Q. The pressure and temperature of an ideal gas in a closed vessel are $720\ kPa$ and $40℃$ respectively. If $\frac{1}{4}$ th of the gas is released from the vessel and the temperature of the remaining gas is raised to $353℃,$ the final pressure of the gas is
A) $1440\ kPa$
B) $1080\ kPa$
C) $720\ kPa$
D) $540\ kPa$
Explanation
(b) $P_{1} = 720kPa,\ T_{1} = 40℃ = 273 + 40 = 313K$
$P \propto mT \Rightarrow \frac{P_{2}}{P_{1}} = \frac{m_{2}}{m_{1}}\frac{T_{2}}{T_{1}} = \frac{3}{4} \times \frac{626}{313} = 1.5$
$\Rightarrow P_{2} = 1.5P_{1} = 1.5 \times 720 = 1080kPa$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.