If you only heat the cylinder, the ammonia boils, during which time
- The whole apparatus remains at saturation temperature,
- The vapor phase expands simply because more ammonia is being added to it, and
- Some of the heat-energy goes into pushing up the piston to accommodate more gas.
After all the liquid has boiled off, the vapor heats up and expands as suggested by the ideal gas law.
If you pull on the piston, the pressure drops and more liquid boils to restore the vapor pressure. But this takes energy out of the system and lowers its temperature, which makes a new equilibrium pressure and temperature lower than what you started with. If you pull on the piston after all the liquid has boiled, the vapor cools down further.
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?
Best Answer
Vapor pressure is something which applies at equilibrium for a liquid vaporizing into an established gas phase. If the gas phase is not established, the liquid must first cavitate. And if the gas phase is already established, the liquid can still exist in a metastable state for quite a while before reaching equilibrium.
So if you take an indestructible piston-cylinder with no nucleation sites and fill it with degassed water, the water will remain as pure liquid with no vapor phase. But this isn't really a challenge, since atmospheric pressure keeps the water pressurized at 1 bar. If you push the piston further, the water will mostly just push back. Sometime well beyond 10 000 bar, the water will solidify into a denser form of ice, maybe ice VII or ice X, and then the piston will give way a bit more. Conversely, you could also pull the piston all you want, and the water will just pull back. This way, you could even get negative pressures – one paper reports getting down to -277 bar this way, but theoretically you could pull at ten times that tension before cavitation lets the piston-cylinder expand.
If you start with a degassed saturated mixture of liquid water and water vapor, that's not a complication. Pushing the piston in that case could condense the vapor phase entirely. Conversely, pulling the piston could vaporize the liquid phase entirely.
You would see something similar if the water had air bubbles. Pushing would make water vapor in the bubbles condense into the liquid phase, while simultaneously dissolving more air into the liquid phase. Pulling would encourage water in the liquid phase to vaporize to maintain vapor pressure in the bubbles.
Some takeaways: