Equipartition Theorem and $C_p$, $C_v$

Every quadratic degree of freedom in a system carries on average $\tfrac12 k_BT$ of energy. This single rule gives the molar heat capacities of all classical ideal gases.

Concept

Equipartition: $E$ per molecule $= \tfrac{f}{2}k_BT$, $f$ = active degrees of freedom (DOF).

Degrees of freedom:

• Monatomic (He, Ar): $f = 3$ (translation only).
• Diatomic at moderate T (N₂, O₂, H₂): $f = 5$ (3 translation + 2 rotation).
• Diatomic at high T (vibration active): $f = 7$.
• Linear triatomic (CO₂): $f = 5$ (rot) + vibration if active.
• Non-linear triatomic (H₂O): $f = 6$.

Molar internal energy: $U = \tfrac{f}{2}RT$ per mole.

Heat capacities: $C_v = \tfrac{f}{2}R,\quad C_p = C_v + R = \tfrac{f+2}{2}R$

Ratio of specific heats: $\gamma = \frac{C_p}{C_v} = 1 + \frac{2}{f}$

So $\gamma_{mono} = 5/3$, $\gamma_{di} = 7/5$, $\gamma_{tri,nonlinear} = 4/3$.

Derivation

$C_p - C_v = R$ (Mayer's relation): when 1 mole is heated by $\Delta T$:

• At constant V: $Q = n C_v \Delta T = \Delta U$.
• At constant P: $Q = n C_p \Delta T = \Delta U + P\Delta V$. With $P\Delta V = nR\Delta T$: $nC_p\Delta T = nC_v\Delta T + nR\Delta T \Rightarrow C_p - C_v = R$

JEE Worked Example

Problem: A mixture of 1 mole He and 2 moles N₂ — find effective $\gamma$.

Solution: $C_v$ mixture (per mole): $C_v = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \frac{1\cdot(3R/2)+2\cdot(5R/2)}{3} = \frac{13R/2}{3} = \frac{13R}{6}$ $C_p = C_v + R = \frac{19R}{6}$ $\gamma = \frac{19}{13} \approx 1.46$

Traps

• Each quadratic DOF gives $\tfrac12 k_BT$ — vibration counts as 2 (KE + PE) per mode.
• Rotational DOF for linear molecules = 2 (no rotation about molecular axis with negligible moment of inertia).
• $C_p - C_v = R$ is the molar relation, not specific heat per unit mass.
• $\gamma$ for mixtures is not the average of $\gamma$'s; compute $C_v$ first.
• At very low T, rotational DOFs freeze out; quantum effects matter.

Key Takeaways

• $C_v = fR/2$, $C_p = (f+2)R/2$, $\gamma = 1 + 2/f$.
• Mayer's relation: $C_p - C_v = R$.
• Mix gases by adding $n_i C_{v,i}$, then divide.
• $\gamma$ determines adiabatic behavior: $PV^\gamma =$ const.