Question

The normal boiling point of liquid propanol is 371 K. Assuming that its molar heat of vaporization is constant at 43.9 kJ/mol, the boiling point of C3H7OH when the external pressure is 1.33 atm is

Answer 1

Given :

Normal boiling point of propanol = 371 K

Molar heat of vaporization Delta Hv =43.9 kJ /mol

Boiling point of propanol = ? ( unknown)

External pressure = 1.33 atm

Here we use Classius clayperon equation ,

Ln ( P2/P1) = - Delta Hvap / R x ( 1 / T2 – 1 / T1)

Let , T1= 371 K = normal boiling point. We know, liquid boils when its vapor pressure = atmospheric pressure so P1 = atmospheric pressure = 1 atm

T2 = unknown , P2 = 1.33 atm (given)

Delta Hv is heat of vaporization in J / mol

R is gas constant = 8.314 J / ( K mol )

Value of delta H in J /mol = 43.9 E 3 J /mol

Lets plug these value in .

Ln ( 1.33 / 1.0 ) = (- 43.9 E 3 / 8.314 ) x ( 1/ T2 – 1 / 371)

T2 = 371.89

T2 = 372 K

So the boiling temperature or point of the propanol at 1.33 atm = 372 K