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

To determine the temperature at which the average translational kinetic energy of a gas becomes twice the initial kinetic energy, we need to use the relationship between kinetic energy and temperature, governed by the kinetic theory of gases.

The average translational kinetic energy \(K\) of a gas molecule is directly proportional to its absolute temperature \(T\). This can be described by the equation:

\(K \propto T\)

Given that the initial temperature \(T_1 = -78^\circ\text{C}\), we first convert it to Kelvin, the unit of absolute temperature used in physics:

\(T_1 = -78 + 273 = 195 \text{ K}\)

Let \(K\) be the initial average kinetic energy. At the new temperature \(T_2\), the kinetic energy increases to \(2K\). Thus,

\(\frac{K_2}{K_1} = \frac{T_2}{T_1} = 2\)

Therefore,

\(T_2 = 2 \times 195 = 390 \text{ K}\)

We convert \(T_2\) from Kelvin back to Celsius:

\(T_2 = 390 - 273 = 117^\circ\text{C}\)

Thus, the temperature at which the average translational kinetic energy of the molecules of the gas becomes \(2K\) is \(117^\circ\text{C}\).

Therefore, the correct answer is \(117^\circ\text{C}\).

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:

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