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(B) From the ideal gas equation,$PV = nRT$,we have $n = \frac{PV}{RT}$.For container $X$: $n_X = \frac{P_X V_X}{R T_X}$.For container $Y$: $n_Y = \fra

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$m/3 \, g$

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(B) From the ideal gas equation,$PV = nRT$,we have $n = \frac{PV}{RT}$.For container $X$: $n_X = \frac{P_X V_X}{R T_X}$.For container $Y$: $n_Y = \frac{P_Y V_Y}{R T_Y}$.Given $P_X = 3P_Y$,$V_X = 3V_Y$,and $T_X = 3T_Y$.Substituting these values: $n_X = \frac{(3P_Y)(3V_Y)}{R(3T_Y)} = 3 	imes \frac{P_Y V_Y}{R T_Y} = 3n_Y$.Since $n = \frac{	ext{mass}}{	ext{molar mass}}$,and the gas is the same,the molar mass is constant.Therefore,$\frac{m_X}{M} = 3 	imes \frac{m_Y}{M}$.Given $m_X = m$,we get $m = 3m_Y$.Thus,$m_Y = \frac{m}{3} \, g$.

The same gas is filled in two containers $X$ and $Y$. If the pressure,volume,and absolute temperature of the gas in container $X$ are three times those in container $Y$,and the mass of the gas in $X$ is $m \, g$,what will be the mass of the gas in $Y$?

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