Question:
What will be the average value of energy along one degree of freedom for an ideal gas in thermal equilibrium at a temperature T ? (kB is Boltzmann constant)
What will be the average value of energy along one degree of freedom for an ideal gas in thermal equilibrium at a temperature T ? (kB is Boltzmann constant)
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Total energy is $\frac{f}{2} k_B T$. For "one" degree of freedom, $f=1$, so the energy is $\frac{1}{2} k_B T$.
Updated On: Jan 21, 2026
- kBT
- (1/2) kBT
- (3/2) kBT
- (2/3) kBT
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The Correct Option is B
Solution and Explanation
Step 1: According to the **Law of Equipartition of Energy**, the total internal energy of a system is shared equally among all its degrees of freedom.
Step 2: The energy associated with each independent degree of freedom for each molecule is exactly $\frac{1}{2} k_B T$.
Step 2: The energy associated with each independent degree of freedom for each molecule is exactly $\frac{1}{2} k_B T$.
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