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Standard VI

Physics

Thermal Equilibrium

Two chambers,...

Question

Two chambers, one containing $m_{1} g$ of a gas at $P_{1}$ pressure and other containing $m_{2} g$ of a gas at $P_{2}$ pressure are put in communication with each other. If temperature remains constant, the common pressure reached will be:

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Solution

$P V = n R T$
$∴ P_{1} V_{1} = n_{1} R T$
$\Rightarrow V_{1} = \frac{m_{1}}{M} \frac{R T}{P_{1}}$
$M =$ Molecule $w t$ at gas.
$T =$ Common temperature.
$V_{2} = \frac{m_{2}}{M} \frac{R T}{P_{2}}$
$V_{1} + V_{2} = \frac{(m + m_{1})}{M} \frac{R T}{P_{e f f}} = V_{e f f}$
$\frac{m_{1}}{P_{1}} + \frac{m_{2}}{P_{2}} = \frac{m_{1} + m_{2}}{P_{e f f}}$
$P_{e f f} = \frac{(m_{1} + m_{2}) P_{1} P_{2}}{P_{2} m_{1} + P_{1} m_{2}}$

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Two chambers, one containing m1g of a gas at P1 pressure and other containing m2g of a gas at P2 pressure are put in communication with each other. If temperature remains constant, the common pressure reached will be:

PV=nRT∴P1V1=n1RT⇒V1=m1MRTP1M= Molecule wt at gas.T= Common temperature.V2=m2MRTP2V1+V2=(m+m1)MRTPeff=Veffm1P1+m2P2=m1+m2PeffPeff=(m1+m2)P1P2P2m1+P1m2

Two chambers, one containing $m_{1} g$ of a gas at $P_{1}$ pressure and other containing $m_{2} g$ of a gas at $P_{2}$ pressure are put in communication with each other. If temperature remains constant, the common pressure reached will be: