Question

The normal boiling point of liquid octane is 399 K. Assuming that its molar heat of vaporization is constant at 36.3 kJ/mol, the boiling point of C8H18 when the external pressure is 1.30 atm is __________K.

Answer 1

Solution:-

Using the Clausius-Clapeyron equation we can clculate the boiling temperature at external pressure, when we know the normal boiling point of a liquid (Octane).
Equation,
         ln(P2/P1) = - (dHvap /R) * (1/T2- 1/T1)

ln(1.30atm/1.00atm) = -(36300J/mole / 8.314J/moleK) * (1/T2(K) - 1/399K)
There fore T2 = 409 K
A:- The boiling point of octane at expternal pressure, 1.30 atm is 409 K