is related to the internal, disordered motion of the molecules inside the bullet — the random jiggling of its atoms — and not to the ordered motion of the bullet as a whole through space. A fast-moving object is not automatically a hot object.
With this careful distinction in place — between the motion of a body and the internal energy stored in its molecules, and between the old caloric fluid and the modern idea of heat as energy — we are ready to build up the laws of thermodynamics step by step. We begin in the next section with the most basic question of all: what does it mean for two bodies to be in thermal equilibrium, and how does that lead us to a proper definition of temperature?
11.2 Thermal Equilibrium
You have already met the idea of “equilibrium” in mechanics. There, a body is in equilibrium when the net external force and the net external torque acting on it are both zero — the body has no tendency to start moving or to start rotating.
In thermodynamics, the same word “equilibrium” is used, but in a different sense. Here we are not talking about forces and motion at all. Instead, we watch the macroscopic variables of the system — quantities such as pressure, volume, temperature, mass, and composition that we met in the previous section.
We say a system is in an equilibrium state when these macroscopic variables do not change with time. If you measure them now, and again after an hour, you get the same values — provided nothing from outside disturbs the system.
For example, a gas sealed inside a closed, rigid container that is completely insulated from its surroundings — with fixed values of pressure, volume, temperature, mass, and composition that stay unchanged in time — is in a state of thermodynamic equilibrium.
The role of the wall
Whether a system stays in equilibrium depends on its surroundings and, in particular, on the nature of the wall that separates the system from those surroundings. This turns out to be a surprisingly important point.
Consider two gases, A and B, held in two separate containers. Experiments show that for a fixed mass of gas, pressure and volume can be treated as its two independent variables — fix these two and the state of the gas is settled. Let gas A be described by the pair \((P_A, V_A)\) and gas B by the pair \((P_B, V_B)\).
Now bring the two containers close together, so that they share a common wall. What happens next depends entirely on what that wall is made of. There are two possibilities.
Case 1 — The adiabatic wall
Suppose the two gases are placed side by side but separated by an adiabatic wall — an insulating wall (it may even be free to move) that does not allow heat to pass through it. To keep the experiment clean, the whole assembly is also surrounded by similar adiabatic walls, so that no heat leaks to the outside either. This arrangement is shown in Fig. 11.1(a).
Figure to come
Fig. 11.1(a) – Two gas chambers A and B side by side, separated by an insulating (adiabatic) wall, the whole set-up enclosed by insulating walls; no heat crosses between A and B.
In this case it is found that any pair of values \((P_A, V_A)\) can coexist with any pair of values \((P_B, V_B)\). The two gases do not influence each other in the least. A may be hot while B is cold, or the other way round — the adiabatic wall simply keeps them from interacting.
Case 2 — The diathermic wall
Now replace the adiabatic wall with a diathermic wall — a conducting wall that does allow heat to flow from one side to the other. This is shown in Fig. 11.1(b).
Figure to come
Fig. 11.1(b) – The same chambers A and B, now separated by a conducting (diathermic) wall; arrows show heat flowing across the wall until both settle to a common state.
With a diathermic wall in place, something new happens. The macroscopic variables of A and B begin to change on their own — that is, spontaneously. Heat flows across the wall, and the pressures and volumes adjust themselves, moving from the original values \((P_A, V_A)\) and \((P_B, V_B)\) to new values \((P_A', V_A')\) and \((P_B', V_B')\).
After some time, this changing stops. The variables settle to steady values and no more heat flows between the two gases. When this settled condition is reached, we say that system A is in thermal equilibrium with system B.
11.3 Zeroth Law of Thermodynamics
At the end of the previous section we were left with a question. We had seen that two systems in thermal equilibrium share the same temperature — but we relied on our everyday sense of “hot” and “cold” to say so. Can we arrive at the idea of temperature more carefully, without just trusting our senses? The Zeroth Law of Thermodynamics provides the clue.
A simple experiment with three systems
Imagine three systems: A, B, and a third system C. Arrange them as follows. Keep A and B separated from each other by an adiabatic wall, so that no heat can pass directly between them. But let each of them be in contact with C through a conducting (diathermic) wall, so that heat can flow between A and C, and between B and C. This arrangement is shown in Fig. 11.2(a).
Figure to come
Fig. 11.2(a) – Systems A and B side by side, separated from each other by an adiabatic wall, but each in contact with a third system C above them through a conducting (diathermic) wall.
Because A is in thermal contact with C, and B is in thermal contact with C, their macroscopic variables change until both A and B reach thermal equilibrium with C. Once this happens, nothing changes further as long as this arrangement is kept.
The crucial second step
Now perform a clever switch. Replace the adiabatic wall between A and B with a conducting wall, so that heat could now flow directly between A and B if it wanted to. At the same time, insulate C from both A and B using an adiabatic wall, so C is now cut off. This new arrangement is shown in Fig. 11.2(b).
Figure to come
Fig. 11.2(b) – The adiabatic wall between A and B is replaced by a conducting wall, while C is now separated from both A and B by an adiabatic wall.
What do we observe? The states of A and B do not change any further. Even though heat is now free to flow between A and B, none does. This means A and B were already in thermal equilibrium with each other — even though they had never been in direct contact before.
Stating the law
This simple but important observation is the content of the Zeroth Law of Thermodynamics.
From the law to the meaning of temperature
The Zeroth Law tells us something deep. When two systems A and B are in thermal equilibrium, there must be some physical quantity that has the same value for both of them. This shared thermodynamic quantity is what we call temperature, denoted by \(T\).
We can now see the logic clearly. Suppose A and B are each separately in thermal equilibrium with C. Then A and C share the same temperature, so \(T_A = T_C\), and B and C share the same temperature, so \(T_B = T_C\). Putting these together:
\[T_A = T_C \quad \text{and} \quad T_B = T_C \;\Rightarrow\; T_A = T_B\]
So A and B must be at the same temperature, which is exactly why they are found to be in thermal equilibrium with each other. The Zeroth Law thus gives temperature a firm scientific footing: it is the marker that is equal for all bodies in mutual thermal equilibrium.
Looking ahead
We have now arrived at the concept of temperature in a formal way, through the Zeroth Law. A natural next question is how to attach actual numbers to it — how to build a temperature scale — but that task of thermometry was taken up in the previous chapter on the thermal properties of matter.
Having secured the ideas of thermal equilibrium and temperature, we are ready to look closely at the two ways energy is transferred into or out of a system — heat and work — and at the quantity they change, the internal energy. That is the subject of the next section.
11.4 Heat, Internal Energy and Work
The Zeroth Law led us to a clear idea of temperature, a marker of the “hotness” of a body. Temperature also decides the direction in which heat flows when two bodies are placed in thermal contact: heat always flows from the body at higher temperature to the body at lower temperature. The flow stops once the two temperatures become equal — that is, once the bodies reach thermal equilibrium. (How numbers are actually assigned to temperature, through thermometry, was dealt with in the earlier chapter on the thermal properties of matter.)
We now build up three closely related ideas that lie at the heart of thermodynamics: internal energy, heat, and work. Getting the differences between them exactly right is essential — many exam mistakes come from confusing heat with internal energy.
Internal energy
Every bulk system — a gas, a liquid, a solid — is made of an enormous number of molecules. These molecules are always in motion and also exert forces on one another. Each moving molecule has kinetic energy, and because of the forces between them, the molecules also have potential energy.
The internal energy of a system, written \(U\), is simply the total of all these molecular kinetic and potential energies added together.
There is one careful condition. In thermodynamics we do not count the kinetic energy of the system moving as a whole. We add up the molecular energies in the frame of reference in which the centre of mass of the system is at rest — that is, a frame moving along with the system. So internal energy includes only the energy of the disordered, random motion of the molecules, not the ordered motion of the body through space.
Consider a closed box of gas sitting on a table, as in Fig. 11.3(a). The random darting, spinning, and jiggling of its molecules make up its internal energy. Now imagine the very same box being carried across the room at some speed, as in Fig. 11.3(b). The molecules now also share the ordered motion of the box, but this extra bulk kinetic energy is not added to \(U\) — internal energy stays the same as when the box was at rest.
Figure to come
Fig. 11.3(a) – A stationary box of gas with molecules moving randomly in all directions; caption notes translational, rotational and vibrational motions all count toward U. (b) The same box moving as a whole with velocity v; the box’s bulk kinetic energy is excluded from U.
Although we used the molecular picture to understand internal energy, thermodynamics treats \(U\) simply as a macroscopic variable of the system. The most important property of \(U\) is this: it depends only on the state of the system, not on how that state was reached.
This makes internal energy a state variable.
So the internal energy of a fixed mass of gas is completely fixed once we know its state — described by definite values of pressure, volume, and temperature. It does not matter whether the gas was heated, compressed, or reached that state by some roundabout route; for a given \(P\), \(V\), and \(T\), the internal energy has one definite value. Pressure, volume, temperature, and internal energy are therefore all state variables of the gas (this idea is developed further in Section 11.7).
If we neglect the small forces between molecules — as we may for an ideal gas — the potential energy part drops out, and the internal energy of the gas becomes just the sum of the kinetic energies of the random molecular motions. As you will see in the next chapter, this random motion is not only translational (molecules moving from place to place inside the container); it can also be rotational (molecules spinning) and vibrational (atoms within a molecule vibrating). All of these count toward \(U\).
Two ways to change the internal energy
How can we change the internal energy of a system? Take the standard example of a fixed mass of gas in a cylinder fitted with a movable piston, shown in Fig. 11.4. Experience shows there are exactly two ways to change the state of the gas, and hence its internal energy.
Figure to come
Fig. 11.4 – A gas in a cylinder with a movable piston; (a) the cylinder in contact with a hotter body, heat flowing in; (b) a weight on the piston doing work by pushing it down. Both change U.
The first way is to place the cylinder in contact with a body hotter than the gas. The temperature difference drives a flow of energy — heat — from the hotter body into the gas, raising the gas’s internal energy. This is energy transfer as heat, shown in Fig. 11.4(a).
The second way is to push the piston down, doing mechanical work on the gas, which also raises its internal energy. This is energy transfer as work, shown in Fig. 11.4(b), and it needs no temperature difference at all.
Both processes can also run in reverse. If the surroundings are colder, heat flows out of the gas to the surroundings. And the gas can itself push the piston up, doing work on the surroundings. In short, heat and work are two different modes of transferring energy to or from a system, and either one can change its internal energy.
Heat and work are not stored — they are “in transit”
Here lies the subtle point that students most often get wrong. Heat is certainly a form of energy, but it is energy in transit — energy on the move between a system and its surroundings because of a temperature difference. It is not something a body “contains.”
Similarly, work in thermodynamics is energy transferred by mechanical means, such as a moving piston, without needing any temperature difference.
This “energy in transit” idea is not just a play on words; the distinction is of basic importance. The state of a thermodynamic system is characterised by its internal energy, not by heat. That is why a statement such as “a gas in a given state has a certain amount of heat” is as meaningless as saying “a gas in a given state has a certain amount of work.” Neither heat nor work is stored in the gas.
By contrast, “a gas in a given state has a certain amount of internal energy” is perfectly meaningful, because \(U\) is a genuine property of the state. And statements such as “a certain amount of heat is supplied to the system” or “a certain amount of work is done by the system” are also meaningful — because these describe energy being transferred, which is exactly what heat and work are.
To summarise: heat and work are not state variables. They are modes of transferring energy to a system, and the effect of that transfer is a change in the system’s internal energy — and internal energy is a state variable.
In everyday language we often blur heat and internal energy together, and elementary books sometimes ignore the difference. For a proper understanding of thermodynamics, however, keeping them apart is crucial — and it prepares us directly for the next section, where heat, work, and internal energy are tied together by the First Law of Thermodynamics.
11.5 First Law of Thermodynamics
In the previous section we saw that the internal energy \(U\) of a system can be changed in exactly two ways — by supplying heat to it, or by doing work on it (or the system doing work on its surroundings). The First Law of Thermodynamics simply ties these three quantities together using the principle of conservation of energy.
Setting up the three quantities
Let us give clear names and, most importantly, clear sign conventions to the three quantities involved. These sign conventions are exam-critical — reversing one is the single most common mistake students make.
\(\Delta Q\) = the heat supplied to the system by the surroundings
\(\Delta W\) = the work done by the system on the surroundings
\(\Delta U\) = the change in the internal energy of the system
Energy cannot appear from nowhere or vanish into nothing — this is the general principle of conservation of energy. Apply it to our system: the heat we pour into the system must be accounted for. Part of it goes into raising the system’s own internal energy, and the rest leaves the system as work done on the surroundings. In symbols,
\[\Delta Q = \Delta U + \Delta W \tag{11.1}\]
This is the First Law of Thermodynamics. It says nothing mysterious — it is just energy conservation, carefully written for a system that can exchange energy with its surroundings as both heat and work.
In words: the heat you put in splits into two parts — one part is stored inside as extra internal energy, and the other part is spent doing work on the outside world. It need not be an equal split; how it divides depends on the process.
An alternative form and the idea of path independence
We can rearrange Equation (11.1) into a very useful form:
\[\Delta Q - \Delta W = \Delta U \tag{11.2}\]
Why is this form so instructive? Because a system can move from one state to another by many different routes, and this equation tells us which quantities care about the route and which do not.
Suppose we take a gas from an initial state \((P_1, V_1)\) to a final state \((P_2, V_2)\). There is more than one way to do it. Along one path, we could first change the volume from \(V_1\) to \(V_2\) at constant pressure \(P_1\) (reaching \((P_1, V_2)\)), and then change the pressure from \(P_1\) to \(P_2\) at constant volume \(V_2\). Along another path, we could first change the pressure at constant volume, and then change the volume at constant pressure. Both routes start and end at the same two states, as shown in Fig. 11.4a.
Figure to come
Fig. 11.4a – A P–V diagram showing two different paths from state (P1, V1) to state (P2, V2): one going first horizontally then vertically, the other first vertically then horizontally, illustrating that ΔU is the same for both while ΔQ and ΔW differ.
Since internal energy \(U\) is a state variable, its change \(\Delta U\) depends only on the initial and final states — not on the path taken. So \(\Delta U\) is the same for both routes.
However, \(\Delta Q\) and \(\Delta W\) are not state variables; they generally depend on the path chosen. The remarkable thing the First Law tells us is that although \(\Delta Q\) and \(\Delta W\) each depend on the path, their combination \(\Delta Q - \Delta W\) does not — it always equals \(\Delta U\), which is path-independent.
A special case makes this vivid. If a process is arranged so that the internal energy does not change at all (\(\Delta U = 0\)) — for instance, the isothermal expansion of an ideal gas, which we will study in Section 11.8 — then Equation (11.2) gives
\[\Delta Q = \Delta W\]
This means all the heat supplied to the system is used up entirely in doing work on the surroundings; none of it is stored as internal energy.
Work done at constant pressure
Now take the familiar case of a gas in a cylinder with a movable piston. When the gas expands, it pushes the piston out and therefore does work.
We can find this work easily. Force equals pressure times area (\(F = P A\)), and when the piston moves through a small displacement, area times displacement equals the change in volume. So, if the gas expands against a constant external pressure \(P\), the work done by it is
\[\Delta W = P\,\Delta V\]
where \(\Delta V\) is the change in volume of the gas (in \(\text{m}^3\)), \(P\) is the pressure (in \(\text{Pa} = \text{N m}^{-2}\)), and \(\Delta W\) is in joules (J).
Substituting this into Equation (11.1), we get the First Law in the form most often used for a gas expanding at constant pressure:
\[\Delta Q = \Delta U + P\,\Delta V \tag{11.3}\]
Applying Eq. (11.3): boiling one gram of water
Let us use Equation (11.3) to find how the internal energy changes when \(1\ \text{g}\) of water turns from liquid into vapour (steam) at atmospheric pressure.
The measured latent heat of vaporisation of water is \(2256\ \text{J g}^{-1}\). So for \(1\ \text{g}\) of water, the heat supplied is
\[\Delta Q = 2256\ \text{J}\]
At atmospheric pressure, \(1\ \text{g}\) of water occupies a volume of about \(1\ \text{cm}^3\) in the liquid phase and about \(1671\ \text{cm}^3\) in the vapour phase — a huge expansion. The work done by the system in expanding against the constant atmospheric pressure is
\[\Delta W = P\,(V_g - V_l)\]
Here the liquid volume \(V_l = 1\ \text{cm}^3\) is tiny compared with the vapour volume \(V_g = 1671\ \text{cm}^3\), so \((V_g - V_l)\) is very nearly equal to \(V_g\). Taking \(P = 1.013 \times 10^5\ \text{Pa}\) and converting the volume to \(\text{m}^3\) (\(1671\ \text{cm}^3 = 1671 \times 10^{-6}\ \text{m}^3\)),
\[\Delta W = 1.013 \times 10^5 \times (1671 \times 10^{-6}) = 169.2\ \text{J}\]
Now Equation (11.3), rearranged as \(\Delta U = \Delta Q - \Delta W\), gives
\[\Delta U = 2256 - 169.2 = 2086.8\ \text{J}\]
The conclusion is striking: of the \(2256\ \text{J}\) supplied, only about \(169\ \text{J}\) goes into doing work as the water expands into steam, while the large remainder — over \(2086\ \text{J}\) — goes into increasing the internal energy of the water as it changes from the liquid to the vapour phase. Most of the latent heat is stored internally, chiefly in pulling the water molecules apart against their mutual attraction.
Looking ahead
The First Law tells us how heat, work, and internal energy balance in any process. But the amount of heat needed to change a substance’s temperature depends on the substance and on the conditions under which heat is supplied. To measure that, we introduce the idea of specific heat capacity in the next section.
11.6 Specific Heat Capacity
The First Law told us how heat, work, and internal energy are balanced. A very practical question now arises: how much heat must we supply to raise a substance’s temperature by a given amount? The answer depends both on the substance and on how much of it we have — and this is captured by the ideas of heat capacity and specific heat capacity.
Heat capacity
Suppose an amount of heat \(\Delta Q\) supplied to a substance raises its temperature from \(T\) to \(T + \Delta T\). The heat capacity \(S\) of the substance is defined as the heat needed per unit rise in temperature (this idea was introduced in Chapter 10):
\[S = \frac{\Delta Q}{\Delta T} \tag{11.4}\]
Heat capacity clearly depends on how much substance there is — a large pot of water needs far more heat than a spoonful to warm by the same amount. So \(S\) is proportional to the mass. It can also depend on the temperature at which the heating is done, because a different amount of heat may be needed for a unit rise at different temperatures.
Specific heat capacity
To get a quantity that is a characteristic of the material itself — independent of how much of it we have — we divide the heat capacity by the mass \(m\) (in kg):
\[s = \frac{S}{m} = \frac{1}{m}\frac{\Delta Q}{\Delta T} \tag{11.5}\]
This \(s\) is the specific heat capacity of the substance.
It depends on the nature of the substance and on its temperature, but not on the amount present.
Molar specific heat capacity
Sometimes it is more convenient to specify the amount of substance in moles (\(\mu\)) rather than in kilograms. Dividing the heat capacity by the number of moles gives the molar specific heat capacity:
\[C = \frac{S}{\mu} = \frac{1}{\mu}\frac{\Delta Q}{\Delta T} \tag{11.6}\]
Like \(s\), the molar specific heat \(C\) is independent of the amount of substance. It depends on the nature of the substance, its temperature, and — importantly — the conditions under which the heat is supplied. As we shall see shortly for gases, these conditions can make a large difference.
Predicting the specific heat of solids
We can actually predict the molar specific heat of a solid using the law of equipartition of energy, which you will meet in detail in Chapter 12. In brief, this law says that a system in thermal equilibrium shares its energy equally among all the independent ways (“modes”) in which it can store energy, each such quadratic mode carrying an average energy of \(\frac{1}{2}k_B T\), where \(k_B\) is Boltzmann’s constant.
Consider a solid made of \(N\) atoms, each vibrating about its fixed mean position. A vibration in one dimension stores energy in two ways — as kinetic energy and as potential energy — so its average energy is \(2 \times \frac{1}{2}k_B T = k_B T\). Since each atom can vibrate in three dimensions, its average energy is \(3k_B T\).
For one mole of the solid there are \(N_A\) (Avogadro number) atoms, so the total internal energy is
\[U = 3 k_B T \times N_A = 3RT\]
using the fact that Boltzmann’s constant times Avogadro’s number equals the universal gas constant, \(k_B N_A = R\).
For a solid, the volume change on heating is negligible, so at constant pressure \(\Delta Q = \Delta U + P\,\Delta V \approx \Delta U\). Therefore the molar specific heat of a solid is predicted to be
\[C = \frac{\Delta Q}{\Delta T} = \frac{\Delta U}{\Delta T} = 3R \tag{11.7}\]
Numerically, \(3R \approx 3 \times 8.31 = 24.9\ \text{J mol}^{-1}\,\text{K}^{-1}\).
Table 11.1 lists measured specific and molar heat capacities of some solids at atmospheric pressure and ordinary room temperature.
Table 11.1 — Specific and molar heat capacities of some solids at room temperature and atmospheric pressure
| Substance | Specific heat (\(\text{J kg}^{-1}\,\text{K}^{-1}\)) | Molar specific heat (\(\text{J mol}^{-1}\,\text{K}^{-1}\)) |
|---|---|---|
| Aluminium | 900.0 | 24.4 |
| Carbon | 506.5 | 6.1 |
| Copper | 386.4 | 24.5 |
| Lead | 127.7 | 26.5 |
| Silver | 236.1 | 25.5 |
| Tungsten | 134.4 | 24.9 |
As the table shows, the measured molar specific heats generally agree with the predicted value \(3R \approx 24.9\ \text{J mol}^{-1}\,\text{K}^{-1}\) at ordinary temperatures. Carbon is a clear exception (only \(6.1\)). The good agreement also breaks down for all solids at low temperatures.
The special case of water
The old unit of heat was the calorie. Originally, one calorie was defined as the amount of heat needed to raise the temperature of \(1\ \text{g}\) of water by \(1\,^\circ\text{C}\).
But more precise measurements revealed a subtlety: the specific heat of water is not perfectly constant — it varies slightly with temperature over the range \(0\) to \(100\,^\circ\text{C}\), as shown in Fig. 11.5.
Figure to come
Fig. 11.5 – A graph of the specific heat capacity of water (in cal g⁻¹ °C⁻¹) versus temperature from 0 to 100 °C, showing a shallow dip near the middle of the range.
Because of this variation, the calorie had to be pinned to a specific temperature interval.
Since heat is just a form of energy, it is preferable to measure it in the SI energy unit, the joule. In SI units, the specific heat capacity of water is
\[s_{\text{water}} = 4186\ \text{J kg}^{-1}\,\text{K}^{-1} = 4.186\ \text{J g}^{-1}\,\text{K}^{-1}\]
This also fixes the conversion \(1\ \text{cal} = 4.186\ \text{J}\). Historically, this conversion factor was called the mechanical equivalent of heat — the amount of work needed to produce one calorie of heat. But now that we use the joule for heat, work, and every other form of energy alike, this term is superfluous and need not be used.
Two specific heats for gases
As already noted, the specific heat of a substance depends on the conditions under which heat is supplied. For gases this matters enormously, because a gas expands strongly when heated. We therefore define two separate molar specific heats for a gas: one measured at constant volume (\(C_v\)) and one at constant pressure (\(C_p\)).
For an ideal gas, these two molar specific heats are linked by a beautifully simple relation:
\[C_p - C_v = R \tag{11.8}\]
where \(R\) is the universal gas constant.
Proving \(C_p - C_v = R\)
We begin with the First Law written for one mole of gas (Eq. 11.3):
\[\Delta Q = \Delta U + P\,\Delta V\]
At constant volume, \(\Delta V = 0\), so no work is done and all the heat goes into internal energy. Hence
\[C_v = \left(\frac{\Delta Q}{\Delta T}\right)_v = \left(\frac{\Delta U}{\Delta T}\right)_v = \frac{\Delta U}{\Delta T} \tag{11.9}\]
The subscript \(v\) (which just labels the quantity held fixed) can be dropped in the last step, because the internal energy \(U\) of an ideal gas depends only on its temperature — not on its volume or pressure.
At constant pressure, both terms survive:
\[C_p = \left(\frac{\Delta Q}{\Delta T}\right)_p = \left(\frac{\Delta U}{\Delta T}\right)_p + P\left(\frac{\Delta V}{\Delta T}\right)_p \tag{11.10}\]
Again, since \(U\) depends only on \(T\), the first term \(\left(\frac{\Delta U}{\Delta T}\right)_p\) is just \(\frac{\Delta U}{\Delta T} = C_v\).
Now use the ideal gas equation for one mole, \(PV = RT\). Holding \(P\) constant and looking at how \(V\) changes with \(T\) gives \(P\,\Delta V = R\,\Delta T\), that is,
\[P\left(\frac{\Delta V}{\Delta T}\right)_p = R \tag{11.11}\]
Putting Eqs. (11.9), (11.10), and (11.11) together, Eq. (11.10) becomes \(C_p = C_v + R\), which is exactly the desired relation, \(C_p - C_v = R\).
Looking ahead
We have now seen how much heat different substances need. Next, in Section 11.7, we look more carefully at the state variables that describe a system and at the equation of state that connects them.
11.7 Thermodynamic State Variables and Equation of State
We have already used quantities like pressure, volume, temperature, and internal energy to describe a gas. In this section we look more carefully at these describing quantities — the state variables — and at the relation that ties them together, the equation of state.
State variables describe equilibrium states
Every equilibrium state of a thermodynamic system is completely described by the specific values of a few macroscopic variables, called state variables (recall from Section 11.4 that a state variable depends only on the current state, not on how that state was reached).
For a gas, an equilibrium state is completely specified once we give its pressure, volume, temperature, and mass — together with its composition, if it is a mixture of gases. Fix these, and the state is settled.
But there is an important catch: a system is not always in an equilibrium state, and when it is not, state variables cannot describe it.
Consider a gas that is suddenly allowed to expand into a vacuum — a so-called free expansion. Imagine a box divided by a partition, with gas on one side and vacuum on the other; when the partition is suddenly removed, the gas rushes out to fill the whole box, as shown in Fig. 11.6(a). During this rapid expansion the pressure of the gas is not the same everywhere inside the box.
Figure to come
Fig. 11.6(a) – A box divided by a partition, gas on one side and vacuum on the other; the partition suddenly removed and gas rushing to fill the whole box (free expansion).
A similar thing happens in an explosive chemical reaction — for instance, a mixture of petrol vapour and air ignited by a spark, as shown in Fig. 11.6(b). During the explosion, the temperature and pressure are not uniform from point to point.
Figure to come
Fig. 11.6(b) – A container of petrol vapour and air being ignited by a spark, with a violent, non-uniform burst; temperature and pressure vary from point to point.
In both cases, because pressure and temperature have no single well-defined value throughout, we cannot assign the system one value of \(P\) or one value of \(T\) — so state variables simply do not apply. Only after the disturbance dies down, when the gas has settled to a uniform temperature and pressure and reached thermal and mechanical equilibrium with its surroundings, can we describe it by state variables once again.
In short, thermodynamic state variables describe the equilibrium states of a system.
The equation of state
The various state variables of a system are not all independent of one another. The relation that connects them is called the equation of state.
For an ideal gas, this relation is the familiar ideal gas equation:
\[PV = \mu R T\]
where \(P\) is the pressure (in \(\text{Pa}\)), \(V\) is the volume (in \(\text{m}^3\)), \(\mu\) is the number of moles of gas, \(R\) is the universal gas constant (\(8.31\ \text{J mol}^{-1}\,\text{K}^{-1}\)), and \(T\) is the absolute temperature (in \(\text{K}\)). (This is the general, \(\mu\)-mole form of the \(PV = RT\) relation we used for one mole while proving \(C_p - C_v = R\).)
The value of the equation of state is that it removes one variable from independent choice. For a fixed amount of gas (given \(\mu\)), there are therefore only two independent state variables — we may pick \(P\) and \(V\), or \(T\) and \(V\), and the third is then fixed by the equation.
If we plot the pressure against the volume of a gas while holding the temperature constant, the resulting curve is called an isotherm.
For an ideal gas, since \(PV = \text{constant}\) at fixed \(T\), an isotherm is a smooth hyperbola-shaped curve, as sketched in Fig. 11.6a. Real gases, whose molecules interact more strongly, obey more complicated equations of state, so their isotherms deviate from this simple shape.
Figure to come
Fig. 11.6a – A P–V graph showing a single smooth hyperbola-like curve labelled “isotherm (T constant)”, pressure falling as volume rises.
Extensive and intensive variables
Thermodynamic state variables come in two kinds — extensive and intensive — and telling them apart is a useful skill.
Extensive variables reflect the “size” or amount of the system; intensive variables do not. There is a simple test to classify any variable. Imagine a system in equilibrium and mentally divide it into two equal halves. Then ask what happens to the variable:
- If its value gets halved in each part, it is an extensive variable.
- If its value stays unchanged in each part, it is an intensive variable.
For instance, split a container of gas into two equal halves: each half has half the volume, half the mass, and half the internal energy — so \(V\), \(M\), and \(U\) are extensive. But each half is still at the same pressure and the same temperature as before — so \(P\) and \(T\) are intensive.
This classification is more than a curiosity — it is a handy consistency check for thermodynamic equations. Every valid equation must have the same character (extensive or intensive) on both sides. Take the First Law in the form
\[\Delta Q = \Delta U + P\,\Delta V\]
Here \(\Delta U\) is extensive and \(\Delta V\) is extensive, while \(P\) is intensive. The product of an intensive variable \(P\) and an extensive quantity \(\Delta V\) is extensive, so the whole right-hand side is extensive — and therefore the left-hand side, \(\Delta Q\), must be extensive too, which it is.
Looking ahead
We now know how to describe the equilibrium states of a system and how its state variables are linked. The next natural step is to study the processes by which a system moves from one equilibrium state to another — beginning with the idealised quasi-static process in Section 11.8.
11.8 Thermodynamic Processes
So far we have described the states of a system. We now turn to the processes by which a system moves from one equilibrium state to another — and to a special idealised way of carrying out such a process that makes it easy to analyse.
11.8.1 Quasi-static Process
Consider a gas that is in both thermal and mechanical equilibrium with its surroundings. Being in mechanical equilibrium means the pressure of the gas equals the external pressure pushing on it; being in thermal equilibrium means its temperature equals that of the surroundings.
Now suppose we suddenly disturb this balance — for example, by abruptly lifting the weight sitting on the movable piston, so the external pressure drops all at once. The piston accelerates outward, and the gas rushes to expand.
During such a sudden change, the gas passes through states that are not equilibrium states. In these non-equilibrium states, the pressure and temperature are not the same everywhere in the gas, so they are not even well-defined for the system as a whole.
The same trouble arises with heat. If there is a large (finite) temperature difference between the gas and its surroundings, heat rushes across quickly, and again the gas is thrown into non-equilibrium states while it settles. Eventually, of course, the gas will calm down and reach a final equilibrium state with a well-defined temperature and pressure equal to those of the surroundings. The free expansion of a gas into vacuum and the explosive chemical reaction mentioned in Section 11.7 are further examples where the system passes through such non-equilibrium states.
The problem is that non-equilibrium states are very difficult to deal with — we cannot even assign them a single pressure or temperature, so our state variables and equations do not apply to them.
The idealised, infinitely slow process
To get around this difficulty, we imagine an idealised process in which the system is in an equilibrium state at every stage. For this to be possible, the process must be carried out infinitely slowly. This is called a quasi-static process — the word “quasi-static” means “nearly static.”
The trick is slowness. If we change the system’s variables (\(P\), \(V\), \(T\)) slowly enough, the system is never pushed out of balance: it stays in thermal and mechanical equilibrium with its surroundings throughout the change. At every stage, the pressure of the system differs from the external pressure by only an infinitesimally small amount, and likewise the temperature of the system differs from that of the surroundings only infinitesimally, as suggested in Fig. 11.7.
Figure to come
Fig. 11.7 – A gas at state (P, T) in a cylinder, surrounded by a reservoir at temperature T + ΔT and an external pressure P + ΔP, the differences being only infinitesimal, illustrating a quasi-static process.
How would we actually take a gas from a state \((P, T)\) to a new state \((P', T')\) quasi-statically? To change the pressure, we alter the external pressure by a tiny amount, let the gas settle so its pressure matches, then repeat — continuing infinitely slowly until the pressure reaches \(P'\). To change the temperature, we place the gas in contact with a series of reservoirs whose temperatures differ from it by only infinitesimal steps, moving gradually from \(T\) to \(T'\).
A quasi-static process is, of course, a hypothetical idealisation — no real process is truly infinitely slow. But it is a very useful one. In practice, any process that is sufficiently slow and avoids sudden effects — no accelerated piston, no large temperature gradients — is a reasonable approximation to an ideal quasi-static process. From here on, unless we say otherwise, we shall deal only with quasi-static processes.
Special kinds of quasi-static processes
Among quasi-static processes, four special types occur again and again, each defined by one quantity being held fixed (or one condition imposed).
A process in which the temperature of the system is kept constant throughout is called an isothermal process. A good example is the slow expansion of a gas in a metallic cylinder placed inside a large reservoir kept at a fixed temperature. (Heat flows between the reservoir and the gas, but because the reservoir has an enormous heat capacity, its own temperature hardly changes — so the gas stays at the reservoir’s fixed temperature.)
A process in which the pressure is kept constant is called an isobaric process.
A process in which the volume is kept constant is called an isochoric process.
Finally, if the system is insulated from its surroundings so that no heat flows in or out, the process is called an adiabatic process.
These four special processes are summarised in Table 11.2.
Table 11.2 — Some special thermodynamic processes
| Type of process | Feature |
|---|---|
| Isothermal | Temperature constant |
| Isobaric | Pressure constant |
| Isochoric | Volume constant |
| Adiabatic | No heat flow between system and surroundings (\(\Delta Q = 0\)) |
We now consider these processes one by one in more detail, beginning with the isothermal process in the next section.
11.8.2 Isothermal Process
We begin our detailed study of thermodynamic processes with the isothermal process — a quasi-static process in which the temperature of the gas is held constant throughout (recall its definition from Section 11.8.1).
Pressure and volume in an isothermal process
For an ideal gas at fixed temperature \(T\), the equation of state \(PV = \mu RT\) has a constant right-hand side (since \(\mu\), \(R\), and \(T\) are all fixed). Therefore
\[PV = \text{constant}\]
This tells us that at constant temperature, the pressure of a given mass of gas varies inversely with its volume — squeeze it into half the volume and its pressure doubles. This is exactly Boyle’s Law, which you met in an earlier class.
Work done in an isothermal process
Now let us calculate the work done by an ideal gas as it expands isothermally from an initial state \((P_1, V_1)\) to a final state \((P_2, V_2)\), all at the same temperature \(T\).
Consider any intermediate stage, where the gas has pressure \(P\) and its volume increases by a small amount from \(V\) to \(V + \Delta V\) (with \(\Delta V\) very small). The small work done by the gas in this step is
\[\Delta W = P\,\Delta V\]
To get the total work, we let \(\Delta V \to 0\) and add up all these tiny contributions across the whole process — which is exactly what integration does:
\[W = \int_{V_1}^{V_2} P\,dV\]
The pressure \(P\) changes as the gas expands, so we cannot pull it out of the integral directly. Instead, we use the ideal gas equation to write \(P = \dfrac{\mu RT}{V}\). Since \(\mu\), \(R\), and \(T\) are constant in an isothermal process, they can come out of the integral:
\[W = \int_{V_1}^{V_2} \frac{\mu RT}{V}\,dV = \mu RT \int_{V_1}^{V_2} \frac{dV}{V} = \mu RT \,\ln\!\left(\frac{V_2}{V_1}\right) \tag{11.12}\]
Here we used the standard result that the integral of \(\dfrac{1}{V}\) is the natural logarithm \(\ln V\), so evaluating between the limits gives \(\ln V_2 - \ln V_1 = \ln(V_2/V_1)\). In this formula \(W\) is in joules (J), \(\mu\) is the number of moles, \(R = 8.31\ \text{J mol}^{-1}\,\text{K}^{-1}\), \(T\) is in kelvin (K), and \(\ln\) denotes the natural logarithm (base \(e\)).
Figure to come
Fig. 11.7a – A P–V graph showing a single isotherm falling from (P1, V1) to (P2, V2), with the area under the curve between V1 and V2 shaded to represent the work done by the gas.
The work done is just the area under the isotherm between \(V_1\) and \(V_2\), as shaded in Fig. 11.7a.
Heat and internal energy in an isothermal process
There is a crucial simplification for an ideal gas. Its internal energy depends only on its temperature. So if the temperature does not change during an isothermal process, the internal energy does not change either:
\[\Delta U = 0 \quad \text{(isothermal, ideal gas)}\]
Now apply the First Law, \(\Delta Q = \Delta U + \Delta W\). With \(\Delta U = 0\), it reduces to
\[Q = W\]
So in an isothermal process, all the heat supplied to the gas is converted entirely into work done by the gas — none of it is stored as internal energy.
Finally, look at the sign of the work in Equation (11.12). If the gas expands, \(V_2 > V_1\), so \(\ln(V_2/V_1) > 0\) and \(W > 0\): the gas does positive work and, by \(Q = W\), absorbs heat. If the gas is compressed, \(V_2 < V_1\), so \(\ln(V_2/V_1) < 0\) and \(W < 0\): work is done on the gas by the surroundings, and an equal amount of heat is released by the gas.
We next study the adiabatic process, in which the gas is thermally insulated so that no heat can flow in or out.
11.8.3 Adiabatic Process
In the isothermal process, heat could flow freely so that the temperature stayed fixed. We now look at the opposite extreme — the adiabatic process — in which the system is thermally insulated so that no heat flows in or out at all (recall its definition from Section 11.8.1). In symbols, \(\Delta Q = 0\) throughout.
What happens to internal energy
Since no heat enters or leaves, the First Law \(\Delta Q = \Delta U + \Delta W\) becomes simply \(0 = \Delta U + \Delta W\), that is,
\[\Delta U = -\Delta W\]
So any work done by the gas comes entirely out of its own internal energy. When the gas expands adiabatically (\(\Delta W > 0\)), its internal energy must fall — and for an ideal gas, a fall in internal energy means a fall in temperature. Conversely, if work is done on the gas (compression), its internal energy and temperature rise.
The adiabatic relation between P and V
For a quasi-static adiabatic process of an ideal gas, pressure and volume are linked by a relation that we quote here without proof (you will derive it in higher courses):
\[PV^{\gamma} = \text{constant} \tag{11.13}\]
Here \(\gamma\) (gamma) is the ratio of the specific heats at constant pressure and constant volume:
\[\gamma = \frac{C_p}{C_v}\]
Because Equation (11.13) says \(PV^{\gamma}\) stays constant, if an ideal gas changes adiabatically from state \((P_1, V_1)\) to state \((P_2, V_2)\), we can equate the value of \(PV^{\gamma}\) at the two states:
\[P_1 V_1^{\gamma} = P_2 V_2^{\gamma} \tag{11.14}\]
Figure 11.8 shows the \(P\)–\(V\) curves of an ideal gas: two adiabatic curves connecting two isotherms. Notice that each adiabatic curve is steeper than an isotherm passing through the same point.
Figure to come
Fig. 11.8 – A P–V graph showing two isotherms (labelled) and two steeper adiabatic curves connecting them, illustrating that adiabatics fall more steeply than isotherms.
Work done in an adiabatic process
Let us calculate the work done by an ideal gas in an adiabatic change from state \((P_1, V_1, T_1)\) to state \((P_2, V_2, T_2)\). As before, the work is the area under the \(P\)–\(V\) curve:
\[W = \int_{V_1}^{V_2} P\, dV \tag{11.15}\]
From Equation (11.13), \(P = \dfrac{\text{constant}}{V^{\gamma}}\), so
\[W = \text{constant} \times \int_{V_1}^{V_2} \frac{dV}{V^{\gamma}} = \text{constant} \times \left[\frac{V^{-\gamma + 1}}{1 - \gamma}\right]_{V_1}^{V_2}\]
using the standard integration result \(\int V^{-\gamma}\, dV = \dfrac{V^{\,1-\gamma}}{1-\gamma}\). Evaluating between the limits,
\[W = \frac{\text{constant}}{1 - \gamma}\left[\frac{1}{V_2^{\,\gamma - 1}} - \frac{1}{V_1^{\,\gamma - 1}}\right]\]
Now, from Equation (11.14), the “constant” equals \(P_1 V_1^{\gamma}\) (or equally \(P_2 V_2^{\gamma}\)). Substituting the appropriate one into each term,
\[W = \frac{1}{1 - \gamma}\left[\frac{P_2 V_2^{\gamma}}{V_2^{\,\gamma - 1}} - \frac{P_1 V_1^{\gamma}}{V_1^{\,\gamma - 1}}\right] = \frac{1}{1 - \gamma}\left[P_2 V_2 - P_1 V_1\right]\]
because \(\dfrac{V^{\gamma}}{V^{\gamma - 1}} = V\). Finally, using the ideal gas equation \(P_1 V_1 = \mu R T_1\) and \(P_2 V_2 = \mu R T_2\),
\[W = \frac{\mu R (T_1 - T_2)}{\gamma - 1} \tag{11.16}\]
where \(W\) is in joules (J), \(\mu\) is the number of moles, \(R = 8.31\ \text{J mol}^{-1}\,\text{K}^{-1}\), \(T_1\) and \(T_2\) are the initial and final temperatures in kelvin (K), and \(\gamma\) is dimensionless.
This result confirms what we argued from the First Law. If the gas does work in expanding (\(W > 0\)), Equation (11.16) requires \(T_1 > T_2\), i.e. \(T_2 < T_1\) — the gas cools. If work is done on the gas in compressing it (\(W < 0\)), then \(T_2 > T_1\) — the temperature of the gas rises.
We turn next to the isochoric process, in which the volume of the gas is held constant.
11.8.4 Isochoric Process
We now consider the isochoric process — a quasi-static process in which the volume of the gas is held constant throughout (recall its definition from Section 11.8.1). The word comes from “iso” (same) and “choric” (space/volume): the volume stays the same.
No work is done
Because the volume does not change, \(\Delta V = 0\). But we saw earlier that the work done by a gas is \(\Delta W = P\,\Delta V\). With no change in volume, this work is zero:
\[\Delta W = P\,\Delta V = 0 \quad \text{(isochoric)}\]
So in an isochoric process the gas neither does work on its surroundings nor has work done on it. The piston (if any) simply does not move.
All the heat goes into internal energy
Now apply the First Law, \(\Delta Q = \Delta U + \Delta W\). Since \(\Delta W = 0\), it becomes simply
\[\Delta Q = \Delta U \quad \text{(isochoric)}\]
This means that the entire heat supplied to the gas goes into increasing its internal energy — and hence its temperature. None of it is “wasted” in doing work.
How much the temperature rises for a given amount of heat is decided by the specific heat of the gas at constant volume, \(C_v\) (Section 11.6). For \(\mu\) moles of gas, the heat supplied at constant volume is
\[\Delta Q = \mu\, C_v\, \Delta T\]
where \(\Delta Q\) is in joules (J), \(\mu\) is the number of moles, \(C_v\) is the molar specific heat at constant volume (in \(\text{J mol}^{-1}\,\text{K}^{-1}\)), and \(\Delta T\) is the temperature rise (in K).
On a \(P\)–\(V\) diagram, an isochoric process is a vertical straight line, since the volume stays fixed while the pressure changes, as shown in Fig. 11.8a.
Figure to come
Fig. 11.8a – A P–V graph showing a vertical straight line at fixed volume, pressure rising from a lower to a higher value, representing an isochoric process.
We turn next to the isobaric process, in which the pressure of the gas is held constant.
11.8.5 Isobaric Process
In the isochoric process the volume was fixed, so no work was done. We now consider the isobaric process — a quasi-static process in which the pressure of the gas is held constant throughout (recall its definition from Section 11.8.1). Here the gas is free to expand or contract, so work is done once again.
Work done in an isobaric process
Since the pressure \(P\) is constant, it can be taken straight out of the work integral. The work done by the gas as its volume changes from \(V_1\) to \(V_2\) is therefore
\[W = \int_{V_1}^{V_2} P\, dV = P(V_2 - V_1)\]
We can rewrite this in terms of temperature using the ideal gas equation. At the constant pressure \(P\), the initial and final states obey \(PV_1 = \mu R T_1\) and \(PV_2 = \mu R T_2\). Subtracting,
\[P(V_2 - V_1) = PV_2 - PV_1 = \mu R T_2 - \mu R T_1\]
which gives the compact result
\[W = P(V_2 - V_1) = \mu R (T_2 - T_1) \tag{11.17}\]
Here \(W\) is in joules (J), \(P\) in pascals (Pa), \(V_1\) and \(V_2\) in \(\text{m}^3\), \(\mu\) is the number of moles, \(R = 8.31\ \text{J mol}^{-1}\,\text{K}^{-1}\), and \(T_1\), \(T_2\) are the initial and final temperatures in kelvin (K).
On a \(P\)–\(V\) diagram, an isobaric process is a horizontal straight line, and the work done is the area of the rectangle beneath it, as shown in Fig. 11.8b.
Figure to come
Fig. 11.8b – A P–V graph showing a horizontal line at constant pressure from V1 to V2, with the rectangular area beneath it shaded to represent the work P(V2 − V1).
Heat, work, and internal energy
In an isobaric process the temperature changes (unlike in the isothermal case), so the internal energy also changes. Applying the First Law, \(\Delta Q = \Delta U + W\): the heat supplied is now shared between two jobs — part of it raises the internal energy of the gas, and the rest is spent doing the expansion work.
How much the temperature rises for a given amount of heat is decided by the specific heat of the gas at constant pressure, \(C_p\) (Section 11.6). For \(\mu\) moles of gas, the heat supplied at constant pressure is
\[\Delta Q = \mu\, C_p\, \Delta T\]
Finally, we consider the cyclic process, in which the system is brought back to its starting state.
11.8.6 Cyclic Process
The processes so far were each defined by holding one quantity fixed — temperature, pressure, or volume — or by preventing heat flow. The cyclic process is defined differently: not by what is held fixed, but by where the system ends up.
In a cyclic process, the system undergoes a series of changes but is finally brought back to its initial state, so that all its state variables regain their starting values.
Internal energy over a cycle
Here the key property of internal energy comes into play. Internal energy \(U\) is a state variable — it depends only on the state of the system, not on the path taken. Since a cyclic process returns the system to exactly the same state it started in, the internal energy returns to its starting value. Therefore, over one complete cycle,
\[\Delta U = 0 \quad \text{(cyclic process)}\]
Heat and work over a cycle
Now apply the First Law, \(\Delta Q = \Delta U + \Delta W\), to a full cycle. Since \(\Delta U = 0\), we are left with
\[\Delta Q = \Delta W \quad \text{(cyclic process)}\]
So over one complete cycle, the total heat absorbed by the system equals the total work done by the system. The system takes in heat and gives out an equal amount of work, again and again, cycle after cycle.
On a \(P\)–\(V\) diagram, a cyclic process appears as a closed loop, since the system returns to its starting point. The net work done by the gas in one cycle equals the area enclosed by this loop, as shown in Fig. 11.8c.
Figure to come
Fig. 11.8c – A P–V graph showing a closed loop representing a cyclic process, with the area enclosed by the loop shaded to indicate the net work done by the gas in one cycle.
Looking ahead
The cyclic process shows how a system can convert heat into work over and over — the very idea behind a heat engine. But a crucial question now arises: can such an engine convert all the heat it absorbs into work? The First Law alone does not forbid it. The answer, as we shall see, is no — and understanding why leads us to the Second Law of Thermodynamics, the subject of the next section.
11.9 Second Law of Thermodynamics
The First Law of Thermodynamics is simply the principle of conservation of energy: energy is never created or destroyed. But the First Law, powerful as it is, is not the whole story. Everyday experience shows there are many processes that would perfectly obey energy conservation and yet are never seen to happen.
A process the First Law allows but nature forbids
Consider a simple example. Nobody has ever seen a book lying quietly on a table suddenly leap up into the air by itself. Yet such a jump would not violate energy conservation at all. The table could cool very slightly, converting a tiny part of its internal energy into an equal amount of mechanical energy, which the book could use to rise to a height whose potential energy exactly matches. Energy would balance perfectly.
But this never happens. Clearly, some additional law of nature forbids it — a law that rules out many processes even though they satisfy the First Law. This additional principle is the Second Law of Thermodynamics.
What the Second Law limits: engines and refrigerators
The Second Law has enormous practical importance, because it sets a fundamental limit on two vital kinds of machines: heat engines and refrigerators.
A heat engine is a device that absorbs heat from a hot source, converts part of it into useful work, and rejects the rest to a cold sink — repeating this in a cycle, as sketched in Fig. 11.8d. Its efficiency is the fraction of the absorbed heat that it manages to turn into work.
Figure to come
Fig. 11.8d – A schematic heat engine: a hot reservoir at temperature T1 supplying heat Q1 to an engine, which delivers work W and rejects heat Q2 to a cold reservoir at temperature T2.
The Second Law says the efficiency of a heat engine can never be \(1\) (i.e. never 100%): some heat must always be rejected to the cold sink and cannot be converted into work.
A refrigerator (or heat pump) does the reverse. Using an external supply of work, it pumps heat from a cold body to a hot body — the opposite of the natural direction, as sketched in Fig. 11.8e. Its effectiveness is measured by its coefficient of performance — roughly, how much heat it extracts from the cold body per unit of work supplied.
Figure to come
Fig. 11.8e – A schematic refrigerator: work W supplied to a device that extracts heat Q2 from a cold reservoir at T2 and delivers heat Q1 to a hot reservoir at T1.
The Second Law says the coefficient of performance of a refrigerator can never be infinite: moving heat from cold to hot always requires some work — it can never be done for free.
These two limitations are captured precisely by two classic statements of the Second Law — one due to Kelvin and Planck (about engines), the other due to Clausius (about refrigerators).
The Kelvin–Planck statement
In plain terms: you cannot build a “perfect” heat engine that takes in heat and turns all of it into work with no other effect. Some heat must always be given up to a colder body. This is exactly why no engine can be 100% efficient.
The Clausius statement
In plain terms: heat will not flow on its own from cold to hot. To force it in that direction — as a refrigerator does — you must supply external work. There is no “perfect” refrigerator that moves heat from cold to hot for free.
The two statements are equivalent
Although the Kelvin–Planck statement talks about engines and the Clausius statement talks about refrigerators, it can be proved that the two are completely equivalent — if a device could violate one, it could be combined with an ordinary machine to violate the other as well. They are simply two faces of the same underlying law.
Looking ahead
The Second Law tells us that real engines must always waste some heat and that heat has a natural one-way direction. This one-wayness is deeply connected to the idea of reversible and irreversible processes, which we examine in the next section.
11.10 Reversible and Irreversible Processes
The Second Law told us that natural processes have a preferred direction — heat flows hot to cold, engines waste some heat, and so on. This “one-wayness” of nature is captured precisely by the ideas of reversible and irreversible processes.
The question of reversibility
Imagine a thermodynamic system going from an initial state \(i\) to a final state \(f\). During this process the system absorbs heat \(Q\) from the surroundings and does work \(W\) on them. Now ask a sharp question: can we reverse this process so as to bring both the system and the surroundings back to their initial states, leaving no other change anywhere else in the universe?
Experience says that for most processes in nature, the answer is no. The spontaneous processes of nature are irreversible — they run in one direction and do not undo themselves.
Everyday examples of irreversibility
Several familiar processes make this clear.
The base of a cooking vessel taken off a stove is hotter than its rim and sides. Heat then flows from the hot base through the whole vessel until it reaches a uniform temperature (and eventually cools to room temperature). This never runs backward: one part of the vessel will not spontaneously cool down and pour its heat back into the base. If it did, it would violate the Second Law of Thermodynamics.
The free expansion of a gas into vacuum is irreversible — the gas will not gather itself back into one half of the container on its own. The combustion of a petrol–air mixture ignited by a spark cannot be un-burned. And cooking gas leaking from a cylinder diffuses throughout the kitchen; those gas molecules will never spontaneously stream back into the cylinder.
Figure to come
Fig. 11.8f – A gas cylinder with gas molecules spreading out to fill an entire room (one large arrow pointing outward), and a crossed-out reverse arrow showing that the gas never spontaneously collects back into the cylinder — illustrating irreversibility.
One more example: if we stir a liquid that is in thermal contact with a reservoir, the work of stirring is converted into heat, raising the internal energy of the reservoir. This cannot be exactly reversed — undoing it would mean converting that heat entirely back into work, which the Second Law forbids.
In short, irreversibility is the rule in nature, not the exception.
Why processes are irreversible: two causes
Irreversibility arises mainly from two causes.
First, many processes pass through non-equilibrium states. A free expansion or an explosive chemical reaction throws the system into states where pressure and temperature are not even well-defined. A process that races through such non-equilibrium states cannot be retraced step-by-step in reverse.
Second, most processes involve dissipative effects — friction, viscosity, and the like. When a moving body slides to a stop, its ordered mechanical energy is lost as heat to the floor and the body. When a rotating blade in a liquid slows and stops due to viscosity, its mechanical energy is converted into internal energy of the liquid. Such conversions of ordered energy into disordered heat cannot be spontaneously undone.
Because dissipative effects like friction and viscosity are present everywhere and can be reduced but never entirely eliminated, almost every real process we deal with is irreversible.
The reversible process — an idealisation
We can now define the opposite, ideal case.
Both conditions are essential: the process must be quasi-static and completely free of dissipation. A good example is the quasi-static isothermal expansion of an ideal gas in a cylinder fitted with a perfectly frictionless movable piston — carried out infinitely slowly and without any friction, it is a reversible process.
A reversible process is, of course, an idealised notion. No real process meets both conditions perfectly, but the concept is extremely useful as a limiting, best-possible case.
Why reversibility matters: the best possible engine
Why is reversibility such a central idea? Because one of the main concerns of thermodynamics is the efficiency with which heat can be converted into work.
The Second Law has already ruled out a perfect heat engine with 100% efficiency. But that leaves a crucial question: what is the highest efficiency possible for an engine working between two reservoirs at temperatures \(T_1\) and \(T_2\)?
It turns out that an engine built entirely from idealised reversible processes achieves the maximum efficiency possible between those two temperatures. Every other engine — one involving irreversibility in any way, as all practical engines do — must have a lower efficiency than this limit.
Looking ahead
This remarkable result points directly to the next section. If a reversible engine is the most efficient possible, what does such an engine look like, and exactly how efficient is it? These questions are answered by the Carnot engine, which we study next.
11.11 Carnot Engine
In the previous section we learned that a reversible engine achieves the highest possible efficiency between two temperatures. Now we build that ideal engine explicitly and find exactly how efficient it can be.
Suppose we have a hot reservoir at temperature \(T_1\) and a cold reservoir at temperature \(T_2\). What is the maximum efficiency possible for a heat engine working between these two reservoirs, and what cycle of processes achieves it?
Why the ideal engine must use only isothermal and adiabatic steps
We expect the ideal engine to be a reversible engine, since irreversibility (from dissipative effects) always lowers efficiency. Recall that a process is reversible only if it is quasi-static and non-dissipative — and that a process is not quasi-static if there is any finite temperature difference between the system and the reservoir.
This has an important consequence. When the engine exchanges heat with a reservoir, it must do so at the same temperature as that reservoir (no finite temperature difference allowed) — that is, isothermally. So two of the steps must be isothermal: one at \(T_1\) absorbing heat \(Q_1\) from the hot reservoir, and one at \(T_2\) releasing heat \(Q_2\) to the cold reservoir.
But to complete a cycle, the gas must also be carried from temperature \(T_1\) down to \(T_2\) and back up from \(T_2\) to \(T_1\). Which reversible process can change the temperature? Since our engine has only two reservoirs, it cannot exchange heat during these temperature changes without introducing finite temperature differences. The only choice is an adiabatic process, in which no heat flows at all.
The Carnot cycle
A reversible heat engine operating between two temperatures is called a Carnot engine, and its cycle of four steps is the Carnot cycle, shown in Fig. 11.9. We take the working substance to be an ideal gas.
Figure to come
Fig. 11.9 – A P–V diagram of the Carnot cycle for an ideal gas: state 1 (P1,V1,T1) → 2 (P2,V2,T1) isothermal expansion, 2 → 3 (P3,V3,T2) adiabatic expansion, 3 → 4 (P4,V4,T2) isothermal compression, 4 → 1 adiabatic compression, forming a closed loop.
Step 1 → 2 (Isothermal expansion at \(T_1\)). The gas expands isothermally from \((P_1, V_1, T_1)\) to \((P_2, V_2, T_1)\), absorbing heat \(Q_1\) from the hot reservoir. Since the process is isothermal, the heat absorbed equals the work done by the gas (Eq. 11.12):
\[W_{1 \to 2} = Q_1 = \mu R T_1 \ln\!\left(\frac{V_2}{V_1}\right) \tag{11.18}\]
Step 2 → 3 (Adiabatic expansion). The gas expands adiabatically from \((P_2, V_2, T_1)\) to \((P_3, V_3, T_2)\), cooling from \(T_1\) to \(T_2\). Using the adiabatic work formula (Eq. 11.16), the work done by the gas is
\[W_{2 \to 3} = \frac{\mu R (T_1 - T_2)}{\gamma - 1} \tag{11.19}\]
Step 3 → 4 (Isothermal compression at \(T_2\)). The gas is compressed isothermally from \((P_3, V_3, T_2)\) to \((P_4, V_4, T_2)\), releasing heat \(Q_2\) to the cold reservoir. Here the work is done on the gas by the surroundings:
\[W_{3 \to 4} = Q_2 = \mu R T_2 \ln\!\left(\frac{V_3}{V_4}\right) \tag{11.20}\]
Step 4 → 1 (Adiabatic compression). The gas is compressed adiabatically from \((P_4, V_4, T_2)\) back to \((P_1, V_1, T_1)\), warming from \(T_2\) to \(T_1\). The work done on the gas is
\[W_{4 \to 1} = \frac{\mu R (T_1 - T_2)}{\gamma - 1} \tag{11.21}\]
Net work and efficiency
Notice that the two adiabatic works, \(W_{2 \to 3}\) and \(W_{4 \to 1}\), are equal in magnitude — one is done by the gas and the other on the gas — so they cancel. The total work done by the gas in one complete cycle is therefore
\[W = W_{1 \to 2} + W_{2 \to 3} - W_{3 \to 4} - W_{4 \to 1} = \mu R T_1 \ln\!\left(\frac{V_2}{V_1}\right) - \mu R T_2 \ln\!\left(\frac{V_3}{V_4}\right) \tag{11.22}\]
The efficiency \(\eta\) of any heat engine is the fraction of the absorbed heat that is turned into useful work.
Using \(\eta = 1 - Q_2/Q_1\) with \(Q_1\) and \(Q_2\) from Eqs. (11.18) and (11.20),
\[\eta = \frac{W}{Q_1} = 1 - \frac{Q_2}{Q_1} = 1 - \frac{T_2}{T_1}\,\frac{\ln(V_3/V_4)}{\ln(V_2/V_1)} \tag{11.23}\]
This still contains volume ratios — but they simplify beautifully using the adiabatic steps.
Since step 2 → 3 is adiabatic, \(TV^{\gamma - 1} = \text{constant}\) gives \(T_1 V_2^{\gamma - 1} = T_2 V_3^{\gamma - 1}\), that is,
\[\frac{V_2}{V_3} = \left(\frac{T_2}{T_1}\right)^{1/(\gamma - 1)} \tag{11.24}\]
Similarly, since step 4 → 1 is adiabatic, \(T_2 V_4^{\gamma - 1} = T_1 V_1^{\gamma - 1}\), that is,
\[\frac{V_1}{V_4} = \left(\frac{T_2}{T_1}\right)^{1/(\gamma - 1)} \tag{11.25}\]
The right-hand sides of Eqs. (11.24) and (11.25) are identical, so \(\dfrac{V_2}{V_3} = \dfrac{V_1}{V_4}\), which rearranges to
\[\frac{V_3}{V_4} = \frac{V_2}{V_1} \tag{11.26}\]
Therefore \(\ln(V_3/V_4) = \ln(V_2/V_1)\), and the volume-ratio factor in Eq. (11.23) becomes exactly 1. The efficiency collapses to a strikingly simple result:
\[\eta = 1 - \frac{T_2}{T_1} \quad \text{(Carnot engine)} \tag{11.27}\]
The Carnot engine reversed: a refrigerator
A Carnot engine is reversible — in fact, it is the only reversible engine possible between two reservoirs at different temperatures. Each step of the Carnot cycle can be run backward. Reversing the whole cycle means taking heat \(Q_2\) from the cold reservoir at \(T_2\), having work \(W\) done on the system, and delivering heat \(Q_1\) to the hot reservoir at \(T_1\). This reversed Carnot engine is a reversible refrigerator.
Carnot’s theorem
We now state a fundamental result, sometimes called Carnot’s theorem.
Proof of part (a). Imagine a reversible (Carnot) engine \(R\) and an irreversible engine \(I\) working between the same hot source and cold sink. Couple them so that \(I\) runs as an engine while \(R\) runs as a refrigerator, as shown in Fig. 11.10.
Figure to come
Fig. 11.10 – An irreversible engine I (absorbing Q1 from the hot reservoir at T1, delivering work W′, rejecting Q1 − W′ to the cold reservoir at T2) coupled to a reversible refrigerator R (driven by work W, taking Q2 from the cold reservoir, returning Q1 to the hot reservoir).
Let \(I\) absorb heat \(Q_1\) from the source, deliver work \(W'\), and reject \(Q_1 - W'\) to the sink. Arrange \(R\) so that it returns the same heat \(Q_1\) to the source, taking heat \(Q_2\) from the sink and requiring work \(W = Q_1 - Q_2\) to run it.
Now suppose, contrary to the theorem, that \(I\) were more efficient than \(R\), i.e. \(\eta_I > \eta_R\). Then for the same \(Q_1\), engine \(I\) would deliver more work: \(W' > W\). Since \(R\) runs as a refrigerator, it draws \(Q_2 = Q_1 - W > Q_1 - W'\) from the sink.
Consider the coupled \(I\)–\(R\) system as a whole. The source is left unchanged (it gives \(Q_1\) to \(I\) and gets \(Q_1\) back from \(R\)). The net heat extracted from the cold sink is \((Q_1 - W) - (Q_1 - W') = W' - W\), and this appears entirely as net work output — with no other change anywhere. But that means the coupled system converts heat drawn from a single (cold) reservoir entirely into work, which directly violates the Kelvin–Planck statement of the Second Law.
Therefore the assumption \(\eta_I > \eta_R\) must be wrong. No engine can be more efficient than a Carnot engine working between the same two temperatures.
A similar argument shows part (b): a reversible engine using one substance cannot be more efficient than one using another. Hence the Carnot efficiency, Eq. (11.27), is independent of the working substance.
A universal temperature scale
The independence of the working substance leads to one more deep result. For a Carnot cycle,
\[\frac{Q_1}{Q_2} = \frac{T_1}{T_2} \tag{11.28}\]
where \(Q_1\) and \(Q_2\) are the heats absorbed and released isothermally at the hot and cold reservoirs. Because this relation holds for any Carnot engine regardless of substance, it can be used to define a truly universal thermodynamic temperature scale — one that does not depend on the properties of any particular material. For an ideal gas as the working substance, this thermodynamic temperature turns out to be the same as the ideal gas temperature scale established earlier.
Looking ahead
With the Carnot engine we have reached the summit of classical thermodynamics: a precise upper limit on how efficiently heat can be turned into work, set purely by the two temperatures involved. The chapter summary and the end-of-chapter exercises that follow will help you consolidate and apply everything from the Zeroth Law through to the Carnot cycle.
11.12 Summary
The Zeroth Law of Thermodynamics states that two systems in thermal equilibrium with a third system separately are in thermal equilibrium with each other. The Zeroth Law leads to the concept of temperature.
The internal energy of a system is the sum of the kinetic energies and potential energies of the molecular constituents of the system. It does not include the overall kinetic energy of the system. Heat and work are two modes of energy transfer to the system. Heat is the energy transfer arising due to a temperature difference between the system and the surroundings. Work is energy transfer brought about by other means, such as moving the piston of a cylinder containing the gas, by raising or lowering some weight connected to it.
The First Law of Thermodynamics is the general law of conservation of energy applied to any system in which energy transfer from or to the surroundings (through heat and work) is taken into account. It states that \[\Delta Q = \Delta U + \Delta W\] where \(\Delta Q\) is the heat supplied to the system, \(\Delta W\) is the work done by the system, and \(\Delta U\) is the change in internal energy of the system.
The specific heat capacity of a substance is defined by \[s = \frac{1}{m}\frac{\Delta Q}{\Delta T}\] where \(m\) is the mass of the substance and \(\Delta Q\) is the heat required to change its temperature by \(\Delta T\). The molar specific heat capacity of a substance is defined by \[C = \frac{1}{\mu}\frac{\Delta Q}{\Delta T}\] where \(\mu\) is the number of moles of the substance. For a solid, the law of equipartition of energy gives \[C = 3R\] which generally agrees with experiment at ordinary temperatures. Calorie is the old unit of heat. One calorie is the amount of heat required to raise the temperature of \(1\ \text{g}\) of water from \(14.5\,^\circ\text{C}\) to \(15.5\,^\circ\text{C}\). \(1\ \text{cal} = 4.186\ \text{J}\).
For an ideal gas, the molar specific heat capacities at constant pressure and volume satisfy the relation \[C_p - C_v = R\] where \(R\) is the universal gas constant.
Equilibrium states of a thermodynamic system are described by state variables. The value of a state variable depends only on the particular state, not on the path used to arrive at that state. Examples of state variables are pressure (\(P\)), volume (\(V\)), temperature (\(T\)), and mass (\(m\)). Heat and work are not state variables. An equation of state (like the ideal gas equation \(PV = \mu RT\)) is a relation connecting different state variables.
A quasi-static process is an infinitely slow process such that the system remains in thermal and mechanical equilibrium with the surroundings throughout. In a quasi-static process, the pressure and temperature of the environment can differ from those of the system only infinitesimally.
In an isothermal expansion of an ideal gas from volume \(V_1\) to \(V_2\) at temperature \(T\), the heat absorbed (\(Q\)) equals the work done (\(W\)) by the gas, each given by \[Q = W = \mu R T \ln\!\left(\frac{V_2}{V_1}\right)\]
In an adiabatic process of an ideal gas \[PV^{\gamma} = \text{constant}\] where \[\gamma = \frac{C_p}{C_v}\] Work done by an ideal gas in an adiabatic change of state from \((P_1, V_1, T_1)\) to \((P_2, V_2, T_2)\) is \[W = \frac{\mu R (T_1 - T_2)}{\gamma - 1}\]
The Second Law of Thermodynamics disallows some processes consistent with the First Law of Thermodynamics. It states:
Kelvin–Planck statement: No process is possible whose sole result is the absorption of heat from a reservoir and the complete conversion of the heat into work.
Clausius statement: No process is possible whose sole result is the transfer of heat from a colder object to a hotter object.
Put simply, the Second Law implies that no heat engine can have efficiency \(\eta\) equal to \(1\), and no refrigerator can have coefficient of performance \(\alpha\) equal to infinity.
A process is reversible if it can be reversed such that both the system and the surroundings return to their original states, with no other change anywhere else in the universe. Spontaneous processes of nature are irreversible. The idealised reversible process is a quasi-static process with no dissipative factors such as friction, viscosity, etc.
A Carnot engine is a reversible engine operating between two temperatures \(T_1\) (source) and \(T_2\) (sink). The Carnot cycle consists of two isothermal processes connected by two adiabatic processes. The efficiency of a Carnot engine is given by \[\eta = 1 - \frac{T_2}{T_1} \quad \text{(Carnot engine)}\] No engine operating between two temperatures can have efficiency greater than that of the Carnot engine.
Sign conventions for heat and work:
- If \(Q > 0\), heat is added to the system.
- If \(Q < 0\), heat is removed from the system.
- If \(W > 0\), work is done by the system.
- If \(W < 0\), work is done on the system.
Table of Physical Quantities
| Quantity | Symbol | Dimensions | Unit | Remark |
|---|---|---|---|---|
| Coefficient of volume expansion | \(\alpha_v\) | \([\text{K}^{-1}]\) | \(\text{K}^{-1}\) | \(\alpha_v = 3\,\alpha_1\) |
| Heat supplied to a system | \(\Delta Q\) | \([\text{ML}^2\text{T}^{-2}]\) | \(\text{J}\) | \(Q\) is not a state variable |
| Specific heat capacity | \(s\) | \([\text{L}^2\text{T}^{-2}\text{K}^{-1}]\) | \(\text{J kg}^{-1}\,\text{K}^{-1}\) | |
| Thermal conductivity | \(K\) | \([\text{MLT}^{-3}\text{K}^{-1}]\) | \(\text{J s}^{-1}\,\text{K}^{-1}\) | \(H = -KA\dfrac{dt}{dx}\) |
11.13 Points to Ponder
The temperature of a body is related to its average internal energy, not to the kinetic energy of motion of its centre of mass. A bullet fired from a gun is not at a higher temperature because of its high speed.
Equilibrium in thermodynamics refers to the situation when the macroscopic variables describing the thermodynamic state of a system do not depend on time. Equilibrium of a system in mechanics means the net external force and torque on the system are zero.
In a state of thermodynamic equilibrium, the microscopic constituents of a system are not in equilibrium (in the mechanical sense).
Heat capacity, in general, depends on the process the system goes through when heat is supplied.
In isothermal quasi-static processes, heat is absorbed or given out by the system even though at every stage the gas has the same temperature as that of the surrounding reservoir. This is possible because of the infinitesimal difference in temperature between the system and the reservoir.
11.14 NCERT Questions
A geyser heats water flowing at the rate of \(3.0\) litres per minute from \(27\,^\circ\text{C}\) to \(77\,^\circ\text{C}\). If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is \(4.0 \times 10^4\ \text{J g}^{-1}\)?
What amount of heat must be supplied to \(2.0 \times 10^{-2}\ \text{kg}\) of nitrogen (at room temperature) to raise its temperature by \(45\,^\circ\text{C}\) at constant pressure? (Molecular mass of \(\text{N}_2 = 28\); \(R = 8.3\ \text{J mol}^{-1}\,\text{K}^{-1}\).)
Explain why:
- Two bodies at different temperatures \(T_1\) and \(T_2\), if brought in thermal contact, do not necessarily settle to the mean temperature \((T_1 + T_2)/2\).
- The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.
- Air pressure in a car tyre increases during driving.
- The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.
A cylinder with a movable piston contains \(3\) moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?
In changing the state of a gas adiabatically from an equilibrium state \(A\) to another equilibrium state \(B\), an amount of work equal to \(22.3\ \text{J}\) is done on the system. If the gas is taken from state \(A\) to \(B\) via a process in which the net heat absorbed by the system is \(9.35\ \text{cal}\), how much is the net work done by the system in the latter case? (Take \(1\ \text{cal} = 4.19\ \text{J}\).)
Two cylinders \(A\) and \(B\) of equal capacity are connected to each other via a stopcock. \(A\) contains a gas at standard temperature and pressure. \(B\) is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following:
- What is the final pressure of the gas in \(A\) and \(B\)?
- What is the change in internal energy of the gas?
- What is the change in the temperature of the gas?
- Do the intermediate states of the system (before settling to the final equilibrium state) lie on its \(P\)-\(V\)-\(T\) surface?
An electric heater supplies heat to a system at a rate of \(100\ \text{W}\). If the system performs work at a rate of \(75\ \text{joules per second}\), at what rate is the internal energy increasing?
A thermodynamic system is taken from an original state \(D\) to an intermediate state \(E\) by the linear process shown in Fig. 11.11. Its volume is then reduced to the original value from \(E\) to \(F\) by an isobaric process. Calculate the total work done by the gas from \(D\) to \(E\) to \(F\).
Figure to come
Fig. 11.11 – A P–V graph (pressure P in N m⁻² on the y-axis, volume V in m³ on the x-axis) showing point D at (2.0, 600), a straight line falling from D to point E at (5.0, 300), and an isobaric line at constant pressure 300 from E back to point F at (2.0, 300).
11.15 Check Your Concepts
Two bodies \(A\) and \(B\) are each separately found to be in thermal equilibrium with a third body \(C\), but \(A\) and \(B\) are never brought into contact with each other. Can we be certain that \(A\) and \(B\) are at the same temperature? Which law of thermodynamics justifies your conclusion?
Explain why the statement “a gas in a given state contains a certain amount of heat” is physically meaningless, whereas “a gas in a given state has a certain amount of internal energy” is a meaningful statement.
A gas is taken from an initial state to a final state by two different paths. Along which quantity — \(\Delta U\), \(\Delta Q\), or \(\Delta W\) — is the change the same for both paths, and which depend on the path chosen? Justify your answer using the First Law of Thermodynamics.
Explain why the molar specific heat of an ideal gas at constant pressure is greater than that at constant volume. State by how much the two differ and name the relation.
On a \(P\)–\(V\) diagram, the adiabatic curve of an ideal gas is steeper than the isotherm passing through the same point. Explain why, and state what happens to the temperature of the gas during an adiabatic expansion.
Classify each of the following as an extensive or an intensive variable: internal energy, temperature, volume, density, mass, pressure. Briefly describe the test you used to decide.
State the Kelvin–Planck and the Clausius statements of the Second Law of Thermodynamics. Explain briefly why the two statements are regarded as equivalent.
List the two conditions a process must satisfy in order to be reversible, and explain why no real process in nature is ever perfectly reversible.
Why does the efficiency of a Carnot engine depend only on the temperatures of the two reservoirs and not on the nature of the working substance?
A fast-moving bullet is not at a higher temperature merely because of its high speed. Explain this statement in terms of internal energy and the motion of the centre of mass of the bullet.
What happens to the efficiency of a Carnot engine as the sink temperature \(T_2\) approaches the source temperature \(T_1\)? What does this imply for engines that work across a small temperature difference?
In an isochoric process no work is done, yet the temperature of the gas can still rise. Explain why no work is done and where the supplied heat goes.
11.16 Practice with Numericals
A system absorbs \(300\ \text{J}\) of heat from its surroundings and does \(120\ \text{J}\) of work on them. Find the change in the internal energy of the system, stating its sign.
One mole of an ideal gas expands isothermally at a temperature of \(400\ \text{K}\) from a volume \(V\) to \(4V\). Taking \(R = 8.31\ \text{J mol}^{-1}\,\text{K}^{-1}\) and \(\ln 4 = 1.386\), find (a) the work done by the gas and (b) the heat absorbed by the gas.
A monatomic ideal gas (\(\gamma = 5/3\)) at a pressure of \(8.0 \times 10^5\ \text{Pa}\) is expanded adiabatically until its volume becomes \(8\) times its original value. Find the final pressure of the gas. (Take \(8^{5/3} = 32\).)
For a certain ideal gas, the molar specific heat at constant pressure is \(C_p = 29.1\ \text{J mol}^{-1}\,\text{K}^{-1}\). Taking \(R = 8.31\ \text{J mol}^{-1}\,\text{K}^{-1}\), find the molar specific heat at constant volume \(C_v\) and the ratio \(\gamma = C_p/C_v\).
A Carnot engine operates between a source at \(600\ \text{K}\) and a sink at \(300\ \text{K}\). (a) Find its efficiency. (b) If it absorbs \(1200\ \text{J}\) of heat per cycle from the source, find the work done and the heat rejected per cycle.
A Carnot engine has an efficiency of \(40\%\) when its sink is maintained at \(27\,^\circ\text{C}\). Find the temperature of the source (in kelvin and in degrees Celsius).
In one complete cycle, a gas absorbs \(800\ \text{J}\) of heat from a hot reservoir and rejects \(500\ \text{J}\) of heat to a cold reservoir. Find (a) the net work done by the gas per cycle, (b) the change in internal energy over the cycle, and (c) the efficiency of the cycle.
\(0.5\) mole of an ideal gas is heated at constant pressure, raising its temperature by \(40\ \text{K}\). Taking \(R = 8.31\ \text{J mol}^{-1}\,\text{K}^{-1}\), find the work done by the gas.
How much heat is required to raise the temperature of \(200\ \text{g}\) of copper from \(30\,^\circ\text{C}\) to \(80\,^\circ\text{C}\)? (Specific heat capacity of copper \(= 386.4\ \text{J kg}^{-1}\,\text{K}^{-1}\).)
A rigid, closed container holds \(1\) mole of an ideal gas. When \(200\ \text{J}\) of heat is supplied at constant volume, the temperature of the gas rises by \(16\ \text{K}\). Find (a) the work done by the gas, (b) the change in its internal energy, and (c) the molar specific heat at constant volume \(C_v\).