Q:

Does a phase change require more energy as pressure is increased? Specifically, the latent heat of vaporization for water.

Thanks

- STeve (age 55)

Thanks

- STeve (age 55)

A:

Steve- Interesting question. I don’t think there’s a general answer that applies to all phase changes, but I know the answer for water evaporation and I believe it would generalize to most evaporation phase changes. Increasing pressure p reduces the latent heat. How do I know?

First, I happen to know the p-T (T is temperature) phase diagram of water (search this site or others to find it) , and at high p there’s a ’critical point’ at which the liquid-vapor change ceases to be a phase transition. So at high p the latent heat goes to zero.

Also, it is well known that increasing p increases the boiling point T. That’s very general for evaporation processes, because the vapor occupies more volume than the condensed phase (solid or liquid). Now it happens that the heat capacity of liquid water is well known, and the heat capacity of the vapor is easy to calculate. The liquid has a bigger heat capacity. So as T is raised, the liquid enthalpy (a little more appropriate than energy) catches up with the vapor enthalpy, reducing the difference. At the critical point, the difference falls to zero and the phases merge.

The reason I think the answer probably generalizes to most other liquid-vapor transitions is that they typically show critical points. The vapor is more compressible so as the pressure is raised it becomes more and more like the liquid, and the latent heat is reduced. Another way to think of it is that the liquid typically has more heat capacity, since the potential energies of interactions between the molecules contribute to its heat capacity, so our second argument would apply typically.

Someone with more sophistication in thermodynamics could probably have given you a sharper answer, but I hope this one will do.

Mike W.

First, I happen to know the p-T (T is temperature) phase diagram of water (search this site or others to find it) , and at high p there’s a ’critical point’ at which the liquid-vapor change ceases to be a phase transition. So at high p the latent heat goes to zero.

Also, it is well known that increasing p increases the boiling point T. That’s very general for evaporation processes, because the vapor occupies more volume than the condensed phase (solid or liquid). Now it happens that the heat capacity of liquid water is well known, and the heat capacity of the vapor is easy to calculate. The liquid has a bigger heat capacity. So as T is raised, the liquid enthalpy (a little more appropriate than energy) catches up with the vapor enthalpy, reducing the difference. At the critical point, the difference falls to zero and the phases merge.

The reason I think the answer probably generalizes to most other liquid-vapor transitions is that they typically show critical points. The vapor is more compressible so as the pressure is raised it becomes more and more like the liquid, and the latent heat is reduced. Another way to think of it is that the liquid typically has more heat capacity, since the potential energies of interactions between the molecules contribute to its heat capacity, so our second argument would apply typically.

Someone with more sophistication in thermodynamics could probably have given you a sharper answer, but I hope this one will do.

Mike W.

*(published on 10/22/2007)*

Q:

"the heat capacity of the vapor is easy to calculate"

Please show me.

Thanks

- Tom Gearing

Denver, CO

Please show me.

Thanks

- Tom Gearing

Denver, CO

A:

Sure. In the vapor, the molecules don't interact significantly. So the heat capacity is the sum of the contributions of the separate molecules. It's a fairly good approximation to treat small molecules at room temperature as rigid rotors. Then for monatomic gases, there are 3 degrees of freedom (3 directions of motion, no rotations), so the equipartition theorem gives a heat capacity of (3/2)k_{B} per atom, where k_{B}is Boltzmann's constant, about 1.4 10^{-23}J/K. For diatomic molecules, there are two rotational modes, giving (5/2)k_{B} net per molecule. For water, there are 3 rotational modes, giving 3k_{B} net per molecule. These are all constant volume heat capacities. For onstant pressure heat capacities one must add another k_{B} per molecule.

You can calculate the number of molecules per unit volume, N/V, in the vapor using the ideal gas law:

p=(N/V)k_{B}T, where p is pressure and T is absolute temperature.

You end up with C_{V}=pV*alpha/T, where alpha is 3/2 for monatomic gases, 5/2 for diatomic gases, and 3 for non-linear but rigid gas molecules. For any ideal gas C_{p}=C_{V}+pV/T.

Of course you have to be careful to use consistent units in any calculation like this.

Mike W.

Lee H

You can calculate the number of molecules per unit volume, N/V, in the vapor using the ideal gas law:

p=(N/V)k

You end up with C

Of course you have to be careful to use consistent units in any calculation like this.

Mike W.

Lee H

*(published on 10/22/2007)*

Q:

Hi,
I came across your site on doing a search on vapor pressures and latent heat of vaporization. It seems that vapor pressure is solely temperature dependent. The Clausius equation confirms this by not having pressure and only has heat of vaporization and temperature.
My question is: why isn't vapor pressure external pressure dependent? When the external pressure rises, gases become more soluble in liquids and thus should also be more likely to return to their liquid state. Thus, their vapor pressure should fall.
Secondly the clausius equation has latent heat. But the answer on the site (to which I am responding) itself says that latent heat is pressure dependent. Thus, for this reason too, the vapor pressure of a liquid should be external pressure dependent.
Could you please point me to references and resources that talk about vapor pressures and these ideas in both conceptual and in equation forms?
Thanks!

- Al (age 27)

- Al (age 27)

A:

Let's distinguish two situations.

1. If there's only one substance around, the pressure (p) in the gas is the same (in equilibrium) as the vapor pressure of the liquid, by definition. So there aren't two different pressures to adjust. There's just a vapor pressure which is a function of temperature (T), following the Clausius-Clapeyron equation () giving dp/dT. It's true that the latent heat (L) which appears in that equation is somewhat dependent on p and T (here we don't separate those dependences, since for any T there's a unique p), and so to be very accurate one should use that T-dependent L. Including that T-dependence slightly changes the approximate version of the equation which assumes fixed L and an ideal gas phase.

2. Sometimes there are other gases present. Now the total pressure depends on how much of those other gases are around, as well as on T. Usually, the gas phase is nearly ideal, so at least you can treat the pressure as the sum of the vapor pressure plus the pressure of the other gases. Here the solubility of these other gases affects the equilibrium pressure, as you surmised. This new p(T) results in a slightly different L(T) than in the first case.

Mike W.

1. If there's only one substance around, the pressure (p) in the gas is the same (in equilibrium) as the vapor pressure of the liquid, by definition. So there aren't two different pressures to adjust. There's just a vapor pressure which is a function of temperature (T), following the Clausius-Clapeyron equation () giving dp/dT. It's true that the latent heat (L) which appears in that equation is somewhat dependent on p and T (here we don't separate those dependences, since for any T there's a unique p), and so to be very accurate one should use that T-dependent L. Including that T-dependence slightly changes the approximate version of the equation which assumes fixed L and an ideal gas phase.

2. Sometimes there are other gases present. Now the total pressure depends on how much of those other gases are around, as well as on T. Usually, the gas phase is nearly ideal, so at least you can treat the pressure as the sum of the vapor pressure plus the pressure of the other gases. Here the solubility of these other gases affects the equilibrium pressure, as you surmised. This new p(T) results in a slightly different L(T) than in the first case.

Mike W.

*(published on 05/15/2009)*