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(D) The average speed of gas molecules is given by the formula $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$.Since the gas is the same,$M$ is constant. Thus,$v

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$8$ times that of $Y$

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(D) The average speed of gas molecules is given by the formula $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$.Since the gas is the same,$M$ is constant. Thus,$v_{avg} \propto \sqrt{T}$.Given $v_X = 2v_Y$,we have $\sqrt{T_X} = 2\sqrt{T_Y}$,which implies $T_X = 4T_Y$.Using the ideal gas equation $PV = nRT$,we have $P = \frac{nRT}{V}$.For flask $X$: $P_X = \frac{1 \cdot R \cdot T_X}{V_X} = \frac{RT_X}{1}$.For flask $Y$: $P_Y = \frac{1 \cdot R \cdot T_Y}{V_Y} = \frac{RT_Y}{2}$.Taking the ratio: $\frac{P_X}{P_Y} = \frac{RT_X}{1} \cdot \frac{2}{RT_Y} = 2 \cdot \frac{T_X}{T_Y}$.Substituting $T_X = 4T_Y$: $\frac{P_X}{P_Y} = 2 \cdot 4 = 8$.Therefore,$P_X = 8P_Y$.

Two flasks $X$ and $Y$ have capacities of $1 \ L$ and $2 \ L$ respectively,and each contains $1 \ mol$ of gas. The temperature of the flasks is maintained such that the average speed of molecules in flask $X$ is twice that in flask $Y$. The pressure in flask $X$ is ...

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