Question

Explain why the vapor pressure of water in equilibrium with liquid water is temperature dependent?

Please show all work with the correct answer. Thank you!!

Answer 1

Note that

for 2 phases to be in equilibrium:

dG(liquid) = dG(vapor)

dG(liquid) = -Sliquid * dT + Vliquid * dP

dG(gas) = -Sgas * dT + Vgas * dP so

so; when we make the equality (in equilibrium, dG = 0, so dG1 = dG2)

-Sgas * dT + Vgas * dP = -Sliquid * dT + Vliquid * dP

when we rearrange;

(Sgas - Sliquid)*dT = (Vgas - Vliquid) * dP

get the derivatives:

dP/dT = (Sgas - Sliquid)/(Vgas - Vliquid)

we can see how T and P have a dependance

Also

Vgas >> Liquid so Vgas can be used alone

dP/dT = (Sgas - Sliquid)/(Vgas - 0)

dP/dT = (Sgas - Sliquid)/(Vgas)

Also,

Sgas - Sliquid = dSvaporization

and

S = -Q/T = -Hvaporization / T

dP/dT = (dSvaporization)/(Vgas)

dP/dT = (dHvaporization)/T * 1/(Vgas)

so

clearly

dP/dT = (dHvaporization)/T * 1/(Vgas)

T is present in the vapor pressure