JEE MAIN - Physics (2023 - 1st February Evening Shift - No. 16)

For three low density gases A, B, C pressure versus temperature graphs are plotted while keeping them at constant volume, as shown in the figure.

JEE Main 2023 (Online) 1st February Evening Shift Physics - Heat and Thermodynamics Question 124 English

The temperature corresponding to the point '$$\mathrm{K}$$' is :

$$-273^{\circ} \mathrm{C}$$
$$-373^{\circ} \mathrm{C}$$
$$-100^{\circ} \mathrm{C}$$
$$-40^{\circ} \mathrm{C}$$

Explanation

In thermodynamics, the pressure-temperature graph for an ideal gas kept at constant volume (isochoric process) is a straight line. This is because for an ideal gas, the pressure is proportional to the temperature (as described by the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature).

It is clear from graph that for all the gases lines of graphs meet at same value.

At $\mathrm{x}$-axis (temperature axis) $\mathrm{P}$ is zero but temperature is negative and it will be equal to 0 K or $$-273^{\circ} \mathrm{C}$$.

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