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# Enthalpy and Internal Energy for Isothermal Expansion

2018-03-17 10:25:43

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 $\Delta U = 0$?

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:

Liquid water. The ideal ga

• 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:

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

Water vapor. The ideal gas law does apply.

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

2018-03-17 11:34:29
• 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 vap

2018-03-17 11:53:54