Question

2 moles of liquid A and 3 moles of liquid B are mixed to from an ideal solution. The vapour pressure of ideal solution is 320 mm Hg. When 1 mole of A & 1 mole of B is further added then new vapour pressure of solution is 328.57 mm Hg. Find the vapour pressure of pure A ($P_A^\circ$) & pure B ($P_B^\circ$):

• A. $P_A^\circ = 200, P_B^\circ = 500$
• B. $P_A^\circ = 500, P_B^\circ = 200$
• C. $P_A^\circ = 300, P_B^\circ = 400$
• D. $P_A^\circ = 200, P_B^\circ = 300$

Hint:

Recognize decimal values like $.57$ as fractions ($4/7$) to facilitate integer arithmetic.

Answer 1

The Correct Option is B

Eq 1: $X_A = 2/5, X_B = 3/5$. $P = 320$.
$0.4 P_A^\circ + 0.6 P_B^\circ = 320 \implies 2 P_A^\circ + 3 P_B^\circ = 1600$.
Eq 2: Add 1 mol each. $n_A = 3, n_B = 4$. $X_A = 3/7, X_B = 4/7$. $P = 328.57 \approx 2300/7$.
$\frac{3}{7} P_A^\circ + \frac{4}{7} P_B^\circ = \frac{2300}{7} \implies 3 P_A^\circ + 4 P_B^\circ = 2300$.
Solve linear system:
Multiply (1) by 3: $6 P_A^\circ + 9 P_B^\circ = 4800$.
Multiply (2) by 2: $6 P_A^\circ + 8 P_B^\circ = 4600$.
Subtract: $P_B^\circ = 200$.
Substitute in (1): $2 P_A^\circ + 600 = 1600 \implies 2 P_A^\circ = 1000 \implies P_A^\circ = 500$.