Two chambers containing m₁ g and m₂ g of a gas at pressures P₁ and P₂ respectively are put in communication with each other, temperature remaining constant. The common pressure reached will be

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

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

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

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

Solution

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

According to Boyle's law, PV== k (a constant )

p $\frac{m}{ρ}$ = k

ρ = $\frac{P m}{k}$ $(∵ V = \frac{m}{ρ})$

$ρ_{1} = \frac{P_{1}}{K}$

and V₁ = $\frac{m_{1}}{ρ_{1}}$

= $\frac{m_{1}}{\frac{P_{1}}{K}}$

V₁ = $\frac{K m_{1}}{P_{1}}$

Simillarly

V₂ = $\frac{K m_{2}}{P_{1}}$

∴ Total volume = V₁ + V₂

= 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

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

⇒ P = $\frac{P_{1} P_{2} (m_{1} + m_{2})}{(m_{1} P_{2} + m_{2} P_{2})}$

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