A container holds 10 kg of gas at a pressure | vedclass

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(B) For an ideal gas at constant temperature,the pressure $P$ is directly proportional to the mass $m$ of the gas in the container $(P \propto m)$.Giv

Vedclass · Accepted Answer

$7.5$

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(B) For an ideal gas at constant temperature,the pressure $P$ is directly proportional to the mass $m$ of the gas in the container $(P \propto m)$.Given:Initial pressure $P_1 = 10^7 \ N/m^2$Initial mass $m_1 = 10 \ kg$Final pressure $P_2 = 2.5 	imes 10^6 \ N/m^2$Using the relation $\frac{P_1}{P_2} = \frac{m_1}{m_2}$:$\frac{10^7}{2.5 	imes 10^6} = \frac{10}{m_2}$Solving for $m_2$:$m_2 = \frac{2.5 	imes 10^6 	imes 10}{10^7} = 2.5 \ kg$The mass of gas removed is $\Delta m = m_1 - m_2 = 10 \ kg - 2.5 \ kg = 7.5 \ kg$.

$A$ container holds $10 \ kg$ of gas at a pressure of $10^7 \ N/m^2$. How much gas in $kg$ must be removed so that the final pressure becomes $2.5 \times 10^6 \ N/m^2$? (Assume temperature remains constant.)

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