Question

Ethanol has a heat of vaporization of 38.56 kJ/mol and a normal boiling point of 78.4 degrees Celsius. What is the vapor pressure of ethanol at 15 degrees Celsius?

Answer 1

Sol.

As the Clausius - Clapeyron equation ,  

ln ( P2 / P1 ) = ( deltaHvap / R ) × ( T2 - T1 ) / ( T2 × T1 )

where P2 and P1 are vapour pressures at temperatures T2 and T1 respectively

deltaHvap is Enthalpy(Heat) of vapourization and R is Gas constant  

Now , at boiling point , vapour pressure is 1 atm

So , P1 = 1 atm , T1 = 78.4 °C = 78.4 + 273.15 K  

= 351.55 K  

T2 = 15 °C = 15 + 273.15 K = 288.15 K  

deltaHvap = 38.56 KJ / mol  

R = 0.008314 KJ / K mol  

So ,

ln ( P2 / 1 atm ) = ( 38.56 KJ / mol / 0.008314 KJ / K mol ) × ( 288.15 K - 351.55 K ) / ( 288.15 K × 351.55 K )

ln ( P2 / 1 atm ) = - 2.9027

P2 / 1 atm = e^-2.9027 = 0.0548  

P2 = 0.0548 × 1 atm = 0.0548 atm  

Therefore , Vapour Pressure of ethanol at 15°C is  

0.0548 atm