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(D) From the ideal gas equation,$PV = nRT = (m/M)RT$. Since the temperature $T$,volume $V$,and molar mass $M$ are constant,we have $P \propto m$. Ther

Vedclass · Accepted Answer

$7.5$

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(D) From the ideal gas equation,$PV = nRT = (m/M)RT$. Since the temperature $T$,volume $V$,and molar mass $M$ are constant,we have $P \propto m$. Therefore,$\frac{P_1}{P_2} = \frac{m_1}{m_2}$. Given $P_1 = 10^7\, N/m^2$,$m_1 = 10\, kg$,and $P_2 = 2.5 	imes 10^6\, N/m^2$. Substituting the values: $\frac{10^7}{2.5 	imes 10^6} = \frac{10}{m_2}$. $4 = \frac{10}{m_2} \implies m_2 = \frac{10}{4} = 2.5\, kg$. The mass of the gas taken out is $\Delta m = m_1 - m_2 = 10 - 2.5 = 7.5\, kg$.

$A$ cylinder contains $10\, kg$ of gas at a pressure of $10^7\, N/m^2$. The quantity of gas taken out of the cylinder,if the final pressure is $2.5 \times 10^6\, N/m^2$,will be ..... $kg$ (Temperature of the gas is constant).

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