(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 ?


marked as duplicate by Community Mar 31 '17 at 4:23

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ ΔU is only required to be zero for an isothermal process undergone by an ideal gas. In this case, we are dealing with liquid water, so the ideal gas law will not apply. For a related question on this site:chemistry.stackexchange.com/questions/37720/… $\endgroup$ – Tyberius Mar 31 '17 at 4:13
  • $\begingroup$ @Tyberius Second answer solved my problem. Thank you. $\endgroup$ – A---B Mar 31 '17 at 4:24