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

Answer: (B)

$117^\circ\text{C}$

Answer 2

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:

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