Reviewed CSCA Physics question · Easy

In a container of ideal gas, the average translational kinetic energy of a single molecule is 6.0 × 10^-21 J. If the absolute temperature of the gas is increased by a factor of 4, what is the new average translational kinetic energy of a single molecule?

1.5 × 10^-21 J
2.4 × 10^-20 J
6.0 × 10^-21 J
1.2 × 10^-20 J

Show the answer and explanation

Correct answer

B. 2.4 × 10^-20 J

Principle or equation

The average translational kinetic energy of an ideal gas molecule is directly proportional to the absolute temperature: KE_avg = (3/2)kT.

Why this answer is correct

Since KE_avg ∝ T, if T is multiplied by 4, KE_avg is also multiplied by 4. New KE_avg = 4 × 6.0 × 10^-21 J = 2.4 × 10^-20 J.

Example

If KE_avg = 5.0 × 10^-21 J at 300 K, then at 900 K (factor of 3) it becomes 1.5 × 10^-20 J.

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