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(D) Given: Initial mass $M_1 = 10 \, kg$,Initial pressure $P_1 = 10^7 \, N/m^2$,Final pressure $P_2 = 2.5 	imes 10^6 \, N/m^2$.Using the ideal gas eq

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$7.5$

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(D) Given: Initial mass $M_1 = 10 \, kg$,Initial pressure $P_1 = 10^7 \, N/m^2$,Final pressure $P_2 = 2.5 	imes 10^6 \, N/m^2$.Using the ideal gas equation $PV = nRT = \frac{M}{M_w} RT$,where $M$ is the mass of the gas.Since $V$,$R$,$T$,and $M_w$ are constant,we have $P \propto M$.Therefore,$\frac{P_2}{P_1} = \frac{M_2}{M_1}$.Substituting the values: $\frac{2.5 	imes 10^6}{10^7} = \frac{M_2}{10}$.$0.25 = \frac{M_2}{10} \Rightarrow M_2 = 2.5 \, kg$.The mass of gas removed is $\Delta M = M_1 - M_2 = 10 - 2.5 = 7.5 \, kg$.

$A$ cylinder contains $10 \, kg$ of gas at a pressure of $10^7 \, N/m^2$. What amount of gas in $kg$ must be removed so that the final pressure becomes $2.5 \times 10^6 \, N/m^2$?

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