The pressure and temperature of an ideal gas | vedclass

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Question

(B) Given: Initial pressure $P_1 = 720 \, kPa$,Initial temperature $T_1 = 40^\circ C = 273 + 40 = 313 \, K$.Final temperature $T_2 = 353^\circ C = 273

Vedclass · Accepted Answer

$1080$

Vedclass · Answer

(B) Given: Initial pressure $P_1 = 720 \, kPa$,Initial temperature $T_1 = 40^\circ C = 273 + 40 = 313 \, K$.Final temperature $T_2 = 353^\circ C = 273 + 353 = 626 \, K$.Since $\frac{1}{4}^{th}$ of the gas is released,the remaining mass $m_2 = \frac{3}{4} m_1$.Using the ideal gas law $PV = nRT = \frac{m}{M} RT$,we have $P \propto mT$ (since $V$ and $M$ are constant).Therefore,$\frac{P_2}{P_1} = \frac{m_2}{m_1} 	imes \frac{T_2}{T_1}$.Substituting the values: $\frac{P_2}{720} = \frac{3}{4} 	imes \frac{626}{313} = \frac{3}{4} 	imes 2 = 1.5$.$P_2 = 1.5 	imes 720 = 1080 \, kPa$.

The pressure and temperature of an ideal gas in a closed vessel are $720 \, kPa$ and $40^\circ C$ respectively. If $\frac{1}{4}^{th}$ of the gas is released from the vessel and the temperature of the remaining gas is raised to $353^\circ C$,the final pressure of the gas is ....... $kPa$.

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