Question

If both the containers are maintained at constant temperature (R = 0.0821 L-atm/K-mol)

Container-I: 300 K, 2 mol H₂, 16.42 lit. Container-II: 400 K, 1 mol H₂, 8.21 lit.

• A. Pressure in container-I is 3 atm before opening the valve.
• B. Pressure after opening the valve is 3.57 atm.
• C. Moles in each container are same after opening the valve.
• D. Pressure in each container are same after opening the valve.

Answer 1

Answer: (A), (B), (D)

A, B, D

Answer 2

Calculate initial pressures using the ideal gas law: $P_1 = \frac{n_1RT_1}{V_1} = \frac{2 \times 0.0821 \times 300}{16.42} = 3$ atm. $P_2 = \frac{n_2RT_2}{V_2} = \frac{1 \times 0.0821 \times 400}{8.21} = 4$ atm. After opening the valve, $n_{total} = 2+1=3$ mol and $V_{total} = 16.42+8.21=24.63$ L. The final temperature is the weighted average of initial temperatures: $T_f = \frac{P_1T_1 + P_2T_2}{P_1 + P_2} = \frac{3 \times 300 + 4 \times 400}{3+4} = \frac{900+1600}{7} = \frac{2500}{7}$ K. The final pressure is $P_f = \frac{n_{total}RT_f}{V_{total}} = \frac{3 \times 0.0821 \times (2500/7)}{24.63} \approx 3.57$ atm. If the final temperature is uniform, the pressure is uniform throughout the combined volume.