Kvant Physics Problem 174

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Problem

It was mistakenly published in the journal under number 164.

Find the heat capacity of an ideal gas in a process in which the temperature of the gas

1. is proportional to the square of its volume;
2. is inversely proportional to its volume.

I. Sh. Slobodetsky

Setup and Assumptions

An ideal gas containing $n$ moles undergoes a quasistatic process in which pressure, volume, and temperature are always well defined and related by the equation of state $pV=nRT$. The task is to determine the heat capacity in the process, defined as

$$C=\frac{\delta Q}{dT},$$

for two different functional dependences between temperature $T$ and volume $V$. The molar heat capacities at constant volume $C_V$ and the gas constant $R$ are treated as constants. The process is assumed to be reversible so that thermodynamic differentials are applicable.

Physical Principles

The first law of thermodynamics for a quasistatic process is written as

$$\delta Q = dU + p,dV.$$

For an ideal gas, the internal energy depends only on temperature and satisfies

$$dU = n C_V dT.$$

The equation of state is

$$p = \frac{nRT}{V}.$$

The heat capacity in a process is defined by

$$C = \frac{\delta Q}{dT}.$$

Derivation

Combining the first law with the expression for internal energy gives

$$\delta Q = nC_V dT + p,dV.$$

Dividing by $dT$ yields

$$C = nC_V + p\frac{dV}{dT}.$$

Substituting the ideal gas pressure,

$$C = nC_V + \frac{nRT}{V}\frac{dV}{dT}.$$

Case 1: $T \propto V^2$

Let

$$T = kV^2,$$

where $k$ is a constant. Differentiation gives

$$\frac{dT}{dV} = 2kV.$$

Expressing $k$ through $T$, since $k = T/V^2$, yields

$$\frac{dT}{dV} = 2\frac{T}{V}.$$

Hence,

$$\frac{dV}{dT} = \frac{V}{2T}.$$

Substitution into the heat capacity expression gives

$$C = nC_V + \frac{nRT}{V}\cdot \frac{V}{2T}.$$

Cancellation of $V$ and $T$ leads to

$$C = nC_V + \frac{nR}{2}.$$

Case 2: $T \propto \frac{1}{V}$

Let

$$T = \frac{k}{V}.$$

Differentiation yields

$$\frac{dT}{dV} = -\frac{k}{V^2}.$$

Using $k = TV$, the derivative becomes

$$\frac{dT}{dV} = -\frac{T}{V}.$$

Hence,

$$\frac{dV}{dT} = -\frac{V}{T}.$$

Substitution into the general expression gives

$$C = nC_V + \frac{nRT}{V}\left(-\frac{V}{T}\right).$$

Cancellation of $V$ and $T$ yields

$$C = nC_V - nR.$$

Result

For the process $T \propto V^2$,

$$C_1 = nC_V + \frac{nR}{2} = n\left(C_V + \frac{R}{2}\right).$$

For the process $T \propto \frac{1}{V}$,

$$C_2 = nC_V - nR = n\left(C_V - R\right).$$

If expressed per mole,

$$\boxed{C_1 = C_V + \frac{R}{2}}, \qquad \boxed{C_2 = C_V - R}.$$

Sanity Checks

Dimensional consistency holds since $C_V$ and $R$ both have units of $\mathrm{J,mol^{-1},K^{-1}}$, making each term in $C$ homogeneous.

In the first case, the additional contribution $\frac{R}{2}$ is positive, corresponding to work done by expansion that increases the heat required per unit temperature change.

In the second case, the term $-R$ reduces the heat capacity, and the process can yield negative total heat capacity per mole if $C_V < R$, which is consistent with a process where compression accompanies heating so strongly that heat must be extracted to raise temperature.

The most sensitive step is the conversion of $dT/dV$ into $dV/dT$, where a sign error would invert the physical interpretation of expansion versus compression contributions in the second case.