Ideal gas: entropy changes

Question 1. Choose a reversible path between the two states. The work received by the gas is $\delta W=-P\,\mathrm dV$. The first law gives

$$\mathrm dU=\delta Q_{\mathrm{rev}}-P\,\mathrm dV, \qquad \delta Q_{\mathrm{rev}}=\mathrm dU+P\,\mathrm dV.$$

For $n$ moles, $\mathrm dU=nc_V\,\mathrm dT$ and $P=nRT/V$. Substituting these expressions and dividing by $T$ gives

$$\mathrm dS=\frac{\delta Q_{\mathrm{rev}}}{T} =nc_V\frac{\mathrm dT}{T}+nR\frac{\mathrm dV}{V}.$$

First change the temperature at constant volume, then change the volume at constant temperature. Adding the two integrals gives

$$\begin{aligned} \Delta S&=nc_V\int_{T_I}^{T_F}\frac{\mathrm dT}{T} +nR\int_{V_I}^{V_F}\frac{\mathrm dV}{V}\\ &=nc_V[\ln T]_{T_I}^{T_F}+nR[\ln V]_{V_I}^{V_F}\\ &=nc_V\ln\frac{T_F}{T_I}+nR\ln\frac{V_F}{V_I}. \end{aligned}$$

Entropy is a state function: this result depends only on the initial and final states. It therefore also applies to an actual irreversible process between these states, even though $\mathrm dS=\delta Q/T$ cannot be used along that process.

Question 2. For a fixed amount of substance, $V=nRT/P$ gives

$$\frac{V_F}{V_I}=\frac{T_F}{T_I}\frac{P_I}{P_F},\qquad \ln\frac{V_F}{V_I}=\ln\frac{T_F}{T_I}-\ln\frac{P_F}{P_I}.$$

Substituting into the result of question 1 and collecting the temperature terms gives

$$\Delta S=n(c_V+R)\ln\frac{T_F}{T_I}-nR\ln\frac{P_F}{P_I} =nc_P\ln\frac{T_F}{T_I}-nR\ln\frac{P_F}{P_I}.$$

Here Mayer's relation was used in the form $c_V+R=c_P$. Similarly, $\mathrm dV/V=\mathrm dT/T-\mathrm dP/P$, hence

$$\mathrm dS=nc_V\frac{\mathrm dT}{T} +nR\left(\frac{\mathrm dT}{T}-\frac{\mathrm dP}{P}\right) =nc_P\frac{\mathrm dT}{T}-nR\frac{\mathrm dP}{P}.$$

Question 3. Setting $\Delta S=0$ and dividing by $nc_V$ gives

$$\ln\frac{T_F}{T_I}+\frac{R}{c_V}\ln\frac{V_F}{V_I}=0.$$

Using $\gamma=c_P/c_V$ and $R/c_V=\gamma-1$, the logarithm rules give

$$\ln\left[\frac{T_F}{T_I}\left(\frac{V_F}{V_I}\right)^{\gamma-1}\right]=0.$$

The argument of the logarithm is therefore 1: $T_FV_F^{\gamma-1}=T_IV_I^{\gamma-1}$, or $TV^{\gamma-1}=\mathrm{const}$. Replacing $T$ by $PV/(nR)$ also yields $PV^\gamma=\mathrm{const}$. Along a reversible adiabatic process, $\delta Q_{\mathrm{rev}}=0$, so $\mathrm dS=0$ at every step: the process is isentropic.

Question 4. For one mole, $nc_V=3R/2$ and $nR=R$. In (a), $T_F/T_I=1$ and $V_F/V_I=2$:

$$\Delta S_a=\frac32R\ln1+R\ln2=R\ln2\simeq5{,}76\ \mathrm{J\,K^{-1}}.$$

In (b), $T_F/T_I=600/300=2$ and $V_F/V_I=1$:

$$\Delta S_b=\frac32R\ln2+R\ln1\simeq8{,}64\ \mathrm{J\,K^{-1}}.$$

For the two successive processes, the changes add: $\Delta S=\Delta S_a+\Delta S_b=(5/2)R\ln2\simeq14{,}4\ \mathrm{J\,K^{-1}}$. The numerical values use $R=8{,}314\ \mathrm{J\,mol^{-1}\,K^{-1}}$.