JEE Physics 2024 Question with Solution

Given: Initial temperature is $T_1 = -78^\circ \text{C}$ and initial average translational kinetic energy is $K_1 = K$.

Find: The temperature at which the average translational kinetic energy becomes $K_2 = 2K$.

For an ideal gas, the average translational kinetic energy is directly proportional to the absolute temperature:

$$K \propto T$$

So,

$$\frac{K_2}{K_1} = \frac{T_2}{T_1}$$

Convert the initial temperature to Kelvin:

$$T_1 = -78 + 273 = 195 \, \text{K}$$

Now use $K_2 = 2K_1$:

$$\frac{T_2}{195} = 2$$

Therefore,

$$T_2 = 2 \times 195 = 390 \, \text{K}$$

Convert back to Celsius:

$$T_2 = 390 - 273 = 117^\circ \text{C}$$

Therefore, the correct option is B, and the required temperature is $117^\circ \text{C}$.

Given: Average translational kinetic energy changes from $K$ to $2K$.

Find: The corresponding temperature.

Since translational kinetic energy is directly proportional to Kelvin temperature, doubling the kinetic energy means doubling the Kelvin temperature.

Initial temperature:

$$-78^\circ \text{C} = 195 \, \text{K}$$

Double it:

$$T_2 = 2 \times 195 = 390 \, \text{K}$$

Convert to Celsius:

$$390 - 273 = 117^\circ \text{C}$$

This works because proportionality is with absolute temperature, not Celsius temperature. Therefore, the correct option is B.