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Answer: Average kinetic energy per molecule \(= \dfrac{3}{2}k_BT\), so it is directly proportional to \(T\).

From the kinetic theory, the pressure of an ideal gas of \(N\) molecules, each of mass \(m\), in volume \(V\) is

\[P = \frac{1}{3}\frac{Nm}{V}\overline{v^2} \quad\Rightarrow\quad PV = \frac{2}{3}N\left(\frac{1}{2}m\overline{v^2}\right)\]

For an ideal gas \(PV = Nk_BT\), where \(k_B\) is Boltzmann’s constant. Comparing,

\[\frac{2}{3}N\left(\frac{1}{2}m\overline{v^2}\right) = Nk_BT \Rightarrow \frac{1}{2}m\overline{v^2} = \frac{3}{2}k_BT\]

Since \(k_B\) is a constant, the average (kinetic) energy per molecule \(\propto T\).

Doubtly AI · 29 Oct 2026 AI-generated · not yet verified Report
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