In this example, $2.50 \mathrm{~mol}$ of an ideal gas with $C_{V, m}=12.47 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ is expanded adiabatically against a constant external pressure of 1.00 bar. The initial temperature and pressure of the gas are $325 \mathrm{~K}$ and $2.50 \mathrm{bar}$, respectively. The final pressure is 1.25 bar. Calculate the final temperature, $q, w, \Delta U$.
Because the process is adiabatic, $q=0$, and $\Delta U=w$. Therefore,
$$
\Delta U=n C_{\mathrm{v}, m}\left(T_f-T_i\right)=-P_{e x t e r n a l}\left(V_f-V_i\right)
$$
Using the ideal gas law,
$$
\begin{aligned}
& n C_{\mathrm{v}, m}\left(T_f-T_i\right)=-n R P_{\text {external }}\left(\frac{T_f}{P_f}-\frac{T_i}{P_i}\right) \\
& T_f\left(n C_{\mathrm{v}, m}+\frac{n R P_{\text {external }}}{P_f}\right)=T_i\left(n C_{\mathrm{v}, m}+\frac{n R P_{\text {external }}}{P_i}\right) \\
& T_f=T_i\left(\frac{C_{\mathrm{v}, m}+\frac{R P_{\text {external }}}{P_i}}{C_{\mathrm{v}, m}+\frac{R P_{\text {external }}}{P_f}}\right) \\
& =325 \mathrm{~K} \times\left(\frac{12.47 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}+\frac{8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times 1.00 \mathrm{bar}}{2.50 \mathrm{bar}}}{12.47 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}+\frac{8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times 1.00 \mathrm{bar}}{1.25 \mathrm{bar}}}\right)=268 \mathrm{~K} \\
&
\end{aligned}
$$
We calculate $\Delta U=w$ from
$$
\begin{aligned}
\Delta U & =n C_{V, m}\left(T_f-T_i\right)=2.5 \mathrm{~mol} \times 12.47 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times(268 \mathrm{~K}-325 \mathrm{~K}) \\
& =-1.78 \mathrm{~kJ}
\end{aligned}
$$