Q 11-12-058JEE MainJEE Main 2024 (31 Jan, Shift 1)Easy
The given figure represents two isobaric processes, at pressures $P_1$ and $P_2$, for the same mass of an ideal gas. Then
Answer: (D) $P_1 > P_2$
For an isobaric process $V = \dfrac{nR}{P}T$, so the slope of the $V$–$T$ line is inversely proportional to the pressure. The $P_2$ line is steeper, so $P_2 < P_1$, i.e. $P_1 > P_2$.
Solution by Sreeraj P, M.Sc Physics