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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 $\Delta U = 0$?

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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).

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  • $\begingroup$ Thanks a lot sir, i was infact wondering that phase change and higher energy of steam could be responsible for that internal energy. But does it mean that ΔU for water alone and steam alone is zero and the ΔU for the system is the difference between internal energies of water and steam? $\endgroup$ – Shahbaaz1104 Sep 22 '15 at 10:13
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    $\begingroup$ you cannot define "$\Delta U$ of water" or "$\Delta U$ of steam" here, your problem is that you don't understand exactly what the term $\Delta U$ applies to. You apply it to a process that a system, or part of a system, is undergoing. There is only one process here, and that is water turning into steam. You cannot break up the process and say the water on the LHS is one process and that the steam on the RHS is another process and that the overall $\Delta U$ is the sum of both processes. $\endgroup$ – orthocresol Sep 22 '15 at 10:19
  • $\begingroup$ Now that water turned to steam and that it is expanding, what is going on inside the system. How is internal energy changing, I find this topic really hard to understand. $\endgroup$ – Shahbaaz1104 Sep 23 '15 at 13:40
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The internal energy per unit mass of the liquid water and the internal energy per unit mass of the vapor are not changing. What is changing is the number of moles of liquid and the number of moles of vapor. The total internal energy of the system is $$U=n_Lu_L+n_Vu_V$$where the n's are the numbers of moles and the u's are the internal energies per mole (Note that there is a discontinuous change in the internal energy per mole in going from liquid to vapor). Similarly for the total volume of the system, we have: $$V=n_Lv_L+n_Vv_V$$ If $\Delta n$ moles of liquid evaporate, the change in total internal energy of the system and total volume of the system will be:$$\Delta U=(u_V-u_L)\Delta n$$ $$\Delta V=(v_V-v_L)\Delta n$$ So, from the first law,$$(u_V-u_L)\Delta n=Q-P(v_V-v_L)\Delta n$$ Rearranging this equation gives: $$Q=[(u_V+Pv_V)-(u_L+Pv_L)]\Delta n=(h_V-h_L)\Delta n$$ where the h's are the enthalpies per mole of saturated liquid and saturated vapor. We define the heat of vaporization as the enthalpy per mole of water vapor minus the enthalpy per mole of liquid water $(h_V-h_L)$. So the amount of heat added is equal to the heat of vaporization times the number of moles evaporated.

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