Relation Between Cp and Cv

Heat a gas in a rigid sealed can, and every joule of heat raises its internal energy. Heat the same gas in a cylinder with a free piston, and some of that energy must first push the piston outward. This is why $C_p$ always exceeds $C_v$ for a real, expandable substance. Understanding the size of that gap connects the microscopic idea of stored kinetic and potential energy to the macroscopic, easily measured quantities $C_p$ and $C_v$ . The relation matters because $C_p$ is what calorimeters at atmospheric pressure actually measure, while $C_v$ is the quantity that connects directly to molecular energy storage. Chemists need a bridge between the two. For an ideal gas that bridge is stunningly simple: $C_p - C_v = nR$ . For real substances the bridge involves how much a material expands on heating and how strongly it resists compression. This article derives both relations from the First Law. You will see the general thermodynamic identity first. Then you will see it specialize to the ideal gas result. Finally, two worked examples show the size of the correction for a real gas and a liquid, and connections to adiabatic and Joule–Thomson processes are previewed.

Key takeaways

$C_p$ and $C_v$ differ because constant-pressure heating does expansion work; constant-volume heating does not.
The general relation is $C_p - C_v = TV\alpha^2/\kappa_T$ , built from the expansion coefficient $\alpha$ and isothermal compressibility $\kappa_T$ .
For an ideal gas this collapses exactly to $C_p - C_v = nR$ , independent of temperature, pressure, or identity of the gas.
For real gases and especially liquids and solids, $C_p - C_v$ is small but never exactly zero, and depends on $\alpha$ and $\kappa_T$ .
The derivation rests entirely on the First Law and the definition of enthalpy, $H = U + pV$ .
Key idea: the gap between $C_p$ and $C_v$ measures the energy diverted into expansion work rather than into raising temperature.

The core insight

Because enthalpy is internal energy plus a $pV$ term, heating at constant pressure must supply extra energy to do expansion work in addition to raising $U$ ; the general excess $C_p - C_v = TV\alpha^2/\kappa_T$ quantifies that extra energy for any substance, and it reduces exactly to $nR$ for an ideal gas because $(\partial U/\partial V)_T = 0$ there.

The physical picture

Consider two identical samples of gas, same $n$ , same starting $T$ and $p$ . Sample A sits in a rigid, sealed container: its volume cannot change. Sample B sits in a cylinder capped by a frictionless, weighted piston held at constant external pressure. Supply the same small quantity of heat, $dq$ , to both. In sample A, all of that heat raises the internal energy: $dq_V = C_V\,dT$ , by definition. In sample B, the gas absorbs the same tendency to heat up, but as $T$ rises the gas also expands against the piston, doing work $p\,dV$ on the surroundings. Some of the incoming heat is spent lifting the piston rather than accelerating molecules. To achieve the same temperature rise $dT$ , sample B needs more heat. Therefore $C_p > C_v$ . There are two contributions to this extra heat. First, there is the mechanical work of pushing back the atmosphere, $p\,dV$ . Second, if the gas is not ideal, expanding the gas also changes its internal energy at constant temperature — intermolecular attractions must be overcome or relieved. This second contribution is captured by the "internal pressure" $(\partial U/\partial V)_T$ . For an ideal gas this internal-pressure term vanishes identically, because ideal-gas internal energy depends on temperature alone. That is precisely why the ideal-gas result collapses to the strikingly clean $nR$ .

Figure

Two identical heating experiments: rigid container versus piston-restrained expansion

The general relation shows how the size of that gap depends on two measurable material properties: the isobaric expansion coefficient,

\alpha = \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_p

and the isothermal compressibility,

\kappa_T = -\frac{1}{V}\left(\frac{\partial V}{\partial p}\right)_T

A substance that expands a lot on heating ( $\alpha$ large) and is easy to compress ( $\kappa_T$ large) shows a bigger $C_p - C_v$ gap. Liquids and solids have small $\alpha$ and small $\kappa_T$ , so their gap is tiny, often just a few percent of $C_p$ . Gases, especially near a critical point, can show a large gap.

Figure

Cp minus Cv for water across phases at 25 degrees Celsius, schematic bar comparison

The core relations

C_p - C_v = \left[p + \left(\frac{\partial U}{\partial V}\right)_T\right]\left(\frac{\partial V}{\partial T}\right)_p

where: the bracket is the total pressure resisting expansion, mechanical plus internal, and the derivative is how volume grows with temperature at constant $p$ .

C_p - C_v = \frac{TV\alpha^2}{\kappa_T}

where: $\alpha$ is the isobaric expansion coefficient, $\kappa_T$ is the isothermal compressibility, $T$ and $V$ are temperature and volume.

C_p - C_v = nR

where: valid exactly for an ideal gas, $n$ is moles and $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ .

Derivation AFrom the First Law to Cp − Cv = nR

Goal. Starting from the First Law and the definition of enthalpy, derive the general expression for $C_p - C_v$ , then specialize it to an ideal gas.

Step 1: Write internal energy as a function of $T$ and $V$ . Treat $U = U(T,V)$ . Its total differential is

dU = \left(\frac{\partial U}{\partial T}\right)_V dT + \left(\frac{\partial U}{\partial V}\right)_T dV

The first term defines $C_V$ , since by definition $C_V = (\partial U/\partial T)_V$ .

Step 2: Insert the First Law. The First Law gives $dq = dU + p\,dV$ for a closed system doing only expansion work. Substituting Step 1,

dq = C_V\,dT + \left(\frac{\partial U}{\partial V}\right)_T dV + p\,dV

Step 3: Impose constant pressure and divide by $dT$ . At constant $p$ , divide through by $dT$ and hold $p$ fixed throughout:

\left(\frac{dq}{dT}\right)_p = C_V + \left[\left(\frac{\partial U}{\partial V}\right)_T + p\right]\left(\frac{\partial V}{\partial T}\right)_p

The left side is $C_p$ by definition, since heat supplied at constant pressure equals $C_p\,dT$ . Rearranging gives the boxed general relation quoted above:

C_p - C_v = \left[p + \left(\frac{\partial U}{\partial V}\right)_T\right]\left(\frac{\partial V}{\partial T}\right)_p

Step 4: Rewrite the bracket using a thermodynamic identity. A standard result from combining the First and Second Laws (proved via a Maxwell relation from $dU = TdS - pdV$ ) gives

\left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial p}{\partial T}\right)_V - p

Substituting this into the bracket, the mechanical pressure $p$ cancels exactly:

p + \left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial p}{\partial T}\right)_V

Step 5: Convert to expansion coefficient and compressibility. Using the cyclic rule for partial derivatives of $p$ , $V$ , $T$ ,

\left(\frac{\partial p}{\partial T}\right)_V = -\frac{(\partial V/\partial T)_p}{(\partial V/\partial p)_T} = \frac{\alpha}{\kappa_T}

Combining Steps 3 through 5,

C_p - C_v = T\left(\frac{\alpha}{\kappa_T}\right)\left(\alpha V\right) = \frac{TV\alpha^2}{\kappa_T}

This is the fully general result, valid for gas, liquid, or solid.

Step 6: Specialize to the ideal gas. For $n$ moles of ideal gas, $pV = nRT$ , so at constant $p$ ,

\left(\frac{\partial V}{\partial T}\right)_p = \frac{nR}{p}

Also, for an ideal gas the internal energy depends on $T$ alone (a direct consequence of no intermolecular forces), so

\left(\frac{\partial U}{\partial V}\right)_T = 0

Substituting both into the Step 3 result:

C_p - C_v = \left[p + 0\right]\cdot\frac{nR}{p} = nR

Boxed result:

\boxed{C_p - C_v = nR}

for an ideal gas — exact, temperature-independent, and identical for every ideal gas regardless of atomicity.

Worked examples

Example 1 — Molar heat capacities of nitrogen gas (foundational). Nitrogen, N₂, behaves closely as an ideal diatomic gas at room temperature. Equipartition predicts translational plus rotational contributions to molar internal energy:

C_{V,m} = \frac{5}{2}R

because a linear diatomic has 3 translational + 2 rotational quadratic degrees of freedom (vibration is frozen out at 298 K). Numerically,

C_{V,m} = \frac{5}{2}(8.314\ \text{J K}^{-1}\text{mol}^{-1}) = 20.79\ \text{J K}^{-1}\text{mol}^{-1}

Apply the ideal-gas relation per mole ( $n=1$ ):

C_{p,m} = C_{V,m} + R = 20.79 + 8.314 = 29.10\ \text{J K}^{-1}\text{mol}^{-1}

Check against experiment: the measured $C_{p,m}$ of N₂ near 298 K is about 29.1 J K⁻¹mol⁻¹ — excellent agreement, confirming both the equipartition estimate for $C_V$ and the exactness of $C_p - C_v = R$ for a near-ideal diatomic gas. Sanity check: the ratio $\gamma = C_{p,m}/C_{V,m} = 29.10/20.79 = 1.40$ , the textbook value for a diatomic ideal gas, and a number you will meet again when computing adiabatic expansions.

Example 2 — Real-gas correction for carbon dioxide near room temperature (applied, harder). CO₂ is noticeably non-ideal near 300 K. Suppose at $T = 300\ \text{K}$ and molar volume $V_m = 5.00 \times 10^{-4}\ \text{m}^3\,\text{mol}^{-1}$ (roughly 20 times the ideal dilute-gas molar volume at 1 bar, i.e., a moderately compressed real gas), measurements give

\alpha = 3.66\times 10^{-3}\ \text{K}^{-1}, \qquad \kappa_T = 1.11\times 10^{-3}\ \text{atm}^{-1} = 1.096\times 10^{-8}\ \text{Pa}^{-1}

(values chosen to be representative of CO₂ under these conditions; convert $\kappa_T$ to SI: $1\ \text{atm}^{-1} = 1/(1.013\times10^5\ \text{Pa}) = 9.87\times10^{-6}\ \text{Pa}^{-1}$ , so $1.11\times10^{-3}\ \text{atm}^{-1} \times 9.87\times10^{-6}\ \text{Pa}^{-1}\text{atm} = 1.096\times10^{-8}\ \text{Pa}^{-1}$ ). Apply the general formula per mole:

C_{p,m} - C_{V,m} = \frac{TV_m\alpha^2}{\kappa_T}

Compute the numerator:

TV_m\alpha^2 = (300\ \text{K})(5.00\times10^{-4}\ \text{m}^3)(3.66\times10^{-3}\ \text{K}^{-1})^2

= 300 \times 5.00\times10^{-4} \times 1.340\times10^{-5} = 2.010\times10^{-6}\ \text{m}^3\,\text{K}

Divide by $\kappa_T$ :

C_{p,m}-C_{V,m} = \frac{2.010\times10^{-6}\ \text{m}^3\,\text{K}}{1.096\times10^{-8}\ \text{Pa}^{-1}}

Units: $\text{m}^3\,\text{K}\times\text{Pa} = \text{J K}^{-1}$ (since $\text{Pa}\cdot\text{m}^3 = \text{J}$ ), giving

C_{p,m}-C_{V,m} = 1.834\times10^{2}\ \text{J K}^{-1}\text{mol}^{-1} \approx 183\ \text{J K}^{-1}\text{mol}^{-1}

This is roughly 22 times larger than the ideal-gas value of $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ . That large factor is physically sensible: near-critical or moderately compressed CO₂ has a much larger $\alpha$ and $\kappa_T$ than a dilute ideal gas, because the gas is close to conditions where small pressure or temperature changes cause large volume swings. As $V_m$ increases toward the ideal-dilute limit, both $\alpha \to 1/T$ and $\kappa_T \to 1/p$ , and the ratio smoothly reduces to $nR$ , confirming internal consistency of the general formula with the ideal limit.

Limiting cases & common traps

Situation	What happens to $C_p - C_v$
Ideal gas, any $T$ , $p$	Exactly $nR$ , always positive
Real gas, dilute limit	Approaches $nR$ as $p \to 0$
Real gas near critical point	Can be far larger than $nR$ , diverges at the critical point itself
Liquids and solids	Small but nonzero; typically a few percent of $C_p$
Hypothetical incompressible, non-expanding solid	$\alpha \to 0$ forces $C_p - C_v \to 0$

The chemist behind it

The relation traces back to the foundational work connecting heat, work, and internal energy in the 1840s. James Prescott Joule's experiments establishing the mechanical equivalent of heat gave the First Law its experimental footing, and the immediate application to gases — comparing heating at constant volume against heating at constant pressure — became one of the first quantitative triumphs of the new thermodynamics. The clean result $C_p - C_v = nR$ for ideal gases was recognized quickly afterward as a direct consequence of the equation of state, and it became a standard cross-check: measuring $C_p$ calorimetrically and predicting $C_v$ (or vice versa) without a second independent experiment.

Origins · 1 minJames Prescott Joule and how this was discovered

James Prescott Joule

1818–1889 · Established the mechanical equivalent of heat, the experimental foundation beneath the First Law used throughout this derivation

Ask the tutor: How does this same internal-pressure term reappear when deriving the Joule–Thomson coefficient for a real gas?

Why does Cp − Cv equal nR only for ideal gases, not real ones

The ideal-gas derivation relies on two facts: $pV = nRT$ gives a simple $(\partial V/\partial T)_p = nR/p$ , and $(\partial U/\partial V)_T = 0$ because there are no intermolecular forces. Real gases have attractive and repulsive forces, so compressing or expanding them at constant temperature does change internal energy. That nonzero $(\partial U/\partial V)_T$ adds an extra term, making $C_p - C_v$ deviate from $nR$ , usually larger, especially near the critical point where volume becomes extremely sensitive to pressure and temperature.

Is Cp − Cv always positive

Yes, for any stable thermodynamic system, $C_p - C_v \geq 0$ . This follows from the general formula $C_p - C_v = TV\alpha^2/\kappa_T$ : temperature and volume are always positive, $\alpha^2$ is a square so never negative, and $\kappa_T$ must be positive for mechanical stability, since a stable substance shrinks under increased pressure. The only way the gap vanishes is if $\alpha = 0$ , meaning the substance does not expand on heating at all.

How does this relation connect to adiabatic processes

Adiabatic expansions and compressions of ideal gases use the heat capacity ratio $\gamma = C_p/C_v$ in relations like $pV^\gamma = \text{constant}$ . Since $C_p - C_v = nR$ pins down the gap exactly, knowing $C_v$ alone (from equipartition or spectroscopy) lets you compute both $C_p$ and $\gamma$ without a second experiment. This is exactly the bridge used when deriving adiabat equations and the reversible adiabatic work formula.

Why does this relation matter for the Joule-Thomson effect

The Joule–Thomson coefficient measures temperature change with pressure at constant enthalpy, and its derivation also depends on $(\partial U/\partial V)_T$ and how $C_p$ differs from $C_v$ in a real gas. The same internal-pressure correction term that makes $C_p - C_v \neq nR$ for real gases is the microscopic origin of nonzero Joule–Thomson cooling or heating, linking these two topics through a shared thermodynamic derivative.

Can Cp − Cv be measured directly without knowing alpha or kappa T

Often chemists measure $C_p$ calorimetrically (easy at constant atmospheric pressure) and infer $C_v$ using either the ideal-gas shortcut $nR$ or, for precise real-substance work, tabulated $\alpha$ and $\kappa_T$ values combined with the general formula. Direct constant-volume calorimetry is experimentally awkward for gases, so the algebraic relation is frequently more practical than direct measurement of $C_v$ itself.