Question

Two chambers, one containing $m_1$ gm of a gas at $P_1$ pressure and other containing $m_2$ gm of a gas at $P_2$ pressure, are put in communication with each other. If temperature remains constant, the common pressure reached will be:

• A. $\frac{P_1{P}_2(m_1+m_2)}{P_2{m}_1+P_1{m}_2}$
• B. $\frac{m_1{m}_2(P_1+P_2)}{(P_2{m}_1+P_1{m}_2)}$
• C. $\frac{P_1{P}_2{m}_1}{P_2{m}_1+P_1{m}_2}$
• D. $\frac{m_1{m}_2{P}_2}{(P_2{m}_1+P_1{m}_2)}$

Answer 1

Answer: (A)

$\frac{P_1{P}_2(m_1+m_2)}{P_2{m}_1+P_1{m}_2}$

Since, Volume is the additive term in the question

$V_1=\frac{m_1}{{\rho }_1}$

Similarly,

$V_2=\frac{m_2}{{\rho }_2}$

Now, total volume is given by, V = $V_1+V_2$

Also, for constant temperature $\rho$ is directly proportional to Pressure.

Hence, V = $\frac{m_1}{{\rho }_1}+\frac{m_2}{{\rho }_2}$

Now, $\rho =\frac{m_1+m_2}{V}$

Thus, eliminating V and $\rho$ from the expression,

P = $\frac{m_1+m_2}{\frac{m_1}{P_1}+\frac{m_2}{P_2}}$

Or, in other words,

P = $\frac{(m_1+m_2)P_1{P}_2}{P_2{m}_1+P_1{m}_2}$