Question

The enthalpy of vaporization of Substance X is 8.00kJmol and its normal boiling point is −100.°C . Calculate the vapor pressure of X at −117.°C .

round to 2 sig figs

Answer 1

We shall employ the Clausius-Clapeyron equation to calculate the vapour pressure of substance X at 117°C.

The Clausius-Clapeyron equation is

ln(P₂/P₁) = ΔH/R(1/T₁ – 1/T₂)

where P₂ is the vapour pressure at absolute temperature T₂ and P₁ is the vapour pressure at absolute temperature T₁; ΔH is the enthalpy of vapourization and R is the gas constant.

Here, T₁ = {273 + (-100)} K = 173 K and T₂ = {273 + (-117)} K = 156 K; P₁ = 760 mmHg (since vapour pressure at normal boiling point is equal to atmospheric pressure = 760 mmHg).

We need to calculate P₂; therefore,

ln(P₂/760 mmHg) = ΔH/R(1/173 – 1/156) = (8.00 kJ/mol)/8.314 J/mol.K * (156 K – 173 K)/(173 K)(156 K)

or ln(P₂/760 mmHg) = (8.00*1000)K/8.314 * (-17/26988 K)

or, ln(P₂/760 mmHg) = -0.606

or, P₂/(760 mmHg) = e^-0.606 = 0.5455

or, P₂ = (760 mmHg)*0.5455 = 414.58

The vapour pressure of X at -117°C is 414.58 mmHg = (414.58 mmHg/760 mmHg)*(1 atm) = 0.5455 ≡ 0.55 atm (since 1 atm = 760 mmHg)

Ans: The vapour pressure of X at -117°C is 414.58 mmHg or 0.55 atm