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Q. An ideal gas initially at pressure 1 bar is being compressed from 30 m3 to 10 m3 volume and its temperature decreases from 320 K to 280 K then find final pressure of gas.

AIIMSAIIMS 2019

Solution:

$P_{1}=1 bar, V_{1}=30 m^{3}$
$V_{2}=10 m^{3}$
$T_{1}=320 K, T_{2}=280 K$
$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}\quad\Rightarrow \quad\quad \frac{1\times30}{320}=\frac{P_{2}\times10}{280}$
$P_{2} = 2.625 bar$