The normal boiling point of acetone, (CH3)2CO, is 56.5°C and the enthalpy of vaporization is 31.3kJ/mol. At what temperature (in °C) would acetone boil at a higher altitude where the atmospheric pressure is 226 mmHg?

95.41°C

Explanation:

T1 = 56.5°C = 329.65K

∇Hᵥₐ = 31.3kJ/mol

p2 = 226mmHg = 0.301bar

p1 = 1bar

R = ideal gas constant = 0.008314kJ/Mol.K

From definition, the normal boiling point is at 1 bar which relates to Clausius-Clapeyron equation relates to vapour pressure to temperature.

In(P₂/P₁) = -∇Hᵥₐ /R (1/T₂ - 1/T₁)

solving for T₂,

R / ∇Hᵥₐ . In(P₂/P₁) = 1/T₂ - 1/T₁

1/T₂ = 1/T₁ - R / ∇Hᵥₐ * In(P₂/P₁)

1/T₂ = (1 / 329.65) - (0.008314 / 31.3) In (0.30 / 1)

1/T₂ = (3.033*10⁻³) - (2.656*10⁻⁴ * -1.204)

1/T₂ = 3.033*10⁻³ - 3.197*10⁻⁴

1/T₂ = 2.713*10⁻³

T₂ = 368.56K

T2 = 95.41°C