## Equipartition Theorem

### Task number: 3945

Using the equipartition theorem determine the specific heat capacity at constant volume for argon and nitrogen. Compare the resulting values with table values.

#### Hint – Equipartition Theorem

The equipartition theorem is a statement from statistical physics and states that in equilibrium for every generalized momentum or generalized coordinate acting in the formula for the energy of the system quadraticaly there is a mean energy of

\[\frac{1}{2}kT,\]where

*k*is the Boltzmann constant and*T*thermodynamic temperature of the system.How is this fact related to a specific heat capacity (or molar heat capacity) and degrees of freedom of gas?

#### Hint 2

Argon is a one-atom gas and nitrogen is a two-atom gas. What is their number of degrees of freedom?

#### Given Values

*c*_{Vi}= ?specific heat capacity of gases at constant volume

*Table values:**M*_{mN}= 28 g mol^{−1}= 0.028 kg mol^{−1}molar mass of nitrogen N _{2}(two-atom molecule)*M*_{mAr}= 36 g mol^{−1}= 0.039 kg mol^{−1}molar mass of argon Ar (one-atom molecule) *R*= 8.31 Jmol^{−1}K^{−1}molar gas constant #### Analysis

**The equipartition theorem**is a statement from statistical physics and states that in equilibrium, every generalized momentum or generalized coordinate acting in the formula for the energy of the system quadratically corresponds to mean energy of 1/2*k**T*, where*k*is the Boltzmann constant and*T*is thermodynamic temperature of the system.This corresponds to the contribution of 1/2

*R*to molar heat capacity*C*_{V}at constant pressure.This contribution belongs to each translational and rotational degree of freedom. It is different with vibrational degrees of freedom. The contribution is twice as big, i.e.

*R*. This is due to the fact that the energy of vibrational motion is composed of kinetic and potential energy, and not only kinetic energy as it is with translational and rotational motion.The task is to determine the degrees of freedom for one-atom molecules of argon and two-atom molecules of nitrogen.

#### Solution

**The equipartition theorem**is a statement from statistical physics according to which to each generalized momentum and each generalized coordinate acting in the formula for energy of the system quadratically corresponds to mean energy of \(\frac{1}{2}kT,\) where*k*is the Boltzmann constant and*T*is thermodynamic temperature.In other words each generalized momentum and generalized coordination acting in the formula for the energy of the system quadratically corresponds to a contribution of \(\frac{1}{2}R\) to molar heat capacity

*C*_{V}.This means that the state contribution corresponds to each translational and rotational degree of freedom. It is different with vibrational degrees of freedom. The contribution is twice as big, i.e.

*R*. This is due to the fact that the energy of vibrational motion constitutes of kinetic and potential energy, and not only kinetic energy as it is with translational and rotational motion.Now we can easily determine heat capacities of given gases.

\[C_{VAr}=3\cdot\frac{1}{2}R=\frac{3}{2}R.\]**Argon**is a one-atom gas which means that its molecules have only three translational degrees of freedom. Molar heat capacity*C*_{V}at constant volume is thereforeRelated specific heat capacity

\[c_{VAr}=\frac{C_{VAr}}{M_{mAr}}=\frac{3R}{2M_{mAr}}.\]*c*_{V}is determined from molar mass*M*_{mAr}as follows:

\[C_{VN}=3\cdot\frac{1}{2}R+2\cdot\frac{1}{2}R+R=\frac{7}{2}R.\]**Nitrogen**is a two-atom gas. It has six degrees of freedom in total. Three of them are translational, two are rotational and one is vibrational. According to the equipartition theorem the molar heat capacity*C*_{VN}at constant volume is given byFor a corresponding specific heat capacity

\[c_{VN}=\frac{C_{VN}}{M_{mN}}=\frac{7R}{2M_{mN}},\]*c*_{VN}we then obtain:where

*M*_{mN}is molar mass of nitrogen.#### Numerical Solution

\[c_{VAr}=\frac{3R}{2M_{mAr}}=\frac{3\cdot{8.31}}{2\cdot{0.039}}\,\mathrm{Jkg}^{-1}\mathrm{K}^{-1}\] \[c_{VAr}=319.6\,\mathrm{Jkg}^{-1}\mathrm{K}^{-1}=0.32\,\mathrm{kJkg}^{-1}\mathrm{K}^{-1}\]#### Answer and Result Discussion

Calculated value Table value for *t*= 20 °C*c*_{VAr}0.32 kJ kg ^{−1}K^{−1}0.32 kJ kg ^{−1}K^{−1}*c*_{VN}1.04 kJ kg ^{−1}K^{−1}0.74 kJ kg ^{−1}K^{−1}By comparing calculated results with table values we can see that while the values for argon are the same, there is a big difference in the values for nitrogen. This is due to the fact that the equipartition theorem is derived within classical physics and thus does not respect the specifics of quantum mechanics that are manifested quite profoundly when dedscribing the vibrational motion at normal temperatures. In reality, the vibrations practically do not contribute to heat capacity under normal conditions. Therefore, to determine molar heat capacity

\[C_{VN}=3\cdot\frac{1}{2}R+2\cdot\frac{1}{2}R=\frac{5}{2}R\]*C*_{VN}of nitrogen, we should use a more precise formulathat involves only the translation and rotation.

If we calculate corresponding specific heat capacity

\[c_{VN}=\frac{C_{VN}}{M_{mN}}=\frac{5R}{2M_{mN}},\]*c*_{VN}we obtain a numerical value of

\[c_{VN}\dot{=}742\,\mathrm{Jkg}^{-1}\mathrm{K}^{-1}\dot{=}0.74\,\mathrm{kJkg}^{-1}\mathrm{K}^{-1}.\]This result is in a good agreement with the table value.