Question

The vapor pressure of benzene is found to obey the empirical equation ln(p/torr) = 16.725 - (3229.86 K)/T - (118345 K^2)/T^2 from 298.2K to its normal boiling point 353.24 K. Given that the molar enthalpy of vaporization at 353.24K is 30.8 kJ/mol and that the molar volume of liquid benzene at 353.24K is 96.0 cm^3/mol, use the above equation to determine the molar volume of the vapor at the boiling temperature. HINT: Use the Clausius Clapeyron equation in some way when solving.

Answer 1

Given lnP= 16.725- 3229.85/T- 118345/T² ,

(dlnP/dT)= 3229.85/T²+2*118345/T³,

dP/dT= P*(3229.85/T²- 236690/T³)

at normal boiling point, lnP= 16.725-3229.85/353.24- 118345/(353.24)²

P= 760 Torr =1 atm

dP/dT=(3229.85/T²- 236690/T³)

at boiling temperature, dP/dT= 3229.85/(353.24²)- 236690/(353.243³)=0.020515 atm/K

but as per Classisus- clayperon equation

dP/dT= latent heat of vaporization/ T*(Vg-Vl)

where Vg= gas volume and Vl = volume of liquid

Latent heat of vaporization =30.68*1000 J/mol, 1 1joules= 101.3 L.atm

Latent heat of vaporization = 30.68*1000*101.3 L.atm

Vl = volume of liquid = 96 cm3/mol = 96*0.001 L/mole=0.096 L/mol

Hence 0.020515atm/K=    30.68*1000*101.3 L.atm/ *353.24mole/(Vg-0.096)

Vg= 428868 L/mole = 428688/0.001 cc/mole = 4.3*10⁸ cc/mole