Carbon tetrachloride,an organic solvent,has a molar heat of vaporization of 29.82 kJ/mol.If CCl₄ has a normal boiling point of 3.50 × 10² K,what is its vapor pressure at 273 K? (R = 8.31 J/mol⋅K)

Question

Quizplus · Accepted Answer

We can use the Clausius-Clapeyron equation to solve this problem:

ln(P2/P1) = -(ΔHvap/R) * (1/T2 - 1/T1)

where P1 and T1 are the known vapor pressure and temperature (3.50 × 10^2 K), respectively, P2 is the vapor pressure at the unknown temperature, T2 (273 K), ΔHvap is the molar heat of vaporization (29.82 kJ/mol), and R is the gas constant (8.31 J/mol*K).

Substituting the values and solving for P2, we get:

ln(P2/760 Torr) = -(29.82 kJ/mol / 8.31 J/mol*K) * (1/273 K - 1/350.0 K)

ln(P2/760 Torr) = -15.85

P2/760 Torr = e^(-15.85)

P2 = 42.2 Torr or mm Hg

Therefore, the vapor pressure of CCl4 at 273 K is 42.2 mm Hg, which corresponds to choice B.

Carbon Tetrachloride,an Organic Solvent,has a Molar Heat of Vaporization of 29.82