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The container shown in Figure has two chambers, separated by a partition, of volumes V1 = 2.0 litre and V2 = 3.0 litre. The chambers contain $\mu _{1}=4.0$ and $\mu _{2}=5.0$ moles of a gas at pressures $p_{1} = 1.00$ atm and $p_{2} = 2.00$ atm. Calculate the pressure after the partition is removed and the mixture attains equilibrium.

$V_{1}$ $V_{2}$

$\mu_{1}$ $\mu_{2}$

$p_{1}$ $p_{2}$

Answers (1)

According to the ideal gas situation, we have PV = nRT

For chamber 1 and chamber 2,

$P_{1}V_{1} = n_{1}RT_{1}$ and $P_{2}V_{2} = n_{2}RT_{2}$

$P_{1}=1atm, P_{2}=2 atm, V_{1}=2L, V_{2}=3L$

$T_{1}=T, T_{2}=T, n_{1}=4, n_{2}=5$

When we remove the partition, $n = n_{1}+n_{2}$ and $V = V_{1} + V_{2}$

According to kinetic theory, Translational gas kinetic energy $=$ $PV = \frac{2}{3}$ . E per mole

So, the translational KE for both cases will be,

$P_{1}V_{1} = \frac{2}{3}. n_{1} E_{1}$ and $P_{2}V_{2} = \frac{2}{3}. n_{2} E_{2}$

Adding, $P_{1}V_{1}+P_{2}V_{2} =\frac{2}{3}. n_{1} E_{1}+ \frac{2}{3}. n_{2} E_{2}$

$n_{1}E_{1}+n_{2}E_{2}=\frac{3}{2}\left ( P_{1}V_{1}+P_{2} V_{2} \right )$

hence by combining the results we get,

$P(V_{1}+V_{2}) = \frac{2}{3}\left [ \frac{3}{2}(P_{1}V_{1}+P_{2}V_{2}) \right ]$

$P = \frac{P_1V_1 + P_{2}V_2}{V_{1} + V_{2}} = 1.6 atm$

Posted by

infoexpert24

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