Heat capacity and work

Push a piston with your thumb and you feel resistance — a gas pushing back does work on its surroundings, and that work has a price tag in heat. Ask a fixed steel cylinder of the same gas to warm up by the same 50 K and it needs noticeably less heat, because it never has to pay for expansion. That single difference — whether the system is allowed to expand while it is heated — is why chemists keep two separate heat capacities on the books: $C_v$ for constant volume and $C_p$ for constant pressure. This article builds both from the first law of thermodynamics, derives the tidy relation $C_p - C_v = R$ for one mole of an ideal gas, and works through the arithmetic of heating a real sample under each condition so the difference stops being abstract and becomes a number you can check.

The core insight

Because heat absorbed at constant volume becomes pure internal energy ( $q_v = \Delta U = C_v\Delta T$ ) while heat absorbed at constant pressure must also fund expansion work ( $q_p = \Delta H = C_p\Delta T$ ), an ideal gas always has $C_p$ greater than $C_v$ , and for one mole the excess is exactly $R$ : $C_p - C_v = R$ .

The core idea

Heat capacity $C$ of a system is the heat needed to raise its temperature by 1 K:

C = \frac{q}{\Delta T}

where: $q$ is heat absorbed (J) and $\Delta T$ is the temperature rise (K). Heat capacity is an extensive property — a bigger sample needs more heat for the same rise — so chemists normalise it. Specific heat capacity $c$ is heat capacity per gram ( $\text{J g}^{-1}\text{K}^{-1}$ ), and molar heat capacity $C_m$ is heat capacity per mole ( $\text{J mol}^{-1}\text{K}^{-1}$ ); both are intensive, independent of how much sample you have. A gas can be heated in two very different ways, and each defines its own molar heat capacity.

Constant volume — $C_v$ . Lock the gas in a rigid container. It cannot expand, so no expansion work is done ( $w = 0$ ), and by the first law every joule of heat becomes internal energy:

q_v = \Delta U = n\,C_v\,\Delta T \quad\Rightarrow\quad C_v = \left(\frac{\partial U}{\partial T}\right)_V

where: $n$ is moles, $C_v$ is molar heat capacity at constant volume, and $\Delta U$ is the change in internal energy.

Constant pressure — $C_p$ . Let the gas sit under a piston held at constant external pressure. As it warms it also expands, pushing the piston out and doing work on the surroundings, so some of the heat goes into that work rather than staying as internal energy. Heat absorbed at constant pressure equals the enthalpy change:

q_p = \Delta H = n\,C_p\,\Delta T \quad\Rightarrow\quad C_p = \left(\frac{\partial H}{\partial T}\right)_P

where: $\Delta H$ is enthalpy change and $C_p$ is molar heat capacity at constant pressure. Because $C_p$ heat has to cover both the temperature rise and the expansion work, while $C_v$ heat only covers the temperature rise, $C_p$ is always the larger of the two for a gas that is free to expand. The next section pins that gap down to an exact number, $R$ , for an ideal gas. The figure below lines the two processes up side by side: the same 50 K rise, the same starting point, but one path (constant $V$ ) needs less heat because there is no piston to push, and the other (constant $P$ ) needs more because part of the heat leaves as work on the piston.

Figure

Two heating paths from 300 kelvin to 350 kelvin, one at constant volume needing less heat and one at constant pressure needing more heat because it also does expansion work

Common traps

Confusing $C_p - C_v = R$ with the ratio $\gamma = C_p/C_v$ from kinetic theory — this unit only asks for the difference relation.
Assuming $q_v = \Delta U$ or $q_p = \Delta H$ hold in general — they hold only when the process is at constant volume or constant pressure respectively, with no work other than $P$ – $V$ expansion work.
Mixing up specific heat capacity (per gram) with molar heat capacity (per mole) — always check which unit a given constant is quoted in before plugging into $q = nC\Delta T$ or $q = mc\Delta T$ .
Applying $C_p - C_v = R$ to solids, liquids, or real (non-ideal) gases — it is derived specifically for one mole of an ideal gas, where $PV = RT$ holds exactly.

Derivation ADeriving Cp minus Cv equals R for an ideal gas

The goal is to show, starting only from the definition of enthalpy and the ideal gas law, that for exactly one mole of an ideal gas the molar heat capacity at constant pressure exceeds the molar heat capacity at constant volume by precisely the gas constant $R$ , with no approximation involved anywhere in the algebra.

Step 1. Start from the definition of enthalpy, $H = U + PV$ , which holds for any system, ideal or not. For one mole of an ideal gas the equation of state is $PV = RT$ , so substituting gives $H = U + RT$ . This is the key simplification that only works for an ideal gas: the awkward $PV$ product has collapsed into the simple term $RT$ .

Step 2. Differentiate both sides with respect to temperature at constant pressure. Since $R$ is a constant, $\left(\dfrac{\partial H}{\partial T}\right)_P = \left(\dfrac{\partial U}{\partial T}\right)_P + R$ .

Step 3. Recognise the left side as $C_p$ by definition. For the term $\left(\dfrac{\partial U}{\partial T}\right)_P$ , use the fact that for an ideal gas the internal energy depends on temperature alone, not on volume or pressure (there are no intermolecular attractions to disturb by expanding). So $\left(\dfrac{\partial U}{\partial T}\right)_P$ equals the same quantity as $\left(\dfrac{\partial U}{\partial T}\right)_V$ , which is $C_v$ by definition.

Step 4. Substitute back: $C_p = C_v + R$ , which rearranges to $C_p - C_v = R$ . The physical reading matches Step 1 of the core idea: the extra $R$ is exactly the work an ideal gas does pushing back the surroundings while expanding at constant pressure, per mole per kelvin of temperature rise. Result: $C_p - C_v = R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ , valid for one mole of any ideal gas regardless of whether it is monatomic, diatomic, or polyatomic.

Worked example

Calculate the heat required to raise the temperature of 2 mol of an ideal monatomic gas from 300 K to 350 K at constant volume, given $C_v = 12.5\ \text{J mol}^{-1}\text{K}^{-1}$ . Then find the heat required for the same temperature change at constant pressure, and confirm the difference equals the expansion work.

Step 1. At constant volume, $q_v = n\,C_v\,\Delta T = (2\ \text{mol})(12.5\ \text{J mol}^{-1}\text{K}^{-1})(350 - 300\ \text{K}) = 2 \times 12.5 \times 50 = 1250\ \text{J} = 1.25\ \text{kJ}$ .

Step 2. Find $C_p$ using $C_p - C_v = R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ : $C_p = 12.5 + 8.314 = 20.814\ \text{J mol}^{-1}\text{K}^{-1}$ .

Step 3. At constant pressure, $q_p = n\,C_p\,\Delta T = 2 \times 20.814 \times 50 = 2081.4\ \text{J} \approx 2.08\ \text{kJ}$ .

Step 4. Since $\Delta U$ depends only on temperature for an ideal gas, it is the same $1250\ \text{J}$ in both processes (same $\Delta T$ ). By the first law, $q_p = \Delta U + w_{\text{by gas}}$ , so the expansion work done by the gas is $w_{\text{by gas}} = q_p - \Delta U = 2081.4 - 1250 = 831.4\ \text{J}$ .

Step 5. Check against theory: expansion work per mole per kelvin should equal $R$ , so for 2 mol over 50 K it should be $n R \Delta T = 2 \times 8.314 \times 50 = 831.4\ \text{J}$ — matches exactly.

The heat needed is 1.25 kJ at constant volume and 2.08 kJ at constant pressure, with the extra 831.4 J at constant pressure accounted for entirely by expansion work against the surroundings.

Competitive stretch

If instead 3.0 kJ of heat is absorbed at constant pressure while the gas does 500 J of expansion work, the first law gives $\Delta U = q + w = 3000 + (-500) = 2500\ \text{J}$ (using $w = -500\ \text{J}$ for work done by the system), and $\Delta H = q_p = 3000\ \text{J}$ directly, since heat absorbed at constant pressure is always the enthalpy change regardless of what $C_p$ or $\Delta T$ happen to be.

Ask the tutor: Want a few more mixed problems converting between $C_v$ and $C_p$ , or checking how $\Delta U$ , $\Delta H$ , and work fit together in a single constant-pressure process — just ask the AI tutor to set some up.

Why does Cp minus Cv equal R and not some other constant

It falls straight out of $H = U + RT$ for one mole of an ideal gas. Differentiating with temperature gives $C_p = C_v + R$ because the only extra term contributed by enthalpy over internal energy is $RT$ , and its derivative with respect to $T$ is just $R$ itself. Physically, $R$ per mole per kelvin is exactly the work an ideal gas does pushing back a constant external pressure as it expands on heating, so the algebra and the physics agree.

Is Cp minus Cv equal to R for real gases too, or only ideal gases

Only ideal gases give the clean, exact relation $C_p - C_v = R$ , because the derivation leans on $PV = RT$ and on internal energy depending on temperature alone. Real gases have intermolecular forces, so their internal energy also depends on volume, and $PV$ is not exactly $RT$ ; for real gases $C_p - C_v$ is close to $R$ but not exactly equal, and the exact expression needs extra correction terms beyond this syllabus.

If Cv is heat capacity at constant volume, why does q v equal ΔU and not just some heat quantity

Because at constant volume the gas cannot expand, so no $P$ – $V$ work is possible ( $w = 0$ ). The first law states $\Delta U = q + w$ , and with $w = 0$ this becomes $\Delta U = q_v$ exactly. It is this absence of expansion work — not anything special about heat capacity itself — that lets heat absorbed equal the internal energy change at constant volume.

Heat capacity and work

The core idea

Common traps

Worked example

Competitive stretch

Before this

Where this leads