Relation between First Law of Thermodynamics and Ideal Gas Law

Question

Thermodynamics has always been a tough thing for me. There are lots of assumptions in this subject (those assumptions, I know, are necessary, I know the science of thermodynamics is a very practical science).
First Law of Thermodynamics states mathematically: $$\Delta U=Q+W$$ (with proper sign conventions must be used). This is just a law of conservation of energy and a very straightforward equation, but when we come to chemical thermodynamics this equation changes its form and becomes: $$\Delta U=Q+p\,\Delta V$$ My intuition says as soon as pressure and volume comes in any equation it becomes specifically for gases. So, my first question is:

Why thermodynamical equations are just for gases?

Let's imagine an isothermal expansion of a gas (that simple piston and gas experiment) under a constant pressure, now work $W$ is $$W=p\,\Delta V$$ but if use ideal gas Law equation i.e. $$pV=nRT$$ $$p\,\Delta V = \Delta nRT + nR\,\Delta T\tag1$$

since the expansion is isothermal therefore $\Delta T = 0$ and I can think that during expansion no atom or molecule has been annihilated therefore $\Delta n = 0$, so after all we get $$p\,\Delta V = 0$$ $$W=0$$

I want to know my mistakes in above consideration.

There is a question in my book:

A swimmer coming out of a pool is covered with a film of water weighing $18\ mathrm g$. How much heat must be supplied to evaporate this water at $298\ \mathrm K$? Calculate the internal energy change of vaporization at $100\ \mathrm{^\circ C}$. $\Delta_\mathrm{vap}H^\circ = 40.66\ \mathrm{kJ\ mol^{-1}}$ for water at $373\ \mathrm K$

My book give its solution like this $$\ce{H2O(l) -> H2O(g)}$$ Amount of substance of $18\ \mathrm g$ of $\ce{H2O(l)}$ is just $1\ \mathrm{mol}$. Since, $\Delta U=Q-p\,\Delta V$, therefore,
$$\Delta U=\Delta H-p\,\Delta V $$. $$\Delta U=\Delta H-\Delta nRT$$ $$\Delta U=40.66 \times 10^3\ \mathrm{J\ mol^{-1}}-1\ \mathrm{mol}\times8.314\ \mathrm{J\ K^{-1}mol^{-1}}\times373\ \mathrm K$$ $$ \Delta U=37.56\ \mathrm{kJ\ mol^{-1}}$$

I have a lot of problems with this solution which goes directly to the foundations of science of thermodynamics. (I must say it's because of these books that science becomes a rotten subject, these books destroy the real essence of science).
How is $\Delta n=1\ \mathrm{mol}$?
Why temperature is taken as $373\ \mathrm K$ and not $298\ \mathrm K$, since the process starts at $298\ \mathrm K$ we should use it?
At $373\ \mathrm K$ the process becomes an isothermal one (latent heat) so $\Delta U$ ought to be zero, if we think the process of vaporization starts from $373\ \mathrm K$.

Any help will be much appreciated. Thank you.

Answer 1

Why thermodynamical equations are just for gases?

They are not. The equation $\Delta U = q + P\Delta V$ applies to any phase (gas, liquid, solid...) when only pV work is done. In the particular form of the equation you present, the pressure is in addition constant during the work.

Gases are (1) an easy way to introduce thermodynamics concepts because they allow for a simplified analysis and may exhibit dramatic behavior, and (2) they provide a link to understanding the behavior of other phases. They also happen to be inherently important for practical and historical reasons during the development in the science.

I want to know my mistakes in above consideration.

There are no mistakes. Consider the ideal gas law $$V=\frac{nRT}{p}$$ If you assume that $p$, $n$ and $T$ are constant, then the dependent variable $V$ will also be constant.

What's probably causing the confusion is that the water is undergoing a phase change. We assume that the liquid does no work, that the change in volume is only due to the formation of vapor. In practice $\pu{\Delta n=+ 1 mol}$ for the gas, $\pu{\Delta n= - 1 mol}$ for the liquid, and $\pu{\pu{\Delta n= 0}}$ for all of the water. We ignore the work done when reducing the amount of liquid because it is small compared to that done when the gas is formed (the change in volume of gas is much greater).

How is Δn=1 ?

See the answer to the previous question. You are converting $\pu{18 g}$ of liquid water into vapor. Since the molecular weight of water is $\pu{18 g/mol}$, you are converting $\pu{1 mol}$ of water.

Why temperature is taken as 373 K and not 298 K, since the process starts at 298 K we should use it?

I agree, if the work is actually performed at a lower temperature than $\pu{373 K}$, so this is an estimate. It is assumed that water is "boiling off" the skin at $\pu{373 K}$ and $\pu{1 atm}$ vapor pressure (the boiling point of water at $\pu{1 atm}$ of pressure is $\pu{373 K}$). Maybe not an accurate portrayal of what is going on, but it gets you to practice the theory.

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I have lot of problems with this solution which goes directly to the foundations of science of thermodynamics. (I must say it's because of these books that science becomes a rotten subject, these books destroy the real essence of science).

The problems more generally are that (1) we have an intuition about the way the world works, based on everyday observations, and this intuition sometimes misinforms us; some of the effort of education is to develop a more accurate intuition; and (2) when teaching science, practice problems are sometimes abstract and don't reflect real life situations except approximately. For instance, when you dry yourself after swimming, you are far from a thermodynamic equilibrium, with air currents, sun heating your skin, and a low water vapor pressure all playing a potential role. Modeling all this is beyond the scope of an introductory course. Probably the most important approximations here (assuming a closed system free of mass flow) are that the enthalpy of vaporization is constant over a broad temperature range and/or (as you rightly pointed out) that the water evaporates at $373 K$.