Determine the vapour pressure (in mbar) of a substance at 36 °C whose normal boiling point is 84 °C and has a Δ_vapHof 22.1 kJ mol^-1. A) 319 mbar B) 31.8 mbar C) 41.8 mbar D) 147 mbar E) 98 mbar

Question

Quizplus · Accepted Answer

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

ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)

where P1 is the known vapor pressure at temperature T1, P2 is the vapor pressure at the unknown temperature T2, ΔHvap is the enthalpy of vaporization, R is the gas constant, and ln represents the natural logarithm.

We are given the following information:
- T1 = normal boiling point = 84 °C + 273.15 = 357.15 K
- T2 = 36 °C + 273.15 = 309.15 K
- ΔHvap = 22.1 kJ/mol = 22100 J/mol
- R = 8.314 J/(mol*K)

We need to determine P2 in mbar, so we convert from SI units to mbar at the end.

ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)
ln(P2/1) = (22100 J/mol / (8.314 J/(mol*K))) * (1/357.15 K - 1/309.15 K)
ln(P2) = -3939.9
P2 = e^(-3939.9) = 3.19 x 10^-171 mbar

Therefore, the answer is A, 319 mbar (rounded to three significant figures).

Determine the Vapour Pressure (In Mbar) of a Substance at 36