Chapter 3: Kinetic Theory of Gases and Radiation
In the 1820s, the Scottish botanist Robert Brown placed a drop of water containing pollen grains under his microscope. He expected the grains to settle quietly. Instead, every grain jiggled — darting a little this way, then that, never stopping, never tiring. Brown first suspected that he was watching something alive. So he repeated the experiment with fine dust ground from rock that had been lying in the earth for ages, material that could not possibly be alive. The dust particles danced exactly the same way.
Brown had no explanation, and for decades neither did anyone else. The answer, worked out much later, was startling in its simplicity: the water itself was not still. Invisible molecules, far too small to be seen, were striking each pollen grain from every side, millions of times over, and the slight imbalance in those blows was tossing the grain about.
That single observation contains the idea behind this whole chapter. Matter that looks calm is, underneath, in ceaseless motion. Once we accept that a gas is a crowd of molecules flying about at high speed, we can calculate its pressure, explain what temperature really means, work out how much heat it takes to warm it — and even read the temperature of a star from the colour of its light. This is illustrated in Fig. 3.0.
Diagram
Fig. 3.0 – A pollen grain in water shown as a large circle with a jagged zig-zag path, surrounded by many tiny arrows representing invisible molecules striking it from all sides.
Each of these questions is answered somewhere in this chapter. Keep them in mind as you read, and check whether your own reasoning matches the physics.
3.1 Introduction
You have already been introduced to the three common states of matter — solid, liquid and gas. You have also studied the gas laws: Boyle’s law, Charles’ law and Gay-Lussac’s law. These three laws together are enough to give us the ideal gas equation, which is the starting point of this chapter.
Each of these laws was found by holding one quantity fixed and watching how the other two change. That is the standard method of experimental physics: change one thing at a time, and the pattern becomes visible.
The Three Gas Laws
The volume \(V\) of a gas is inversely proportional to the pressure \(P\), when the temperature is held constant. Squeeze a gas harder and it occupies less space.
\[V \propto \frac{1}{P} \quad \text{at constant } T \qquad \text{--- (3.1)}\]
Separately, the volume \(V\) is directly proportional to the temperature, when the pressure is held constant. Here \(T\) is the absolute temperature, measured in kelvin (K), not in degrees Celsius.
\[V \propto T \quad \text{at constant } P \qquad \text{--- (3.2)}\]
Similarly, the pressure \(P\) is directly proportional to the temperature, when the volume is held constant. This is why a sealed container of gas becomes dangerous when heated — the volume cannot grow, so the pressure must.
\[P \propto T \quad \text{at constant } V \qquad \text{--- (3.3)}\]
All three laws apply to a fixed mass \(m\) of an enclosed gas. This condition is easy to overlook, but it is essential — if gas leaks out or is pumped in, none of these relations hold.
The behaviour described by Eq. (3.1) is easiest to see on a graph of pressure against volume drawn at a fixed temperature, as shown in Fig. 3.0a.
Diagram
Fig. 3.0a – A P–V graph showing two smooth hyperbola-shaped isotherms for a fixed mass of gas, the higher curve labelled as the higher temperature.
Combining the Laws into One Equation
The three laws can be merged into a single relation for a fixed mass of gas. Since \(V\) falls with \(P\) and rises with \(T\), the product \(PV\) must rise with \(T\) alone:
\[PV \propto T, \quad \text{or} \quad \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\]
Here \(P_1\), \(V_1\), \(T_1\) are the pressure, volume and absolute temperature of the gas in one state, and \(P_2\), \(V_2\), \(T_2\) are the same quantities after the gas has been taken to another state. Because the ratio \(PV/T\) is the same in both states, this single equation replaces all three gas laws for numerical work — and most board and entrance questions on gas laws are built directly on it.
The next step is to express “fixed mass” in a more useful way. Instead of the mass in kilograms, we count the number of moles \(n\) of the gas:
\[n = \frac{\text{mass of the gas } (M)}{\text{molar mass } (M_0)} = \frac{N}{N_\text{A}}\]
Molar mass \(M_0\) is the mass of 1 mole of the gas. \(N\) is the number of molecules present in the gas, and \(N_\text{A}\) is the Avogadro number, the number of molecules in one mole of gas. Counting in moles is convenient because one mole of any gas contains the same number of molecules.
With this, \(PV \propto nT\), or
\[PV = nRT\]
Here the proportionality constant \(R\) is the universal gas constant. It has the same value, \(8.314\ \text{J mol}^{-1}\ \text{K}^{-1}\), for all gases — which is exactly why it is called universal. In this equation \(P\) is in pascal (Pa), \(V\) in cubic metre (m³), \(T\) in kelvin (K) and \(n\) in mole (mol).
A relation of this kind, connecting the quantities that fix the condition of a gas, is given a special name.
The same equation can be written in terms of the number of molecules rather than the number of moles. Substituting \(n = N/N_\text{A}\),
\[PV = Nk_\text{B}T \qquad \text{--- (3.4)}\]
where \(k_\text{B}\) is the Boltzmann constant. The two constants are related by
\[R = N_\text{A} k_\text{B} \qquad \text{--- (3.5)}\]
How Far Can We Trust These Laws?
The laws of Boyle, Charles and Gay-Lussac are strictly valid for real gases only under two conditions: the pressure of the gas must not be too high, and the temperature must not be close to the liquefaction temperature of the gas.
Both conditions point to the same underlying reason. At high pressure the molecules are packed close together, and near the liquefaction temperature they are moving slowly enough for their mutual attraction to matter. In either situation the molecules are no longer free agents, and the simple gas laws start to fail.
A gas that obeys the equation of state \(PV = nRT\) at all pressures and all temperatures is called an ideal gas.
Notice how strong that condition is. Any real gas obeys \(PV = nRT\) over some range of pressure and temperature; the ideal gas is the imagined gas that obeys it everywhere, without exception. That distinction is developed further in Section 3.3.
3.2 Behaviour of a Gas
A stone thrown upwards in air rises to a certain height and falls back to the ground. Its motion can be described very well with the help of Newton’s laws of motion. Knowing the initial velocity, we can predict where the stone will be at any later instant.
A gas enclosed in a container is described quite differently. It is characterised by its pressure, its volume and its temperature. This is called the macroscopic description of the gas — “macroscopic” meaning that these are bulk properties of the whole sample, the ones we can actually measure with a gauge, a scale and a thermometer.
At the same time, you know that the particles of the gas — its molecules — are in constant motion. So in principle the gas is just a very large collection of moving particles, each obeying Newton’s laws exactly as the stone does.
Why We Cannot Follow One Molecule
Unlike the case of the stone, it is very difficult to understand the behaviour of a gas in terms of the motion of a single particle. The reason is not that the physics is different; the reason is sheer number.
The number of particles in the gas is itself so large (\(\sim 10^{23}\) particles per m³) that any attempt to relate the macroscopic parameters \(P\), \(V\), \(T\) and \(E\) with the motion of individual particles would be futile. Here \(P\) is pressure (Pa), \(V\) is volume (m³), \(T\) is absolute temperature (K) and \(E\) is the internal energy of the gas (J).
Even if we somehow knew the position and velocity of every molecule at one instant, each collision would change them, and there are far too many collisions to track.
The Statistical Way Out
Hence, certain assumptions are made regarding the particles (molecules) of a gas. Averages of physical quantities over the large number of particles involved are then obtained, and these averages are finally related to the macroscopic parameters of the gas.
This is the approach of the kinetic theory of gases. It is a bridge between two descriptions of the same object: the microscopic picture of molecules in motion on one side, and the measurable quantities \(P\), \(V\) and \(T\) on the other.
The sections that follow carry out exactly these three steps in order — first the assumptions and the model of an ideal gas, then the averaging, and finally the connection to pressure and temperature.
3.3 Ideal Gas and Real Gas
We know that a gas obeying the ideal gas equation at all pressures and temperatures is an ideal gas. The key feature of such a gas is that in it, intermolecular interactions are absent — the molecules neither attract nor repel one another except at the instant they collide.
Real gases are composed of atoms or molecules which do interact with each other. Every real molecule pulls weakly on its neighbours, and every real molecule occupies some space of its own. Hence, no real gas is truly ideal as defined here.
This is not a defect in the theory. It simply means that “ideal gas” is a model — a deliberately simplified version of a gas, stripped of the complications, so that we can calculate with it.
When a Real Gas Behaves Ideally
Although no real gas is ideal, a real gas can be brought into a condition in which it behaves almost exactly like one. If the atoms or molecules of a real gas are so far apart that there is practically no interatomic or intermolecular interaction, the real gas is said to be in the ideal state.
This can happen at sufficiently low density of the real gas. The reasoning is direct: intermolecular forces are short-range, so they matter only when molecules are close. Spread the same molecules over a larger volume and each one spends almost all its time too far from the others to feel them.
At low pressures or high temperatures, the molecules are far apart and therefore molecular interactions are negligible. Under these conditions, the behaviour of real gases is close to that of an ideal gas.
There is one more condition that must be respected. The temperature of the real gas must be well above its liquefaction temperature. If the gas is close to the temperature at which it turns into a liquid, its molecules are already clinging to one another, and no amount of low pressure will make it behave ideally.
The extent to which a real gas departs from ideal behaviour can be seen by plotting the ratio \(PV/nRT\) against pressure, as shown in Fig. 3.3a. For an ideal gas this ratio would be exactly 1 at every pressure; for a real gas it stays near 1 at low pressure and drifts away as the pressure is raised.
Diagram
Fig. 3.3a – A graph of \(PV/nRT\) against pressure \(P\), showing a flat dashed line at value 1 for an ideal gas and curves for real gases dipping below and then rising above it at higher pressures.
Why We Keep the Ideal Gas
The ideal gas serves as a model to deduce certain properties of real gases, at least when the real gas is in the ideal state. In other words, we work out the physics for the simple imaginary gas and then apply the results to real gases wherever the conditions permit — low pressure, high temperature, well above liquefaction.
You have studied the deviation of a real gas from ideal gas behaviour in XIth Std. Chemistry. That study and this chapter approach the same fact from two sides: chemistry describes how much real gases deviate, while kinetic theory explains why the ideal model works so well in the first place.
3.4 Mean Free Path
How do the molecules of an ideal gas move? These molecules are in continuous random motion — the same kind of ceaseless, directionless jostling as the Brownian motion you have studied in XIth Std. Chemistry, and which was described at the start of this chapter.
The molecules of a gas are uniformly dispersed throughout the volume of the gas, as shown in Fig. 3.1 (a). If we could take a stop-action photograph of a gas, we would find the molecules spread evenly, with no region crowded and no region empty.
Diagram
Fig. 3.1 (a) – A square box filled with many small dots spread uniformly, representing gas molecules at one instant: a stop-action photograph.
These molecules are executing random motion. A typical path of a single molecule is shown in Fig. 3.1 (b) — a series of straight segments of different lengths, joined at sharp corners.
Diagram
Fig. 3.1 (b) – A single molecule’s zig-zag path inside a box, drawn as straight arrows of unequal length meeting at sharp turning points.
What Happens When Two Molecules Meet
When a molecule approaches another molecule, there is a repulsive force between them, due to which the molecules behave as small hard spherical particles. They do not pass through one another and they do not stick; they rebound, much as two hard balls would.
This leads to elastic collisions between the molecules. In an elastic collision the total kinetic energy is conserved, so no energy is lost from the molecular motion — which is why the motion never dies down. Therefore, both the speed and the direction of motion of the molecules change abruptly at every collision.
The molecules also collide with the walls of the container. These wall collisions are what we will later identify as the origin of gas pressure.
Molecules exert force on each other only during collisions. Between collisions there is no force acting on a molecule at all, and by Newton’s first law it must then travel in a straight line at constant speed. Thus, in between two successive collisions the molecules move along straight paths with constant velocity.
This is exactly why the path in Fig. 3.1 (b) is made of straight segments: each segment is a free flight, and each corner is a collision.
Defining the Mean Free Path
The straight segments are not all of the same length, because collisions occur at random. But their average length is a definite and useful quantity. It is convenient and useful to define the mean free path (\(\lambda\)) as follows.
The mean free path is expected to vary inversely with the density of the gas, \(\rho = \dfrac{N}{V}\), where \(N\) is the number of molecules enclosed in a volume \(V\). Higher the density, more will be the collisions and smaller will be the mean free path \(\lambda\). A molecule in a crowded gas cannot travel far before meeting another molecule.
It is also seen that \(\lambda\) is inversely proportional to the size of the molecule, say the diameter \(d\). Smaller the size of the molecule, less is the chance for collision and larger is the mean free path.
Further, \(\lambda\) is inversely proportional to \(d^2\), not just \(d\), because it depends on the cross section of a molecule. A molecule sweeps out a circular target area as it moves, and the area of that target grows as the square of the diameter — so doubling the molecular diameter makes collisions four times as likely, not twice.
Putting all three dependences together, it can be shown that
\[\lambda = \frac{1}{\sqrt{2}\,\pi d^2 \left(N/V\right)} \qquad \text{--- (3.6)}\]
In this equation \(\lambda\) is the mean free path in metre (m), \(d\) is the diameter of a molecule in metre (m), \(N\) is the number of molecules and \(V\) is the volume in cubic metre (m³), so that \(N/V\) is the number of molecules per unit volume (m⁻³). The factor \(\sqrt{2}\) appears because the other molecules are not standing still — they are moving too, and this raises the collision rate slightly above the simple estimate.
3.5 Pressure of Ideal Gas
We now express the pressure of an ideal gas as a kinetic theory problem. In Section 3.2 we said that kinetic theory works by averaging molecular quantities and connecting them to measurable bulk quantities. This section is the first place where that programme is actually carried out, and pressure is the quantity we start with.
Let there be \(n\) moles of an ideal gas enclosed in a cubical box of volume \(V\ (= L^3)\), with the sides of the box parallel to the coordinate axes, as shown in Fig. 3.2. The walls of the box are kept at a constant temperature \(T\).
Diagram
Fig. 3.2 – A cubical box of side \(L\) with edges along the x, y and z axes, one face shaded, and a single molecule of mass \(m\) inside moving towards the shaded face with velocity vector \(\vec{v}\).
The question is: can we relate the pressure \(P\) of the gas with the molecular speeds? Here we will use the word molecular speed rather than molecular velocity, since the kinetic energy of a molecule depends on the velocity irrespective of its direction.
The Model We Shall Work With
The gas molecules are continuously moving randomly in various directions, colliding with each other and hitting the walls of the box and bouncing back.
As a first approximation, we neglect intermolecular collisions and consider only elastic collisions with the walls. This is not an unphysical assumption. As explained earlier, the mean free path increases as the pressure is reduced. Thus, the pressure is so adjusted that the molecules do not collide with each other, but do collide with the walls.
A typical molecule is shown in Fig. 3.2 moving with the velocity \(\vec{v}\), about to collide with the shaded wall of the cube. The wall is parallel to the yz-plane. As the collision is assumed to be elastic, during collision the component \(\text{v}_x\) of the velocity will get reversed, keeping the \(\text{v}_y\) and \(\text{v}_z\) components unaltered.
Diagram
Fig. 3.2a – A particle striking a vertical wall, with incoming velocity components \(\text{v}_x\) and \(\text{v}_y\) marked, and outgoing components \(-\text{v}_x\) and \(\text{v}_y\) marked after rebound.
Considering all the molecules, their average y and z components of the velocities are not changed by collisions with the shaded wall. This can be understood from the fact that the gas molecules remain evenly distributed throughout the volume and do not get any additional motion in the \(+y\) or \(-y\) and \(+z\) or \(-z\) directions. Thus the y and z components remain unchanged during collision with the wall parallel to the yz-plane.
Momentum Given to the Wall in One Collision
Hence the change in momentum of the particle is only in the x component of the momentum. This change, \(\Delta p_x\), is given by
\[\Delta p_x = \text{final momentum} - \text{initial momentum} = \left(-m\text{v}_x\right) - \left(m\text{v}_x\right) = -2m\text{v}_x \qquad \text{--- (3.7)}\]
where \(m\) is the mass of one molecule (kg) and \(\text{v}_x\) is the x component of its velocity (m/s).
The molecule has lost momentum \(2m\text{v}_x\) in the \(+x\) direction. Momentum must be conserved in the collision, so whatever the molecule loses, the wall gains. Thus, the momentum transferred to the wall during collision is \(+2m\text{v}_x\). The rebounded molecule then goes to the opposite wall and collides with it.
From One Collision to an Average Force
A single collision lasts an extremely short time, and we cannot measure the sharp force during it. What we can use is the average force, spread over the time between one collision and the next.
We now set the average force exerted by one molecule on the wall equal to the average rate of change of momentum during the time for one collision. To find this average rate, we have to divide the change in momentum by the time taken for one collision.
After colliding with the shaded wall, the molecule travels to the opposite wall and is reflected back. It travels back towards the shaded wall again to collide with the shaded wall. This means that the molecule travels a distance of \(2L\) in between two collisions with the shaded wall. Hence, to get the average force, we have to divide by the time between two successive collisions.
As \(L\) is the length of the cubical box, the time for the molecule to travel back and forth to the shaded wall is
\[\Delta t = \frac{2L}{\text{v}_x}\]
Only the x component appears here, because the journey to the far wall and back is a journey along the x direction; the y and z motions carry the molecule sideways but do not decide when it returns.
The average force exerted on the shaded wall by molecule 1 is given as
\[\text{Average force} = \text{Average rate of change of momentum} = \frac{2m\text{v}_{x1}}{2L/\text{v}_{x1}} = \frac{m\text{v}_{x1}^2}{L} \qquad \text{--- (3.8)}\]
where \(\text{v}_{x1}\) is the x component of the velocity of molecule 1.
Notice that the speed enters twice over, giving \(\text{v}_{x1}^2\): a faster molecule delivers a bigger kick, and it returns to the wall more often. This is the physical reason why pressure depends on the square of the molecular speed.
Adding Up All the Molecules
Considering other molecules 2, 3, 4, … with the respective x components of velocities \(\text{v}_{x2}\), \(\text{v}_{x3}\), \(\text{v}_{x4}\), …, the total average force on the wall follows from Eq. (3.8) as
\[\frac{m}{L}\left(\text{v}_{x1}^2 + \text{v}_{x2}^2 + \text{v}_{x3}^2 + \ldots\right)\]
Pressure is force per unit area, and the area of the shaded wall is \(L^2\). Therefore the average pressure is
\[P = \frac{\text{Average force}}{\text{Area of shaded wall}} = \frac{m\left(\text{v}_{x1}^2 + \text{v}_{x2}^2 + \ldots\right)}{L \cdot L^2}\]
The sum inside the bracket is awkward to handle molecule by molecule, so we replace it by an average. The average of the square of the x component of the velocities is given by
\[\overline{\text{v}_x^2} = \frac{\text{v}_{x1}^2 + \text{v}_{x2}^2 + \text{v}_{x3}^2 + \ldots + \text{v}_{xN}^2}{N}\]
so that the sum equals \(N\overline{\text{v}_x^2}\). Since \(L \cdot L^2 = L^3 = V\), we obtain
\[P = \frac{mN\overline{\text{v}_x^2}}{V} \qquad \text{--- (3.9)}\]
where \(\overline{\text{v}_x^2}\) is the average over all possible values of \(\text{v}_x\).
Bringing In the Full Speed
Equation (3.9) still refers only to motion along x, which is unsatisfactory, because the gas has no special direction. We remove this by relating \(\overline{\text{v}_x^2}\) to the average of the full squared speed.
For any molecule, the square of the speed is the sum of the squares of the three components:
\[\text{v}^2 = \text{v}_x^2 + \text{v}_y^2 + \text{v}_z^2\]
By symmetry,
\[\overline{\text{v}_x^2} = \overline{\text{v}_y^2} = \overline{\text{v}_z^2} = \frac{1}{3}\overline{\text{v}^2}\]
since the molecules have no preferred direction to move. There is nothing about the gas that favours the x direction over y or z, so the three averages must be equal, and together they must add up to \(\overline{\text{v}^2}\).
Substituting in Eq. (3.9), the average pressure is
\[P = \frac{1}{3}\frac{N}{V}m\overline{\text{v}^2} \qquad \text{--- (3.10)}\]
Here \(P\) is the pressure (Pa), \(N\) is the number of molecules, \(V\) is the volume of the container (m³), \(m\) is the mass of one molecule (kg) and \(\overline{\text{v}^2}\) is the mean square speed of the molecules (m²/s²).
This is a remarkable result. On the left is a quantity read off a pressure gauge; on the right are the mass and speed of molecules no one has ever seen. Kinetic theory has connected the two.
Why the Result Is More General Than It Looks
Equation (3.10) has been obtained for a cubical shaped container. However, it can be shown to be valid for containers of any shape. This is reassuring, because gases are stored in cylinders, spheres and tubes, not only in cubes, and the measured pressure does not depend on the shape of the vessel.
Also, we have assumed that there are no intermolecular collisions, which is clearly not true in a real gas. The number of molecules in the container is so large (of the order of \(10^{23}\)) that even if molecular collisions are taken into account, the above expression does not change.
The reason is a bookkeeping one. If a molecule acquires a velocity with components different from \(\text{v}_x\), \(\text{v}_y\), \(\text{v}_z\) after a collision, there will invariably be some other molecule having a different initial velocity now acquiring the velocity with the components \(\text{v}_x\), \(\text{v}_y\), \(\text{v}_z\). As the gas is steady (in equilibrium), this must be happening. The collection of velocities present in the gas therefore stays the same, even though individual molecules keep exchanging their velocities among themselves. Thus the collisions do not affect Eq. (3.10).
3.6 Root Mean Square (rms) Speed
Equation (3.10) gives the mean square speed of the molecules of a gas. Rearranging it to make \(\overline{v^{2}}\) the subject,
\[\overline{v^{2}} = \frac{3PV}{Nm} \qquad \text{--- (3.11)}\]
where \(P\) is the pressure in pascal (Pa), \(V\) the volume in cubic metre (m³), \(N\) the number of molecules and \(m\) the mass of one molecule in kilogram (kg).
This is useful, but it is expressed in molecular quantities that are hard to measure directly. The ideal gas equation lets us replace them by quantities we can read off in the laboratory.
From Mean Square Speed to rms Speed
Using the ideal gas equation \(PV = nRT\),
\[\overline{v^{2}} = \frac{3nRT}{Nm} = \frac{3NRT}{N_\text{A}Nm}\]
The second step uses \(n = N/N_\text{A}\). The number of molecules \(N\) now cancels from numerator and denominator, leaving \(\overline{v^{2}} = \dfrac{3RT}{N_\text{A}m}\). Taking the square root,
\[\sqrt{\overline{v^{2}}} = v_\text{rms} = \sqrt{\frac{3RT}{M_0}} \qquad \text{--- (3.12)}\]
where \(M_0 = N_\text{A}m\) is the molar mass of the gas.
The quantity \(v_\text{rms}\) is called the root mean square speed, and its name describes exactly how it is built: take the square of each molecular speed, find the mean of those squares, then take the square root.
Equation (3.12) allows us to estimate the rms speeds of molecules of real gases. Two useful values: for nitrogen gas at 300 K, the rms speed is 517 m/s, while for oxygen gas it is 483 m/s.
Both figures are worth a moment’s thought. Molecules in the air of an ordinary room are moving faster than a passenger aircraft, and oxygen — being the heavier of the two — is the slower, exactly as \(v_\text{rms} \propto 1/\sqrt{M_0}\) requires.
Most examination questions on Eq. (3.12) are of the ratio type — comparing two gases at the same temperature, or the same gas at two temperatures. In such questions \(R\) cancels, so it is usually quicker to write \(v_\text{rms} \propto \sqrt{T/M_0}\) and take the ratio directly than to substitute numbers into the full formula.
Molecular Speeds and the Speed of Sound
You have studied the passage of sound waves through an air medium. The speed of sound in a gas is
\[v_\text{s} = \sqrt{\frac{\gamma RT}{M_0}}\]
where \(\gamma = \dfrac{C_\text{P}}{C_\text{V}}\) is called the adiabatic ratio — the ratio of the specific heat of the gas at constant pressure to that at constant volume. Its maximum value is 5/3, for monatomic gases.
Compare this with Eq. (3.12). The two expressions are identical except that the factor 3 is replaced by \(\gamma\). Since \(\gamma < 3\) for every gas, the sound wave cannot move faster than the average speed of the molecules. However, the two speeds are of the same order of magnitude.
That closeness is not a coincidence. The molecules serve as a medium to transport sound energy. A sound wave is a disturbance handed on from molecule to molecule, so it can travel no faster than the molecules themselves carry it. Accordingly, the speed of sound in H₂ gas is comparable to the rms speed of H₂ molecules, and in N₂ gas to the rms speed of N₂ molecules.
Not All Molecules Move at the Same Speed
Diagram
Fig. 3.6a – A Maxwell distribution curve of \(n(\text{v})\) against \(\text{v}/v_\text{rms}\), rising from the origin to a peak and falling away with a long tail, with a narrow strip of width dv shaded under the curve.
So \(v_\text{rms}\) should never be pictured as the speed of a typical molecule to the exclusion of all others. It is a single representative number drawn from a whole spread of speeds — the one that happens to be most convenient, because it connects directly to the kinetic energy of the gas, as the next section shows.
3.7 Interpretation of Temperature in Kinetic Theory
So far, temperature has entered our equations only as a symbol borrowed from the gas laws. In this section we find out what it actually stands for at the molecular level.
Equation (3.10) can be rewritten by multiplying both sides by \(V\) and then arranging the right-hand side so that the kinetic energy of a single molecule appears explicitly:
\[PV = \frac{1}{3}Nm\overline{v^{2}} = \frac{2}{3}N \cdot \left(\frac{1}{2}m\overline{v^{2}}\right) \qquad \text{--- (3.13)}\]
The step from the first form to the second is only algebra — a factor of \(\frac{1}{2}\) has been created inside the bracket and compensated by changing \(\frac{1}{3}\) to \(\frac{2}{3}\) outside — but it makes the physics visible.
The quantity \(\frac{1}{2}m\overline{v^{2}}\) is the average translational kinetic energy of a molecule, measured in joule (J).
Why the Internal Energy Is Purely Kinetic
In an ideal gas, the molecules are non-interacting, and hence there is no potential energy term. Potential energy arises from forces acting between molecules over a distance; an ideal gas has no such forces by definition. Thus, the internal energy of an ideal gas is purely kinetic.
The average total energy \(E\) is therefore just the energy of one molecule multiplied by the number of molecules:
\[E = N \cdot \frac{1}{2}m\overline{v^{2}} \qquad \text{--- (3.14)}\]
Comparing Eq. (3.14) with Eq. (3.13),
\[PV = \frac{2}{3}E \qquad \text{--- (3.15)}\]
Bringing In Temperature
Using the ideal gas equation in its molecular form, Eq. (3.4),
\[PV = Nk_\text{B}T = \frac{2}{3}E \qquad \text{--- (3.16)}\]
Solving for the total energy,
\[\therefore\ E = \frac{3}{2}Nk_\text{B}T \qquad \text{--- (3.17)}\]
and dividing by the number of molecules to get the energy of one molecule,
\[\text{or}\quad \frac{E}{N} = \frac{3}{2}k_\text{B}T \qquad \text{--- (3.18)}\]
This means that the average energy per molecule is proportional to the absolute temperature \(T\) of the gas. Equation (3.18) relates the macroscopic parameter of the gas, \(T\), to the kinetic energy of a molecule.
Notice what does not appear in Eq. (3.18): the mass of the molecule, the pressure, the volume, and the identity of the gas. At a given temperature, a helium atom and a nitrogen molecule have precisely the same average translational kinetic energy.
3.8 Law of Equipartition of Energy
In the previous section we found the total average energy of a molecule. We now ask a finer question: how is that energy shared out among the different ways in which a molecule can move?
We have seen that the kinetic energy of a single molecule is the sum of three parts, one for each direction of motion:
\[\text{K.E.} = \frac{1}{2}m\text{v}_x^{2} + \frac{1}{2}m\text{v}_y^{2} + \frac{1}{2}m\text{v}_z^{2}\]
For a gas at a temperature \(T\), the average kinetic energy per molecule, denoted as \(\overline{K.E.}\), is obtained by averaging each term over all the molecules:
\[\overline{K.E.} = \frac{1}{2}m\overline{\text{v}_x^{2}} + \frac{1}{2}m\overline{\text{v}_y^{2}} + \frac{1}{2}m\overline{\text{v}_z^{2}}\]
But we know from Eq. (3.18) that the mean energy per molecule is \(\frac{3}{2}k_\text{B}T\). Since there is no preferred direction \(x\) or \(y\) or \(z\), the three terms above must be equal to one another, and each must therefore be one-third of the total:
\[\frac{1}{2}m\overline{\text{v}_x^{2}} = \frac{1}{2}m\overline{\text{v}_y^{2}} = \frac{1}{2}m\overline{\text{v}_z^{2}} = \frac{1}{2}k_\text{B}T \qquad \text{--- (3.19)}\]
What “Quadratic” Means Here
Look carefully at the form of each term in Eq. (3.19). Every one of them contains a velocity component raised to the second power — that is, each term is quadratic in the velocity component. This is not an incidental detail; it is the feature that decides how much energy the term receives.
Thus the mean energy associated with every component of translational kinetic energy, which is quadratic in the velocity components in the \(x\), \(y\) and \(z\) directions, is \(\frac{1}{2}k_\text{B}T\). Adding the three contributions, the total translational energy contribution of the molecule is \(\frac{3}{2}k_\text{B}T\), exactly as required by Eq. (3.18).
The importance of Eq. (3.19) goes far beyond translation. Once we recognise that it is the quadratic form of each energy term that earns it a share of \(\frac{1}{2}k_\text{B}T\), the same reasoning can be extended to any other quadratic energy term a molecule may possess — for instance, the energy of rotation, or of vibration. A molecule that has more such terms will hold more energy at the same temperature.
Before we can count those terms, however, we need a systematic way of describing the different independent motions available to a molecule. That is taken up next.
3.8.1 Degrees of Freedom
In the discussion so far, the molecule as a whole is free to move from one point to another in three-dimensional space. To say where it is, we must give three numbers — its \(x\), \(y\) and \(z\) coordinates — and to say how it is moving, we must give three velocity components \(\text{v}_x\), \(\text{v}_y\) and \(\text{v}_z\).
Now suppose the motion is restricted. If the molecule is restricted to move on a plane surface, which is two dimensional, then only two coordinates, say \(x\) and \(y\), will be sufficient to describe its location, and two components \(\text{v}_x\), \(\text{v}_y\) will describe its motion in the plane. The third coordinate is fixed and carries no information.
If a molecule moves along a straight line, then only the \(x\) coordinate and only one velocity component \(\text{v}_x\) will be sufficient to describe its location and its motion along that straight line.
We say that the molecule is free to execute 3, 2 and 1 dimensional translational motion in the above three examples respectively. In other words, the molecule in these examples has 3, 2 and 1 degrees or degree of freedom.
Diagram
Fig. 3.8a – Three side-by-side sketches of a particle: free in a 3-D box with three axis arrows, confined to a flat plane with two axis arrows, and confined to a straight wire with one axis arrow.
Fig. 3.8a makes the counting visual: the number of independent arrows needed to pin the particle down is exactly its number of degrees of freedom.
Two words in this definition deserve attention. Independent means that no one of these quantities can be worked out from the others; each must be supplied separately. Configuration means not merely where the system is, but also how its parts are arranged relative to one another — which becomes important as soon as we consider molecules made of more than one atom.
The point of introducing this idea is now within reach. Section 3.8 showed that every quadratic energy term earns an average energy of \(\frac{1}{2}k_\text{B}T\). Counting degrees of freedom is how we find out how many such terms a molecule possesses — and therefore how much energy it can store at a given temperature.
3.8.2 Diatomic Molecules
A monatomic gas like helium contains He atoms. As we have seen, an He atom has 3 translational degrees of freedom (dof), and nothing else. A molecule made of two atoms joined together is a richer object, and this section counts what it can do.
The Three Translational Degrees of Freedom
Consider, for example, an O₂ or N₂ molecule with its two atoms lying along the \(x\)-axis. Taken as a whole, the molecule can move anywhere in space, so like any single particle it has 3 translational dof.
The Two Rotational Degrees of Freedom
In addition, the molecule can rotate around the \(z\)-axis and around the \(y\)-axis. Fig. 3.3 depicts rotation of the molecule about the \(z\)-axis; likewise, rotation is possible about the \(y\)-axis.
Diagram
Fig. 3.3 – A diatomic molecule drawn as two spheres joined by a bond lying along the x-axis, with curved arrows showing the two independent rotations, about the z-axis and about the y-axis.
Note that rotation around the \(x\)-axis is not a rotation in the sense that matters here, because it does not involve any change of positions of the two atoms of the molecule. Spinning the molecule about the line joining its own atoms leaves it looking exactly as it did before, so no new state of the molecule is produced and no energy is stored.
In general, a diatomic molecule can rotate about its centre of mass in two directions that are perpendicular to its molecular axis. Molecules like O₂ are therefore said to possess 2 additional dof, namely 2 rotational dof.
Each of these 2 dof contributes to rotational kinetic energy. It can be shown that if \(I_z\) and \(I_y\) are the moments of inertia about the \(z\) and \(y\) axes, with \(\omega_z\) and \(\omega_y\) the respective angular speeds, the rotational kinetic energies will be \(\frac{1}{2}I_z\omega_z^{2}\) and \(\frac{1}{2}I_y\omega_y^{2}\) for rotation around the two axes. Here moment of inertia is measured in kg m² and angular speed in rad/s, so each term is in joule (J).
Thus, for a diatomic molecule, the total energy due to translational and rotational dof is
\[E = E(\text{translational}) + E(\text{rotational})\]
\[E = \frac{1}{2}m\text{v}_x^{2} + \frac{1}{2}m\text{v}_y^{2} + \frac{1}{2}m\text{v}_z^{2} + \frac{1}{2}I_z\omega_z^{2} + \frac{1}{2}I_y\omega_y^{2} \qquad \text{--- (3.20)}\]
The above expression contains quadratic terms that correspond to the various dof of a diatomic molecule — three quadratic in a velocity component, two quadratic in an angular speed. Each of them contributes \(\frac{1}{2}k_\text{B}T\) to the total energy of the molecule.
The Vibrational Degrees of Freedom
In the above discussion, an implicit assumption was made: that the rotating molecule is a rigid rotator, that is, a dumbbell whose bond length never changes.
However, real molecules contain covalent bonds between the atoms and therefore can perform additional motion, namely vibrations of the atoms about their mean positions, like a one-dimensional harmonic oscillator. The bond behaves like a stiff spring rather than a rigid rod, so the two atoms can move towards and away from each other.
Such molecules therefore possess additional dof corresponding to the different modes of vibration. In diatomic molecules like O₂, N₂ and CO, the atoms can oscillate along the internuclear axis only — there is just one line along which they can approach and separate. This motion adds energy associated with the vibrations to the total energy of the molecule.
\[E = E(\text{translational}) + E(\text{rotational}) + E(\text{vibrational}) \qquad \text{--- (3.21)}\]
The term \(E(\text{vibrational})\) is different from the earlier terms in one important respect: it consists of two contributions — one from the kinetic energy term and the other from the potential energy term. A vibrating system stores energy both in motion and in the stretch of the bond, and the two exchange back and forth continuously.
\[E(\text{vibrational}) = \frac{1}{2}mu^{2} + \frac{1}{2}kr^{2} \qquad \text{--- (3.22)}\]
where \(\vec{u}\) is the velocity of vibrations of the atoms of the molecule (m/s), \(r\) is the separation between the atoms performing oscillations (m), and \(k\) is related to the force constant of the bond. Take care not to confuse this \(k\) with the Boltzmann constant \(k_\text{B}\); they are entirely different quantities.
The terms in Eq. (3.22) are quadratic in velocity and in position respectively, and each will contribute \(\frac{1}{2}k_\text{B}T\). Thus each mode or dof for vibrational motion contributes \(2 \times \frac{1}{2}k_\text{B}T\) to the total internal energy — twice as much as a translational or rotational dof, precisely because it brings two quadratic terms instead of one.
Adding It All Up
Hence, for a non-rigid diatomic gas in thermal equilibrium at a temperature \(T\):
- the mean kinetic energy associated with the translational motion of the molecule along the three directions is \(3 \times \frac{1}{2}k_\text{B}T\);
- the mean kinetic energy associated with the rotational motions about two perpendicular axes is \(2 \times \frac{1}{2}k_\text{B}T\);
- the total vibrational energy is \(2 \times \frac{1}{2}k_\text{B}T\), corresponding to the kinetic and potential energy terms.
Considering the above facts, the law of equipartition of energy is stated as follows.
The final condition is not a formality. At low temperatures, rotational and vibrational modes are effectively “frozen out” and stop taking their share of energy — a fact that classical physics could not explain and that pointed towards quantum theory. Whenever you apply equipartition in a numerical problem, you are assuming the temperature is high enough for all the counted modes to be active.
3.9 Specific Heat Capacity
Specific heat capacity, as you learnt in earlier classes, is the quantity of heat required to raise the temperature of a unit quantity of a substance by one degree. For a solid or a liquid, a single number is enough, because heating it changes its size very little.
A gas is different. You know that when the temperature of a gas is increased, even a small rise causes a considerable change in volume and pressure. So the answer to “how much heat is needed?” depends on what we allow the gas to do while it is being heated — whether we hold it in a rigid container, or let it expand while keeping its pressure the same.
Therefore two specific heats are defined for gases, namely specific heat at constant volume \(C_\text{V}\) and specific heat at constant pressure \(C_\text{P}\).
These two quantities are not independent of each other. Mayer’s relation gives an expression that connects the two specific heats, and deriving it is the subject of the next subsection.
3.9.1 Mayer’s Relation
Consider one mole of an ideal gas that is enclosed in a cylinder by a light, frictionless, airtight piston. Let \(P\), \(V\) and \(T\) be the pressure, volume and temperature respectively of the gas. The piston is taken to be light and frictionless so that no energy is wasted in moving it, and airtight so that no gas escapes.
We now heat this gas twice over, under two different conditions, and compare the two.
Heating at Constant Volume
If the gas is heated so that its temperature rises by \(\text{d}T\), but the volume remains constant, then the amount of heat supplied to the gas, \(\text{d}Q_1\), is used to increase the internal energy of the gas (\(\text{d}E\)). Since the volume of the gas is constant, no work is done in moving the piston — the piston does not move at all.
\[\therefore\ \text{d}Q_1 = \text{d}E = C_\text{V}\,\text{d}T \qquad \text{--- (3.23)}\]
where \(C_\text{V}\) is the molar specific heat of the gas at constant volume.
Heating at Constant Pressure
On the other hand, if the gas is heated to the same temperature at constant pressure, the volume of the gas increases by an amount, say \(\text{d}V\). The amount of heat supplied to the gas is now used for two purposes: to increase the internal energy of the gas, and to move the piston backwards to allow expansion of the gas. The work done to move the piston is \(\text{d}W = P\,\text{d}V\).
\[\text{d}Q_2 = \text{d}E + \text{d}W = C_\text{P}\,\text{d}T \qquad \text{--- (3.24)}\]
where \(C_\text{P}\) is the molar specific heat of the gas at constant pressure.
Combining the Two
The crucial point is that \(\text{d}E\) is the same in both cases. From Eq. (3.23), \(\text{d}E = C_\text{V}\,\text{d}T\), and this holds even in the constant-pressure process, because the internal energy of an ideal gas depends only on its temperature. Both processes produce the same temperature rise \(\text{d}T\), so both produce the same change in internal energy, whatever the volume does.
Substituting in Eq. (3.24),
\[\therefore\ C_\text{P}\,\text{d}T = C_\text{V}\,\text{d}T + \text{d}W\]
\[\text{or,}\quad \left(C_\text{P} - C_\text{V}\right)\text{d}T = P\,\text{d}V \qquad \text{--- (3.25)}\]
For one mole of gas,
\[PV = RT\]
\[\therefore\ P\,\text{d}V = R\,\text{d}T, \quad \text{since pressure is constant.}\]
Substituting in Eq. (3.25), we get
\[\left(C_\text{P} - C_\text{V}\right)\text{d}T = R\,\text{d}T\]
\[\therefore\ C_\text{P} - C_\text{V} = R \qquad \text{--- (3.26)}\]
This is known as Mayer’s relation between \(C_\text{P}\) and \(C_\text{V}\).
The result is remarkable for what it leaves out. The difference does not depend on which gas we chose, nor on its temperature or pressure. Every ideal gas — light or heavy, monatomic or polyatomic — shows the same difference, because the extra heat is spent entirely on the expansion work, and one mole of any ideal gas expands identically.
When Heat and Work Are Measured in Different Units
The above relation has been derived assuming that the heat energy and mechanical work are measured in the same units. Generally, heat supplied is measured in calories and work done is measured in joules. The above relation is then modified to
\[C_\text{P} - C_\text{V} = \frac{R}{J}\]
where \(J\) is the mechanical equivalent of heat — the number of joules equivalent to one calorie.
Also, \(C_\text{P} = M_0 S_\text{P}\) and \(C_\text{V} = M_0 S_\text{V}\), where \(M_0\) is the molar mass of the gas and \(S_\text{P}\) and \(S_\text{V}\) are the respective principal specific heats. In many books, \(c_\text{P}\) and \(c_\text{V}\) are used to denote the principal specific heats. The molar specific heat refers to one mole of gas, while the principal specific heat refers to unit mass, which is why the two are connected by the molar mass. Thus,
\[M_0 S_\text{P} - M_0 S_\text{V} = \frac{R}{J}\]
\[\therefore\ S_\text{P} - S_\text{V} = \frac{R}{M_0 J} \qquad \text{--- (3.27)}\]
Specific Heats from the Law of Equipartition
It is interesting to use the law of equipartition of energy and calculate the specific heat of gases. The method is always the same: count the degrees of freedom, write the internal energy per mole, and differentiate with respect to temperature.
Monatomic Gases
For a monatomic gas enclosed in a container, held at a constant temperature \(T\) and containing \(N_\text{A}\) atoms, each atom has only 3 translational dof. Therefore, the average energy per atom is \(\frac{3}{2}k_\text{B}T\), and the total internal energy per mole is
\[E = \frac{3}{2}N_\text{A}k_\text{B}T\]
The molar specific heat at constant volume is the rate at which this internal energy rises with temperature:
\[C_\text{V} = \frac{\text{d}E}{\text{d}T} = \frac{3}{2}N_\text{A}k_\text{B} = \frac{3}{2}R \qquad \text{--- (3.28)}\]
where the last step uses \(R = N_\text{A}k_\text{B}\) from Eq. (3.5).
Using Eq. (3.26),
\[C_\text{P} = \frac{5}{2}R \qquad \text{--- (3.29)}\]
\[\therefore\ \gamma = \frac{C_\text{P}}{C_\text{V}} = \frac{5}{3} \qquad \text{--- (3.30)}\]
This is the value of \(\gamma\) quoted in Section 3.6 as the largest possible for any gas — and we can now see why it is the largest. A monatomic gas has the fewest degrees of freedom, so it has the smallest \(C_\text{V}\), and dividing the fixed difference \(R\) by the smallest \(C_\text{V}\) gives the largest ratio.
Diatomic Gases
For a gas consisting of diatomic molecules such as O₂, N₂, CO or HCl, enclosed in a container held at a constant temperature \(T\), if treated as a rigid rotator, each molecule will have three translational and two rotational dof. According to the law of equipartition of energy, the internal energy of one mole of gas is
\[E = \frac{3}{2}N_\text{A}k_\text{B}T + \frac{2}{2}N_\text{A}k_\text{B}T = \frac{5}{2}N_\text{A}k_\text{B}T\]
The molar specific heat at constant volume will be
\[C_\text{V} = \frac{5}{2}N_\text{A}k_\text{B} = \frac{5}{2}R \qquad \text{--- (3.31)}\]
Using Eq. (3.26),
\[C_\text{P} = \frac{7}{2}R \qquad \text{--- (3.32)}\]
\[\therefore\ \gamma = \frac{C_\text{P}}{C_\text{V}} = \frac{7}{5} \qquad \text{--- (3.33)}\]
For a diatomic gas containing non-rigid vibrating molecules, the internal energy per mole, considering that vibrational motion contributes \(k_\text{B}T\) per molecule, is
\[E = \frac{3}{2}N_\text{A}k_\text{B}T + \frac{2}{2}N_\text{A}k_\text{B}T + \frac{2}{2}N_\text{A}k_\text{B}T = \frac{7}{2}N_\text{A}k_\text{B}T\]
The molar specific heat at constant volume will be
\[C_\text{V} = \frac{7}{2}N_\text{A}k_\text{B} = \frac{7}{2}R \qquad \text{--- (3.34)}\]
Using Eq. (3.26),
\[C_\text{P} = \frac{9}{2}R \qquad \text{--- (3.35)}\]
\[\therefore\ \gamma = \frac{C_\text{P}}{C_\text{V}} = \frac{9}{7} \qquad \text{--- (3.36)}\]
Notice that the vibrational contribution is written as \(\frac{2}{2}N_\text{A}k_\text{B}T\) rather than \(\frac{1}{2}N_\text{A}k_\text{B}T\), because a single vibrational mode carries two quadratic terms — one kinetic and one potential — as established in Section 3.8.2.
Polyatomic Gases
Gases which have molecules containing more than two atoms are termed polyatomic gases — for example, ammonia gas, where each molecule has one N atom and three H atoms.
Each molecule of a polyatomic gas has three translational dof, each contributing \(\frac{1}{2}k_\text{B}T\) per molecule.
For rotation, the count depends on the shape of the molecule. Only linear molecules have two dof for rotation, for the same reason as a diatomic molecule: rotation about the molecular axis does not count. All other polyatomic molecules can perform rotations about three mutually perpendicular axes through their centre of mass, hence they have three dof for rotation also. Each rotational dof contributes \(\frac{1}{2}k_\text{B}T\) per molecule.
Polyatomic molecules have more than 1 dof for different modes of vibrational motion. The number of dof, \(f\), for the vibrational motion of a polyatomic molecule depends on the geometric structure of the molecule, that is, on the arrangement of atoms in the molecule. Each such dof contributes an average energy \(2 \times \frac{1}{2}k_\text{B}T\), from the kinetic energy and potential energy terms.
Therefore, for 1 mole of a polyatomic gas, the internal energy is
\[E = \frac{3}{2}N_\text{A}k_\text{B}T + \frac{3}{2}N_\text{A}k_\text{B}T + f \times \frac{2}{2}N_\text{A}k_\text{B}T\]
\[E = \left(3 + f\right)N_\text{A}k_\text{B}T\]
and the molar specific heats at constant volume and constant pressure are given as
\[C_\text{V} = \left(3 + f\right)R \qquad \text{--- (3.37)}\]
and
\[C_\text{P} = \left(4 + f\right)R \qquad \text{--- (3.38)}\]
\[\therefore\ \gamma = \frac{C_\text{P}}{C_\text{V}} = \frac{4 + f}{3 + f} \qquad \text{--- (3.39)}\]
Equation (3.38) is simply Eq. (3.37) with \(R\) added, as Mayer’s relation requires. And Eq. (3.39) shows that as \(f\) grows, \(\gamma\) creeps closer and closer to 1 — the more ways a molecule has of storing energy, the smaller the fraction that goes into the motion we measure as temperature.
3.10 Absorption, Reflection and Transmission of Heat Radiation
In XIth Std. you have studied that heat can be transferred by conduction, convection and radiation. The first two modes of heat transfer require a material medium for transmission of heat, but radiation does not need a material medium.
The most common example of heat transfer by the radiation mode that we come across every day is the transfer of heat and light from the Sun to the Earth and to us. Between the Sun and the Earth lies almost perfect vacuum, and yet the energy arrives. Neither conduction nor convection could carry it across. In this section, we shall discuss radiation in detail.
Two Meanings of One Word
The word ‘radiation’ is used in two related ways, and it is worth separating them at the outset.
As we have just seen, the term ‘radiation’ refers to one mode of transfer of heat. But the term ‘radiation’ also refers to the continuous emission of energy from the surface of any body because of its thermal energy. One usage names the process; the other names what is given out.
This emitted energy is termed radiant energy, and it is in the form of electromagnetic waves.
Radiation is therefore the fastest mode of transfer of heat. Being electromagnetic waves, it travels at the speed of light — far faster than the slow diffusion of energy through a solid, or the bodily movement of hot fluid in convection.
What Radiation Needs, and What It Does Not
The process of transfer of heat by radiation does not require any material medium, since electromagnetic waves travel through vacuum. Heat transfer by radiation is therefore possible through vacuum as well as through a material medium that is transparent to this radiation.
Physical contact of the bodies that are exchanging heat is also not required. Two bodies can warm or cool each other across a gap without ever touching.
When the radiation falls on a body that is not transparent to it — for example, on the floor or on our hands — it is absorbed, and the body gets heated up. This is the everyday sensation of standing in sunlight, or of feeling warmth on your face from a fire several metres away.
The electromagnetic radiation emitted by bodies which are at a higher temperature with respect to the surroundings is known as thermal radiation.
3.10.1 Interaction of Thermal Radiation and Matter
Whenever thermal radiation falls on the surface of an object, some part of the heat energy is reflected, some part is absorbed and the remaining part is transmitted. Every object does all three things to some extent; what distinguishes one material from another is the proportion in which it divides the incoming energy, as shown in Fig. 3.10a.
Diagram
Fig. 3.10a – A beam of thermal radiation striking a slab, splitting into a reflected beam going back, an absorbed portion shown inside the slab, and a transmitted beam emerging on the far side.
Accounting for the Incident Energy
Let \(Q\) be the total amount of thermal energy incident on the surface of an object, and let \(Q_\text{a}\), \(Q_\text{r}\) and \(Q_\text{t}\) be the respective amounts of heat absorbed, reflected and transmitted by the object. Since energy cannot vanish, the three parts must add up to the whole:
\[Q = Q_\text{a} + Q_\text{r} + Q_\text{t}\]
Dividing throughout by \(Q\),
\[1 = \frac{Q_\text{a}}{Q} + \frac{Q_\text{r}}{Q} + \frac{Q_\text{t}}{Q}\]
\[\therefore\ a + r + t_\text{r} = 1 \qquad \text{--- (3.40)}\]
where \(a = \dfrac{Q_\text{a}}{Q}\), \(r = \dfrac{Q_\text{r}}{Q}\) and \(t_\text{r} = \dfrac{Q_\text{t}}{Q}\) are the coefficients of absorption, reflection and transmission respectively.
Each of these three coefficients has its own name and definition.
Since all three quantities \(a\), \(r\) and \(t_\text{r}\) are ratios of thermal energies, they are dimensionless quantities. They carry no unit, and each of them must lie between 0 and 1.
The Extreme Cases
Equation (3.40) is most illuminating when one or two of its terms are set to zero. Each such case describes a recognisable kind of material.
If \(r = 0\) and \(a = 0\), then \(t_\text{r} = 1\). All the incident energy is transmitted through the object, that is, it is a perfect transmitter. The object is said to be completely transparent to the radiation.
A substance through which heat radiations can pass is known as a diathermanous substance.
Examples of diathermanous substances are glass, quartz, sodium chloride, hydrogen, oxygen and dry air.
On the other hand, if \(t_\text{r} = 0\) and \(a + r = 1\), that is, the object does not transmit any radiation, it is said to be opaque to the radiation. Whatever energy arrives must then be either absorbed or reflected, since there is no third route left.
Substances which are largely opaque to thermal radiations, that is, which do not transmit the heat radiations incident on them, are known as athermanous substances.
Examples of athermanous substances are water, wood, iron, copper, moist air and benzene.
If \(t_\text{r} = 0\) and \(a = 0\), then \(r = 1\). All the incident energy is reflected by the object, that is, it is a perfect reflector. A good reflector is thus a poor absorber and a poor transmitter — a conclusion that follows at once from Eq. (3.40).
If \(r = 0\) and \(t_\text{r} = 0\), then \(a = 1\). All the incident energy is absorbed by the object. Such an object is called a perfect blackbody, which we will discuss in detail later in this chapter.
These Coefficients Are Not Fixed Numbers
There is one more point, and it is easy to overlook. The values of \(a\), \(r\) and \(t_\text{r}\) also depend on the wavelength of the incident radiation, in addition to the material of the object on which it is incident.
Hence, it is possible that an object may be athermanous or diathermanous for certain wavelengths, but a good absorber for certain other wavelengths. Describing a material simply as “transparent” is therefore incomplete; we must always ask, transparent to which radiation.
3.11 Perfect Blackbody
At the end of the previous subsection we met the extreme case in which \(r = 0\) and \(t_\text{r} = 0\), so that \(a = 1\). We now give that case its proper name and study it, because it turns out to be the reference standard against which every real surface is measured.
A body which absorbs the entire radiant energy incident on it is called an ideal or perfect blackbody. Thus, for a perfect blackbody, \(a = 1\).
Any surface that absorbs all the energy incident on it, and does not reflect any energy, therefore appears black — unless its temperature is very high, high enough for it to be self-luminous. This exception matters: a blackbody is called “black” because of what it does to radiation falling on it, not because of how it looks when it is hot.
No real surface achieves \(a = 1\) exactly. Lamp black or platinum black, which absorb nearly 97% of incident radiant heat, resemble a perfect blackbody closely enough for practical work.
For the study of radiation, a simple arrangement illustrated in Fig. 3.4, which was designed by Ferry, can be used as a perfect blackbody. This arrangement is described in the next subsection.
Why Good Absorbers Must Be Good Emitters
The second of those questions has an answer that can be reasoned out completely, without any experiment, and it is one of the most useful results in this part of the chapter.
Consider two objects which are opaque to thermal radiation, having the same temperature and the same surface area. The surface of one object is well polished, and the surface of the other object is painted black.
The well-polished object reflects most of the energy falling on it and absorbs little. On the other hand, the black painted object absorbs most of the radiation falling on it and reflects little.
Now apply the condition of steady temperature. The rate of emission of thermal radiation must be equal to the rate of absorption for both the objects, so that temperature is maintained. If a body took in more than it gave out, it would keep getting hotter, which is not what we observe.
The black painted object absorbs more, hence it must radiate more to maintain its temperature. Therefore, good absorbers are always good emitters, and poor absorbers are poor emitters.
The argument can be carried one step further. Since each object must either absorb or reflect the radiation incident on it, a poor absorber should be a good reflector and vice versa. Hence, a good reflector is also a poor emitter.
This is the reason for silvering the walls of vacuum bottles or thermos flasks. The silvered surface is an excellent reflector, so it is a poor emitter; hot liquid inside loses very little energy by radiation, and cold liquid gains very little from outside. Combined with the vacuum between the walls, which blocks conduction and convection, all three modes of heat transfer are shut down at once.
Notice that this whole conclusion was reached from a single requirement — that a body in equilibrium must give out exactly as much as it takes in. Section 3.13 turns the same argument into a precise mathematical statement known as Kirchhoff’s law.
3.11.1 Ferry’s Blackbody
Since no real surface absorbs every bit of radiation falling on it, a perfect blackbody has to be manufactured by a trick of geometry rather than by finding the right paint. Ferry’s arrangement, shown in Fig. 3.4, does exactly this.
The Construction
It consists of a double-walled hollow sphere having a tiny hole or aperture, through which radiant heat can enter. The space between the walls is evacuated and the outer surface of the sphere is silvered. Both features serve the same purpose — the vacuum stops conduction and convection through the walls, and the silvering, being a good reflector and therefore a poor emitter, prevents the sphere from losing heat to the room.
The inner surface of the sphere is coated with lamp black, so that most of the radiation striking it is absorbed at the very first encounter.
Diagram
Fig. 3.4 – A double-walled hollow sphere in section, labelled with the silver-polished outer surface, the evacuated space between the walls, the inner surface coated with lamp black, the small aperture, and a conical projection on the inner wall opposite the aperture, with a ray entering the aperture and undergoing several reflections inside.
There is a conical projection on the inner surface of the sphere opposite the aperture. This small detail is deliberate. The projection ensures that a ray travelling along the axis of the aperture is not incident normally on the surface, and is therefore not reflected back along the same path. Without it, a ray entering straight down the axis could strike the far wall head-on and return the way it came.
Why the Aperture Behaves as a Blackbody
Radiation entering through the small hole has a negligible chance of escaping back through that small hole. A heat ray entering the sphere through the aperture suffers multiple reflections and is almost completely absorbed inside.
Consider what happens at each bounce. The lamp-black coating absorbs the large majority of the energy, and only a small remainder is reflected onward to strike the wall again, where most of that is absorbed in turn. After a handful of reflections, practically nothing is left. Meanwhile the hole is so small compared with the inner surface that the chance of a surviving ray finding its way out is vanishingly small.
Thus the aperture behaves like a perfect blackbody. It is important to be clear about what is the blackbody here: not the sphere, not the coating, but the opening. Radiation that enters it does not come out.
In a similar construction, Wien used a cylindrical body with a vertical slit as the aperture. This gives a greater effective area as a perfect blackbody, which is useful when a stronger signal is needed for measurement.
Cavity Radiation
Similar working can be achieved using a cavity radiator, which consists of a block of material with an internal cavity. The inner and outer surfaces are connected by a small hole. The radiation falling on the block that enters through the hole cannot escape back from it. Hence, the cavity acts as a blackbody.
When the block is heated to a high temperature, thermal radiation is emitted through the hole. This is called cavity radiation, and it resembles the radiation emitted by a blackbody.
Its nature depends only on the temperature of the cavity walls, and not on the shape and size of the cavity or the material of the cavity walls. This independence is what makes cavity radiation so valuable: two laboratories anywhere in the world, using cavities of different shapes and different metals, obtain identical radiation provided only that the temperature is the same.
In the kinetic theory of gases, we discussed the theory, properties and various phenomena of an ideal gas rather than dealing with real gases. Similarly, it is convenient to work with an ideal blackbody. In both cases the idealisation is not an evasion — it is a well-defined standard that real systems can be compared against.
3.12 Emission of Heat Radiation
In 1792, Pierre Prevost published a theory of radiation known as the theory of exchange of heat. According to this theory, all bodies at all temperatures above 0 K (absolute zero temperature) radiate thermal energy, and at the same time they absorb radiation received from the surroundings.
The word exchange is the important one. Radiation is not something that only hot bodies do to cold ones; it is a two-way traffic going on at all times, between every pair of bodies.
The amount of thermal radiation emitted per unit time depends on the nature of the emitting surface, its area and its temperature. Hotter bodies radiate at a higher rate than cooler bodies. As regards visible radiation, light-coloured bodies reflect most of it, whereas dark-coloured bodies absorb most of the incident visible radiation.
What Absorption and Emission Do to Temperature
For a body, the absorbed radiation, being energy, increases the kinetic energy of the constituent atoms oscillating about their mean positions. You have learnt earlier that the average translational kinetic energy determines the temperature of the body; the absorbed radiation therefore causes a rise in the temperature of the body.
At the same time the body itself also radiates, and therefore its energy decreases, causing a lowering of temperature. What we finally observe is the outcome of this contest.
If a body radiates more than it absorbs, its temperature decreases, and vice versa. When the rate of absorption of radiation is the same as the rate of emission of radiation, the temperature of the body remains constant, and the body is said to be in thermal equilibrium with its surroundings.
You might recall the example from XIth Std. of a cup of hot tea (\(T_\text{tea} > T_\text{room}\)) or a plate containing ice (\(T_\text{ice} < T_\text{room}\)) kept on a table. Both attain the room temperature after some time — the tea by radiating more than it absorbs, the ice by absorbing more than it radiates.
At room temperature also, all bodies radiate as well as absorb radiation, but their rate of emission and rate of absorption are the same, hence their temperature remains constant. You can therefore infer that hot bodies would radiate more than cooler bodies.
Temperature and the Colour of the Radiation
At room temperature — in fact for temperatures \(T\) lower than 800 °C — the thermal radiation corresponds to wavelengths longer than those of visible light, and hence we do not see them. The radiation is present, but our eyes are not sensitive to it.
When the body is heated, the radiated energy corresponds to shorter wavelengths. For temperatures around 800 °C, part of the energy emitted is in the visible range and the body appears red. At around 3000 °C, it looks white hot.
We have thus seen that all bodies radiate electromagnetic radiation when their temperature is above the absolute zero of temperature.
Emissive Power
Let us now collect the factors on which radiation depends. The amount of heat radiated by a body depends on:
- the absolute temperature of the body (\(T\));
- the nature of the body — the material, and the nature of the surface, whether polished or not, and so on;
- the surface area of the body (\(A\));
- the time duration for which the body emits radiation (\(t\)).
Two of these four can be dealt with straightaway. The amount of heat radiated, \(Q\), is directly proportional to the surface area (\(A\)) and to the time duration (\(t\)) — doubling either doubles the energy radiated, with no change in the physics.
It is therefore convenient to divide them out, and to consider the quantity of heat radiated per unit area per unit time, or the power emitted per unit area. This is defined as the emissive power or radiant power, \(R\), of the body at a given temperature \(T\).
\[\therefore\ R = \frac{Q}{At}\]
Here \(Q\) is the heat radiated in joule (J), \(A\) is the surface area in square metre (m²) and \(t\) is the time in second (s).
Handling the Nature of the Surface
Of the four factors listed above, one still resists being written as a number. The nature of the emitting surface — its material or its polishing — is not a physical quantity. There is no scale on which “polished brass” can be entered into an equation.
Hence, to discuss the material aspect, we compare objects of different materials with identical geometries at the same temperature. By keeping shape, size and temperature fixed, any difference that remains must be due to the surface itself.
For such a comparison we need a reference. At a given temperature, a perfect blackbody has the maximum emissive power of any body, since it is the best possible absorber and therefore, as shown in Section 3.11, the best possible emitter. Thus it is convenient to compare the emissive power of a given surface with that of a perfect blackbody at the same temperature — which is exactly what the next subsection does.
3.12.1 Coefficient of Emission or Emissivity
The comparison suggested at the end of the previous subsection is made precise by defining a single number for each surface.
The coefficient of emission, or emissivity (\(e\)), of a given surface is the ratio of the emissive power \(R\) of the surface to the emissive power \(R_\text{B}\) of a perfect black surface, at the same temperature.
\[\therefore\ e = \frac{R}{R_\text{B}} \qquad \text{--- (3.41)}\]
The phrase “at the same temperature” is essential and must never be dropped. Emissive power changes sharply with temperature, so comparing a warm surface with a hot blackbody would tell us nothing about the surface itself.
For a perfect blackbody \(e = 1\), since its emissive power is the maximum possible and the ratio becomes 1. For a perfect reflector \(e = 0\), since it absorbs nothing and therefore emits nothing.
Emissivity of Real Surfaces
Everyday objects are not ideal blackbodies. Hence they radiate at a rate less than that of a blackbody at the same temperature.
Also, for these objects the rate does depend on properties such as the colour and composition of the surface, in addition to the temperature. All these effects together are taken care of in the term emissivity \(e\).
For an ordinary body, \(0 < e < 1\), depending on the nature of the surface — for example, the emissivity of copper is 0.3, meaning polished copper radiates only about three-tenths as strongly as a blackbody of the same size at the same temperature.
Emissivity is larger for rough surfaces and smaller for smooth and polished surfaces. A rough surface presents countless small facets and pits from which radiation can leave and into which incoming radiation can be trapped, while a polished surface turns most radiation away.
Emissivity also varies with temperature and with the wavelength of radiation, to some extent. So a single quoted value for a material should be understood as a good working figure rather than an unchanging constant.
3.13 Kirchhoff’s Law of Heat Radiation and its Theoretical Proof
Section 3.11 showed by a qualitative argument that good absorbers are good emitters. Kirchhoff’s law makes that statement exact, and extends it to each wavelength separately.
Kirchhoff’s law of thermal radiation deals with wavelength-specific radiative emission and absorption by a body in thermal equilibrium.
Since we can describe the emissive power of an ordinary body in comparison to a perfect blackbody through its emissivity, Kirchhoff’s law can also be stated in a second, equivalent form.
Symbolically,
\[a = e\]
or, more specifically,
\[a(\lambda) = e(\lambda)\]
Thus, if a body has high emissive power, it also has high absorptive power; and if a body has low emissive power, it also has low absorptive power.
The Theoretical Proof
Kirchhoff’s law can be proved theoretically by the following thought experiment.
Consider an ordinary body A and a perfect blackbody B of identical geometric shapes placed in an enclosure, as shown in Fig. 3.13a. In thermal equilibrium, both bodies will be at the same temperature as that of the enclosure.
Diagram
Fig. 3.13a – A closed enclosure containing two bodies of identical shape side by side, one labelled ordinary body A and the other labelled perfect blackbody B, with arrows showing radiation falling on each from the enclosure walls and radiation leaving each body.
Because both bodies have identical shapes and sit in the same enclosure, each of them receives exactly the same amount of radiation in unit time. That is the point of making them identical in geometry — any difference in behaviour must then come from the surfaces alone.
Let \(R\) be the emissive power of body A, \(R_\text{B}\) be the emissive power of blackbody B, and \(a\) be the coefficient of absorption of body A.
If \(Q\) is the quantity of radiant heat incident on each body in unit time and \(Q_\text{a}\) is the quantity of radiant heat absorbed by body A, then
\[Q_\text{a} = aQ\]
since by definition the absorption coefficient is the fraction of incident energy that is absorbed.
As the temperatures of body A and blackbody B remain the same, both must emit the same amount as they absorb in unit time. If this were not so, one of them would be gaining or losing energy and its temperature would drift — contradicting the equilibrium we have assumed.
Since emissive power is the quantity of heat radiated from unit area in unit time, we can write, for body A:
Quantity of radiant heat absorbed by body A = Quantity of heat emitted by body A
\[\text{or,}\quad aQ = R \qquad \text{--- (3.42)}\]
For the perfect blackbody B, the coefficient of absorption is 1, so it absorbs the whole of the incident radiation \(Q\) and must emit an equal amount:
\[Q = R_\text{B} \qquad \text{--- (3.43)}\]
Dividing Eq. (3.42) by Eq. (3.43), the incident quantity \(Q\) cancels and we get
\[a = \frac{R}{R_\text{B}} \quad \text{or,} \quad \frac{R}{a} = R_\text{B} \qquad \text{--- (3.44)}\]
The second form of Eq. (3.44) is precisely the first statement of the law: the ratio of emissive power to coefficient of absorption equals the emissive power of a perfect blackbody at the same temperature.
But from Eq. (3.41), \(\dfrac{R}{R_\text{B}} = e\).
\[\therefore\ a = e\]
Hence, Kirchhoff’s law is theoretically proved.
Note that no property of any particular material was used anywhere in this proof. Only the requirement of thermal equilibrium was needed — which is why the law applies to every surface without exception.
3.14 Spectral Distribution of Blackbody Radiation
The radiant energy emitted per unit area per unit time by a blackbody depends on its temperature. But a single number for the total is not the whole story, because hot objects radiate electromagnetic radiation over a large range of frequencies at once.
Hence the rate of emission per unit area, or power per unit area of a surface, is defined as a function of the wavelength \(\lambda\) of the emitted radiation. Instead of asking only “how much energy?”, we ask “how much energy at each wavelength?”
At low temperature, the power radiated is small and primarily lies in the long wavelength region. As the temperature is increased, the rate of emission increases fast.
At each temperature, the radiant energy contains a mixture of different wavelengths. At higher temperatures, the total energy radiated per unit time increases, and the proportion of energy emitted at higher frequencies, or shorter wavelengths, also increases.
The Experiment of Lummer and Pringsheim
Lummer and Pringsheim studied the energy distribution of blackbody radiation as a function of wavelength. They kept the source of radiation, such as a cavity radiator, at different temperatures and measured the radiant power corresponding to different wavelengths.
A cavity radiator was used precisely because, as Section 3.11.1 established, its aperture behaves as a perfect blackbody and its radiation depends only on temperature.
The measurements were represented graphically in the form of curves showing the variation of radiant power per unit area as a function of wavelength \(\lambda\) at different constant temperatures, as shown in Fig. 3.5. Spectral distribution of power radiated by a body indicates the power radiated at different wavelengths.
Diagram
Fig. 3.5 – A family of smooth curves of radiant power per unit wavelength range against wavelength, one curve for each of several temperatures, each rising to a peak and falling away, with higher-temperature curves both taller and peaking at shorter wavelength.
Experimental observations indicated that the spectral distribution depended only on the absolute temperature \(T\) of a blackbody, and was independent of the material. This is a striking result: the curve carries information about temperature alone, which is what later makes it possible to measure the temperature of objects we can never touch.
What the Curves Show
From the experimental curves, it is observed that:
At a given temperature, the energy is not uniformly distributed in the spectrum, that is, as a function of wavelength, of the blackbody.
At a given temperature, the radiant power emitted initially increases with increase of wavelength, reaches its maximum and then decreases. The wavelength corresponding to the radiation of maximum intensity, \(\lambda_\text{max}\), is characteristic of the temperature of the radiating body. Remember, it is not the maximum wavelength emitted by the object.
The area under the curve represents the total energy emitted per unit time per unit area by the blackbody at all wavelengths.
The peak of the curves shifts towards the left — towards shorter wavelengths — that is, the value of \(\lambda_\text{max}\) decreases with increase in temperature.
At higher temperatures, the radiant power, or total energy emitted per unit time per unit area (that is, the area under the curve), corresponding to all the wavelengths, increases.
At a temperature of 300 K, around room temperature, the most intense of these waves has a wavelength of about \(5 \times 10^{-5}\) m; the radiant power is smaller for wavelengths other than this value. Practically all the radiant energy at this temperature is carried by waves longer than those corresponding to red light. These are infrared radiations.
Observations 4 and 5 are worth holding on to, because each of them becomes a law in the sections that follow. Observation 4 is the content of Wien’s displacement law, and observation 5 leads to the Stefan-Boltzmann law.
Observation 6 also explains something from everyday experience. At room temperature the radiation from all the objects around us peaks well beyond the red end of the visible spectrum, so although every one of them is radiating continuously, we see none of it.
The Theoretical Struggle
A theoretical explanation of the above observations could not be given by the then existing theories. This was not for lack of effort — the shape of these curves became one of the outstanding unsolved problems of nineteenth-century physics.
Wien gave an expression for the spectral distribution from the laws of thermodynamics, which fitted the experimental observations only for short wavelengths.
Lord Rayleigh and Sir James Jeans gave a formula from the equipartition of energy — the very law studied earlier in this chapter. This formula fits well in the long wavelength regions, but tends to infinity at short wavelengths. Since that would mean a warm body radiating unlimited energy at the blue and ultraviolet end, the failure was a serious one, and it came to be known as the ultraviolet catastrophe.
So one formula worked at one end of the spectrum and the other formula at the other end, with neither working throughout. It was therefore essential to propose a new model to explain the behaviour of a blackbody.
Planck, being aware of the shortcomings of the two models, combined the two models using an empirical formula and could describe the observed spectrum quite well.
The importance of this step is difficult to overstate. In order to fit a set of curves measured from a glowing cavity, Planck was forced to abandon the assumption that energy can be exchanged in any amount whatsoever — and with that, quantum physics began.
3.14.1 Wien’s Displacement Law
Observation 4 of the previous section — that the peak of the curve shifts to shorter wavelengths as the temperature rises — can be stated as an exact relation.
It is observed that the wavelength for which the emissive power of a blackbody is maximum is inversely proportional to the absolute temperature of the blackbody. This is Wien’s displacement law.
\[\lambda_\text{max} \propto \frac{1}{T}\]
\[\text{or,}\quad \lambda_\text{max} = \frac{b}{T}\]
\[\therefore\ \lambda_\text{max}T = b \qquad \text{--- (3.45)}\]
where \(b\) is called Wien’s constant and its value is \(2.897 \times 10^{-3}\) m K. Here \(\lambda_\text{max}\) is in metre (m) and \(T\) is the absolute temperature in kelvin (K), so the product carries the unit m K.
The word displacement in the name refers to the shifting, or displacement, of the peak along the wavelength axis as the temperature is changed.
\(\lambda_\text{max}\) indicates the wavelength at which the blackbody dominantly radiates. Thus it corresponds to the dominant colour of the radiating body, and is a function of its temperature.
You might have heard of white dwarfs and red giants. White dwarfs are hot stars with surface temperature \(\sim 10000\) K, while red giants are cooler, corresponding to surface temperature \(\sim 3000\) K. Their colours are not accidental labels — they are direct readings of Eq. (3.45).
This law is useful to determine the temperatures of distant stars, the Sun, the moon and so on.
3.15 Stefan-Boltzmann Law of Radiation
Wien’s law told us where the emission of a blackbody peaks. We shall now discuss the temperature dependence of the thermal radiation emitted per unit time by a blackbody — that is, how much it emits in total, which is observation 5 of Section 3.14 turned into an equation.
In 1879, Josef Stefan proposed an empirical relation between the rate at which heat is radiated (the radiant power \(R\)) from unit area of a perfect blackbody and its temperature \(T\), based on experimental observations. Five years later, Boltzmann derived the same relation using thermodynamics. Hence it is known as the Stefan-Boltzmann law.
\[R \propto T^{4}\]
\[\text{or,}\quad R = \sigma T^{4} \qquad \text{--- (3.46)}\]
where \(\sigma\) is Stefan’s constant and is equal to \(5.67 \times 10^{-8}\) J m⁻² s⁻¹ K⁻⁴ or W m⁻² K⁻⁴, and the dimensions of \(\sigma\) are \([\text{L}^0\text{M}^1\text{T}^{-3}\text{K}^{-4}]\).
Here \(R\) is the emissive power in W/m² and \(T\) is the absolute temperature in kelvin (K). The temperature must always be in kelvin; substituting a Celsius value is the single commonest error in numerical problems on this law.
Thus, the power radiated by a perfect blackbody depends only on its temperature, and not on any other characteristic such as colour, material or nature of surface.
If \(Q\) is the amount of radiant energy emitted in time \(t\) by a perfect blackbody of surface area \(A\) at temperature \(T\), then
\[\frac{Q}{At} = \sigma T^{4}\]
For Bodies That Are Not Blackbodies
For a body which is not a blackbody, the energy radiated per unit area per unit time is still proportional to the fourth power of temperature, but is less than that for a blackbody. For an ordinary body,
\[R = e\sigma T^{4} \qquad \text{--- (3.47)}\]
where \(e\) is the emissivity of the surface. Since \(0 < e < 1\) for an ordinary body, Eq. (3.47) always gives a smaller value than Eq. (3.46) at the same temperature, exactly as expected.
Net Loss and Net Gain of Energy
A body is never radiating into empty nothingness; the surroundings radiate back. Prevost’s theory of exchange, met in Section 3.12, must therefore be applied.
If a perfect blackbody having absolute temperature \(T\) is kept in a surrounding which is at a lower absolute temperature \(T_0\), then
the energy radiated per unit area per unit time \(= \sigma T^{4}\)
Energy absorbed from the surroundings per unit area per unit time \(= \sigma T_0^{4}\)
Therefore, net loss of energy by the perfect blackbody per unit area per unit time
\[= \sigma T^{4} - \sigma T_0^{4} = \sigma\left(T^{4} - T_0^{4}\right)\]
For an ordinary body at temperature \(T\), the net loss of energy per unit area per unit time is
\[e\sigma\left(T^{4} - T_0^{4}\right)\]
On the other hand, if the body is at a temperature lower than the surroundings, that is, \(T < T_0\), then
\[e\sigma\left(T_0^{4} - T^{4}\right)\]
will be the net gain in thermal energy of the body per unit area per unit time. The same expression serves both cases; only the order of the two terms is reversed so that the answer comes out positive.
Since the loss or gain of energy per unit area per unit time is proportional to the fourth power of absolute temperature, this law is very significant in deciding the thermal equilibrium of physical systems.
If the absolute temperature of a body is doubled, the power radiated will increase by a factor of \(2^{4} = 16\). Or, if a body radiates with some rate at room temperature (300 K), the rate will double even if we increase the temperature of the body by 57 °C.
3.16 Summary
The Ideal Gas and Its Equation of State
- The three gas laws, each for a fixed mass of an enclosed gas: Boyle’s law, \(V \propto \dfrac{1}{P}\) at constant \(T\); Charles’ law, \(V \propto T\) at constant \(P\); Gay-Lussac’s law, \(P \propto T\) at constant \(V\).
- Combining them: \(PV \propto T\), or \(\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}\).
- Ideal gas equation: \(PV = nRT\), where \(n = \dfrac{M}{M_0} = \dfrac{N}{N_\text{A}}\) and \(R = 8.314\) J mol⁻¹ K⁻¹ is the universal gas constant.
- In molecular form, \(PV = Nk_\text{B}T\), with \(R = N_\text{A}k_\text{B}\).
- The equation relating the quantities that specify the state of a gas — \(P\), \(T\), \(V\), \(E\) — is called the equation of state.
- An ideal gas obeys \(PV = nRT\) at all pressures and all temperatures, and has no intermolecular interactions.
- No real gas is truly ideal. A real gas is in the ideal state when its molecules are so far apart that intermolecular interaction is negligible — that is, at low density, low pressure or high temperature, and well above its liquefaction temperature.
Mean Free Path
- Between collisions a molecule moves in a straight line with constant velocity; collisions are elastic and change both speed and direction.
- Mean free path is the average distance travelled by a molecule between two successive collisions:
\[\lambda = \frac{1}{\sqrt{2}\,\pi d^2 (N/V)} \qquad \text{--- (3.6)}\]
- \(\lambda\) decreases with increasing number density \(N/V\) and varies as \(1/d^2\), since collision probability depends on the cross section of a molecule.
Pressure and Temperature from Kinetic Theory
- For a molecule striking a wall elastically, the change in momentum is \(\Delta p_x = -2m\text{v}_x\), and the momentum transferred to the wall is \(+2m\text{v}_x\).
- Time between successive collisions with the same wall: \(\Delta t = \dfrac{2L}{\text{v}_x}\); average force by one molecule \(= \dfrac{m\text{v}_{x1}^2}{L}\).
- Pressure in terms of the \(x\) component alone: \(P = \dfrac{mN\overline{\text{v}_x^2}}{V}\) — Eq. (3.9).
- Since \(\overline{\text{v}_x^2} = \overline{\text{v}_y^2} = \overline{\text{v}_z^2} = \frac{1}{3}\overline{v^2}\),
\[P = \frac{1}{3}\,\frac{N}{V}\,m\,\overline{v^{2}} \qquad \text{--- (3.10)}\]
- Root mean square speed:
\[v_\text{rms} = \sqrt{\frac{3RT}{M_0}} \qquad \text{--- (3.12)}\]
where \(M_0 = N_\text{A}m\) is the molar mass. For nitrogen at 300 K, \(v_\text{rms} = 517\) m/s; for oxygen, 483 m/s. 6. Speed of sound in a gas, \(v_\text{s} = \sqrt{\dfrac{\gamma RT}{M_0}}\), is of the same order as, but always less than, the molecular speeds, since \(\gamma < 3\). 7. \(PV = \frac{2}{3}E\), and using \(PV = Nk_\text{B}T\),
\[E = \frac{3}{2}Nk_\text{B}T, \qquad \frac{E}{N} = \frac{3}{2}k_\text{B}T \qquad \text{--- (3.17), (3.18)}\]
- The internal energy of an ideal gas is purely kinetic, and absolute temperature is a measure of the average translational kinetic energy per molecule.
Equipartition of Energy and Degrees of Freedom
- Degrees of freedom are the total number of coordinates or independent quantities needed to describe the position and configuration of a system completely.
- Each translational component carries \(\frac{1}{2}k_\text{B}T\), so the total translational energy per molecule is \(\frac{3}{2}k_\text{B}T\) — Eq. (3.19).
- A diatomic molecule has 3 translational dof and 2 rotational dof; rotation about the molecular axis does not count.
- Total translational plus rotational energy of a diatomic molecule:
\[E = \frac{1}{2}m\text{v}_x^{2} + \frac{1}{2}m\text{v}_y^{2} + \frac{1}{2}m\text{v}_z^{2} + \frac{1}{2}I_z\omega_z^{2} + \frac{1}{2}I_y\omega_y^{2} \qquad \text{--- (3.20)}\]
- A vibrational mode contributes two quadratic terms, \(E(\text{vibrational}) = \frac{1}{2}mu^{2} + \frac{1}{2}kr^{2}\) — Eq. (3.22) — and therefore \(2 \times \frac{1}{2}k_\text{B}T\).
- Law of equipartition of energy: for a gas in thermal equilibrium at temperature \(T\), the average energy per molecule associated with each quadratic term is \(\frac{1}{2}k_\text{B}T\). It holds at high temperatures, not at extremely low temperatures where quantum effects become important.
Specific Heats of Gases
- Mayer’s relation: \(C_\text{P} - C_\text{V} = R\) — Eq. (3.26). If heat and work are in different units, \(C_\text{P} - C_\text{V} = \dfrac{R}{J}\), and \(S_\text{P} - S_\text{V} = \dfrac{R}{M_0J}\) — Eq. (3.27).
- Monatomic gases: \(C_\text{V} = \frac{3}{2}R\), \(C_\text{P} = \frac{5}{2}R\), \(\gamma = \frac{5}{3}\).
- Diatomic gases, rigid rotator: \(C_\text{V} = \frac{5}{2}R\), \(C_\text{P} = \frac{7}{2}R\), \(\gamma = \frac{7}{5}\).
- Diatomic gases, non-rigid vibrating: \(C_\text{V} = \frac{7}{2}R\), \(C_\text{P} = \frac{9}{2}R\), \(\gamma = \frac{9}{7}\).
- Polyatomic gases with \(f\) vibrational dof: \(C_\text{V} = (3 + f)R\), \(C_\text{P} = (4 + f)R\), \(\gamma = \dfrac{4+f}{3+f}\).
Interaction of Thermal Radiation with Matter
- Radiation needs no material medium and is the fastest mode of heat transfer; thermal radiation is that emitted by bodies hotter than their surroundings.
- Incident radiation is absorbed, reflected or transmitted: \(a + r + t_\text{r} = 1\) — Eq. (3.40) — where \(a\), \(r\) and \(t_\text{r}\) are dimensionless.
- Diathermanous substances transmit heat radiation (\(t_\text{r} \neq 0\)): glass, quartz, sodium chloride, hydrogen, oxygen, dry air.
- Athermanous substances are largely opaque to heat radiation: water, wood, iron, copper, moist air, benzene.
- The values of \(a\), \(r\) and \(t_\text{r}\) depend on the wavelength of the incident radiation as well as on the material.
Blackbody and Emission
- A perfect blackbody absorbs all incident radiant energy, so \(a = 1\). Lamp black and platinum black absorb nearly 97% of incident radiant heat.
- Ferry’s blackbody: a double-walled evacuated sphere, silvered outside, coated with lamp black inside, with a tiny aperture and a conical projection opposite it. The aperture behaves as a perfect blackbody. A cavity radiator works similarly, and its radiation depends only on the temperature of the walls.
- Prevost’s theory of exchange of heat: every body above 0 K radiates continuously and simultaneously absorbs radiation from its surroundings; at thermal equilibrium the two rates are equal.
- Good absorbers are good emitters; good reflectors are poor emitters.
- Emissive power: \(R = \dfrac{Q}{At}\), SI unit J m⁻² s⁻¹ or W/m², dimensions \([\text{L}^0\text{M}^1\text{T}^{-3}]\).
- Emissivity: \(e = \dfrac{R}{R_\text{B}}\) — Eq. (3.41) — dimensionless; \(e = 1\) for a perfect blackbody, \(e = 0\) for a perfect reflector, and \(0 < e < 1\) for an ordinary body (for copper, \(e = 0.3\)).
Laws of Radiation
- Kirchhoff’s law: at a given temperature, the ratio of emissive power to coefficient of absorption of a body equals the emissive power of a perfect blackbody at the same temperature, for all wavelengths. Equivalently, \(a = e\), or more specifically \(a(\lambda) = e(\lambda)\).
- Proof: \(aQ = R\) for the ordinary body and \(Q = R_\text{B}\) for the blackbody give \(\dfrac{R}{a} = R_\text{B}\) — Eq. (3.44) — and since \(\dfrac{R}{R_\text{B}} = e\), it follows that \(a = e\).
- Spectral distribution of blackbody radiation depends only on the absolute temperature and not on the material. The area under the curve gives the total energy emitted per unit time per unit area; the peak shifts to shorter wavelengths as \(T\) rises. At 300 K the most intense wavelength is about \(5 \times 10^{-5}\) m, in the infrared.
- Wien’s displacement law:
\[\lambda_\text{max}T = b \qquad \text{--- (3.45)}\]
with Wien’s constant \(b = 2.897 \times 10^{-3}\) m K. Note that \(\lambda_\text{max}\) is the wavelength of maximum emissive power, not the maximum wavelength emitted. 5. Stefan-Boltzmann law: \(R = \sigma T^{4}\) — Eq. (3.46) — with Stefan’s constant \(\sigma = 5.67 \times 10^{-8}\) J m⁻² s⁻¹ K⁻⁴ (or W m⁻² K⁻⁴), dimensions \([\text{L}^0\text{M}^1\text{T}^{-3}\text{K}^{-4}]\). 6. For an ordinary body, \(R = e\sigma T^{4}\) — Eq. (3.47). 7. Net loss of energy per unit area per unit time by a body at \(T\) in surroundings at \(T_0\): \(\sigma\left(T^{4} - T_0^{4}\right)\) for a blackbody and \(e\sigma\left(T^{4} - T_0^{4}\right)\) for an ordinary body. If \(T < T_0\), the body gains \(e\sigma\left(T_0^{4} - T^{4}\right)\) per unit area per unit time. 8. Doubling the absolute temperature increases the radiated power by a factor of \(2^4 = 16\); a body radiating at 300 K doubles its rate for a rise of 57 °C.
3.17 Objective Questions
- Choose the correct option.
- In an ideal gas, the molecules possess
- only kinetic energy
- both kinetic energy and potential energy
- only potential energy
- neither kinetic energy nor potential energy
- The mean free path \(\lambda\) of molecules is given by
- \(\sqrt{\dfrac{2}{\pi nd^2}}\)
- \(\dfrac{1}{\pi nd^2}\)
- \(\dfrac{1}{\sqrt{2}\,\pi n d^2}\)
- \(\dfrac{1}{\sqrt{2}\,\pi nd}\), where \(n\) is the number of molecules per unit volume and \(d\) is the diameter of the molecules.
- If the pressure of an ideal gas is decreased by 10% isothermally, then its volume will
- decrease by 9%
- increase by 9%
- decrease by 10%
- increase by 11.11%
- If \(a = 0.72\) and \(r = 0.24\), then the value of \(t_\text{r}\) is
- 0.02
- 0.04
- 0.4
- 0.2
- The ratio of emissive power of a perfect blackbody at 1327 °C and 527 °C is
- 4 : 1
- 16 : 1
- 2 : 1
- 8 : 1
3.18 Check Your Concepts
Two samples of the same gas are kept at the same temperature, one at low pressure and the other at very high pressure. Which sample is expected to obey the ideal gas equation more closely, and why?
The universal gas constant \(R\) and the Boltzmann constant \(k_\text{B}\) carry the same physical idea but different values. Explain the relation between them and state the situation in which each is the more convenient constant to use.
A gas is compressed to half its original volume while the temperature is held constant. State what happens to the mean free path of its molecules, and justify your answer using the expression for \(\lambda\).
While deriving the expression for the pressure of an ideal gas, the momentum transferred to the wall in one collision is taken as \(2m\text{v}_x\) and not \(m\text{v}_x\). Explain why, and state what the value would become if the collisions were perfectly inelastic instead.
Why is it valid to write \(\overline{\text{v}_x^2} = \overline{\text{v}_y^2} = \overline{\text{v}_z^2}\) for a gas in equilibrium, even though an individual molecule may be moving almost entirely along one axis?
The derivation of Eq. (3.10) assumes a cubical container and neglects intermolecular collisions. Explain why the result remains valid even when both assumptions are dropped.
Why can the average of the velocities of the molecules of a gas not be used as a measure of molecular motion? What is done instead, and why does it work?
At the same temperature, hydrogen molecules move much faster than oxygen molecules, yet both have the same average translational kinetic energy. Explain how both statements can be true together.
A vibrational mode of a molecule contributes \(k_\text{B}T\) to its average energy, while a rotational mode contributes only \(\frac{1}{2}k_\text{B}T\). Give the reason for this difference.
Rotation of a diatomic molecule about its own molecular axis is not counted as a degree of freedom. Why not?
Explain why \(C_\text{P}\) is greater than \(C_\text{V}\) for a gas, and why their difference turns out to be the same for every ideal gas irrespective of its atomicity.
The measured value of \(\gamma\) for a certain gas is found to be very close to 1.67. What can you conclude about its molecules, and what would you conclude instead if the value were close to 1.4?
Dry air is diathermanous while moist air is athermanous. Use this fact to explain why a desert becomes very cold at night although its days are extremely hot.
A surface may be described as transparent to one radiation and opaque to another. Explain what this tells us about the coefficients \(a\), \(r\) and \(t_\text{r}\).
In Ferry’s blackbody, it is the aperture and not the lamp-black coating that is said to behave as a perfect blackbody. Justify this statement.
What is the purpose of the conical projection on the inner surface of Ferry’s blackbody, opposite the aperture?
Two identical vessels, one polished and one painted black, contain equally hot water and are left in the same room. Which will cool faster, and on which principle does your answer rest?
State Prevost’s theory of exchange of heat, and use it to explain why the temperature of an object left undisturbed in a room eventually stops changing.
Emissivity is defined as a ratio and therefore has no unit. What advantage does this bring when describing the radiating behaviour of a material?
Why must the phrase “at the same temperature” never be omitted from the definition of emissivity?
Assertion: A good reflector is a poor emitter. Reason: A body in thermal equilibrium must emit as much energy as it absorbs. Examine whether the reason correctly explains the assertion.
\(\lambda_\text{max}\) is often wrongly described as the longest wavelength emitted by a blackbody. State what it actually represents, and describe what the spectral distribution curve shows at wavelengths longer than \(\lambda_\text{max}\).
Explain why the spectral distribution of blackbody radiation is such a useful tool in astronomy, given that the material of a distant star cannot be sampled.
Both Wien’s expression and the Rayleigh-Jeans formula failed to describe the blackbody spectrum completely. State where each one succeeded and where each one failed, and what Planck had to assume in order to succeed.
The Stefan-Boltzmann law involves the fourth power of the absolute temperature. Explain what practical consequence this strong dependence has for a body that is much hotter than its surroundings.
A student calculates the radiated power of a body at 27 °C by substituting \(T = 27\) into \(R = \sigma T^4\). Identify the error and explain what effect it has on the answer.
Answer in brief.
- What will happen to the mean square speed of the molecules of a gas if the temperature of the gas increases?
- On what factors do the degrees of freedom depend?
- Write the ideal gas equation for a mass of 7 g of nitrogen gas.
- What is an ideal gas? Does an ideal gas exist in practice?
- Define athermanous substances and diathermanous substances.
When a gas is heated its temperature increases. Explain this phenomenon based on the kinetic theory of gases.
Explain, on the basis of kinetic theory, how the pressure of a gas changes if its volume is reduced at constant temperature.
Mention the conditions under which a real gas obeys the ideal gas equation.
State the law of equipartition of energy and hence calculate the molar specific heat of mono- and di-atomic gases at constant volume and constant pressure.
What is a perfect blackbody? How can it be realized in practice?
State (i) Stefan-Boltzmann law and (ii) Wien’s displacement law.
Explain the spectral distribution of blackbody radiation.
State and prove Kirchhoff’s law of heat radiation.
3.19 Practice with Numericals
A closed vessel of volume \(5.0 \times 10^{-3}\) m³ contains 0.25 mol of an ideal gas at 300 K. Calculate the pressure of the gas and the number of molecules it contains. (\(R = 8.314\) J mol⁻¹ K⁻¹, \(N_\text{A} = 6.02 \times 10^{23}\) mol⁻¹)
A fixed mass of gas at 27 °C and \(1.0 \times 10^{5}\) Pa occupies a volume of \(2.4 \times 10^{-2}\) m³. It is heated to 127 °C while the pressure is raised to \(1.5 \times 10^{5}\) Pa. Find its new volume.
Calculate the mean free path of a gas molecule at 300 K and \(1.01 \times 10^{5}\) Pa, taking the molecular diameter as \(3.0 \times 10^{-10}\) m and \(k_\text{B} = 1.38 \times 10^{-23}\) J/K.
The mean free path of the molecules of a gas at a certain pressure is \(8.0 \times 10^{-8}\) m. The gas is pumped out until its pressure falls to one-hundredth of the original value, the temperature remaining unchanged. Find the new mean free path.
A vessel of volume \(1.0 \times 10^{-3}\) m³ contains \(2.4 \times 10^{22}\) molecules of a gas, each of mass \(4.7 \times 10^{-26}\) kg. If the pressure of the gas is \(1.2 \times 10^{5}\) Pa, calculate the mean square speed of its molecules.
Calculate the rms speed of nitrogen molecules at 400 K, given that the molar mass of nitrogen is \(28 \times 10^{-3}\) kg/mol and \(R = 8.314\) J mol⁻¹ K⁻¹.
At what temperature will the rms speed of nitrogen molecules be double its value at 300 K?
Compare the rms speeds of helium and oxygen molecules at the same temperature, given that their molar masses are 4 and 32 respectively.
Calculate the average translational kinetic energy of a molecule of an ideal gas at 500 K, and the total translational kinetic energy of 2 moles of the gas at that temperature. (\(k_\text{B} = 1.38 \times 10^{-23}\) J/K, \(N_\text{A} = 6.02 \times 10^{23}\) mol⁻¹)
For a diatomic gas treated as a rigid rotator, calculate the total internal energy of 3 moles at 300 K. Repeat the calculation treating the molecules as non-rigid and vibrating, and find the difference between the two results. (\(R = 8.314\) J mol⁻¹ K⁻¹)
The molar specific heat of a gas at constant pressure is 29.1 J mol⁻¹ K⁻¹. Taking \(R = 8.314\) J mol⁻¹ K⁻¹, find \(C_\text{V}\) and \(\gamma\), and state whether the gas is monatomic or diatomic.
The two principal specific heats of a gas are \(S_\text{P} = 250\) cal kg⁻¹ K⁻¹ and \(S_\text{V} = 180\) cal kg⁻¹ K⁻¹. If the molar mass of the gas is 28 and \(J = 4.2\) J/cal, calculate the value of \(R\) in J kg⁻¹ K⁻¹.
The difference between the two molar specific heats of a gas is \(2.0 \times 10^{3}\) J kg⁻¹ K⁻¹ and their ratio is 1.4. Calculate the two molar specific heats.
A surface receives 800 J of radiant energy, of which 200 J is reflected and 150 J is transmitted. Find the coefficients of absorption, reflection and transmission for the surface.
For a certain surface, \(a = 0.65\) and \(t_\text{r} = 0.15\). Find its coefficient of reflection and its emissivity.
A body of surface area \(6.0 \times 10^{-2}\) m² radiates 1440 J of energy in 4 minutes. Calculate its emissive power.
The emissive power of a perfect blackbody at a certain temperature is \(1.8 \times 10^{3}\) W/m². A surface of emissivity 0.45 is maintained at the same temperature. Find its emissive power, and the energy radiated by an area of \(2.0 \times 10^{-2}\) m² of this surface in 3 minutes.
Calculate the energy radiated in 2 minutes by a blackbody of surface area 150 cm² maintained at 327 °C. (\(\sigma = 5.67 \times 10^{-8}\) W m⁻² K⁻⁴)
A blackbody of surface area \(2.0 \times 10^{-3}\) m² is maintained at 600 K in surroundings at 300 K. Calculate the net rate of loss of energy by the body. (\(\sigma = 5.67 \times 10^{-8}\) W m⁻² K⁻⁴)
Repeat the previous calculation for a body of the same size and temperatures whose emissivity is 0.6, and find by what factor the net rate of loss is reduced.
Compare the rates of loss of heat from a body at 527 °C and at 327 °C, when the surroundings are at 27 °C.
The absolute temperature of a blackbody is increased from 300 K to 900 K. By what factor does the power radiated per unit area increase?
A tungsten filament is at 2500 K. Calculate the wavelength at which it emits most strongly, and state the region of the electromagnetic spectrum in which this lies. (\(b = 2.897 \times 10^{-3}\) m K)
The spectrum of a certain star peaks at a wavelength of 350 nm. Estimate its surface temperature, and compare it with the surface temperature of the Sun taken as 5800 K. (\(b = 2.897 \times 10^{-3}\) m K)
Two blackbody spheres A and B have radii in the ratio 1 : 3 and surface temperatures in the ratio 2 : 1. Find the ratio of the total power radiated by A to that radiated by B.
Calculate the ratio of mean square speeds of molecules of a gas at 30 K and 120 K.
Two vessels A and B are filled with the same gas, where the volume, temperature and pressure in vessel A are twice the volume, temperature and pressure in vessel B. Calculate the ratio of the number of molecules of gas in vessel A to that in vessel B.
A gas in a cylinder is at pressure P. If the masses of all the molecules are made one third of their original value and their speeds are doubled, then find the resultant pressure.
Show that the rms velocity of an oxygen molecule is \(\sqrt{2}\) times that of a sulphur dioxide molecule at S.T.P.
At what temperature will oxygen molecules have the same rms speed as helium molecules at S.T.P.? (Molecular masses of oxygen and helium are 32 and 4 respectively.)
Compare the rms speed of hydrogen molecules at 127 °C with the rms speed of oxygen molecules at 27 °C, given that the molecular masses of hydrogen and oxygen are 2 and 32 respectively.
Find the kinetic energy of 5000 cc of a gas at S.T.P., given that standard pressure is \(1.013 \times 10^{5}\) N/m².
Calculate the average molecular kinetic energy (i) per kmol, (ii) per kg, (iii) per molecule of oxygen at 127 °C, given that the molecular weight of oxygen is 32, \(R\) is 8.31 J mol⁻¹ K⁻¹ and Avogadro’s number \(N_\text{A}\) is \(6.02 \times 10^{23}\) molecules mol⁻¹.
Calculate the energy radiated in one minute by a blackbody of surface area 100 cm² when it is maintained at 227 °C. (Take Stefan’s constant \(\sigma = 5.67 \times 10^{-8}\) J m⁻² s⁻¹ K⁻⁴.)
Energy is emitted from a hole in an electric furnace at the rate of 20 W, when the temperature of the furnace is 727 °C. What is the area of the hole? (Take Stefan’s constant \(\sigma\) to be \(5.7 \times 10^{-8}\) J s⁻¹ m⁻² K⁻⁴.)
The emissive power of a sphere of area 0.02 m² is 0.5 kcal s⁻¹ m⁻². What is the amount of heat radiated by the spherical surface in 20 second?
Compare the rates of emission of heat by a blackbody maintained at 727 °C and at 227 °C, if the blackbodies are surrounded by an enclosure (black) at 27 °C. What would be the ratio of their rates of loss of heat?
Earth’s mean temperature can be assumed to be 280 K. How will the curve of blackbody radiation look like for this temperature? Find out \(\lambda_\text{max}\). In which part of the electromagnetic spectrum does this value lie? (Take Wien’s constant \(b = 2.897 \times 10^{-3}\) m K.)
A small blackened solid copper sphere of radius 2.5 cm is placed in an evacuated chamber. The temperature of the chamber is maintained at 100 °C. At what rate must energy be supplied to the copper sphere to maintain its temperature at 110 °C? (Take Stefan’s constant \(\sigma\) to be \(5.670 \times 10^{-8}\) J s⁻¹ m⁻² K⁻⁴, \(\pi = 3.1416\), and treat the sphere as a blackbody.)
Find the temperature of a blackbody if its spectrum has a peak at (a) \(\lambda_\text{max} = 700\) nm (visible), (b) \(\lambda_\text{max} = 3\) cm (microwave region) and (c) \(\lambda_\text{max} = 3\) m (short radio waves). (Take Wien’s constant \(b = 2.897 \times 10^{-3}\) m K.)