Question

At 90 °C, the vapor pressure of 1,2 dimethylbenzene is 20 kPa, and that of 1,3-dimethylbenzene is 18 kPa. What is the composition of a liquid mixture that boils at 90 °C when the pressure is 19 kPa? What is x₂ and x₃? What is y₂ and y₃?

Answer 1

To find the composition of the liquid mixture that boils at 90 °C when the pressure is 19 kPa, we can use the Raoult's law equation. The composition of the liquid mixture is approximately 54% 1,2-dimethylbenzene and 46% 1,3-dimethylbenzene. In the vapor phase, the mole fractions are approximately 48.6% for 1,3-dimethylbenzene and 51.4% for 1,2-dimethylbenzene.

Step-by-step explanation:

To find the composition of the liquid mixture that boils at 90 °C when the pressure is 19 kPa, we can use the Raoult's law equation: P = x1P1 + x2P2. In this case, x1 and x2 represent the mole fractions of 1,2-dimethylbenzene and 1,3-dimethylbenzene respectively, and P1 and P2 represent their respective vapor pressures at 90 °C. We are given that P1 = 20 kPa, P2 = 18 kPa, and the total pressure, P, is 19 kPa.

Let's rearrange the equation to solve for x2:

19 kPa = (1 - x2)(20 kPa) + x2(18 kPa)

Simplifying this equation gives us x2 = 0.46. Since x2 represents the mole fraction of 1,3-dimethylbenzene, the mole fraction of 1,2-dimethylbenzene, x1, can be calculated as 1 - x2 = 1 - 0.46 = 0.54.

To find the vapor phase mole fractions, y2 and y3, we can use Dalton's law of partial pressures. The vapor pressure of a component is proportional to its mole fraction in the liquid phase. Since the total vapor pressure is given as 19 kPa, we can write:

y2 = x2 * (P / P2) = 0.46 * (19 kPa / 18 kPa) = 0.486

y3 = x3 * (P / P3) = (1 - x2) * (19 kPa / 20 kPa) = 0.514

So, the composition of the liquid mixture is approximately 54% 1,2-dimethylbenzene and 46% 1,3-dimethylbenzene. In the vapor phase, the mole fractions are approximately 48.6% for 1,3-dimethylbenzene and 51.4% for 1,2-dimethylbenzene.