Two containers C_{1} and C_{2} of volumes V a | vedclass

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(D) Let $n_{1}$ and $n_{2}$ be the number of moles of gas present in containers $C_{1}$ and $C_{2}$ respectively,before the valve is opened.Using the

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$A$ long time after the valve is opened,the number of moles of gas in the hot container will be thrice that of the cold one.

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(D) Let $n_{1}$ and $n_{2}$ be the number of moles of gas present in containers $C_{1}$ and $C_{2}$ respectively,before the valve is opened.Using the ideal gas equation $pV = nRT$:$n_{1} = \frac{pV}{R(300)}$$n_{2} = \frac{5p(4V)}{R(400)} = \frac{20pV}{400R} = \frac{pV}{20R}$When the valve is opened,gas flows until the pressure in both $C_{1}$ and $C_{2}$ becomes equal to $p_{0}$.Let $n_{1}'$ and $n_{2}'$ be the new number of moles:$n_{1}' = \frac{p_{0}V}{R(300)}$$n_{2}' = \frac{p_{0}(4V)}{R(400)} = \frac{p_{0}V}{100R}$Since the total number of moles is conserved $(n_{1} + n_{2} = n_{1}' + n_{2}')$:$\frac{pV}{300R} + \frac{20pV}{400R} = \frac{p_{0}V}{300R} + \frac{p_{0}V}{100R}$$\frac{p}{300} + \frac{p}{20} = p_{0} \left( \frac{1}{300} + \frac{3}{300} \right)$$\frac{p + 15p}{300} = p_{0} \left( \frac{4}{300} \right)$$16p = 4p_{0} \Rightarrow p_{0} = 4p$Now,the ratio of moles in the hot container $(C_{2})$ to the cold container $(C_{1})$ is:$\frac{n_{2}'}{n_{1}'} = \frac{p_{0}(4V) / R(400)}{p_{0}V / R(300)} = \frac{4/400}{1/300} = \frac{1/100}{1/300} = 3$Thus,the number of moles in the hot container is thrice that of the cold one.

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