Question

At 27°C temperature, the mean kinetic energy of the atoms of an ideal gas is \( E_1 \). If the temperature is increased to 327°C, then the mean kinetic energy of the atoms will be

• A. \( 2E_1 \)
• B. \( \frac{E_1}{2} \)
• C. \( \frac{3}{2} E_1 \)
• D. \( \sqrt{2} E_1 \)

Hint:

The mean kinetic energy of an ideal gas is directly proportional to its absolute temperature. So, doubling the temperature doubles the kinetic energy.

Answer 1

The Correct Option is A

The mean kinetic energy of the atoms of an ideal gas is directly proportional to the absolute temperature, according to the kinetic theory of gases: \[ E \propto T \] where \( E \) is the mean kinetic energy and \( T \) is the absolute temperature in Kelvin. The temperature in Kelvin can be calculated as: \[ T_1 = 27^\circ C + 273 = 300 \, \text{K}, \quad T_2 = 327^\circ C + 273 = 600 \, \text{K} \] Now, the ratio of the mean kinetic energies at temperatures \( T_1 \) and \( T_2 \) is: \[ \frac{E_2}{E_1} = \frac{T_2}{T_1} = \frac{600}{300} = 2 \] Therefore, the new mean kinetic energy \( E_2 \) is: \[ E_2 = 2E_1 \] Thus, the correct option is: \[ 2E_1 \]