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 find the temperature where a gas's average translational kinetic energy doubles its initial value, we use the kinetic theory of gases.

The average translational kinetic energy, denoted as \(K\), is directly proportional to the absolute temperature, \(T\), as expressed by \(K \propto T\).

The initial temperature is given as \(T_1 = -78^\circ\text{C}\). Converting this to Kelvin, the absolute temperature scale used in physics, yields:

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

Let the initial average kinetic energy be \(K\). At a new temperature, \(T_2\), the kinetic energy becomes \(2K\). The ratio of kinetic energies is equal to the ratio of absolute temperatures:

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

Solving for \(T_2\):

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

Converting \(T_2\) back to Celsius:

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

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

The correct answer is \(117^\circ\text{C}\).