Kinetic Energy And Temperature

Question

When the absolute temperature of a fixed sample of ideal gas is raised from 300 K to 600 K, by what factor does the average translational kinetic energy of each molecule increase?

RankGuru · Accepted Answer

This question rests on the temperature dependence of molecular energy set out in NCERT Class 11, Chapter 13 (Kinetic Theory), where the mean translational kinetic energy of a molecule is $\overline{KE} = \frac{3}{2}k_BT$. Because this energy is directly proportional to the absolute temperature, the ratio of energies equals the ratio of kelvin temperatures. Here the temperature rises from 300 K to 600 K, a ratio of $\frac{600}{300} = 2$, so the average kinetic energy simply doubles. The option of a fourfold increase mistakenly assumes energy scales with the square of temperature. The option that energy stays the same ignores the heating altogether. The option involving the square root of two arises from confusing energy with molecular speed, which does scale as the square root of temperature. A plausibility check clarifies the distinction: since $\overline{KE} \propto T$ but $v_{rms} \propto \sqrt{T}$, doubling temperature doubles energy while multiplying speed only by $\sqrt{2}$, and the factor of two for energy is the physically correct answer.

Kinetic Energy And Temperature

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