Question

the vapor pressure of pure benzene at 20 ° C is equal to 74.9 torr and that of toluene pure is 21.6 torr. assuming ideality, find the partial vapor pressure of each component, the total vapor pressure and the molar fractions of the steam in equilibrium with its solution composed of 1,200 mol of benzene and 1.3 mol of toluene

Answer 1

solution composed of 1,200 mol of benzene and 1.3 mol of toluene :

P(total) = p_T + p_B

p_T (partial pressure of Toluene) = x_T p_T* (mole fraction * V. pressure of pure liquid)

p_B (partial pressure of Benzene) = x_B p_B* (mole fraction * V. pressure of pure liquid)

x_B = 1200 mol / 1201.3 mol = 0.9989

x_T = 1.3 mol / 1201.3 mol = 0.0011

p_B (partial pressure of Benzene) = 0.9989 _*74.9 torr = 74.82 torr

p_T (partial pressure of Toluene) = x_T p_T* = 0.0011*21.6 torr =0.0024 torr

total vapor pressure = p_T + p_B = 74.844 torr

Composition of steam : (Using dalton's law of partial pressure )

y_B (mole fraction of benzene in steam )= p_B /P(total) =0.9997

y_T (mole fraction of Toluene in steam )= p_T /P(total) = 1 -y_B = 0.0003.