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(C) For an open container,the pressure $P$ and volume $V$ remain constant.According to the ideal gas law,$PV = \mu RT$,where $\mu$ is the number of mo

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$\frac{2x}{3}$

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(C) For an open container,the pressure $P$ and volume $V$ remain constant.According to the ideal gas law,$PV = \mu RT$,where $\mu$ is the number of moles and $R$ is the gas constant.Since $P$ and $V$ are constant,$\mu T = 	ext{constant}$.Therefore,$\mu_1 T_1 = \mu_2 T_2$.Given $\mu_1 = x$,$T_1 = T$,and $T_2 = 3T$.Substituting these values: $x 	imes T = \mu_2 	imes (3T)$.Solving for $\mu_2$: $\mu_2 = \frac{x}{3}$,which is the amount of gas remaining in the container.The amount of gas that exits the container is the initial amount minus the remaining amount: $\Delta \mu = x - \frac{x}{3} = \frac{2x}{3}$.

An open container contains $x$ moles of gas at $T$ $K$. If the temperature is tripled,then the amount of gas that exits the container is:

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