In the figure,the volume of container X is tw | vedclass

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(C) Since the containers are connected,the pressure $P$ in both containers must be equal,i.e.,$P_X = P_Y$.Using the ideal gas equation $PV = nRT = \fr

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$\frac{m}{4}$

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(C) Since the containers are connected,the pressure $P$ in both containers must be equal,i.e.,$P_X = P_Y$.Using the ideal gas equation $PV = nRT = \frac{m}{M}RT$,we get $P = \frac{mRT}{MV}$.For both containers,$P = \frac{mRT}{MV}$ is constant,so $\frac{m_X T_X}{V_X} = \frac{m_Y T_Y}{V_Y}$.Given $V_X = 2V_Y$,$T_X = 200 \, K$,$T_Y = 400 \, K$,and $m_X = m$.Substituting these values: $\frac{m 	imes 200}{2V_Y} = \frac{m_Y 	imes 400}{V_Y}$.Simplifying: $\frac{100m}{V_Y} = \frac{400m_Y}{V_Y}$.$100m = 400m_Y \Rightarrow m_Y = \frac{100m}{400} = \frac{m}{4}$.

In the figure,the volume of container $X$ is twice the volume of container $Y$. Both contain an ideal gas. The temperature of $X$ is $200 \, K$ and the temperature of $Y$ is $400 \, K$. If the mass of the gas in $X$ is $m$,what will be the mass of the gas in $Y$?

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