Kvant Physics Problem 174

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Verdicts: UNKNOWN + UNKNOWN
Solve time: 11m23s
Source on kvant.digital

Problem

This problem was mistakenly published under number 164 in the journal.

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

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

I. Sh. Slobodetskii

Setup and Assumptions

An ideal gas containing $n$ moles undergoes a quasistatic thermodynamic process in which the temperature $T$ is prescribed as a function of the volume $V$. The heat capacity of the process is defined as

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

where $\delta Q$ is the infinitesimal heat supplied to the gas during an infinitesimal change of state along the given process.

The gas is assumed ideal, so its internal energy depends only on temperature and is given by

$$U = n C_V T,$$

where $C_V$ is the molar heat capacity at constant volume. The process is assumed quasistatic so that the equation of state

$$PV = nRT$$

holds at every intermediate state. No phase transitions or chemical reactions occur.

Two cases are considered. In the first case the temperature is proportional to the square of the volume,

$$T = a V^2,$$

where $a$ is a positive constant. In the second case the temperature is inversely proportional to the volume,

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

The unknown is the heat capacity $C$ of the process in each case.

Physical Principles

The first law of thermodynamics for a quasistatic process is

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

For an ideal gas the internal energy differential is

$$dU = n C_V dT.$$

The equation of state is

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

The heat capacity along a process is defined by

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

Derivation

Substituting the first law and the expression for $dU$ into the definition of heat capacity gives

$$C = \frac{dU + P,dV}{dT} = nC_V + P\frac{dV}{dT}.$$

Using the ideal gas law for pressure,

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

Case 1: $T = aV^2$

Differentiation yields

$$\frac{dT}{dV} = 2aV,$$

so

$$\frac{dV}{dT} = \frac{1}{2aV}.$$

Expressing $a$ through $T$ and $V$ from $T = aV^2$ gives $a = \frac{T}{V^2}$, hence

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

Substituting into the expression for heat capacity,

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

Cancellation of $T$ and $V$ leads to

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

Case 2: $T = \frac{a}{V}$

Differentiation yields

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

Using $T = \frac{a}{V}$ gives $a = TV$, so

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

and therefore

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

Substituting into the heat capacity expression,

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

Cancellation of $T$ and $V$ yields

$$C = nC_V - nR.$$

Result

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

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

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

$$C = n(C_V - R).$$

For one mole of gas ($n = 1$), the molar heat capacities are

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

For a monatomic ideal gas, where $C_V = \frac{3R}{2}$,

$$C_1 = 2R = 16.6\ \mathrm{J,mol^{-1}K^{-1}}, \qquad C_2 = \frac{R}{2} = 4.16\ \mathrm{J,mol^{-1}K^{-1}}.$$

Sanity Checks

The expression $C = nC_V + nR T/V \cdot dV/dT$ has units of $\mathrm{J,K^{-1}}$ because $C_V$ has units $\mathrm{J,mol^{-1}K^{-1}}$ multiplied by $n$, and the second term reduces to $nR$ times a dimensionless factor, ensuring consistency.

In both cases the correction to $nC_V$ is proportional to $nR$, the only available combination of gas parameters that carries heat capacity units independent of state variables, confirming dimensional closure.

For the first process the term $+nR/2$ increases the heat capacity relative to the constant-volume value because the temperature rise is accompanied by expansion that requires additional heat input against external pressure. For the second process the term $-nR$ reduces the heat capacity because the system contracts while being heated, so part of the supplied energy is recovered as work done on the surroundings, reducing the net heat required per unit temperature increase.