The temperature of a gas is $-78^{\circ} \mathrm{C}$ and the average translational kinetic energy of its molecules is $\mathrm{K}$. The temperature at which the average translational kinetic energy of the molecules of the same gas becomes $2 \mathrm{~K}$ is :

<p>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:</p> <p>$K_{avg} = \frac{3}{2}kT$</p> <p>Where:</p> <ul> <li>$K_{avg}$ is the average kinetic energy,</li> <li>$k$ is the Boltzmann's constant, and</li> <li>$T$ is the temperature in Kelvin.</li> </ul> <p>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 $(T_2)$.</p> <p>Using the direct proportionality relation, we can set up the following equation:</p> <p>$K \propto T$</p> <p>$2K = K_2 = \frac{3}{2}kT_2$</p> <p>Given that at $T_1$, the kinetic energy is $K$, and at $T_2$, it's $2K$, we can use the ratio as follows:</p> <p>$\frac{K_2}{K_1} = \frac{2K}{K} = 2 = \frac{T_2}{T_1}$</p> <p>Thus, we have:</p> <p>$T_2 = 2T_1 = 2 \times 195K = 390K$</p> <p>To find the temperature in Celsius, we convert $390K$ back to Celsius:</p> <p>$T_{2(Celsius)} = 390 - 273 = 117^{\circ}C$</p> <p>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:</p> <p>Option D $117^{\circ} \mathrm{C}$</p>