Why does Heat of Vap go down? 2

[ChEMatt]

Never noticed this but it sparked some discussion today. Why does the heat of vaporization of water go down as the pressure goes up? Seems counter-intuitive.

Onwards,

Matt

 

[Latexman]

Not really. At the critical pressure (and temperature), [Δ]H_v = 0, right?

Good luck,
Latexman

 

[25362]

As Latexman says the heat of vaporization decreases monotonically from the triple point to the critical point.

The answer to ChEMatt's question is found in the Clausius-Clapeyron equation for the latent heat of evaporation of pure substances:

[Δ]H = T [Δ]V (dP ^sat/dT)​

where [Δ]V is the molar volume difference between the saturated vapor and the saturated liquid.

As T increases, (dP ^sat/dT) drops over the range of interest, so does [Δ]V, resulting in a narrower [Δ]H as T and P increase.

It should be noted that at low pressures the latent heat of vaporization is about constant since the slope (d ln P ^sat/d(1/T) in the Clausius-Clapeyron equation for the heat of vaporization:

[Δ]H = -R [d lnP ^sat/d (1/T)]​

is about constant.

 

[DickRussell]

Another way to look at it is that as the V/L system approaches the critical point the properties of the two phases must become the same (volumes, enthalpies, compressibilities, viscosity, etc). Since the V & L enthalpies become the same, the heat of vaporization must become zero. Also, the surface tension must go to zero.

 

[ChEMatt]

DickRussell, that's pretty much what I came up with after some more thought.

25362, thank you for your valuable post as well!

Onwards,

Matt