# What is the effect of an increase in pressure on latent heat of vaporization?

What is latent heat of vaporization ($L_v$) in the first place? Wikipedia seems to indicate that it is the energy used in overcoming intermolecular interactions, without taking into account at all any work done to push back the atmosphere to allow for an increase in volume when a liquid boils.

If that is so, then would it be correct that $L_v$ decreases as boiling point rises, because at the higher boiling point, less energy is required to overcome the weaker intermolecular interactions?

Otherwise, would it then be correct to say that $L_v$ increases as boiling point rises, assuming constant-volume?

Let's say there is a beaker containing 1 kg of liquid water at the boiling point, and an identical beaker, containing an identical amount of water at the same temperature. However, this second beaker is perfectly sealed such that the volume of its contents, both liquid and gaseous, will not change.

Since both liquids are at the boiling point, applying heat should cause boiling to occur. Compared to the first beaker, then, would the second beaker (in theory) require more or less heat for its 1 kg of liquid water to completely boil?

Thanks!

• second beaker will consume more energy due to pressure built up
– user41400
Feb 26, 2014 at 16:53

The name of the property is itself a clue here : enthalpy of vaporization. By nature, enthalpy does take into account the work required to push against the atmosphere.

You can see the impact of increasing the pressure on the enthalpy of vaporization on a Mollier diagram. Increasing the pressure has the overall effect of reducing the enthalpy of vaporization, until it becomes zero at the critical point. At this stage, there is no longer a phase change associated with vaporization.

• Why does increasing pressure decrease the enthalpy, however? Is it because, with more pressure, boiling point increases (hence, the temperature of the liquid can increase and further work against the intermolecular attractions holding molecules together)?
– Muno
Dec 15, 2015 at 2:14

Let be simple.

Latent heat refers to the heat required to overcome molecular bonds.

Latent heat of vapourisation of water at 1 bar, $100^\circ C$ is $2257 \frac{kJ}{kg}$. Which means, that much heat is required to break inter-molecular forces and turn into gasoeus phase.

As pressure on molecules increases they require more heat to overcome the pressure force acting or to escape and thus latent heat required is more.

While for latent heat of condensation, $335\frac{kJ}{kg}$ is to be removed, so that liquid water turns into ice at 1 bar 0 degree c.

As pressure increases, pressure acting helps in binding the molecules thus even removal of lesser amount heat would also do.

Thus as presure increases at 100 degree latent heat of vapourisation also increases while as pressure increases latent heat of condensation decreases.

• This could benefit from some improved formatting (e.g. capitalization). Jun 24, 2015 at 19:01

When temperature of the water in the beaker increases, the bond b/w molecules releases and water starts vaporizing. when temperature reaches 100 degree, then water starts boiling but temperature in the beaker will never increases still all the water converted to steam. That is called latent heat of vaporization. But in the case of second beaker its top is fully closed , so steam can not escape from the beaker, so pressure in the beaker will start increase. Due to the pressure exerted by the steam over the water surface, the remaining water molecule cannot escape or cannot converted in to steam ( Reason-: vapor pressure on the water surface increased). So, the heat given to the beaker rises the temperature of the water and then the water molecules gets more energy and starts evaporate. Again the pressure in the beaker will increase( Because it is in closed condition , steam cannot escape to the atmosphere) and that high pressure resist the remaining water molecule to evaporate. So, if we heat the beaker again then that heat is absorbed by the remaining water, again the temperature of water will increase and the process will go on...This is actually happens in our pressure cooker ...

You will need more energy to convert water to steam in sealed one. The open one will be free to expand, while the sealed one have limited volume. When the sealed one evaporates, the molecules of steam pushes molecules of air in the container increasing the pressure. As the boiling point increases so as the enthalpy too, thus the difference in final and initial enthalpy is greater.

Sure, as pressure increases so does boiling point and the amount of work needed to be done against the atmosphere you would assume. But this is not the case. Work done by gas = change in volume x atmospheric pressure.

If the atmospheric pressure is indeed higher the amount of expansion of the vaporized gas required in order to reach pressure equilibrium would be less.

It is easier to imagine the scenario as a fixed volume of substance in a container that can freely expand with no resistance.

In conclusion if pressure increases, volume change decreases. So therefore Latent heat of vaporization remains constant.

I think pressure does make it more difficult to vaporize water. However, heat reduces the strength of hydrogen bonds. What Whelp's answer didn't consider is why is the latent heat of vaporization at the critical point Zero?

If you add all the heat required to raise the temperature of 1 liter of water from 100 C to the critical point, you'll find that it is about half the amount of heat required to vaporize water. That's calculated at 4200 J / degree at atmospheric pressure. However, if you add the heat required at different pressures and temperatures using an isobaric specific heat chart, you'll notice that the heat required does in fact increase sharply as the pressure increases, but the total amount of heat is still below what is required. High temperature weakens hydrogen bonds more than pressure increases them.