JEE Physics 2023 Question with Solution

Given: Pressure versus temperature graphs for low density gases $A$, $B$, $C$ are plotted at constant volume.

Find: The temperature corresponding to point $K$.

For an isochoric process, the ideal gas law gives:

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

So,

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

If Celsius temperature is $t$, then absolute temperature is:

$$T = t + 273$$

Hence,

$$P = \frac{nR}{V}(t+273)$$

At point $K$, pressure is zero. Therefore,

$$0 = \frac{nR}{V}(t+273)$$

Since $\frac{nR}{V} \neq 0$, we get:

$$t+273 = 0$$

Thus,

$$t = -273^\circ \text{C}$$

Therefore, the temperature corresponding to point $K$ is $-273^\circ \text{C}$. The solution working gives this value, although the solution incorrectly labels it as Option 4 / D.