Question

A certain substance has a heat of vaporization of 70.78 kJ/mol. At what Kelvin temperature will the vapor pressure be 6.50 times higher than it was at 331 K?

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)

P2 = 6.5P1; T1 = 331K; T2 = ?

substitute

ln(6.5P1/P1) = -70780/8.314*(1/T2- 1/331)

ln(6.5) = -8513.350(1/T2- 1/331)

ln(6.5) /-8513.350 + 1/331 = 1/T2

0.00280 = 1/T2

T2 = 0.00280 ^-1

T2 = 357.142 K