thermodynamics - Enthalpy and Internal Energy for Isothermal Expansion

Considering a system with constant atmospheric pressure , i.e a massless piston sitting in a cylinder containing water. At constant temperature say $\rm 100\,^{\circ} C$, the water turns into vapour and pushes the piston up. Now shouldnt the $\Delta U=0$ as $Q = -p(V_2 - V_1)$, considering the Ideal Gas Behaviour And then as $\Delta H = \Delta U + \Delta n\cdot R\cdot T$, then shouldn't $\Delta H = \Delta n\cdot R\cdot T$?

Then according to the question in image below, why isn't $\Delta U = 0$?

Answer

You are right that an in in ideal gas, internal energy is a function of temperature only, and that in this problem, temperature is not changing. However, I think you are confused about how broadly the ideal gas law applies to this problem.

The question states that the ideal gas law applies to the water vapor. But the question is about a phase change of water. Let's break down some of the components of the problem. In the question we have:

1. 
   

   Liquid water. The ideal gas law does not apply to liquid water.

   
   


2. 
   

   Water vapor. The ideal gas law does apply.

   
   


3. 
   

   A phase change of liquid water to water vapor. $\ce{H2O(l) <=> H2O(g)}$ The ideal gas law does not apply to the process of the phase change, simply because processes are not gases and cannot be modeled by the ideal gas law.

   
   
   

Thus only one of three "components" of the problem is an ideal gas.

As a look at any reasonable steam table will tell you, the internal energy of water vapor is higher than the internal energy of liquid water. This difference is the heat of vaporization (at constant volume).