JEE Physics 2024 Question with Solution

Given: Two isobaric processes for the same mass of an ideal gas are shown on a $V$-$T$ graph.

Find: The correct relation between $P_1$ and $P_2$.

For an ideal gas,

$$PV = nRT$$

Rearranging,

$$V = \left( \frac{nR}{P} \right) T$$

So, in a $V$-$T$ graph, the slope is

$$\text{Slope} = \frac{V}{T} = \frac{nR}{P}$$

Hence, slope is inversely proportional to pressure:

$$\text{Slope} \propto \frac{1}{P}$$

The line corresponding to $P_2$ is steeper than the line corresponding to $P_1$. Therefore,

$$(\text{Slope})_2 > (\text{Slope})_1 \implies P_2 < P_1$$

Thus,

$$P_1 > P_2$$

Therefore, the correct option is D.

Given: Two isobaric lines are drawn on a $V$-$T$ graph for the same gas.

Find: Which pressure is greater.

In an isobaric $V$-$T$ graph, a steeper line means larger $\frac{V}{T}$. Since

$$V = \left( \frac{nR}{P} \right) T$$

a larger slope means smaller pressure. Therefore, the steeper line corresponds to lower pressure.

Since the line for $P_2$ is steeper, $P_2$ is lower than $P_1$. Therefore, the correct option is D, that is, $P_1 > P_2$.