The thermodynamics of heating water in a rigid container

When the pressure on the liquid surface is less than the vapor pressure of the liquid at a given temperature, the liquid will start to evaporate. This is common sense.

The problem is more difficult when the liquid and its vapor are heated inside a rigid container, with the specific volume of the mixture less than the critical specific volume:

According to my professor's notes, the level of the liquid in the container would rise. Why? What is the physics (or the thermodynamics) behind this? If a mixture is heated in a constant volume what happens to it? I know that it will not boil completely because it would attain equilibrium pressure with its vapor soon enough, but how do we fix the state when it would stop boiling? Why is it that if the specific volume of the mixture is less than the critical specific volume, the level of the liquid will rise when heated?

Edit:

Hi, sorry for bumping this old question! It's been six years and I have a master's degree now but still haven't been able to solve this question! However, I asked around a lot and I received a very crude answer to this, but it did make sense nonetheless.

Suppose we take the T-v diagram of water. For the first case, let us assume the specific volume of the mixture is above the critical specific volume at that temperature. We can mark an arbitrary point under the vapor dome and mark it as state 1. Suppose now we heat the mixture at constant volume, the temperature of the mixture should increase because of conservation of energy. We can mark state 2 under the dome at the same volume but at a higher temperature. If we keep heating the mixture, after some time state 2 will lie on the saturated vapor line. This essentially means that we started with a mixture of water and vapor, but ended with all vapor.

If we do the same exercise but this time the specific volume of the mixture being less than the critical specific volume, we will find that we end up with all liquid at one point.

While this does make sense intuitively, I still haven't found any luck in mathematically proving this. If anybody can work this out, I'd be very relieved!

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?

I wonder how everybody answered without correcting the question.if specific volume is less than critical specific volume then the liquid level will actually rise not fall!!!!! The increased pressure will compress the gaseous molecules and convert them to liquid.

If you add heat to the system, it will be absorbed by liquid molecules changing to the gas phase. That reduces the number of liquid molecules (which reduces the liquid volume) and increases the number of gas molecules (which increases the pressure). The higher pressure will compress the liquid (probably a small amount) to a higher density. Both effects serve to reduce the liquid volume.

• It is not obvious that increasing the temperature will force molecules from the liquid to become vapor: the temperature of the vapor goes up, and so does the pressure. What guarantees that the saturated vapor pressure increases faster than the pressure increase in the vapor due to the ideal gas law? That's the bit I am stuck on... and your answer doesn't address it. Intuitively I agree, but I don't think your argument is solid. – Floris Aug 24 '14 at 4:39
• The only ways you have to absorb heat are to increase the temperature of the system or to convert liquid to gas. Clearly increasing the temperature will increase the pressure of the gas. To increase the volume of liquid you have to have the thermal expansion of the liquid very high, the compressibility of the liquid low, and the heat of vaporization high. I admit I don't have a proof that there cannot be a liquid like this, but usually more will boil than the expansion. – Ross Millikan Aug 24 '14 at 4:59
• As I said - intuitively I agree with everything you say. But I think this question is asking for the mathematical proof. I admit I am struggling with it myself. – Floris Aug 24 '14 at 5:17

The liquid will not boil and when temperature attains critical temperature, two phases becomes indistinguishable.

Definition of vapor pressure: Evaporation and consequently the resulting vapor contributes to the total pressure of the container. The contribution from water vapor is called partial pressure of water vapor or vapor pressure. So the amount of water vapor is usually expressed as vapor pressure.

The maximum amount of vapor that can be present in the air, is dependent only on the temperature and very very little on the pressure(not on pressure). Don't mix up with the phase diagram of water. It is only for the case where the whole container has liquid water and no vacuum/air. Here is the question to ask yourself, if you blindly follow the phase diagram.

If water is to become vapor only at 100 Celcius and 1 bar why is there evaporation and vapor at room temperature?.

To understand evaporation of water in a container with vacuum/air, you need to look at clausius clapeyron curve, which says that maximum vapor that can evaporate is dependent on temperature only. So raising the temperature will make more liquid to evaporate and raise the pressure in the container. But this does not affect evaporation as it is independent of pressure. The increased evaporation with increasing temperature will bring down the volume of the liquid.

The concept that evaporation is dependent on temperature and not pressure is key to this understanding.