Question

Two vessels are filled with ideal gases A and B and are connected through a pipe of zero volume as shown in figure. The stop cock is opened and the gases are allowed to mix homogeneously and the temperature is kept constant. The partial pressures of A and B respectively (in atm) are

• A. 8.0, 5
• B. 9.6, 4
• C. 6.4, 4
• D. 4.8, 2

Hint:

For mixing of non-reacting gases at constant temperature, the new partial pressure is simply the initial pressure scaled by the volume fraction: \(P_{\text{new}} = P_{\text{old}} \times \frac{V_{\text{old}}}{V_{\text{total}}}\).

Answer 1

The Correct Option is D

Step 1: Calculate Total Volume When the stopcock is opened, the total volume available for each gas is the sum of individual volumes. \[ V_{\text{total}} = V_A + V_B = 12\text{ L} + 8\text{ L} = 20\text{ L} \]
Step 2: Calculate Partial Pressure of Gas A Using Boyle's Law (\(P_1 V_1 = P_2 V_2\)) for Gas A: Initial state: \(P_A = 8\text{ atm}, V_A = 12\text{ L}\). Final state: \(P'_A = ?, V_{\text{total}} = 20\text{ L}\). \[ P'_A = \frac{P_A V_A}{V_{\text{total}}} = \frac{8 \times 12}{20} = \frac{96}{20} = 4.8\text{ atm} \]
Step 3: Calculate Partial Pressure of Gas B Using Boyle's Law for Gas B: Initial state: \(P_B = 5\text{ atm}, V_B = 8\text{ L}\). Final state: \(P'_B = ?, V_{\text{total}} = 20\text{ L}\). \[ P'_B = \frac{P_B V_B}{V_{\text{total}}} = \frac{5 \times 8}{20} = \frac{40}{20} = 2.0\text{ atm} \] Final Answer:
4.8, 2.