Question

The normal boiling point of liquid ethanol is 351 K. Assuming that its molar heat of vaporization is constant at 45.1 kJ/mol, the boiling point of C₂H₅OH when the external pressure is 0.602 atm is ? K.

Answer 1

We know that the clausius-clapeyron equation is given by :

log (P₂/P₁) =(ΔHvap / 2.303*R) ( 1/T₁ - 1/T₂)

Where : P₂ is pressure at temperature T₂ and P₁ is pressure at temperature T₁

ΔHvap is enthalpy of vapourisation and R is universal gas constant (8.314 J/mol-K )

Substituting all the values we get :

log (0.602 atm / 1 atm ) = ( 45100 J mol^-1 / 2.303 * 8.314 J mol^-1 K^-1 ) ( 1/351 K - 1/T₂)

( 1/351 K - 1/T₂) = -4.22 / 45100 K

On solving , we get T₂ = 339.839 K

Therefore the boiling point of liquid ethanol at 0.602 atm pressure is 339.839 K.