 Solution

In the one-mole Van der Waals equation, $b>0$ represents excluded volume: the volume available to molecular motion is $V-b$, rather than $V$. Subtracting $b$ in the denominator therefore increases the repulsive contribution to pressure. The $a>0$ term represents intermolecular attraction, which reduces the outward momentum flux at a wall and hence appears as the negative correction $-a/V^2$. The two minus signs do not mean that both physical effects lower pressure.

For a reversible isothermal process along a homogeneous gas branch with $V_2>V_1>b$, the work done by the gas isTo find the internal energy change without guessing a caloric law, combine $dU=TdS-pdV$ with the Helmholtz free energy Maxwell relation $(\partial S/\partial V{)}_T=(\partial p/\partial T{)}_V$. This givesThereforeThe attractive contribution $-a/V$ becomes less negative as the gas expands: separating attracting particles requires energy. This differs from the ideal gas result $\Delta U=0$ at fixed temperature. The required heat is $Q=\Delta U+W_{\mathrm{by}}=RT\ln [(V_2-b)/(V_1-b)]$. Below the critical temperature, an equilibrium path crossing a liquid–gas coexistence region needs the equilibrium coexistence construction rather than the unstable homogeneous Van der Waals isotherm.