Example 2.80. Calculate the change in entropy $S$ of an ideal gas of constant mass $m$ when: a) $P=$ const and the volume changes from $V_{1}$ to $V_{2}$; b) $V=$ const and the pressure changes from $P_{1}$ to $P_{2}$; c) $T=$ const and the volume changes from $V_{1}$ to $V_{2}$.
Solution. a) In an isobaric process in an ideal gas, the elementary amount of heat $d Q$ is related to the mass of the gas $m$, the molar mass $\mu$, the specific heat at constant pressure $C_{p}$, and the temperature, thus $d Q=\frac{m}{\mu} C_{p} d T$. From formula (2.56), we obtain $d Q=\frac{\mu P d V}{m R}$, hence the change in entropy

$$
\begin{gathered}
S_{1}-S_{2}=\int_{V_{1}}^{V_{2}} \frac{m}{\mu} C_{p} \frac{\mu P d V}{m R} \frac{m}{\mu R} \frac{1}{P V}= \\
=\frac{m}{\mu} C_{p} \int_{V_{1}}^{V_{2}} \frac{d V}{V}=\frac{m}{\mu} C_{p} \ln \frac{V_{2}}{V_{1}}
\end{gathered}
$$

b) In an isochoric process, from the Mendeleev-Clapeyron equation (2.56), it follows that $\frac{P_{2}}{P_{1}}=\frac{T_{2}}{T_{1}}$, and $d Q=\frac{m}{\mu} C_{v} d T$, where $C_{v}$ is the specific heat at constant volume. According to formula (2.56), we get

$$
S_{1}-S_{2}=\int_{V_{1}}^{V_{2}} \frac{m}{\mu} C_{v} \frac{d T}{T}=\frac{m}{\mu} C_{v} \ln \frac{T_{2}}{T_{1}}=\frac{m}{\mu} C_{v} \ln \frac{P_{2}}{P_{1}}
$$

c) In an isothermal process, $d Q=P(V) d V$, substituting $P=\frac{m}{\mu} \frac{R T}{V}$ from equation (2.56), we obtain for the change in entropy

$$
S_{1}-S_{2}=\int_{V_{1}}^{V_{2}} \frac{m}{\mu} \frac{R T}{V} d V=\frac{m}{\mu} R \ln \frac{V_{2}}{V_{1}}
$$