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Reviewed CSCA Physics question · Easy

In a sample of an ideal gas, the average translational kinetic energy of a single molecule is 7.5 × 10^-21 J. If the absolute temperature of the gas is tripled, what is the new average translational kinetic energy of a single molecule?

  1. 2.5 × 10^-21 J
  2. 7.5 × 10^-21 J
  3. 2.25 × 10^-20 J
  4. 6.75 × 10^-20 J
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Correct answer

C. 2.25 × 10^-20 J

Principle or equation

The average translational kinetic energy of a molecule in an ideal gas is directly proportional to the absolute temperature: KE_avg = (3/2)kT. Therefore, if the temperature is tripled, the average kinetic energy is also tripled.

Why this answer is correct

Initial KE_avg = 7.5 × 10^-21 J. New temperature = 3 × old temperature, so new KE_avg = 3 × 7.5 × 10^-21 J = 2.25 × 10^-20 J.

Example

If the average kinetic energy is 5.0 × 10^-21 J at 300 K, at 600 K it becomes 1.0 × 10^-20 J.

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