Question

The temperature of a gas is \(-78^\circ\text{C}\) 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:

• A. \(-39^\circ\text{C}\)
• B. \(117^\circ\text{C}\)
• C. \(127^\circ\text{C}\)
• D. \(-78^\circ\text{C}\)

Answer 1

The Correct Option is B

The translational kinetic energy of the molecules of an ideal gas is directly proportional to the temperature in Kelvin. Thus, we have the relation:

\[ K \propto T. \]

Given that the initial temperature is \( T_1 = -78^\circ \text{C} = 195 \, \text{K} \), and the final temperature is \( T_2 \) when the kinetic energy becomes \( 2K \), we know:

\[ \frac{T_2}{T_1} = 2. \]

Thus, the new temperature is:

\[ T_2 = 2 \times 195 = 390 \, \text{K}. \]

Converting back to Celsius:

\[ T_2 = 390 - 273 = 117^\circ \text{C}. \]