Question

Initially, flask A contained oxygen gas at 27°C and 950 mm of Hg, and flask B contained neon gas at 27°C and 900 mm. Finally, the two flasks were joined by means of a narrow tube of negligible volume equipped with a stopcork and gases were allowed to mixup freely. The final pressure in the combined system was found to be 910 mm of Hg.

What is the correct relationship between volumes of the two flasks ?

• A. VB = 3VA
• B. VB = 4VA
• C. VB = 5VA
• D. VB = 4.5VA

Answer 1

Answer: (B)

VB = 4VA

Answer 2

Let $V_A$ and $V_B$ be the volumes of flask A and B, respectively. According to Boyle's Law, the pressure of a gas is inversely proportional to its volume at constant temperature. When the gases mix, the final total pressure ($P_{final}$) is the sum of the partial pressures of each gas in the combined volume ($V_A + V_B$). The partial pressure of gas A in the final mixture is $P_{A,final} = \frac{P_{A,initial} \times V_A}{V_A + V_B}$. The partial pressure of gas B in the final mixture is $P_{B,final} = \frac{P_{B,initial} \times V_B}{V_A + V_B}$. By Dalton's Law of Partial Pressures, $P_{final} = P_{A,final} + P_{B,final}$.

Substituting the given values: $910 = \frac{950 \times V_A}{V_A + V_B} + \frac{900 \times V_B}{V_A + V_B}$ Multiplying by $(V_A + V_B)$: $910 (V_A + V_B) = 950 V_A + 900 V_B$ $910 V_A + 910 V_B = 950 V_A + 900 V_B$ Rearranging the terms: $910 V_B - 900 V_B = 950 V_A - 910 V_A$ $20 V_B = 40 V_A$ $V_B = 2 V_A$

Re-checking with the options, if we assume $V_B = 4V_A$: $910 (V_A + 4V_A) = 950 V_A + 900 (4V_A)$ $910 (5V_A) = 950 V_A + 3600 V_A$ $4550 V_A = 4550 V_A$ This confirms that $V_B = 4V_A$ is the correct relationship.