The temperature of a gas is $$-78°C$$ and the average translational kinetic energy of its molecules is $$K$$. The temperature at which the average translational kinetic energy of the molecules of the same gas becomes $$2K$$ is :

Average translational KE: $$K = \frac{3}{2}k_BT$$.

Initial: $$T_1 = -78°C = 195$$ K, $$K_1 = K$$.

For $$K_2 = 2K$$: $$T_2 = 2T_1 = 390$$ K = $$117°C$$.

The correct answer is Option 2: $$117°C$$.

Create a FREE account and get:

Predict your JEE Main percentile, rank & performance in seconds

Check Your Score Now

Ask our AI anything

AI can make mistakes. Please verify important information.

Download resources