QJEEM24Ph0607: JEE Main 2024 Physics - Class 11 JEEM24Ph0607 Solution: The given figure represents two isobaric processes for the same …

The two isobaric processes depicted in the figure show changes in the volume of an ideal gas with temperature under constant pressure conditions.

Isobaric processes follow the equation $PV = nRT$ , where P is pressure, V is volume, n is the number of moles of the gas, R is the ideal gas constant, and T is temperature.

When we rearrange the equation in terms of V (Volume), we get $V = \left(\frac{nR}{P}\right)T$ . The slope of the volume-temperature graph in such a scenario is given by $\frac{nR}{P}$ . This implies that the slope is inversely proportional to the pressure (i.e., slope $\propto \frac{1}{P}$ ).

From this relationship, if one process has a higher slope compared to another, its corresponding pressure must be lower. Observing the given figure and applying this understanding, if Process 2 has a greater slope than Process 1, it indicates that P_2 < P_1 .