Question

The vapour pressures of pure liquids a and b are 400 and 600 mmhg, respectively at 298 k. On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquid b is 0.5 in the mixture. The vapour pressure of the final solution, the mole fractions of components a and b in vapour phase, respectively are -

Answer 1

The vapour pressure of the final solution is 500 mmHg

The mole fractions of components a in vapour phase is 0.4

The mole fractions of components b in vapour phase is 0.6

Given:

Vapour pressure of a = 400 mmHg $\bold{=P^o_A}$

Vapour pressure of b = 600 mmHg $\bold{=P^o_B}$

Mole fraction of b = 0.5 $\bold{=x_B}$

To find:

The vapour pressure of the final solution = ?

The mole fractions of components a in vapour phase = ?

The mole fractions of components b in vapour phase = ?

Solution:

The vapour pressure of the final solution is given by the formula:

$\bold{P_s=P^o_Ax_A+P^o_Bx_B\longrightarrow(1)}$

The mole fraction of A is given by the formula:

$\bold{x_A+x_B=1}$

$\bold{x_A=1-x_B=1-0.5}$

$\bold{\therefore x_A=0.5}$

Now, on substituting the known values in equation (1), we get,

$\bold{P_s=400(0.5)+600(0.5)}$

$\bold{P_s=200+300}$

$\bold{\therefore P_s=500 \ mmHg}$

The mole fractions of components a in vapour phase:

$\bold{y_A=\frac{P_A}{P_s}=\frac{200}{500}}$

$\bold{\therefore y_A=0.4}$

The mole fractions of components b in vapour phase:

$\bold{y_B=\frac{P_B}{P_s}=\frac{300}{500}}$

$\bold{\therefore y_B=0.6}$