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(C) The ideal gas equation is given by $PV = nRT$,where $n = \frac{m}{M}$ ($M$ is the molar mass of the gas).Since the containers are connected by a t

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$m/4$

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(C) The ideal gas equation is given by $PV = nRT$,where $n = \frac{m}{M}$ ($M$ is the molar mass of the gas).Since the containers are connected by a thin tube,the pressure $P$ in both containers must be the same at equilibrium.For container $X$: $V_X = 2V$,$T_X = 200\,K$,and mass $m_X = m$.Thus,$P(2V) = \left(\frac{m}{M}\right) R(200) \implies P = \frac{mR(200)}{2VM} = \frac{100mR}{VM}$.For container $Y$: $V_Y = V$,$T_Y = 400\,K$,and mass $m_Y = m_Y$.Thus,$P(V) = \left(\frac{m_Y}{M}\right) R(400) \implies P = \frac{400m_Y R}{VM}$.Equating the expressions for $P$: $\frac{100mR}{VM} = \frac{400m_Y R}{VM}$.Solving for $m_Y$: $100m = 400m_Y \implies m_Y = \frac{100m}{400} = \frac{m}{4}$.

$A$ container $X$ has a volume double that of container $Y$,and both are connected by a thin tube. Both contain the same ideal gas. The temperature of $X$ is $200\,K$ and that of $Y$ is $400\,K$. If the mass of the gas in $X$ is $m$,then the mass of the gas in $Y$ will be:

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