Question

A ternary solution prepared by mixing 2 mol of A, 3 mol of B, 5 mol of C is allowed to evaporate till equilibrium is achieved, the vapor collected is allowed to condense and the condensed liquid is again allowed to reach equilibrium, the mole fraction of C in this vapor phase is- [P°=30kPa, P° = 10kPa, P° = 20kPa]

• A. $\frac{10}{19}$
• B. $\frac{1}{2}$
• C. $\frac{20}{41}$
• D. $\frac{3}{10}$

Answer 1

Answer: (C)

$\frac{20}{41}$

Answer 2

To solve this problem, we need to follow a two-step process:

1. Calculate the composition of the vapor phase (V1) in equilibrium with the initial liquid (L1).
2. Assume V1 condenses to form a new liquid (L2), then calculate the composition of the vapor phase (V2) in equilibrium with L2.

Step 1: First Evaporation (L1 to V1)

Given initial moles:

• $n_A = 2$ mol
• $n_B = 3$ mol
• $n_C = 5$ mol

Total moles $n_{total} = 2 + 3 + 5 = 10$ mol

Mole fractions in the initial liquid (L1):

• $x_A = \frac{n_A}{n_{total}} = \frac{2}{10} = 0.2$
• $x_B = \frac{n_B}{n_{total}} = \frac{3}{10} = 0.3$
• $x_C = \frac{n_C}{n_{total}} = \frac{5}{10} = 0.5$

Given pure vapor pressures:

• $P_A^\circ = 30$ kPa
• $P_B^\circ = 10$ kPa
• $P_C^\circ = 20$ kPa

According to Raoult's Law, the partial pressure of each component in the vapor phase (V1) is:

• $P_A = x_A P_A^\circ = 0.2 \times 30 = 6$ kPa
• $P_B = x_B P_B^\circ = 0.3 \times 10 = 3$ kPa
• $P_C = x_C P_C^\circ = 0.5 \times 20 = 10$ kPa

The total pressure of the first vapor phase ($P_{T1}$) is:

$P_{T1} = P_A + P_B + P_C = 6 + 3 + 10 = 19$ kPa

According to Dalton's Law of Partial Pressures, the mole fraction of each component in the first vapor phase ($y_i$) is:

• $y_A = \frac{P_A}{P_{T1}} = \frac{6}{19}$
• $y_B = \frac{P_B}{P_{T1}} = \frac{3}{19}$
• $y_C = \frac{P_C}{P_{T1}} = \frac{10}{19}$

Step 2: Condensation and Second Evaporation (L2 to V2)

The vapor collected (V1) is condensed to form a new liquid (L2). Therefore, the mole fractions in L2 are the same as the mole fractions in V1:

• $x'_A = y_A = \frac{6}{19}$
• $x'_B = y_B = \frac{3}{19}$
• $x'_C = y_C = \frac{10}{19}$

This new liquid (L2) is allowed to reach equilibrium, meaning it evaporates to form a second vapor phase (V2).

Using Raoult's Law again for L2:

• $P'_A = x'_A P_A^\circ = \frac{6}{19} \times 30 = \frac{180}{19}$ kPa
• $P'_B = x'_B P_B^\circ = \frac{3}{19} \times 10 = \frac{30}{19}$ kPa
• $P'_C = x'_C P_C^\circ = \frac{10}{19} \times 20 = \frac{200}{19}$ kPa

The total pressure of the second vapor phase ($P_{T2}$) is:

$P_{T2} = P'_A + P'_B + P'_C = \frac{180}{19} + \frac{30}{19} + \frac{200}{19} = \frac{180 + 30 + 200}{19} = \frac{410}{19}$ kPa

The mole fraction of C in this second vapor phase ($y'_C$) is:

$y'_C = \frac{P'_C}{P_{T2}} = \frac{\frac{200}{19}}{\frac{410}{19}} = \frac{200}{410} = \frac{20}{41}$

The final answer is $\frac{20}{41}$.

Explanation of the solution:

1. Calculate initial liquid mole fractions ($x_i$): Divide moles of each component by total moles.
2. Calculate partial pressures in first vapor ($P_i$): Use Raoult's Law ($P_i = x_i P_i^\circ$).
3. Calculate total pressure of first vapor ($P_{T1}$): Sum of partial pressures.
4. Calculate mole fractions in first vapor ($y_i$): Use Dalton's Law ($y_i = P_i / P_{T1}$).
5. Define second liquid mole fractions ($x'_i$): These are equal to the mole fractions of the first vapor ($x'_i = y_i$).
6. Calculate partial pressures in second vapor ($P'_i$): Use Raoult's Law again ($P'_i = x'_i P_i^\circ$).
7. Calculate total pressure of second vapor ($P_{T2}$): Sum of new partial pressures.
8. Calculate mole fraction of C in second vapor ($y'_C$): Use Dalton's Law ($y'_C = P'_C / P_{T2}$).