A mixture of two miscible liquids A and B is distilled under equilibrium conditions at 1 ~atm pressure. The mole fraction of A in solution and vapour phase are 0.30 and 0.60 respectively. Assuming ideal behaviour of the solution and the vapour, calculate the ratio of the vapour pressure of pure A to that of pure B

Step-by-step solution

In solution, x_A=0.30 ; x_B=0.70 In vapour phase, x_A^ =0.60 ; x_B^ =0.40 Using Dalton's law and Raoult's law aligned & x_A^ =0.60= p_A P = p_A p_A+p_B = 0.30 p_A^ 0.30 p_A^ +0.70 p_B^ & x_B^ =0.40= p_B P = p_B p_A+p_B = 0.70 p_B^ 0.30 p_A^ +0.70 p_B^ & x_A^ x_B^ = 0.60 0.40 = 0.30 p_A^ 0.70 p_B^ & p_A^ p_B^ = 0.60 0.70 0.40 0.30 = 7 2 =3.5 aligned