Kinetic Theory • Topic 3 of 3

Degrees of Freedom & Specific Heat

A molecule does not only fly in a straight line — depending on its structure it can also rotate and vibrate. The number of independent ways in which a molecule can store energy is called its number of degrees of freedom ($f$). This idea links the microscopic motion of molecules to the heat capacities we measure in the lab.

  • A monatomic molecule (e.g. helium, argon) is essentially a point and can only move along three axes: it has $f=3$ (all translational).
  • A diatomic molecule (e.g. $\text{O}_2$, $\text{N}_2$) has 3 translational plus 2 rotational degrees of freedom: $f=5$ at ordinary temperatures.
  • A polyatomic (non-linear) molecule has 3 translational plus 3 rotational degrees: $f=6$ (vibrational modes add more at high temperatures).

Law of equipartition of energy. In thermal equilibrium, the total energy of a molecule is shared equally among all its degrees of freedom, and each degree of freedom carries an average energy of $\frac{1}{2}k_BT$ per molecule (or $\frac{1}{2}RT$ per mole). So a molecule with $f$ degrees of freedom has average energy $\frac{f}{2}k_BT$.

Internal energy and specific heats. For one mole of an ideal gas the internal energy is $U=\frac{f}{2}RT$. The molar specific heat at constant volume is $C_v=\frac{dU}{dT}=\frac{f}{2}R$, and at constant pressure $C_p=C_v+R$ (Mayer's relation). Their ratio is the important adiabatic constant $\gamma=\frac{C_p}{C_v}=1+\frac{2}{f}$. Putting in the values of $f$:

  • Monatomic ($f=3$): $C_v=\frac{3}{2}R$, $C_p=\frac{5}{2}R$, $\gamma=\frac{5}{3}\approx1.67$.
  • Diatomic ($f=5$): $C_v=\frac{5}{2}R$, $C_p=\frac{7}{2}R$, $\gamma=\frac{7}{5}=1.40$.
  • Polyatomic ($f=6$): $C_v=3R$, $C_p=4R$, $\gamma=\frac{4}{3}\approx1.33$.

The smaller value of $\gamma$ for more complex molecules simply reflects that they have more ways to soak up energy.

Mean free path. Even at high speeds, a molecule does not travel far before bumping into another. The mean free path ($\lambda$) is the average distance a molecule travels between two successive collisions: $\lambda=\frac{1}{\sqrt{2}\,\pi d^2 n}$, where $d$ is the molecular diameter and $n=\frac{N}{V}$ is the number of molecules per unit volume. The mean free path is larger at low pressure (fewer molecules to hit) and shrinks as the gas is compressed. For air at ordinary conditions $\lambda$ is only about $10^{-7}\ \text{m}$, yet that is hundreds of times the molecular size.

Adiabatic constant gamma for monatomic, diatomic and polyatomic gasesgammamono1.67di1.40poly1.33gamma = 1 + 2/f
Solved Examples
1
Worked Example
Find the molar specific heats $C_v$ and $C_p$ and the value of $\gamma$ for a monatomic gas. ($R=8.314\ \text{J}\,\text{mol}^{-1}\,\text{K}^{-1}$.)
Solution
  1. Step 1: A monatomic gas has $f=3$, so $C_v=\frac{f}{2}R=\frac{3}{2}R$.
  2. Step 2: $C_p=C_v+R=\frac{3}{2}R+R=\frac{5}{2}R$; numerically $C_v=12.5$, $C_p=20.8\ \text{J}\,\text{mol}^{-1}\,\text{K}^{-1}$.
  3. Step 3: $\gamma=\frac{C_p}{C_v}=\frac{5/2}{3/2}=\frac{5}{3}\approx1.67$.

Answer: $C_v=\frac{3}{2}R$, $C_p=\frac{5}{2}R$, $\gamma=\frac{5}{3}\approx1.67$.

2
Worked Example
Calculate the internal energy of 2 moles of a diatomic gas at 300 K. ($R=8.314\ \text{J}\,\text{mol}^{-1}\,\text{K}^{-1}$.)
Solution
  1. Step 1: For a diatomic gas $f=5$, so $U=n\times\frac{f}{2}RT=n\times\frac{5}{2}RT$.
  2. Step 2: Substitute: $U=2\times\frac{5}{2}\times8.314\times300$.
  3. Step 3: Compute: $U=5\times8.314\times300=1.25\times10^{4}\ \text{J}$.

Answer: $U\approx1.25\times10^{4}\ \text{J}$.

3
Worked Example
The value of $\gamma$ for a gas is 1.4. How many degrees of freedom does each molecule have?
Solution
  1. Step 1: Use $\gamma=1+\frac{2}{f}$, so $\frac{2}{f}=\gamma-1$.
  2. Step 2: Substitute: $\frac{2}{f}=1.4-1=0.4$, so $f=\frac{2}{0.4}$.
  3. Step 3: Compute: $f=5$ (a diatomic gas).

Answer: $f=5$ degrees of freedom (diatomic).

4
Worked Example
Using the law of equipartition, find the average energy of a diatomic molecule at temperature $T$.
Solution
  1. Step 1: Each degree of freedom carries $\frac{1}{2}k_BT$ of energy.
  2. Step 2: A diatomic molecule has $f=5$ (3 translational + 2 rotational).
  3. Step 3: Total average energy $=\frac{f}{2}k_BT=\frac{5}{2}k_BT$.

Answer: Average energy $=\frac{5}{2}k_BT$.

5
Worked Example
Calculate the mean free path of a gas molecule given molecular diameter $d=3\times10^{-10}\ \text{m}$ and number density $n=2.7\times10^{25}\ \text{m}^{-3}$.
Solution
  1. Step 1: Use $\lambda=\frac{1}{\sqrt{2}\,\pi d^2 n}$.
  2. Step 2: Compute $\pi d^2=3.14\times(3\times10^{-10})^2=2.83\times10^{-19}$; then $\sqrt{2}\,\pi d^2 n=1.414\times2.83\times10^{-19}\times2.7\times10^{25}$.
  3. Step 3: Denominator $=1.08\times10^{7}$, so $\lambda=\frac{1}{1.08\times10^{7}}\approx9.3\times10^{-8}\ \text{m}$.

Answer: $\lambda\approx9.3\times10^{-8}\ \text{m}$.

6
Worked Example
For a gas, $C_p-C_v=R$ and $\frac{C_p}{C_v}=1.67$. Find $C_v$ in terms of $R$.
Solution
  1. Step 1: From $\gamma=\frac{C_p}{C_v}=1.67$, write $C_p=1.67\,C_v$.
  2. Step 2: Substitute into $C_p-C_v=R$: $1.67\,C_v-C_v=0.67\,C_v=R$.
  3. Step 3: So $C_v=\frac{R}{0.67}=\frac{3}{2}R$ (a monatomic gas).

Answer: $C_v=\frac{3}{2}R$.

Key Points

  • Degrees of freedom $f$: monatomic $=3$, diatomic $=5$, polyatomic (non-linear) $=6$ at ordinary temperatures.
  • Law of equipartition: each degree of freedom has average energy $\frac{1}{2}k_BT$ per molecule ($\frac{1}{2}RT$ per mole).
  • Internal energy of one mole: $U=\frac{f}{2}RT$; molar specific heats $C_v=\frac{f}{2}R$ and $C_p=C_v+R$ (Mayer's relation).
  • Adiabatic constant $\gamma=\frac{C_p}{C_v}=1+\frac{2}{f}$: $\frac{5}{3}$ (mono), $\frac{7}{5}$ (di), $\frac{4}{3}$ (poly).
  • Mean free path: $\lambda=\frac{1}{\sqrt{2}\,\pi d^2 n}$; larger at low pressure (smaller $n$).
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The number of degrees of freedom of a monatomic gas molecule is:
Explanation: A monatomic molecule has only 3 translational degrees of freedom.
Q2.According to the law of equipartition, the average energy per degree of freedom per molecule is:
Explanation: Each degree of freedom carries $\frac{1}{2}k_BT$ of average energy.
Q3.The value of $\gamma=\frac{C_p}{C_v}$ for a diatomic gas is:
Explanation: With $f=5$, $\gamma=1+\frac{2}{5}=\frac{7}{5}=1.40$.
Q4.The mean free path of a gas molecule:
Explanation: $\lambda=\frac{1}{\sqrt{2}\,\pi d^2 n}$; higher pressure means larger $n$, so $\lambda$ decreases.
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