Question

The vapor pressure of an organic solvent is 50mmHg at 25 C and 200mmHg at 45 C. The solvent is
the only species in a closed flask at 35 C and is present in both liquid and vapor states. The volume
of gas above the liquid is 150 mL.

a) Estimate the amount of the solvent (mol) contained in the gas phase.

b) What assumptions did you make? How would you answer change if the species dimerized (one molecule results from two molecules of the species combining?

Answer 1

(a): Given at temperature, T₁ = 25 C = 25+273 = 298 K

Vapor pressure, P₁  = 50 mmHg

at temperature, T₂ = 45 C = 45+273 = 318 K

Vapor pressure, P₂  = 200 mmHg

Applying the relation:

ln (P₂ / P₁) = (Hvap / R) x (1/T₁ - 1/T₂)

=> ln (P₂ / P₁) = (Hvap / R) x (T₂ - T₁) / T₁xT₂  

=> Hvap =  ln (P₂ / P₁) x T₁xT₂ x R / (T₂ - T₁)

=> Hvap = ln(200 / 50) x 298K x 318K x 8.314 J.mol-1.K-1 / (318K - 298K)

=> Hvap = 54611 J = 54.61 kJ

Given temperature, T₃ = 35 C = 35+273 = 308 K

Now we need to calculate vapor pressure, P₃ at temperature T₃.

Again applying the relation:

ln (P₃ / P₁) = (Hvap / R) x (1/T₁ - 1/T₃)

=> ln (P₃ / 50 mmHg) = (54611 J / 8.314 J.mol-1.K-1) x (308K - 298K) / 308Kx298K

=> ln (P₃ / 50 mmHg) = 0.71565

=> P₃ / 50 mmHg = 2.0455

=> P₃ = 2.0455 x 50 mmHg = 102.3 mmHg

or P₃ = 0.1346 atm

Given the volume of the gas above the liquid, V = 150 mL = 0.150 L

Now we can calculate the number of moles in gas phase from ideal gas equation as

n = PV / RT

=> n = 0.1346 atm x 0.150 L / 0.0821 L.atm.mol-1.K-1 x 308 K = 7.98x10^-4 mol (answer)

(b): We made the assumption that there is neither any association or dissociation of the organic solvent.