# Change in internal not zero for an Isothermal process? [duplicate]

(All energy quantities are in $\pu {kJ/mol}$.)

One molar $H_2O$ is condensed isothermally and reversibly at $\pu{373K}$. Given $\Delta_{vap}H^\circ_{100^\circ}[H_2O] = \pu{40.656 kJ/mol}$. Find $q, w, \Delta_r U, \Delta_r H$.

Since the process is isothermal $\Delta U = 0$ and $\Delta H_p = q_p = -w_p = -40.656$.

I got $\Delta H_p = q_p$ part correct but

For other two author used $\Delta H - \Delta n(g)RT = \Delta U = 37.55$ and then first law to get $w = 3.101$.

Why $\Delta U$ is not zero ?