Prove Cp - Cv = R for ideal gas

Question

Prove that for an ideal gas, $C_p - C_v = R$ , where $C_p$ is the molar heat capacity at constant pressure, $C_v$ is the molar heat capacity at constant volume, and $R$ is the universal gas constant.

Solution — Step by Step

The first law: $\Delta Q = \Delta U + P\Delta V$ (heat added = increase in internal energy + work done by gas).

For 1 mole of ideal gas:

At constant volume: $Q_v = C_v \Delta T$ (all heat goes to increasing internal energy, since $\Delta V = 0$ means no work done)
At constant pressure: $Q_p = C_p \Delta T$

At constant volume, $\Delta U = Q_v = C_v \Delta T$ (since $\Delta W = P\Delta V = 0$ ).

Since internal energy of an ideal gas depends only on temperature (not volume or pressure), this same $\Delta U = C_v \Delta T$ applies to any process for an ideal gas — including the constant pressure process.

At constant pressure:

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

So: $C_p \Delta T - C_v \Delta T = P\Delta V$

(C_p - C_v)\Delta T = P\Delta V

For 1 mole of ideal gas: $PV = RT$ .

At constant pressure: $P\Delta V = R\Delta T$ .

Substituting:

(C_p - C_v)\Delta T = R\Delta T

Dividing both sides by $\Delta T$ (assuming $\Delta T \neq 0$ ):

\boxed{C_p - C_v = R}

Why This Works

The difference between $C_p$ and $C_v$ arises because at constant pressure, the gas must expand as it’s heated. This expansion requires the gas to do work against the external pressure ( $W = P\Delta V$ ). At constant volume, no such work is needed — all the heat goes into increasing the internal energy.

The work done per mole per degree of temperature rise at constant pressure is exactly $R$ (from the ideal gas law). So $C_p$ is always $R$ greater than $C_v$ — it takes more heat to raise the temperature by 1 K at constant pressure than at constant volume.

Alternative Method

Using enthalpy: At constant pressure, $Q_p = \Delta H$ (heat at constant pressure equals enthalpy change). So $C_p = (\partial H/\partial T)_p$ and $C_v = (\partial U/\partial T)_v$ . Since $H = U + PV = U + RT$ (for 1 mole ideal gas), $C_p = C_v + R$ .

Typical values: for a monoatomic ideal gas (He, Ne, Ar), $C_v = \frac{3}{2}R$ and $C_p = \frac{5}{2}R$ . For diatomic gases (N₂, O₂, H₂) at room temperature, $C_v = \frac{5}{2}R$ and $C_p = \frac{7}{2}R$ . The ratio $\gamma = C_p/C_v$ is 5/3 for monoatomic and 7/5 for diatomic. These are frequently asked in JEE.

Common Mistake

Students sometimes write $\Delta U = C_p \Delta T$ for the constant pressure process (using $C_p$ instead of $C_v$ ). This is wrong. The internal energy change is ALWAYS $\Delta U = nC_v\Delta T$ for an ideal gas — regardless of whether the process is at constant pressure, constant volume, or neither. $C_v$ characterises how internal energy changes with temperature for an ideal gas; $C_p$ only appears in expressions for heat added at constant pressure.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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