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Vapour Pressure vs Heat of Evaporation

Vapour Pressure vs Heat of Evaporation

Vapour Pressure vs Heat of Evaporation

(OP)
Hi there,

I have a general theory question.
Am I right to assume a liquid with a relatively high vapour pressure will have a relatively low heat of evaporation? Are they directly related?
Say some liquid has a (high) 40 kPa vapour pressure at 25degC it means the liquid has weak inter-molecular forces and many of the molecules can transition from the liquid to gas phase because they do not require much energy to do so. This is occuring at room temperature where the air temperature remains relatively constant. Therefore the amount of energy the liquid takes in from the air (heat of evap) must be quite low when in comparison to a liquid that has a low vapour pressure?

Thanks for any feedback.

RE: Vapour Pressure vs Heat of Evaporation

The Clasius Clayperon eqn sets out how the latent heat of vaporisation is related to vapor pressure:

dP/dT = ΔH/(T.ΔV), where dP/dT is the rate of change of saturation vapor pressure with temp, T is the mean phase transition temp and ΔH and ΔV are the molar latent heat of vaporisation and change in molar phase volume respectively. So, from this, we see it is the rate of change of sat vap pressure which is directly proportional to the phase change enthalpy.

RE: Vapour Pressure vs Heat of Evaporation

(OP)
Ok thanks for that,

It does make sense, you need more energy to get larger change in vapour pressures over a certain temperature range..

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