Question

The molar enthalpy of vaporization of carbon disulfide is 26.74 kJ/mol, and its normal boiling point is 46°C. What is the vapor pressure of CS2 at 0°C?

a)447 torr
b)4160 torr
c)313 torr
d)139 torr
e)5.47 torr

Answer 1

Use the integrated from of Clausius-Clayperon relation.
Knowing the vapor pressure p0 at temperature T0 and the enthalpy of vaporization, you can relate vapor pressure to temperature by:
ln(p/p0) = (?h/R)·(1/T0 - 1/T)
or
p = p0·e^{ (?h/R)·(1/T0 - 1/T) }
(?h enthalpy of vaporization, R universal gas constant)

Here you've got the normal boiling point as reference point. That is the at which vapor pressure equals standard atmospheric pressure, .i.e.
p0 = 1 atm = 101325Pa
at
T0 = 46°C = 319.15K

So the vapor pressure at T=0°C = 273.15K is
p = 1atm · e^{ (26740J/mol/8.314472J/molK)·(1/319.15 - 1/273.15) }
= 0.183atm

Converting: 0.183 atm* 760= 139.08 torr

Answer D