Specific heat at constant pressure vs constant volume — why Cp > Cv

Question

Explain why the specific heat at constant pressure ( $C_p$ ) is always greater than the specific heat at constant volume ( $C_v$ ) for an ideal gas. Derive the relation $C_p - C_v = R$ .

(NCERT Class 11, Chapter 12 — asked almost every year in boards)

Solution — Step by Step

$C_v$ = heat needed to raise the temperature of 1 mole of gas by 1 K at constant volume. Since volume is fixed, no work is done. All heat goes into increasing internal energy.

$C_p$ = heat needed to raise the temperature of 1 mole of gas by 1 K at constant pressure. Here the gas expands as it heats up, so it does work on the surroundings in addition to increasing its internal energy.

At constant volume, $W = 0$ , so:

Q_v = \Delta U = nC_v\Delta T

For 1 mole with $\Delta T = 1$ K: $Q_v = C_v$ .

At constant pressure, the gas does work $W = P\Delta V = nR\Delta T$ (from ideal gas law). So:

Q_p = \Delta U + W = nC_v\Delta T + nR\Delta T = n(C_v + R)\Delta T

For 1 mole with $\Delta T = 1$ K: $Q_p = C_v + R$ .

But $Q_p = C_p$ by definition, so:

\boxed{C_p - C_v = R}

At constant pressure, the gas must do extra work ($P\Delta V$) to expand against the surroundings while also raising its temperature. This extra energy requirement means you need to supply more heat than at constant volume — hence $C_p > C_v$.

The difference is exactly $R = 8.314$ J mol $^{-1}$ K $^{-1}$ for any ideal gas, regardless of whether it’s monatomic, diatomic, or polyatomic.

Why This Works

The internal energy of an ideal gas depends only on temperature, not on pressure or volume. So $\Delta U$ for a 1 K rise is the same whether the process is at constant $P$ or constant $V$ . The only difference is the work term: at constant volume, work is zero; at constant pressure, work equals $nR\Delta T$ .

This is why $C_p - C_v = R$ is universal for ideal gases. For real gases, the relation becomes $C_p - C_v = T V \alpha^2 / \kappa_T$ , which reduces to $R$ in the ideal limit.

Alternative Method

Using enthalpy: $H = U + PV$ . For an ideal gas, $PV = nRT$ , so $H = U + nRT$ .

Differentiating: $dH = dU + nR\,dT$ .

Per mole: $C_p\,dT = C_v\,dT + R\,dT$ , giving $C_p = C_v + R$ .

The ratio $\gamma = C_p/C_v$ is crucial for adiabatic processes. For monatomic gases, $\gamma = 5/3$ . For diatomic (at room temperature), $\gamma = 7/5$ . Memorise these — they appear in sound speed, adiabatic expansion, and Poisson’s law problems.

Common Mistake

Students sometimes state ” $C_p > C_v$ because gas expands at constant pressure.” That is the right idea, but the reasoning must be precise: the gas does work $P\Delta V$ during expansion, which requires additional heat input beyond what goes into internal energy. Simply saying “gas expands” without connecting it to the extra work done is incomplete and will cost marks in board exams.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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