Question

What will be the average value of energy along one degree of freedom for an ideal gas in thermal equilibrium at a temperature $\mathrm{T}$ ? (${\mathrm{k}}_{\mathrm{B}}$ is Boltzmann constant)

• A. ${\mathrm{K}}_{\mathrm{B}}\mathrm{T}$
• B. $\frac{2}{3}{\mathrm{K}}_{\mathrm{B}}\mathrm{T}$
• C. $\frac{3}{2}{\mathrm{K}}_{\mathrm{B}}\mathrm{T}$
• D. $\frac{1}{2}{\mathrm{K}}_{\mathrm{B}}\mathrm{T}$

Answer 1

The correct option is D

$\frac{1}{2}{\mathrm{K}}_{\mathrm{B}}\mathrm{T}$

Thermal Equilibrium:

• Thermal equilibrium is a relationship between two entities or closed systems in which only energy transfers are permitted and take place across a heat-permeable barrier, and in which the transfers have continued until the states of the bodies stop to change.
• In thermal equilibrium, at temperature $\mathrm{T}$, the energy associated where each degree of freedom of 1 mole of a gas is equal to $\frac{1}{2}{\mathrm{K}}_{\mathrm{B}}\mathrm{T}$ .
• The average energy can be calculated using the given formula as: $\mathrm{U}=\mathrm{n}.\frac{1}{2}{\mathrm{K}}_{\mathrm{B}}\mathrm{T}$ (where n is the degree of freedom).
• As the degree of freedom for an ideal gas in thermal equilibrium is one, so we have $\mathrm{n}=1.$
• Substituting the value of $\mathrm{n}$ in the above equation we have: $\mathrm{U}=1.\frac{1}{2}{\mathrm{K}}_{\mathrm{B}}\mathrm{T}$ = $\frac{1}{2}{\mathrm{K}}_{\mathrm{B}}\mathrm{T}$.

Therefore, among the above-given options, option (d) $\frac{1}{2}{\mathrm{K}}_{\mathrm{B}}\mathrm{T}$ is correct.