Relation Between Cp and Cv: MDCAT Physics notes

Why $C_p$ is greater than $C_v$

Heat one mole of an ideal gas through $\Delta T$ in two ways.

• Constant volume: no work is done, so $Q_v = C_v\Delta T = \Delta U$.
• Constant pressure: the gas expands and does work $P\Delta V$, so $Q_p = C_p\Delta T = \Delta U + P\Delta V$.

The rise in internal energy $\Delta U$ is the same in both cases because it depends only on the temperature change. So the first law at constant pressure for one mole reads:

$$C_p\Delta T = C_v\Delta T + P\Delta V$$

Derivation of $C_p - C_v = R$

For one mole of ideal gas $PV = RT$. At constant pressure, $P\Delta V = R\Delta T$. Substituting:

$$C_p\Delta T = C_v\Delta T + R\Delta T \quad\Rightarrow\quad C_p - C_v = R$$

$R = 8.314$ J mol$^{-1}$ K$^{-1}$. This is per mole. For $n$ moles the total heat capacities differ by $nR$; for example, for 4 moles $C_p - C_v = 4R$ (in J K$^{-1}$).

Internal energy in terms of $C_p$

For any process of an ideal gas, $\Delta U = nC_v\Delta T$. If only $C_p$ is given, use $C_v = C_p - R$:

$$\Delta U = n(C_p - R)\Delta T$$

Values for ideal gases

$\gamma$ is a ratio of molar quantities, so it does not depend on the number of moles. Two moles or ten moles of helium both have $\gamma \approx 1.67$.

Identifying a gas

Given $C_p$ and $C_v$, find $\gamma$. A ratio near 1.67 means monatomic; near 1.4 means diatomic. Example: $C_p = 29.1$, $C_v = 20.8$ J mol$^{-1}$ K$^{-1}$ gives $\gamma = 1.4$ and $C_p - C_v = 8.3 \approx R$, so the gas is diatomic.

Finding one from the other

• $C_p = C_v + R$ and $C_v = C_p - R$.
• $C_v = C_p/\gamma$. Example: $C_p = 21$ J mol$^{-1}$ K$^{-1}$ and $\gamma = 1.5$ gives $C_v = 14$ J mol$^{-1}$ K$^{-1}$.

Solids and liquids

A solid (for example aluminium or copper) expands very little on heating, so the work $P\Delta V$ is negligible. Hence for solids and liquids $C_p \approx C_v$. For gases the difference $R$ is always significant.

Key formulas

• $C_p - C_v = R$ (per mole)
• $\gamma = C_p/C_v$
• $\Delta U = nC_v\Delta T$, $Q_p = nC_p\Delta T$
• Monatomic: $C_v = \tfrac32R$, $C_p = \tfrac52R$; diatomic: $C_v = \tfrac52R$, $C_p = \tfrac72R$

Common MDCAT traps

• $\Delta U$ always uses $C_v$, even when heating is at constant pressure. Using $nC_p\Delta T$ for $\Delta U$ is wrong.
• For $n$ moles the difference is $nR$, not $R$; but $\gamma$ is independent of $n$.
• If $C_v = \tfrac52R$ then $C_p = \tfrac72R$, not $\tfrac52R$; add $R$.
• $C_p = C_v$ applies to solids, not to noble gases like He, Ne or Ar.
• First law at constant pressure is $C_p\Delta T = C_v\Delta T + P\Delta V$, with $P\Delta V$, not $V\Delta P$.

Quick revision

• $C_p > C_v$ because at constant pressure the gas also does work.
• $C_p - C_v = R = 8.314$ J mol$^{-1}$ K$^{-1}$.
• $\gamma$ = 1.67 (monatomic), 1.4 (diatomic).
• For solids $C_p \approx C_v$.
• Internal energy change of an ideal gas: $\Delta U = nC_v\Delta T$.