A gas follows the equation of state $P(V-b)=RT$. What is the change in internal energy for an isothermal expansion from $V_1$ to $V_2$?

Solution

1. The given equation of state is $P(V-b)=RT$.
2. For this specific gas, the internal energy $U$ is a function of temperature $T$ only, such that $(\frac{\partial U}{\partial V}{)}_T=0$.
3. Since the expansion is isothermal, the temperature remains constant ($dT=0$).
4. Therefore, the change in internal energy $$\begin{array}{rl}\Delta U& ={\int }_{V_1}^{V_2}(\frac{\partial U}{\partial V}{)}_{T}dV\\ & =0\end{array}$$.
5. Hence the answer is (B).