We can use the first law of thermodynamics to help us calculate how the entropy of an ideal gas changes. For infinitesimal changes, the first law is $d{E}_{\text{th}}=dW+dQ$. Using $dQ$ in $dS=dQ/T$ gives

Chapter 19 found explicit expressions for the thermal energy of and work done on an ideal gas. In differential form, the expressions are $d{E}_{\text{th}}=n{C}_{\text{V}}dT$ and $dW=-pdV=-(nRT/V)dV$. With these substitutions, Equation 20.55 becomes

We can integrate from entropy $S_{\text{i}}$ at initial temperature $T_{\text{i}}$ and volume $V_{\text{i}}$ to $S_{\text{f}}$ at $T_{\text{f}}$ and $V_{\text{f}}$. The integration is like the one we did above for the entropy associated with a temperature change, which leads to

You learned in Chapter 19 that heating a gas at constant pressure requires more heat, for the same temperature change, than heating the gas at constant volume. The second term in Equation 20.57 vanishes for a constant-volume process because $\text{ln}(1)=0$. A constant-pressure process has $V_{\text{f}}/V_{\text{i}}=T_{\text{f}}/T_{\text{i}}$, so the two logarithms are the same. Using this along with $C_{\text{P}}=C_{\text{V}}+R$, which we proved in Chapter 19, leads to