Question

Consider a subject that boils at -34 degrees celcius, at 98 kpa. At that temperature and pressure 1kg of liquid occupies 0.0015 m^3 and 1kg of vapor occupies 1.16m^3. At 80kpa the fluid boils at -38 degrees celcius, estimate the enthalpy of vaporization of t his substance at 98 kpa and estimate the molar mass.

Answer 1

The assumption made is

we have clapeyron eqn

substituing in above equation from assumption

whence

on integration

to estimate molar mass

use

no. of moles (n) = mass/ molar mass =m/M

1.16-0.0015 = (1/M)*8.314*(-34+273)/98

1.1585 *M= 20.2759

M=17.50 kg/kmol=17.50g/mol

no. of moles (n) = mass/ molar mass =m/M