Question

The vapor pressure, p, of a compound varies with temperature as follows

T / oC 0 20 40 50 70 80 90 100

p/ kPa 1.92 6.38 17.7 27.7 62.3 89.3 124.9 170.9

What are the (a) normal boiling point and (b) the enthalpy of vaporization for this compound.

Answer 1

we know that

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

now

consider

T2 = 100 + 273 = 373 K

P2 = 170.9

T1 = 0 + 273 = 273 K

P1 = 1.92

so

using this data

ln ( 170.9 / 1.92) = (dHvap / 8.314) ( 1/273 - 1/373)

dHvap = 38000 J

so

dHvap = 38 kJ

so

the enthalpy of vaporization is 38 kJ

2)

now

at boiling point pressure = 101.325 kPa

so

P2 = 101.325

T2 = ?

P1 = 1.92

T1 = 273 K

dHvap = 38 x 1000

so

ln ( 101.325 / 1.92) = ( 38 x 1000 / 8.314) ( 1/273 - 1/T2)

solving we get

T2 = 357.74 K

T2 = 357.74 - 273 = 84.74

so

the boiling point is 84.74 C