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(C) Using the ideal gas equation,$PV = \mu RT = \frac{m}{M}RT$,where $m$ is the mass of the gas.Since $V$ and $M$ are constant,we have $P \propto mT$.

Vedclass · Accepted Answer

$11.7$

Vedclass · Answer

(C) Using the ideal gas equation,$PV = \mu RT = \frac{m}{M}RT$,where $m$ is the mass of the gas.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: $P_1 = 20 \ atm$,$T_1 = 27^{\circ}C = 300 \ K$.After releasing half the gas,$m_2 = \frac{1}{2}m_1$.The temperature is raised by $50^{\circ}C$,so $T_2 = (27 + 50)^{\circ}C = 77^{\circ}C = 350 \ K$.Using the ratio: $\frac{P_2}{P_1} = \frac{m_2}{m_1} 	imes \frac{T_2}{T_1}$.$\frac{P_2}{20} = \frac{1}{2} 	imes \frac{350}{300} = \frac{1}{2} 	imes \frac{7}{6} = \frac{7}{12}$.$P_2 = 20 	imes \frac{7}{12} = \frac{140}{12} \approx 11.67 \ atm$.Thus,the final pressure is approximately $11.7 \ atm$.

The gas in a vessel is subjected to a pressure of $20 \ atm$ at a temperature of $27^{\circ}C$. If one-half of the gas is released from the vessel and the temperature of the remaining gas is raised by $50^{\circ}C$,the final pressure of the gas in the vessel is ....... $atm$.

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