Q 11-12-093JEE MainJEE Main 2022 (27 Jun, Shift 2)Easy
For a perfect gas, two pressures $P_1$ and $P_2$ are shown in figure. The graph shows
Answer: (A) $P_1 > P_2$
For a fixed amount of gas, $V = \dfrac{nR}{P}T$, so the slope of the $V$-$T$ line is $\dfrac{nR}{P}$.
The line for $P_2$ is steeper, so $P_2$ is smaller: $P_1 > P_2$.
Solution by Sreeraj P, M.Sc Physics