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Molecular Speeds - Vrms, Vavg, Vmps

Two chamber c...

Question

Two chamber containing $m_{1}$ and $m_{2}$ grams 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 C $\frac{P_{1} P_{2} (m_{1} + m_{2})}{P_{2} m_{1} + P_{1} m_{2}}$
According to Boyle's law, PV $=$ kT(a constant)
or $P \frac{m}{ρ} =$ kT or $ρ = \frac{P m}{k T}$

or $ρ = \frac{P}{K}$ (where $\frac{k T}{m} = K$ 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 volume $= V_{1} + V_{2} = K (\frac{m_{1}}{P_{1}} + \frac{m_{2}}{P_{2}})$

Let P be the common pressure and $ρ$ be the common density of mixture. Then.
$ρ = \frac{m_{1} + m_{2}}{V_{1} + V_{2}} = \frac{m_{1} + m_{2}}{K (\frac{m_{1}}{P_{1}} + \frac{m_{2}}{P_{2}})}$

$∴ P = K_{ρ} = \frac{m_{1} + m_{2}}{\frac{m_{1}}{P_{1}} + \frac{m_{2}}{P_{2}}} = \frac{P_{1} P_{2} (m_{1} + m_{2})}{(m_{1} P_{2} + m_{2} P_{1})}$

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