Question

The average kinetic energy of a molecule in air at room temperature of 20\(^\circ\)C

• A. \( 6 \times 10^{-22} \) J
• B. \( 7.06 \times 10^{-21} \) J
• C. \( 6.07 \times 10^{-21} \) J
• D. \( 6.70 \times 10^{-21} \) J

Hint:

To find the average kinetic energy of molecules, remember to use the Boltzmann constant and convert the temperature to Kelvin.

Answer 1

The Correct Option is C

The average kinetic energy of a molecule is given by the equation: \[ K.E. = \frac{3}{2} k_B T \] where: - \( k_B \) is the Boltzmann constant, \( 1.38 \times 10^{-23} \text{ J/K} \), - \( T \) is the temperature in Kelvin. The temperature in Kelvin is: \[ T = 20 + 273 = 293 \text{ K} \] Substituting the values: \[ K.E. = \frac{3}{2} \times (1.38 \times 10^{-23}) \times 293 = 6.07 \times 10^{-21} \text{ J} \] Thus, the correct answer is (C).