Question

At T(K) root mean square (rms) velocity of argon (molar mass 40 g mol⁻¹) is 20 ms⁻¹. The average kinetic energy of the same gas at T(K) (in J mol⁻¹) is

• A. 8
• B. 16
• C. 4
• D. 2

Hint:

The formula \(KE_{avg, mole} = \frac{1}{2} M v_{rms}^2\) is a very useful shortcut that directly relates molar kinetic energy to the rms speed, bypassing the need to calculate the temperature. It is analogous to the macroscopic kinetic energy formula \(KE = \frac{1}{2}mv^2\), but applied on a molar scale. Remember to use molar mass M in kg/mol.

Answer 1

The Correct Option is A

Step 1: Relation between KE and \( v_{rms} \): We know \( KE_{\text{per mole}} = \frac{1}{2} M v_{rms}^2 \). (Derived from \( v_{rms} = \sqrt{\frac{3RT}{M}} \) and \( KE = \frac{3}{2}RT \)).
Step 2: Calculation: Molar mass \( M = 40 \text{ g/mol} = 0.04 \text{ kg/mol} \). \( v_{rms} = 20 \text{ ms}^{-1} \). \[ KE = \frac{1}{2} \times 0.04 \times (20)^2 \] \[ KE = 0.02 \times 400 = 8 \text{ J mol}^{-1} \]