Question

the enthalpy of vaporization of acetone is 32.0kj/mol. the normal boiling point of acetone is 56.5°c. what is the vapor pressure of acetone at 25.0°c?

Answer 1

Sol.

As the Clausius - Clapeyron equation is :

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

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

deltaHvap = Enthalpy of vapourization   

R = Gas constant  

Here for acetone , deltaHvap = 32 KJ / mol  

R = 0.008314 KJ / K mol

At normal boiling point , the vapour pressure is equal to 1 atm

So ,

P1 = 1 atm

T1 = 56.5°C = 56.5 + 273.15 K = 329.65 K

Also , T2 = 25°C = 25 + 273.15 K = 298.15 K   

Therefore ,

ln ( P2 / 1 atm ) = ( 32 KJ / mol / 0.008314 KJ / K mol )  

× ( 298.15 K - 329.65 K ) / ( 298.15 K × 329.65 K )

ln ( P2 / 1 atm ) = - 1.2335

P2 / 1 atm = e^-1.2335 = 0.2913  

P2 = 0.2913 × 1 atm = 0.2913 atm

Hence , Vapour pressure of acetone at 25°C is   0.2913 atm