How can specific heat be predicted from the law of equipartition of energy?

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) According to the law of equipartition of energy,each degree of freedom contributes $\frac{1}{2} k_B T$ to the internal energy of a system.
For a gas with $f$ degrees of freedom,the internal energy $U$ is given by $U = f \cdot \frac{1}{2} n R T$.
The molar specific heat at constant volume is $C_V = \frac{dU}{dT} = \frac{f}{2} R$.
The molar specific heat at constant pressure is $C_P = C_V + R = (\frac{f}{2} + 1) R$.
This classical prediction assumes that all degrees of freedom are active at all temperatures.
However,experimental observations show that specific heat varies with temperature,approaching zero as $T \to 0 \ K$,which indicates that degrees of freedom become 'frozen' or ineffective at low temperatures.
This limitation of classical mechanics is explained by quantum mechanics,where a minimum energy threshold is required to excite a degree of freedom.

Explore More

Similar Questions

$1 \, mole$ of a gas requires $40 \, calories$ of heat to raise its temperature from $20^{\circ}C$ to $30^{\circ}C$ at constant pressure. How many calories of heat are required to raise the temperature of the same gas by the same amount at constant volume? $(R = 2 \, cal \, mol^{-1} K^{-1})$

Difficult
View Solution

If the degree of freedom of a gas is $f,$ then the ratio of two specific heats ${C_P}/{C_V}$ is given by

$Assertion :$ The ratio of $\frac{C_p}{C_v}$ for an ideal diatomic gas is less than that for an ideal monoatomic gas (where $C_p$ and $C_v$ have usual meaning).
$Reason :$ The atoms of a monoatomic gas have less degrees of freedom as compared to molecules of the diatomic gas.

Explain the difference between the expressions $C_P - C_V = R$,$C_P - C_V = \frac{R}{J}$,and $C_P - C_V = \frac{r}{J}$.

Difficult
View Solution

Specific heats of an ideal gas at constant pressure and volume are denoted by $C_p$ and $C_v$ respectively. If $\gamma = \frac{C_p}{C_v}$ and $R$ is the universal gas constant,then $C_v$ is equal to:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo