Toluene is an additive in many gasoline. The ΔH vap of toluene = 37 KJ/mol and its normal boiling point is 111 degrees C. What is the vapour pressure of toluene (in kPa) at 30 degrees Celsius?

Answer 1

The Clausius-Clapeyron equation can be used to estimate the vapor pressure as a function of temperature, so can be used in this case to calculate vapor pressure at 30ºC given the vapor pressure at 111ºC and the ∆H_vap.

There are several varieties of this equation. The one below is commonly used.

ln (P₂/P₁) = ∆H_vap/R (1/T₁ - 1/T₂)

Convert temperatures to K so

T₁ = 111 + 273 = 384K

T₂ = 30 + 273 = 303K

R = 8.3145 J/mol-K or 0.0083145 kJ/mol-K

∆H_vap = 37,000 J/mol or 37 kJ/mol (NOTE: be sure units of R and ∆H are consistent)

P₁ = 1 atm (since that is the pressure for the "normal" bp)

P₂ = ?

Plug these values into the equation and solve for P₂ which will be in atmospheres. Convert to kPa using conversion factor of 1 atm = 101.325 kPa

(I'll leave the math to you)