A certain substance has a heat of vaporization of 55.94kJ/mol. At what Kelvin temperature will the vapor pressure be 8.00 times higher than it was at 319 K

Answer 1

You can think of P_VP as the equilibrium constant for the subs(l) --> subs(g) equation (i.e. a phase change).

P_VP = e^‾ΔG/RT with ΔG = ΔH - TΔS for vaporization. We make the assumption that the enthalpy change and the entropy change do not vary with temperature (not true, but reasonable over a small range of temperatures.

We want to go from T₁ to T₂ so that P_VP(T₂)/P_VP(T₁) = 8

We take the natural log of both sides of the ratio and obtain the Clausius-Clapeyron Equation

ln(P_VP(T₂)/P_VP(T₁)) = -ΔG(T₂)/RT₂ + ΔG(T₁)/RT₁ = -ΔH_Vap(1/T₂ - 1/T₁) or ΔH_vap(1/T₁ - 1/T₂)

because of the assumptions of constancy with T and that the entropy terms cancel.

Now, it's a matter of solving for T₂ given that the LHS = ln(8) and you know everything on the RHS except T₂

Some things to watch out for:

Heat of vaporization must be in Joules/mole

T's must be in Kelvin

R = 8.314 J/mole K

Good luck!