Question

Propane has a normal boiling point of −42.0∘C and a heat of vaporization (ΔHvap) of 19.04 kJ/mol. What is the vapor pressure of propane at 20.0 ∘C?

Express the pressure in torrs to three significant figures.

Answer 1

Ans :- P₂ = 6.19 x 10³ torr

Explanation :-

According to Clausius-Clayperon equation, we have

log (P₂/P₁) = ΔH_vap / 2.303 R [1/T₁ - 1/T₂] .........................(1)

Here,

T₁ = -42.0 ^oC =(-42.0 + 273)K = 231.0 K
T2 = 20.0 ^oC =(20.0 + 273)K = 293.0 K
P₁ = 760 torr (This is normal pressure at boiling takes place)
ΔH = 19.04 KJ/mol = 19040.0 J/mol and

P₂ = ?

Now, from equation (1), we have
log (P₂/P₁) = ΔH_vap / 2.303 R [1/T₁ - 1/T₂]
log (P₂/760) = (19040.0/ 2.303 x 8.314) [ 1/231.0 - 1/293.0]
log (P₂/760) = (994.4) [ 0.000916]
log (P₂/760)= 0.91087

P₂/760 = 10^0.91087

P₂/760 = 8.145

P₂ = 8.145 x 760

P₂ = 6190.2 torr

P₂ = 6.19 x 10³ torr