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(B) Using the ideal gas equation $PV = nRT = (m/M)RT$,where $m$ is the mass and $M$ is the molar mass. Since $V$ and $M$ are constant,we have $P \prop

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$6$

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(B) Using the ideal gas equation $PV = nRT = (m/M)RT$,where $m$ is the mass and $M$ is the molar mass. Since $V$ and $M$ are constant,we have $P \propto mT$.Let the initial state be $(P_1, m_1, T_1)$ and the final state be $(P_2, m_2, T_2)$.Given: $P_1 = 10 \ atm$,$T_1 = 27 + 273 = 300 \ K$,$m_1 = m$.After removing half the mass: $m_2 = m/2$.Final temperature: $T_2 = 87 + 273 = 360 \ K$.Using the ratio: $\frac{P_2}{P_1} = \frac{m_2}{m_1} 	imes \frac{T_2}{T_1}$$\frac{P_2}{10} = \frac{m/2}{m} 	imes \frac{360}{300}$$\frac{P_2}{10} = \frac{1}{2} 	imes \frac{6}{5}$$\frac{P_2}{10} = \frac{3}{5}$$P_2 = 10 	imes \frac{3}{5} = 6 \ atm$.

$A$ vessel is filled with an ideal gas at a pressure of $10 \ atm$ and temperature $27^\circ C$. Half of the mass of the gas is removed from the vessel and the temperature of the remaining gas is increased to $87^\circ C$. Then the pressure of the gas in the vessel will be ...... $atm$.

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