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Vapour Pressure

Two chambers ...

Question

Two chambers containing $m_{1}$ gm and $m_{2}$ gm of a gas at pressure $P_{1}$ 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} (m_{1} + m_{2})}{(P_{2} m_{1} + P_{1} m_{2})}

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B

\frac{P_{1} P_{2} m_{1}}{P_{2} m_{1} + P_{1} m_{2}}

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C

\frac{m_{1} m_{2} (P_{1} + P_{2})}{P_{2} m_{1} + P_{1} m_{2}}

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D

\frac{m_{1} m_{2} P_{2}}{P_{2} m_{1} + P_{1} m_{2}}

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Solution

The correct option is A $\frac{P_{1} P_{2} (m_{1} + m_{2})}{(P_{2} m_{1} + P_{1} m_{2})}$
According to Boyle's law, $P V = k$ (a constant)
or $P \frac{m}{ρ} = k$ or $ρ = \frac{P m}{k} (∵ V = \frac{m}{ρ})$
or $ρ = \frac{P}{K} (∵ \frac{k}{m} = K$ a constant)
So, $ρ_{1} = \frac{P_{1}}{K}$ and $V_{1} = \frac{m_{1}}{ρ_{1}} = \frac{m_{1}}{P_{1} / K} = \frac{K m_{1}}{P_{1}}$
Similarly, $V_{2} = \frac{K m_{2}}{P_{2}}$
$∴$ Total number $= V_{1} + V_{2} = K (\frac{m_{1}}{P_{1}} + \frac{m_{2}}{P_{2}})$
Let $P$ be the common pressure and $ρ$ be the common density of mixture. Then
$ρ = \frac{m_{1} + m_{2}}{V_{1} + V_{2}} = \frac{m_{1} + m_{2}}{K (\frac{m_{1}}{P_{1}} + \frac{m_{2}}{P_{2}})}$
$∴ P = K ρ = \frac{m_{1} + m_{2}}{\frac{m_{1}}{P_{2}} + \frac{m_{2}}{P_{2}}} = \frac{P_{1} P_{2} (m_{1} + m_{2})}{(m_{1} P_{2} + m_{2} P_{1})}$ ..

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