Question

Refrigerators make use of the heat absorption required to vaporize a volatile liquid. A fluorocarbon liquid being investigated to replace a chlorofluorocarbon has a molar heat of vaporization of 26.0 kJ mol⁻¹ at 250K and 750 torr. If 1.50 mol is vaporized at this temperature and pressure, what is

(a) q

(b) w

(c) ΔH

(d) ΔU

Answer 1

Molar heat of vaporization is the amount of heat needed to boil 1 mole of a substance at its boiling point.

here 1.50 moles used, therefore it will be 1.50x 26.0 = 39 kJmol^-1

a) q or the amount of heat absorbed = 39  kJmol^-1

b) workdone or PV = nRT

n=1.5 moles

R= 8.314 JK-1 mole

T=250 K

So nRT = 1.50 x 8.314 x 250 =3117.75 J

c) we know U= q - pV or (pV=q-nRT)

=39 kJ - 3.117 kJ = 35.89 kJ/mol

therefore H=U+PV = 35.89 kJ + 3.11kJ = 39 kJ/mol

conversion of joule into kJ, divide by 1000.