Physics

Physics

Problem

Which of the following expressions gives the average translational kinetic energy KavgK_{\text{avg}} of a single atom in an ideal monatomic gas at absolute temperature TT? a) kBTk_B T b) 52kBT\dfrac{5}{2}k_B T c) 32kBT\dfrac{3}{2}k_B T d) 32RT\dfrac{3}{2}RT
Intuitive explanation

The key insight is that temperature in a gas is fundamentally a measure of random molecular motion. For a monatomic ideal gas, all the thermal energy goes into translational kinetic energy (no rotation or vibration), so there's a direct proportional relationship between TT and KavgK_{\text{avg}}. The equipartition theorem tells us that each translational degree of freedom (motion in xx, yy, and zz directions) contributes equally to the average energy, giving you a simple factor involving Boltzmann's constant kBk_B.

Intuitive explanation

The key insight is that temperature in a gas is fundamentally a measure of random molecular motion. For a monatomic ideal gas, all the thermal energy goes into translational kinetic energy (no rotation or vibration), so there's a direct proportional relationship between TT and KavgK_{\text{avg}}. The equipartition theorem tells us that each translational degree of freedom (motion in xx, yy, and zz directions) contributes equally to the average energy, giving you a simple factor involving Boltzmann's constant kBk_B.

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