The average translational kinetic energy of gas molecules is KE_avg = (3/2) k_B T. If temperature doubles, what happens to KE_avg?

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Multiple Choice

The average translational kinetic energy of gas molecules is KE_avg = (3/2) k_B T. If temperature doubles, what happens to KE_avg?

Explanation:
The essential idea is that the average translational kinetic energy of gas molecules scales linearly with temperature. For an ideal gas, KE_avg = (3/2) k_B T, where k_B is a constant. If the temperature doubles, T becomes 2T, so KE_avg becomes (3/2) k_B (2T) = 2 × [(3/2) k_B T]. That is, the average translational kinetic energy doubles. The energy distribution shifts to higher speeds, but the overall proportionality to T remains the same.

The essential idea is that the average translational kinetic energy of gas molecules scales linearly with temperature. For an ideal gas, KE_avg = (3/2) k_B T, where k_B is a constant. If the temperature doubles, T becomes 2T, so KE_avg becomes (3/2) k_B (2T) = 2 × [(3/2) k_B T]. That is, the average translational kinetic energy doubles. The energy distribution shifts to higher speeds, but the overall proportionality to T remains the same.

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