Question

Vessel A contains an ideal gas at a pressure $5 \times 10^5$ Pa and is connected with a heat source which maintains its temperature at 300K. Another vessel B which has four times greater inner volume contains the same gas at a pressure $10^5$ Pa and is connected to a heat source which maintains its temperature at 400K. What will be the pressure of entire system if two vessels are now connected by a narrow tube tap:-

• A. 1.8 x 10^5 Pa
• B. 1.5 x 10^5 Pa
• C. 2.0 x 10^5 Pa
• D. 2.5 x 10^5 Pa

Answer 1

Answer: (A)

1.8 x 10^5 Pa

Answer 2

1. Calculate initial moles using $PV=nRT$: $n_A = \frac{P_A V_A}{R T_A} = \frac{(5 \times 10^5) V_A}{R \times 300}$ $n_B = \frac{P_B V_B}{R T_B} = \frac{(10^5) (4V_A)}{R \times 400} = \frac{(4 \times 10^5) V_A}{R \times 400}$
2. Simplify moles: $n_A = \frac{5000}{3} \frac{V_A}{R}$ $n_B = 1000 \frac{V_A}{R}$
3. Total volume $V_{total} = V_A + V_B = V_A + 4V_A = 5V_A$.
4. Calculate partial pressures in the total volume, assuming gas A occupies volume $V_A$ at $T_A$ and gas B occupies volume $V_B$ at $T_B$ (a common simplification for finding total pressure): $P_{fA} = \frac{n_A R T_A}{V_{total}} = \frac{\frac{(5 \times 10^5) V_A}{R \times 300} R \times 300}{5V_A} = \frac{(5 \times 10^5) V_A}{5V_A} = 10^5$ Pa. $P_{fB} = \frac{n_B R T_B}{V_{total}} = \frac{\frac{(4 \times 10^5) V_A}{R \times 400} R \times 400}{5V_A} = \frac{(4 \times 10^5) V_A}{5V_A} = 0.8 \times 10^5$ Pa.
5. Sum partial pressures (Dalton's Law): $P_{final} = P_{fA} + P_{fB} = 10^5 \text{ Pa} + 0.8 \times 10^5 \text{ Pa} = 1.8 \times 10^5$ Pa.