Question

The heat of vaporization of water at its normal boiling point is ΔHvap = 40.656 kJ/mol. Estimate the vapor pressure of water at 25°C. Answer in atm.

Answer 1

Recall that in equilibrium; especially in vapor-liquid equilibriums, we can use Clasius Clapyeron combination equation in order to relate two points in the same equilibrium line.

The equation is given as:

ln(P2/P1) = -dHvap/R*(1/T2-1/T1)

Where

P2,P1 = vapor pressure at point 1 and 2

dH = Enthalpy of vaporization, typically reported in kJ/mol, but we need to use J/mol

R = 8.314 J/mol K

T1,T2 = Saturation temperature at point 1 and 2

Therefore, we need at least 4 variables in order to solve this.

Substitute all known data:

ln(P2/P1) = -dHvap/R*(1/T2-1/T1)

Change negative signs

ln(P2/P1) = dHvap/R*(1/T1-1/T2)

P1 = 1 atm, T1 = 100°C = 373 K

P2 = x atm, T2 = 25°C = 298K

ln(P2/1) = 40656/8.314*(1/373 - 1/298)

P2 = exp(40656/8.314*(1/373 - 1/298)) = 0.03690

P2 = 0.03690 atm