Question

At 90 degree C, the vapour pressure of toluene is 400torr and that of sigma-xylene is 150torr. What is the composition of the liquid mixture that boils at 90 degree C, when the pressure is 0.50atm ? What is the composition of vapour produced?

Answer 1

Completely Miscible Liquids. They can be handled by Raoult's Law, i.e.

$$y_i P = x_i P_i^0$$

where $P =$ Total pressure of vapors in equilibrium with the liquid solution, $P_i^0 =$ vapor pressure of component i in pure state, $y_i =$ mole fraction of $i^{\text{th}}$ component in vapor state, $x_i =$ mole fraction of $i^{\text{th}}$ component in liquid state.

This most fundamental expression may be arranged in many useful forms. e.g. for binary solutions :

$$P = x_a P_a^0 + (1 - x_a) + P_b^0$$

a=toluene, b=xylene 1 atm=760 torr, 0.5 atm =380 torr

$$P = x_a P_a^0 + (1 - x_a) + P_b^0 = 400x_a + 150(1 - x_a) = 400x_a + 150 - 150x_a = 250x_a + 150$$ $$250x_a = 380 - 150$$ $$x_a = 230 / 250 = 0.92$$ $$x_b = 1 - 0.92 = 0.08$$ $$P_a = x_a P_a^0 = 0.92 * 400 = 368 \text{ torr}$$ $$P_b = x_b P_b^0 = 0.08 * 150 = 12 \text{ torr}$$ $$y_a = P_a / P = 368 / 380 = 0.968 \quad (96.8\%)$$ $$y_b = 1 - y_a = 0.032 \quad (3.2\%)$$