JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Heat & Thermodynamics

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What will be the average value energy along one degree of freedom for an ideal gas in thermal equilibrium at a temperature

T

? (

k_{B}

is Boltzmann constant)

(A) $k_{B} T$
(B) $\frac{2}{3} k_{B} T$
(C) $\frac{3}{2} k_{B} T$
(D) $\frac{1}{2} K_{B} T$ ✓ Correct

Answer: (D)

Solution

In thermal equilibrium the energy of a molecule is a sum of quadratic terms, one for each degree of freedom, and averaging any one such term over the Maxwell-Boltzmann distribution gives the same value for all of them. The law of equipartition fixes that common value at

\frac{1}{2} k_{B} T

, independent of the mass of the molecule and of which motion the degree of freedom describes. The average energy along one degree of freedom is

\frac{1}{2} k_{B} T

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Single correct option

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