Question 8
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 :
Solution
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: $$117°C$$.
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