Question

# The normal boiling point of liquid propanol is 371 K. Assuming that its molar heat of...

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

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