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Reviewed CSCA Physics question · Standard

In a sample of an ideal gas, the average translational kinetic energy of a molecule is 5.0 × 10^-21 J. If the number of molecules in the sample is doubled while the temperature remains constant, what is the new average translational kinetic energy of a molecule?

  1. 2.5 × 10^-21 J
  2. 5.0 × 10^-21 J
  3. 1.0 × 10^-20 J
  4. 2.0 × 10^-20 J
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Correct answer

B. 5.0 × 10^-21 J

Principle or equation

The average translational kinetic energy of a molecule in an ideal gas depends only on the absolute temperature: KE_avg = (3/2)kT. It is independent of the number of molecules.

Why this answer is correct

Since the temperature is constant, the average translational kinetic energy per molecule remains the same, 5.0 × 10^-21 J.

Example

If the average kinetic energy is 6.0×10^-21 J at 300 K, it remains 6.0×10^-21 J at the same temperature even if the number of molecules changes.

This published item includes a stored explanation and passed the platform’s publication workflow. It is independent preparation material, not a claim of an official or recalled examination question.

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