A particular compound has an enthalpy of vaporization of 28300 J/mol.At 281 K it has a vapor pressure of 101 mmHg.What is its vapor pressure at 301 K? (R = 8.31 J/K·mol; 1 atm = 760 mmHg)
A) 98.8 mmHg
B) 123 mmHg
C) 45.2 mmHg
D) 226 mmHg
E) 103 mmHg

Question

Quizplus · Accepted Answer

First, we need to use the Clausius-Clapeyron equation to find the vapor pressure at 301 K:
ln(P2/P1) = -ΔHvap/R(1/T2 - 1/T1)
where P1 is the given vapor pressure at 281 K (101 mmHg), T1 is the corresponding temperature (281 K), ΔHvap is the enthalpy of vaporization (28300 J/mol), R is the gas constant (8.31 J/K·mol), P2 is the unknown vapor pressure at 301 K, and T2 is 301 K.
Plugging in the values, we get:
ln(P2/101) = -28300/8.31(1/301 - 1/281)
ln(P2/101) = -28.843
P2/101 = e^-28.843
P2 = 101e^-28.843
P2 = 225.8 mmHg
(rounded to the nearest tenth)
Therefore, the vapor pressure at 301 K is 225.8 mmHg, so the correct answer is D.

A Particular Compound Has an Enthalpy of Vaporization of 28300