Question

Two chamber containing $m_1$ and $m_2$ grams of a gas at pressures $P_1$ and $P_2$ respectively are put in communication with each other, temperature remaining constant. The common pressure reached will be

• A. $\frac{P_{1}P_{2}\left(m_{1} + m_{2}\right)}{P_{2}m_{1} + P_{1}m_{2}}$
• B. $\frac{P_{1}P_{2}m_{1} }{P_{2}m_{1} + P_{1}m_{2}}$
• C. $\frac{m_{1}m_{2}\left(P_{1} + P_{2}\right) }{P_{2}m_{1} + P_{1}m_{2}}$
• D. $\frac{m_{1}m_{2} P_{2} }{P_{2}m_{1} + P_{1}m_{2}}$

Answer 1

Answer: (A)

$\frac{P_{1}P_{2}\left(m_{1} + m_{2}\right)}{P_{2}m_{1} + P_{1}m_{2}}$

Answer 2

According to Boyles law, $PV = kT$ (a constant) or $\quad P \frac{m}{\rho} = kT \quad$ or$\quad\rho = \frac{Pm}{kT}$ or$\quad \rho = \frac{P}{K}\quad$ (where $\frac{kT}{m} = K$ a constant) So, $\rho_{1}= \frac{P_{1}}{K}$ and $V_{1} = \frac{m_{1}}{\rho_{1}} = \frac{m_{1}}{P_{1} /K} = \frac{Km_{1}}{P_{1}}$ Similarly, $V_{2} = \frac{Km_{2}}{P_{2}}$ Total volume $=V_{1} + V_{2} = K \left(\frac{m_{1}}{P_{1}} + \frac{m_{2}}{P_{2}}\right)$ Let $P$ be the common pressure and $\rho$ be the common density of mixture. Then. $\rho = \frac{m_{1} + m_{2}}{V_{1} + V_{2}} = \frac{m_{1} + m_{2}}{K\left(\frac{m_{1}}{P_{1}} + \frac{m_{2}}{P_{2}}\right)}$ $\therefore\quad P = K\rho = \frac{m_{1} + m_{2}}{\frac{m_{1}}{P_{1}} + \frac{m_{2}}{P_{2}}} = \frac{P_{1}P_{2}\left(m_{1}+ m_{2}\right)}{\left(m_{1}P_{2} + m_{2}P_{1}\right)}$