Heat Capacity at Constant Volume

Fill a rigid steel tank with gas and heat it slowly. No piston moves, no expansion work happens, yet the temperature climbs. Every joule you supply goes into raising the internal energy of the molecules. The proportionality constant between that heat and the temperature rise is the heat capacity at constant volume, $C_V$ . This quantity matters because it is the direct experimental window onto internal energy, $U$ , one of the two central state functions of the First Law. Unlike $C_p$ , which mixes in expansion work, $C_V$ isolates the pure thermal response of a system at fixed volume. It is the bridge between molecular structure and measurable thermodynamics. This article defines $C_V$ rigorously. It shows how to compute $\Delta U$ from $C_V$ over a temperature range. It works two full numerical examples. It closes by connecting $C_V$ to the equipartition theorem and previewing its role in $C_p - C_V$ and adiabatic processes.

Key takeaways

$C_V = (\partial U/\partial T)_V$ measures how internal energy rises with temperature at fixed volume.
At constant volume, $q_V = \Delta U$ , so $C_V$ is measured calorimetrically with a bomb calorimeter.
For an ideal gas, $U$ depends only on $T$ , so $\Delta U = \int C_V \, dT$ works for any process, not just constant- $V$ ones.
Monatomic ideal gases have $C_{V,m} = \tfrac{3}{2}R \approx 12.47\ \text{J K}^{-1}\text{mol}^{-1}$ ; diatomics add rotational and vibrational terms.
Real gases and solids have temperature-dependent $C_V$ , often fitted to polynomials in $T$ .
Key idea: $C_V$ converts a temperature change into an internal-energy change, and for an ideal gas that conversion is universal, independent of path or volume.

The core insight

The heat capacity at constant volume, $C_V = (\partial U/\partial T)_V$ , is the exact slope of the internal-energy-versus-temperature curve at fixed volume, so integrating it over a temperature interval gives $\Delta U$ directly — and for an ideal gas, whose $U$ has no volume dependence, that integral gives $\Delta U$ for any process between those two temperatures, not merely a constant-volume one.

The physical picture

Consider a fixed amount of gas sealed in a rigid container. Because the walls do not move, the system can exchange no expansion work with the surroundings: $dw = -p_{\text{ex}}\,dV = 0$ . The First Law, $dU = dq + dw$ , then collapses to $dU = dq_V$ . Any heat added at constant volume converts entirely into internal energy. This is the operational meaning behind the definition

C_V \equiv \left(\frac{\partial U}{\partial T}\right)_V .

Microscopically, internal energy is stored in translational, rotational, vibrational, and (at high enough $T$ ) electronic degrees of freedom. Heating the gas populates higher-energy microstates in each active mode. $C_V$ tells you how efficiently a given input of thermal energy raises the average energy per molecule — a large $C_V$ means many degrees of freedom are absorbing energy, so temperature rises slowly per joule added; a small $C_V$ means fewer active modes, so temperature rises quickly. For an ideal gas, molecules do not interact, so $U$ carries no dependence on volume or pressure — it depends only on $T$ and the number of moles $n$ . This is Joule's law, $(\partial U/\partial V)_T = 0$ for an ideal gas. Consequently $C_V$ for an ideal gas is (to a very good approximation) a function of $T$ alone, and the relation $\Delta U = \int C_V\,dT$ holds regardless of whether the volume actually stayed constant during the process. That generality is the single most useful practical fact in this topic. Real gases, liquids, and solids do have some volume-dependence in $U$ (intermolecular forces), so $C_V$ can depend weakly on $V$ as well as $T$ ; the definition above is always local and exact, even though the ideal-gas simplification is not always available.

Figure

Constant-volume heating: heat becomes internal energy, not work

Figure

U versus T for one mole of an ideal monatomic gas

The core relations

C_V = \left(\frac{\partial U}{\partial T}\right)_V

where: $C_V$ is the constant-volume heat capacity (J K⁻¹), $U$ is internal energy, $T$ is temperature.

\Delta U = \int_{T_1}^{T_2} C_V \, dT \quad \left(= C_V \,\Delta T \text{ if } C_V \text{ is constant}\right)

where: valid at constant $V$ always, and for any path if the system is an ideal gas.

q_V = \Delta U \qquad (\text{no non-expansion work, } dV = 0)

where: heat measured in a bomb calorimeter equals the internal-energy change directly.

C_{V,m}^{\text{ideal monatomic}} = \frac{3}{2}R, \qquad C_{V,m}^{\text{ideal linear diatomic (rigid rotor, T} \gg \theta_{\text{rot}}) } \approx \frac{5}{2}R

where: $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ ; the diatomic value assumes fully active translation and rotation but frozen vibration at room temperature.

Derivation AFrom the First Law to

\Delta U = \int C_V\,dT

Goal. Show that $C_V$ , defined as a partial derivative of $U$ , converts directly into a usable formula for $\Delta U$ over a finite temperature change — and identify exactly when that formula needs only $T$ -dependence.

Step 1. Start from the First Law. For a closed system undergoing an infinitesimal change,

dU = dq + dw .

If the boundary is rigid ( $dV = 0$ ) and there is no non-expansion (electrical, etc.) work, then $dw = -p_{\text{ex}}\,dV = 0$ , so

dU = dq_V .

Step 2. Treat $U$ as a state function of $T$ and $V$ . In general,

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

At constant $V$ , $dV = 0$ , so the second term vanishes and

dU = \left(\frac{\partial U}{\partial T}\right)_V dT \equiv C_V \, dT .

Comparing with Step 1 shows $dq_V = C_V\,dT$ , which is the operational definition: $C_V$ is measured by supplying heat at constant volume and recording $dT$ .

Step 3. Integrate over a finite range. For any process where $V$ is held fixed from $T_1$ to $T_2$ ,

\Delta U = \int_{T_1}^{T_2} C_V(T) \, dT .

If $C_V$ can be treated as constant over the interval (a good approximation for gases over modest temperature ranges), this simplifies to $\Delta U = C_V\,\Delta T$ .

Step 4. Remove the constant-volume restriction for an ideal gas. For an ideal gas, molecular energies depend only on speed and internal quantum states, never on the average intermolecular spacing (there are no intermolecular forces). Hence

\left(\frac{\partial U}{\partial V}\right)_T = 0 \quad \text{(Joule's law for an ideal gas)}.

Substituting into the general expression for $dU$ from Step 2, the $dV$ term vanishes identically — not just when $dV=0$ , but always, for any path. Therefore

dU = C_V(T)\, dT

holds along any trajectory an ideal gas follows, whether $V$ changes or not, whether the process is reversible or irreversible.

Step 5. Conclude.

\boxed{\Delta U = \int_{T_1}^{T_2} C_V(T)\, dT \quad \text{for an ideal gas, valid for any process between } T_1 \text{ and } T_2.}

For a general (non-ideal) system, the same integral gives $\Delta U$ correctly only along a path where $V$ is actually held constant; otherwise the $(\partial U/\partial V)_T\,dV$ term must be added back in.

Worked examples

Example 1 — Foundational: heating argon in a sealed vessel. A rigid 5.00 L vessel contains 2.00 mol of argon gas (treat as ideal, monatomic), initially at 300 K. Electrical heating supplies 3.00 kJ of energy. Find the final temperature and confirm energy bookkeeping. Step 1: Molar heat capacity at constant volume for a monatomic ideal gas.

C_{V,m} = \frac{3}{2}R = \frac{3}{2}(8.314\ \text{J K}^{-1}\text{mol}^{-1}) = 12.47\ \text{J K}^{-1}\text{mol}^{-1}.

Step 2: Total heat capacity for 2.00 mol.

C_V = n C_{V,m} = (2.00\ \text{mol})(12.47\ \text{J K}^{-1}\text{mol}^{-1}) = 24.94\ \text{J K}^{-1}.

Step 3: Since $V$ is fixed, $q_V = \Delta U$ , and $C_V$ is temperature-independent for an ideal monatomic gas, so

\Delta T = \frac{q_V}{C_V} = \frac{3000\ \text{J}}{24.94\ \text{J K}^{-1}} = 120.3\ \text{K}.

Step 4: Final temperature.

T_2 = 300\ \text{K} + 120.3\ \text{K} = 420.3\ \text{K}.

Sanity check: $\Delta U = n C_{V,m}\Delta T = (2.00)(12.47)(120.3) = 3000\ \text{J}$ , matching the heat supplied exactly, as required since $w = 0$ for a rigid container. The pressure also rose (via $pV = nRT$ ), but that never entered the calculation — a signature of the constant- $V$ path.

Example 2 — Applied: comparing constant-volume and constant-pressure heating of the same gas. One mole of an ideal diatomic gas (e.g., N₂, rotations active, vibrations frozen at room temperature) is heated from 298 K to 350 K. Compute $\Delta U$ for this temperature change, and then compare the heat required if the process occurs (a) at constant volume, versus (b) at constant pressure. Use $C_{V,m} = \tfrac{5}{2}R$ and, anticipating the Cp–Cv relation, $C_{p,m} = C_{V,m} + R = \tfrac{7}{2}R$ for an ideal gas. Step 1: Compute $\Delta U$ . Because the gas is ideal, $\Delta U = \int C_V\,dT$ regardless of path, so this single number applies to both cases.

C_{V,m} = \frac{5}{2}(8.314) = 20.785\ \text{J K}^{-1}\text{mol}^{-1}.

\Delta U = n C_{V,m}\,\Delta T = (1.00\ \text{mol})(20.785\ \text{J K}^{-1}\text{mol}^{-1})(52\ \text{K}) = 1080.8\ \text{J}.

Step 2 (a): At constant volume, no expansion work occurs, so all of this appears as heat:

q_V = \Delta U = 1080.8\ \text{J} \approx 1.08\ \text{kJ}.

Step 2 (b): At constant pressure, some heat also does expansion work, so more heat is needed for the same $\Delta T$ :

C_{p,m} = \frac{7}{2}(8.314) = 29.099\ \text{J K}^{-1}\text{mol}^{-1},

q_p = n C_{p,m}\,\Delta T = (1.00)(29.099)(52) = 1513.1\ \text{J} \approx 1.51\ \text{kJ}.

Step 3: Check consistency. The difference $q_p - q_V = 1513.1 - 1080.8 = 432.3\ \text{J}$ should equal the expansion work at constant pressure, $p\Delta V = nR\Delta T = (1.00)(8.314)(52) = 432.3\ \text{J}$ . It matches exactly — confirming that the extra heat needed at constant pressure is precisely the work of pushing back the atmosphere, while $\Delta U$ itself, computed from $C_V$ , is identical in both cases because $U$ depends only on $T$ for an ideal gas.

Limiting cases & common traps

Situation	Behavior of $C_V$	Common error
Ideal monatomic gas	$C_{V,m} = \tfrac{3}{2}R$ , constant with $T$	Forgetting rotational/vibrational contributions for polyatomics
Ideal diatomic gas, room T	$C_{V,m} \approx \tfrac{5}{2}R$ (vibration frozen)	Using $\tfrac{7}{2}R$ (that includes a vibrational mode not yet active)
Very high $T$	Vibrational modes activate, $C_V$ rises toward equipartition limit	Assuming $C_V$ is always constant
Real gas, near critical point	$C_V$ can depend noticeably on $V$ and $T$	Applying ideal-gas formula $\Delta U = \int C_V dT$ to any path uncritically
Solids at low $T$	$C_V \to 0$ as $T \to 0$ (Debye $T^3$ law)	Assuming Dulong–Petit ( $C_{V,m} = 3R$ ) holds at all temperatures

The chemist behind it

James Prescott Joule performed the classic experiments that established the connection between mechanical work, heat, and internal energy — most famously his paddle-wheel measurement of the mechanical equivalent of heat, and his gas-expansion experiments that led to the recognition that $(\partial U/\partial V)_T = 0$ for an ideal gas. That result is exactly what makes $\Delta U = \int C_V\,dT$ path-independent for ideal gases, the practical payoff exploited throughout this article.

Origins · 1 minJames Prescott Joule and how this was discovered

James Prescott Joule

1818–1889 · English physicist whose calorimetric and free-expansion experiments underpinned the First Law and the ideal-gas independence of U from V

Ask the tutor: Want to see how $C_p - C_V = nR$ is derived from this same starting point, and why that difference vanishes for incompressible solids and liquids?

Is Cv always temperature-independent

No. The value $C_{V,m} = \tfrac{3}{2}R$ for monatomic ideal gases is essentially constant because only translation is active. For diatomic and polyatomic gases, $C_V$ increases with temperature as rotational and especially vibrational modes become thermally accessible. For solids, $C_V$ drops toward zero as $T \to 0$ following the Debye $T^3$ law, and only approaches the classical Dulong–Petit value of $3R$ per mole of atoms at high temperature. Treating $C_V$ as constant is a useful approximation only over limited temperature ranges.

How is Cv actually measured in the lab

The standard method is bomb calorimetry: a sample reacts or is heated inside a sealed, rigid, insulated steel vessel submerged in a water bath. Because the vessel volume cannot change, no expansion work is possible, so all heat released or absorbed equals $\Delta U$ directly, by $q_V = \Delta U$ . Measuring the temperature rise of the calorimeter (with known heat capacity) and using $C_V = q_V/\Delta T$ for a calibration heater lets chemists extract $C_V$ or reaction internal energies precisely.

Why does Cv matter more for U than Cp does

$C_p$ measures how enthalpy $H$ responds to temperature at constant pressure, since $C_p = (\partial H/\partial T)_p$ . $C_V$ is the analogous, exact partner for internal energy: $C_V = (\partial U/\partial T)_V$ . Because $U$ is the quantity that appears directly in the First Law, $C_V$ gives the most direct thermodynamic and molecular-level handle on internal energy, and it is the quantity equipartition theory predicts most simply, since equipartition assigns energy, not enthalpy, per degree of freedom.

Does Cv depend on whether the gas is doing work

No — $C_V$ is a property of the substance's internal energy landscape, defined through a partial derivative with $V$ held fixed. It does not care what path a process follows. What changes with the path is whether $q_V$ equals $\Delta U$ : that equality requires $dV = 0$ specifically. For an ideal gas, however, $C_V(T)$ alone still lets you compute $\Delta U$ for a path where $V$ does change, because $U$ has no volume dependence to worry about.

How does Cv connect to what comes next

$C_V$ sets up two major results. First, subtracting it from $C_p$ gives the Cp–Cv relation, $C_p - C_V = nR$ for an ideal gas, explaining why heating at constant pressure always costs more heat than at constant volume. Second, $C_V$ appears directly in the equations for reversible adiabatic expansions and compressions of an ideal gas, where $q = 0$ forces all internal-energy change into work, governed by the ratio $\gamma = C_p/C_V$ .