With a constant volume for the container, the sum of the volume of the liquid and vapor is constant.

Let's start in an equilibrium position, temperature $T_1$. Volume of liquid is $V_l$ and vapor is $V_v$. Vapor pressure = $P_v$. Now we increase the temperature to $T_2$.

The first thing that happens is that the pressure of vapor increases to $P_v\frac{T_2}{T_1}$ before any further evaporation takes place (ideal gas law) and the temperature of the liquid also increases. The question is - do we expect more liquid to evaporate? In other words - **is the increase in saturated vapor pressure faster than the increase in pressure with temperature due to the ideal gas law?**

Now the August equation is a simple representation of the relationship between pressure and temperature, and takes the form

$$log_{10}P = - \frac{B}{T}$$
(this is really a different formulation of the Clausius-Clapeyron equation).

We can rewrite this as
$$P = a e^{-b/T}$$
From this it follows that
$$\frac{dP}{dT}=\frac{abe^{-b/T}}{T^2}=\frac{bP}{T^2}$$
Compare this to the ideal gas law
$$PV = nRT\\
\frac{dP}{dT} = \frac{nR}{V} = \frac{P}{T}$$

The increase in vapor pressure will be greater than the increase in pressure due to the ideal gas law if

$$\frac{bP}{T^2} > \frac{P}{T}\\
b > T$$

So that leaves us the question - what is this factor $b$, and how does it relate to the stated fact that the specific volume is less than the critical specific volume?

Here I have to say - I am not sure. I *do* know that $b = \frac{\Delta H}{R}$, but in principle for a liquid with a very low enthalpy of evaporation, $b$ could be very small. It's been a long time since I did thermodynamics, and I am stuck on this very last part. What does this have to do with the critical specific volume?

I'm nonetheless posting this as an "answer", hoping that it will give somebody else the nudge needed to create a complete answer (or give me a hint so I can finish this myself...). It's something simple - but it's late here.

Anybody want to take a shot at finishing this?

This is what it says on the linked page:

...thermal reversibility requires that all heat transfer is isothermal.

This statement is only true in the context of that web-page, where it is being assumed that the system is in contact with a *thermal reservoir* (i.e. an object whose thermal mass is so large that it's temperature doesn't change while exchanging energy via heat with the system).

We know that a process in which there is energy transfer via heat between two objects at *different* temperatures is irreversible, and so in order for a process to be reversible in this context, the system and the environment must be at the same temperature at all times during the heat transfer. Since the environment's temperature doesn't change, neither can the system's.

Thus:

**A reversible process in which energy is exchanged via heat between a finite system and an infinite reservoir must be isothermal.**

Now, it is possible to design a reversible process in which the system's temperature changes *and* the system exchanges energy via heat with its environment. If you bring a system into contact with a sequence of thermal reservoirs, each at an infinitesimally larger temperature than the last, and you allow the system and reservoir to equilibrate at each step, then the system gains energy via heat, *and* its temperature goes up. Crucially, this heat exchange occurs when the system is in contact with a reservoir *at the same temperature*, so the entire process is reversible.

## Best Answer

For saturated water, you must have vapor and liquid water together, presumably at thermal equilibrium. In addition, the pressure of the saturated water must be the vapor pressure at the specified temperature. The Antoine equation specifies vapor pressure as a function of temperature, and it also says that a higher vapor pressure corresponds to a higher temperature. At higher temperatures (hence, at higher pressures), the density of liquid water is lower, so the specific volume is higher.

For further info, see http://antoine.frostburg.edu/chem/senese/101/liquids/faq/antoine-vapor-pressure.shtml