Question

Please work through and explain thoroughly.

1-propanol and 2- propanol form an ideal solution at all concentrations at 25°C. Given that at 25°C, P*1-prop = 20.9 torr, P*2-prop = 45.2, and the total P = 39.1 torr when the molar fraction of 2- propanol (x2-prop) = 0.75, calculate the composition of both species in the vapor phase, as well as the relative amounts of liquid and vapor phases at an overall composition of 0.80 at 25°C.

Answer 1

1-propanol (P1° = 20.9 Torr at 25 °C) and 2-propanol (P2° = 45.2 Torr at 25 °C) form ideal solutions in all proportions.

We are given, mole fraction of 2-propanol, x2 = 0.75 , P1° = 20.9 Torr at 25 °C , P2° = 45.2 torr

Let x1 and x2 represent the mole fractions of 1-propanol and 2-propanol in a liquid mixture, respectively, and y1 and y2 represent the mole fractions of each in the vapor phase.

The partial pressure of propanol will be its mole fraction multiplied by propanol's vapor pressure.

In the vapor, the mole fraction of propanol will be its partial pressure divided by the sum of the two partial pressures. In each solution, the mole fraction of 1-propanol will be 1 - the mole fraction of 2-propanol. .

We use the mole fractions of each liquid to calculate the partial pressure of that component:

2-propanol's vapor pressure = 45.2 torr X 0.75 = 33.9 torr
1-propanol's vapor pressure = 20.9 torr X 0.25 = 5.225 torr

Mole fraction of 1-propanol in the vapor phase = 5.225 torr 33.9 torr = 0.134
Mole fraction of 2-propanol in the vapor phase = 1-0.134 = 0.866

The relative amounts of liquid and vapor phases of propanol-1 and propanol-2 at an overall composition of 0.80 at 25°C is:

1 - 0.80 = 0.2