Molar heat capacities

Heat a gas by one kelvin and ask how much energy it takes — the honest answer is "it depends." Hold the gas in a sealed rigid can and heat it, and every joule you supply goes straight into raising its internal energy. Let the same gas sit under a piston free to rise, and some of your heat instead does the work of pushing that piston back against the atmosphere, leaving less to actually warm the gas. So the same one-mole, one-kelvin rise costs more heat at constant pressure than at constant volume — and the two costs are named the molar heat capacities $C_v$ and $C_p$ . This section pins down what $C_v$ and $C_p$ mean physically, derives the tidy relation joining them — Mayer's relation, $C_p - C_v = R$ — straight from the first law of thermodynamics and the ideal gas equation, and works through the arithmetic examiners actually ask for. It closes out the Heat and Thermodynamics unit and sets up the next unit's job of computing $C_v$ and $C_p$ from a gas's degrees of freedom.

The core insight

For an ideal gas, $C_p$ always exceeds $C_v$ by exactly the gas constant $R$ , because constant-pressure heating must additionally supply the work of expansion that constant-volume heating never needs.

The core idea

Molar heat capacity is the heat needed to raise the temperature of one mole of a substance by one kelvin. For gases — unlike most solids and liquids — this number splits into two distinct values, because heating can happen in two very different ways.

At constant volume ( $C_v$ ). Lock the gas in a rigid container and heat it. The volume cannot change, so the gas does no work, $W = 0$ . By the first law of thermodynamics, $Q = \Delta U + W$ , every joule of heat goes directly into internal energy:

Q = \Delta U = nC_v\,\Delta T \qquad \implies \qquad C_v = \left(\frac{\Delta U}{n\,\Delta T}\right)_{V}

where: $Q$ is heat supplied (J), $n$ is the number of moles, $\Delta T$ is the temperature rise (K), $\Delta U$ is the change in internal energy (J), and $C_v$ is the molar heat capacity at constant volume ( $\text{J mol}^{-1}\text{K}^{-1}$ ).

At constant pressure ( $C_p$ ). Now let the gas sit under a free piston at fixed pressure and heat it. It expands as it warms, doing work $W = p\,\Delta V$ on the surroundings. The first law still holds, but now heat must cover both the internal-energy rise and the work of pushing the piston:

Q = nC_p\,\Delta T = \Delta U + W

Because part of the supplied heat is "spent" on expansion work rather than warming the gas, reaching the same $\Delta T$ at constant pressure always costs more heat than at constant volume — so $C_p > C_v$ for every ideal gas, without exception. The figure below makes the bookkeeping visible: at constant volume all the heat becomes internal energy; at constant pressure the same heat splits into internal energy plus expansion work.

Figure

Two heat flow diagrams comparing constant volume heating where all heat becomes internal energy and constant pressure heating where heat splits into internal energy and expansion work

The relation joining $C_p$ and $C_v$ — derived below — is:

C_p - C_v = R

where: $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ is the universal gas constant. Known as Mayer's relation, it holds for any ideal gas — monatomic, diatomic, or polyatomic — only the actual numeric values of $C_v$ and $C_p$ differ by gas type (fixed by degrees of freedom, the next unit's topic). One useful consequence: the ratio $\gamma = C_p/C_v$ , which shows up in the adiabatic relation $pV^\gamma = \text{const}$ , is always greater than 1.

Derivation AMayer's relation for an ideal gas

The goal is to show, starting only from the first law of thermodynamics and the ideal gas equation, that the constant-pressure and constant-volume molar heat capacities of any ideal gas differ by exactly the universal gas constant $R$ — a relation independent of which specific gas is being heated.

Step 1. Write the first law for one mole heated at constant pressure through a small rise $\Delta T$ . Heat supplied equals the internal energy change plus the work of expansion: $Q_p = \Delta U + p\,\Delta V$ . By definition, $Q_p = nC_p\,\Delta T$ , so $nC_p\,\Delta T = \Delta U + p\,\Delta V$ .

Step 2. For an ideal gas, internal energy depends only on temperature, so the same $\Delta U$ appears whether the heating happens at constant volume or constant pressure, for the same $\Delta T$ . At constant volume, no work is done, so $\Delta U = nC_v\,\Delta T$ directly from $Q_v = \Delta U$ . Substitute this $\Delta U$ into the constant-pressure equation: $nC_p\,\Delta T = nC_v\,\Delta T + p\,\Delta V$ .

Step 3. Evaluate $p\,\Delta V$ using the ideal gas equation $pV = nRT$ . At constant pressure, differencing gives $p\,\Delta V = nR\,\Delta T$ (the pressure stays fixed, so it multiplies straight through). Substitute: $nC_p\,\Delta T = nC_v\,\Delta T + nR\,\Delta T$ .

Step 4. Divide through by the common factor $n\,\Delta T$ (nonzero for any actual heating). Every term simplifies, leaving $C_p = C_v + R$ , or equivalently $C_p - C_v = R$ . Nothing about the gas's identity entered the derivation beyond $pV = nRT$ , so the relation holds for every ideal gas, with only the numerical values of $C_v$ and $C_p$ themselves depending on the gas.

Worked example

For a certain ideal gas, $C_v = 20.8\ \text{J mol}^{-1}\text{K}^{-1}$ . Two moles of this gas are heated at constant volume from 300 K to 350 K, requiring 2080 J of heat. Find $C_p$ , the ratio $\gamma = C_p/C_v$ , and the heat needed to produce the same temperature rise at constant pressure. (Take $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ .)

Step 1. Check $C_v$ against the given constant-volume data: $C_v = Q/(n\,\Delta T) = 2080/(2 \times 50) = 20.8\ \text{J mol}^{-1}\text{K}^{-1}$ , matching the value given — the numbers are consistent.

Step 2. Apply Mayer's relation to get $C_p$ : $C_p = C_v + R = 20.8 + 8.314 = 29.114\ \text{J mol}^{-1}\text{K}^{-1}$ .

Step 3. Find the ratio: $\gamma = C_p/C_v = 29.114/20.8 \approx 1.40$ . This value is characteristic of a diatomic gas (like $\text{N}_2$ or $\text{O}_2$ ), consistent with $C_v \approx \tfrac{5}{2}R$ and 5 degrees of freedom.

Step 4. Find the constant-pressure heat for the same $\Delta T = 50\ \text{K}$ : $Q_p = nC_p\,\Delta T = 2 \times 29.114 \times 50 = 2911.4\ \text{J}$ .

The heat needed at constant pressure is 2911.4 J, compared to 2080 J at constant volume. The extra $831.4\ \text{J}$ equals $nR\,\Delta T = 2 \times 8.314 \times 50 = 831.4\ \text{J}$ — exactly the work of expansion, confirming the bookkeeping in Mayer's relation.

Common traps

Assuming $C_p = C_v$ , as is often true for solids and liquids — for a gas these are always different, with $C_p$ always the larger.
Mixing up molar heat capacity (per mole, paired with $R$ in $\text{J mol}^{-1}\text{K}^{-1}$ ) with specific heat capacity (per unit mass) — converting between them needs the gas's molar mass.
Thinking $C_p - C_v = R$ applies only to one type of gas — it is a universal ideal-gas relation; only the numeric values of $C_v$ , $C_p$ , and $\gamma$ change between monatomic, diatomic, and polyatomic gases.
Restricting $\Delta U = nC_v\,\Delta T$ to constant-volume processes only — for an ideal gas this formula for $\Delta U$ holds for any process, since internal energy depends only on temperature; it is $Q = nC_v\,\Delta T$ that is restricted to constant volume.

Ask the tutor: want to see how $C_v$ and $C_p$ get their actual numeric values from a gas's degrees of freedom, or work through more Mayer's-relation numericals?

Why does heating at constant pressure need more heat than at constant volume, for the same temperature rise?

Because the gas is free to expand under constant pressure, and expanding against the surroundings requires work. Every joule of heat supplied splits into two jobs: raising the internal energy (which sets the temperature) and pushing the piston back to make room for the expanding gas. At constant volume there is no room to expand, so no work is done, and all the heat goes into internal energy alone. Since the same temperature rise needs the same internal-energy change either way, the constant-pressure case simply needs extra heat on top — exactly $nR\Delta T$ worth, which is the algebraic content of Mayer's relation.

Does Delta U = nC_v Delta T only work for constant-volume heating?

No — this is a common mix-up. For an ideal gas, internal energy depends only on temperature (not on volume or pressure), so $\Delta U = nC_v\Delta T$ is true for any process between the same two temperatures, whether volume, pressure, or both are changing. What is genuinely restricted to constant volume is the heat equation $Q = nC_v\Delta T$ , because only at constant volume does all the heat become internal energy with no work done. At constant pressure, $Q = nC_p\Delta T$ instead, even though $\Delta U$ is still $nC_v\Delta T$ .

How is Mayer's relation connected to gamma and the adiabatic process?

The ratio $\gamma = C_p/C_v$ appears directly in the adiabatic relation $pV^\gamma = \text{const}$ used for rapid, insulated compressions and expansions elsewhere in this unit. Since Mayer's relation guarantees $C_p > C_v$ for every ideal gas, it follows immediately that $\gamma > 1$ always. The actual size of $\gamma$ — around 1.67 for monatomic gases, about 1.40 for diatomic gases — depends on how many degrees of freedom the gas molecules have, which is worked out using equipartition of energy in the kinetic theory unit that follows this one.

Molar heat capacities

The core idea

Worked example

Common traps

Before this

Where this leads