# Why the heat of vaporization decreases with increasing temperature ?

1. Jun 23, 2013

### Jano L.

Heat of vaporization of water slightly decreases with temperature.

https://en.wikipedia.org/wiki/Enthalpy_of_vaporization

Does anybody know of a simple explanation of this fact based on molecular theory? Or any explanation whatsoever. The Wikipedia article says

"
The heat of vaporization diminishes with increasing temperature and it vanishes completely at the critical temperature (Tr=1) because above the critical temperature the liquid and vapor phases no longer co-exist.
"

but this does not explain why the heat decreases.

2. Jun 23, 2013

### Andrew Mason

Vaporization requires breaking the intermolecular bonds. That is more easily done if you begin with higher translational molecular energy ie. higher temperature. So the heat needed to vaporize a unit mass of a substance (e.g. water) decreases as the starting temperature increases. When you get to the critical temperature the molecules are moving so rapidly that the intermolecular forces are never sufficient to make the molecules stick to one another.

AM

3. Jun 24, 2013

### Jano L.

That would make sense if part of the kinetic energy of the molecule changed into increment of potential energy $\epsilon$ needed to overcome the inter-molecular attraction. The higher the temperature, the higher the initial kinetic energy; consequently, less additional energy from heat would be needed to overcome the barrier.

But the evaporation process is different: the molecules have the same kinetic energy in liquid as in vapor phase, so their kinetic energy does not contribute to overcome the potential barrier due to inter-molecular bonds. If the potential difference $\epsilon$ is the same for both temperatures, the same amount of energy should be needed to overcome it.

Furthermore, the supplied heat for evaporation does not only increase the potential energy of the molecule, it also does work; approximately $k_B T$ per molecule if the vapor is close to ideal gas (steam does work on the piston).

So, it seems that the heat of vaporization should be

$$\epsilon + k_B T,$$

i.e. should increase with temperature, if $\epsilon$ does not decrease rapidly.

4. Jun 25, 2013

### Andrew Mason

The total energy of the molecules in the liquid state is negative (since there is a negative binding energy and the translational kinetic energy of the molecules is less than the magnitude of the binding energy). The total energy of molecules in the gaseous state is positive. They may have the same KE but the molecules in the gaseous state have more (less negative) potential energy.

When a molecule escapes, it decreases the energy of the molecules in the liquid that it leaves behind, making it harder for other molecules to escape. So in order to maintain temperature and keep the evaporation process going you have to keep supplying positive heat. The lower the temperature, the more energy the other molecules have to give to the escaping molecule, so the more heat you have to supply to keep the process going.

AM

5. Jun 25, 2013

### jartsa

When million molecules evaporate from a big glass of water at temperature 370 C, the heat content of the glass of water decreases quite a lot.

Now we add million molecules liquid water at temperature 370 c into the glass, which increases the heat content of the glass of water quite a lot.

If the heat content of the glass of water is now only slightly smaller than at the beginning, we say the heat of evaporation of water in this temperature is small.

6. Jun 28, 2013

### Jano L.

But my question is why is that so. The kinetic energy of the water does not change in the process, and if the temperature is lower, the work done per one evaporated molecule is lower. So unless the step in potential energy after evaporation gets substantively higher with lowering temperature, I do not see why more heat should be necessary at lower temperature.

7. Jun 28, 2013

### nst.john

Well if temperature is increasing and temperature is measuring the average kinetic energy of molecules than the reason for this is because the molecules have more initial kinetic energy at higher temperatures so less work needs to be done to give the molecules sufficient kinetic energy to break the intermolecular forces of attraction.

8. Jun 29, 2013

### jartsa

Typically gases have lower specific heats than liquids.

When hot liquid evaporates, lot of heat energy is relased when material with high specific heat turns into material with low specific heat.

9. Jun 29, 2013

### Jano L.

This is similar to Andrew's post in #4. Again, the evaporation does not work like penetration of a barrier. It is not sufficient to supply the missing energy -(U+T) to the molecule. The molecule will keep its kinetic energy, so it has to receive at least energy -U, plus the energy required to do work on the piston $k_B T$.

10. Jun 29, 2013

### snorkack

Look at the densities of liquid and vapour.

If you heat a liquid under its own vapour pressure then the following things happen:
1) The liquid undergoes thermal expansion. True, the vapour pressure increases and compresses the liquid, but liquids are not very compressible, so the thermal expansion of liquid dominates and the density of liquid decreases.
As the density of liquid decreases, the intermolecular attractive forces decrease with distance, and thus the heat of evaporation decreases.
2) The vapour is compressed under increasing vapour pressure. It also undergoes thermal expansion; but the gases are very compressible and the vapour pressure grows rapidly, so the density of vapour increases.
As the density of vapour increases, the intermolecular attractive forces in vapour increase, and start to pull molecules from liquid into vapour, causing further decrease of the heat of evaporation.

11. Jun 29, 2013

### Jano L.

Thanks snorkack, that makes sense! Water expands and saturated vapor gets denser with higher temperature, which can lower the required energy -U needed to get to the other phase.

Now I also realize that the isobaric work done by the vapor $k_B T$ (2400 J/mol) is much lower than -U (41000 J/mol), so $l(T)$ is dominated by the behaviour of -U(T).

Incidentally, under 4 °C water gets denser with increasing temperature. If the above explanation is right, then this should be visible on the plot of $l(T)$. Do you know of such plot for temperatures around 4 °C?

12. Jun 30, 2013

### snorkack

No, the above explanation was slightly simplified.

Water expands a lot on freezing, yet the heat of evaporation of ice is much larger than the heat of evaporation of water, because ice has a large heat of melting.

Water expands on freezing because the strongest bonds between molecules are directional bonds making up a network with many holes. The property of expanding on freezing is shared by a few other substances which share the cause - the strongest bonds are directional ones, which make up a network with holes.

Most other substances have nondirectional bonds, whether van der Waals, ionic or metallic, so for them the strongest bonds are in tight packing.

Generally, at low temperatures the molecules will find the positions where they form strongest bonds and therefore have highest enthalpy of vaporization (breaking all or almost all bonds at the same time). For most substances, this is the densest arrangement; for a few, it is an ordered network with holes but with bonds where they are strongest.

As the temperature is increased, the molecules are unable to stay at the bottoms of their potential wells, so they will undergo thermal movement into and through less well bound positions. Most substances thereby expand; the few that contract by allowing their molecules to fill the holes still break the strongest bonds and replace them with weaker ones.

So, evaporation breaks almost all bonds at once; whereas heating of a substance under vapour pressure gradually loosens the bonds. Naturally, the enthalpy of evaporation falls on heating, mainly because there are fewer and weaker bonds left to break.

13. Jul 2, 2013

### Jano L.

Thank you very much snorkack, I like your explanation.

14. Jun 8, 2016

Yes, but the equilibration of energy that 1/2 kbt per mode is derived from a conservation of momentum.
In fact, the entire Boltzmann's distribution is based on Maxwell's distribution which assumes that during
a collision, energy is conserved. At which time the momentum equilibrates between the modes.
The fact that energy is distributed among the modes is only an abstraction of Newton's Laws.
The collisions are elastic so the total momentum is conserved, and thus via Lagrange and
Hamiton's Equations, the total energy is also conserved. This does not apply to an arbitrary
potential energy at all..

Why would you assume that momentum is distributed through a potential well that is not
homogeneous?

Take for example a ball rolling down an inclined plane. If you have 0 velocity at the top,
you do not have 0 velocity at the bottom. Yet the two balls are in equilibrium..

15. Jun 9, 2016

### Andrew Mason

The equipartition theorem does not apply when quantum effects are significant (e.g. vibrational modes frozen out at lower temperatures). Yet conservation of momentum always applies.
I am not sure I understand your concern but it seems to go beyond the question raised by the OP. I would suggest that you start a new thread.

AM