# Does increasing pressure increase or decrease latent heat of vaporisation?

I found some say increasing pressure increases the latent heat of vaporisation (What is the effect of an increase in pressure on latent heat of vaporization?). This doesn't make sense to me because increasing pressure increases the boiling point of liquid. More work is required to convert water to vapour under increased pressure. That is why water boils at lower temperature at higher altitude. Can someone please confirm whether my understanding is correct?

• The evaporation is not limited to boiling. It can be water evaporating at 20 Deg C under 1 atm versus 10 atm. – Poutnik May 21 '19 at 5:52

Higher temperature is needed for boiling at higher pressure, as saturated vapour pressure, raining exponentially with temperature, must match external pressure.

Heat of evaporation decreases with temperature and is zero at the critical point, as properties of gaseous and liquid phase converge to the common point.

The rate of this decreasing is given by the difference of specific(or molar) heat of the vapour and the liquid.

E.g water evaporation heat at $$20 ^\circ \mathrm{C}$$ is equal to

• the evaporation heat at $$100 ^\circ \mathrm{C}$$

• plus heat to warm liquid water $$20->100 ^\circ \mathrm{C}$$

• minus released heat by cooling water vapour $$100->20 ^\circ \mathrm{C}$$

For evaporation at the same temperature but higher pressure, the evaporation heat does not increase, but is effectively integrated with higher work done on gas.

Last but not least, enthalpy ( of evaporation or any other) is state function at constant pressure.For standard enthalpy, standard pressure is implied and any volume work is implicitly included as difference $$\Delta H - \Delta U$$.

• The temperature dependence of evaporation enthalpy is due temperature dependent average energy difference between molecules in gaseous and liquid state. – Poutnik May 21 '19 at 6:55
• I found from Encyclopedia of World on latent heat. "Latent heat can be thought of as the energy required to break the bonds between the molecules at each change of phase.At warmer temperatures the water molecules contain more kinetic energy and thus less energy is required for evaporation to occur". Im now thinking if volume is constant, increasing pressure turns vapour into liquid and the energy released is equivalent to latent heat of vaporisation. So, latent heat of vaporisation remain the same at increasing pressure. However, the rate of evaporation decreases. Am I right? – Catalyst May 21 '19 at 6:58
• At higher temperature, molecules in gas contain more energy as well. But the mean energy difference between gaseous and liquid molecule decreases with temperature, as liquid bonding energy decreases with temperature. – Poutnik May 21 '19 at 7:04

The main point is to keep clear and separate the difference between boiling point (a temperature) and latent heat (an amount of energy).

As pressure goes up, the boiling point goes up. Think for example of the coexistence line on a phase diagram plotted in the $$p,T$$ plane. This line finishes at the critical point.

Latent heat goes down with temperature, falling to zero at the critical point. This means it goes down with pressure too. The critical point gives you the clue why this is: it's because as you increase $$p$$ and $$T$$ the vapour (at that $$p$$ and $$T$$) is becoming more and more like the liquid (at that $$p$$ and $$T$$). The liquid phase doesn't change much, but the vapour does, because it gets more dense as you go towards the critical point.

You are right that the higher pressure would mean more work to achieve a given amount of expansion, but the volume change (per unit mass, say) is smaller for evaporation at high $$p$$ and overall this is the bigger effect.