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: Understanding the Concept:
This problem connects two concepts from the kinetic theory of gases: the root mean square (rms) velocity and the average kinetic energy of gas molecules. The average kinetic energy of a gas depends only on its temperature, while the rms velocity depends on both temperature and molar mass.
Step 2: Key Formula or Approach:
1. Root Mean Square (rms) Velocity: \(v_{rms} = \sqrt{\frac{3RT}{M}}\), where \(R\) is the ideal gas constant, \(T\) is the absolute temperature, and \(M\) is the molar mass in kg/mol.
2. Average Kinetic Energy per mole: \(KE_{avg, mole} = \frac{3}{2}RT\).
We can see a relationship between these two formulas: from the rms velocity formula, \(v_{rms}^2 = \frac{3RT}{M}\), which means \(M v_{rms}^2 = 3RT\). And from the KE formula, \(RT = \frac{2}{3} KE_{avg, mole}\). We can also relate them directly: \(KE_{avg, mole} = \frac{1}{2} M v_{rms}^2\).
Step 3: Detailed Explanation:
We are given:
- Gas: Argon (Ar)
- Molar mass, \(M = 40 \text{ g mol}^{-1} = 0.040 \text{ kg mol}^{-1}\) (must be in SI units).
- RMS velocity, \(v_{rms} = 20 \text{ m s}^{-1}\).
- Temperature = T(K).
We need to find the average kinetic energy per mole (\(KE_{avg, mole}\)) at this same temperature T.
Using the direct relationship derived in Step 2:
\[ KE_{avg, mole} = \frac{1}{2} M v_{rms}^2 \] Substitute the given values (making sure M is in kg/mol):
\[ KE_{avg, mole} = \frac{1}{2} \times (0.040 \text{ kg mol}^{-1}) \times (20 \text{ m s}^{-1})^2 \] \[ KE_{avg, mole} = \frac{1}{2} \times 0.040 \times 400 \text{ J mol}^{-1} \] \[ KE_{avg, mole} = 0.020 \times 400 = 8 \text{ J mol}^{-1} \] Alternative Method (Finding T first):
First, find the temperature T using the rms velocity formula. (Let's use R \(\approx\) 8.314 J/mol·K, though it's not needed for the final answer).
\[ v_{rms}^2 = \frac{3RT}{M} \implies T = \frac{M v_{rms}^2}{3R} \] \[ T = \frac{(0.040)(20)^2}{3R} = \frac{0.040 \times 400}{3R} = \frac{16}{3R} \] Now, calculate the average kinetic energy per mole:
\[ KE_{avg, mole} = \frac{3}{2}RT = \frac{3}{2}R \left( \frac{16}{3R} \right) \] The \(3R\) terms cancel out.
\[ KE_{avg, mole} = \frac{16}{2} = 8 \text{ J mol}^{-1} \] Both methods give the same result.
Step 4: Final Answer:
The average kinetic energy of the gas is 8 J mol⁻¹. Therefore, option (A) is correct.