QJEEM24Ph4f1e: JEE Main 2024 Physics - Class 11 JEEM24Ph4f1e Solution: The temperature of a gas is − 78 ∘ C and the average…

The average translational kinetic energy ( $K_{avg}$ ) of a molecule is directly proportional to the absolute temperature (T) of the gas, as described by the equation:

$K_{avg} = \frac{3}{2}kT$

Where:

• $K_{avg}$ is the average kinetic energy,
• $k$ is the Boltzmann's constant, and
• $T$ is the temperature in Kelvin.

From the given problem, if the temperature of the gas is $-78^{\circ}C$ , which in Kelvin is $T_1 = -78 + 273 = 195K$ , and the average translational kinetic energy is $K$ , when the energy becomes $2K$ , we need to find the new temperature $(T_2)$ .

Using the direct proportionality relation, we can set up the following equation:

$K \propto T$

$2K = K_2 = \frac{3}{2}kT_2$

Given that at $T_1$ , the kinetic energy is $K$ , and at $T_2$ , it's $2K$ , we can use the ratio as follows:

$\frac{K_2}{K_1} = \frac{2K}{K} = 2 = \frac{T_2}{T_1}$

Thus, we have:

$T_2 = 2T_1 = 2 \times 195K = 390K$

To find the temperature in Celsius, we convert $390K$ back to Celsius:

$T_{2(Celsius)} = 390 - 273 = 117^{\circ}C$

Therefore, the temperature at which the average translational kinetic energy of the molecules of the same gas becomes $2K$ is $117^{\circ}C$ . The correct answer is:

Option D $${117}^{\circ }\mathrm{C}$$