150 moles of benzene are held in a closed container of volume V = 1 m^3 at a pre

Question

150 moles of benzene are held in a closed container of volume V = 1 m^3 at a pressure of 1 atm. Under these conditions, benzene separates into a liquid and a vapour phase in equilibrium with each other. The molar volume of liquid benzene is 9.6 times 10^-5 m^3/mol. (a) If the vapour pressure (P^sat) of benzene is given by the Antoine equation below, with pressure in bar and temperature in K, calculate the boiling temperature of benzene at 1 atm. in(P^sat) = 10.882 - 3823.79/T - 1.461 (b) Calculate the mole fraction of vapour in the system. Assume that the vapour phase behaves as an ideal gas.

Explanation / Answer

let x= moles of liquid benzene , 150-x= moles of benzene vapor

1 atm= 1.01325 bar

from Antoine equation , ln (1.01325)= 10.882- 3823.79/(T-1.461)

3823.79/(T-1.461 )= 10.87

T-1.461= 3823.79/10.87= 351.77

T= 350.31 K =350.31-273= 77.3 deg.c

since the tank contains only benzene, its mole fraction in the vapor phase= 1

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