Cp and Cv of Gases: Mayer's Relation and Why Cp − Cv = R
Class 11 thermodynamics asks you again and again for the heat needed to warm a gas, for the ratio γ = Cp/Cv, or for a quick proof that Cp − Cv = R. All of these depend on one idea: a gas needs more heat to warm up at constant pressure than at constant volume. This article explains why, derives the relation step by step, and gives worked numerical examples.
What are Cv and Cp?
Molar heat capacity is the heat needed to raise the temperature of 1 mole of a substance by 1 K. For a gas the answer depends on how the heating is done, so there are two important values:
- Cv — molar heat capacity at constant volume (rigid container, no expansion).
- Cp — molar heat capacity at constant pressure (gas is free to expand, as in a piston-cylinder or an open vessel).
Their definitions in terms of U (internal energy) and H (enthalpy) are:
Heat at constant V: qv = n Cv ΔT = ΔU | Heat at constant P: qp = n Cp ΔT = ΔH
Here n is the number of moles, ΔT is the temperature change in K (or °C, since the size of the degree is the same), and Cv, Cp are in J mol⁻¹ K⁻¹.
Mayer's relation: Cp − Cv = R
For n moles (total heat capacities): Cp − Cv = nR
Derivation in four lines
- By definition, H = U + PV.
- For 1 mole of ideal gas, PV = RT, so H = U + RT.
- Differentiate with respect to T: dH/dT = dU/dT + R.
- For an ideal gas, U and H depend only on temperature, so dH/dT = Cp and dU/dT = Cv. Hence Cp = Cv + R.
The physical reason (no calculus needed)
At constant volume, all the heat goes into raising the internal energy of the gas, that is, making the molecules move faster. At constant pressure, the gas also expands and pushes the surroundings back. This expansion work costs energy, and that energy must be supplied as extra heat. For 1 mole of ideal gas heated by 1 K at constant pressure, the work is PΔV = RΔT = R × 1 = R. So the extra heat per mole per kelvin is exactly R.
Sign convention note: some books write ΔU = q + w (w is work done on the system), others write ΔU = q − w (w is work done by the system). The final result is the same either way: qp = ΔU + PΔV, where PΔV is the expansion work done by the gas.
Values of R and the matching units
| Unit system | R | So Cp − Cv = |
|---|---|---|
| SI | 8.314 J mol⁻¹ K⁻¹ | 8.314 J mol⁻¹ K⁻¹ |
| Litre-atmosphere | 0.0821 L atm mol⁻¹ K⁻¹ | 0.0821 L atm mol⁻¹ K⁻¹ |
| Calorie | 1.987 cal mol⁻¹ K⁻¹ (≈ 2 cal) | ≈ 2 cal mol⁻¹ K⁻¹ |
Cp and Cv must be in the same unit as R. Many school problems give Cv in cal and ask for Cp: add about 2 cal, not 8.314.
Cv, Cp and γ for ideal gases (rigid-molecule model)
From the equipartition idea, each translational or rotational degree of freedom contributes ½R to Cv. Vibrations are ignored at ordinary temperatures in this simple model.
| Gas type | Cv | Cp = Cv + R | γ = Cp/Cv | Example |
|---|---|---|---|---|
| Monatomic | (3/2)R = 12.47 | (5/2)R = 20.79 | 5/3 = 1.67 | He, Ne, Ar |
| Diatomic / linear | (5/2)R = 20.79 | (7/2)R = 29.10 | 7/5 = 1.40 | N₂, O₂, H₂, CO |
| Non-linear polyatomic | 3R = 24.94 | 4R = 33.26 | 4/3 = 1.33 | H₂O vapour, NH₃, CH₄ |
All Cv and Cp values are in J mol⁻¹ K⁻¹, using R = 8.314. Real measured values differ somewhat, especially for polyatomic gases, because vibrations contribute at higher temperature.
Worked example 1 — Cp from Cv
Nitrogen at room temperature has Cv ≈ 20.8 J mol⁻¹ K⁻¹. Find Cp and γ.
Cp = Cv + R = 20.8 + 8.314 = 29.114 ≈ 29.1 J mol⁻¹ K⁻¹
γ = Cp/Cv = 29.114 ÷ 20.8 = 1.3997 ≈ 1.40
Check: this matches the diatomic value 7/5 = 1.40.
Worked example 2 — Heat at constant V and at constant P
2 mol of an ideal monatomic gas is heated from 300 K to 400 K. Find the heat needed (a) at constant volume, (b) at constant pressure, and (c) the work done by the gas in case (b).
ΔT = 400 − 300 = 100 K
Cv = (3/2)(8.314) = 12.471; Cp = (5/2)(8.314) = 20.785 J mol⁻¹ K⁻¹
(a) qv = n Cv ΔT = 2 × 12.471 × 100 = 2494.2 J (= ΔU)
(b) qp = n Cp ΔT = 2 × 20.785 × 100 = 4157.0 J (= ΔH)
(c) Work by gas = qp − ΔU = 4157.0 − 2494.2 = 1662.8 J
Second route: nRΔT = 2 × 8.314 × 100 = 1662.8 J ✓. Also ΔU is the same (2494.2 J) in both cases because U of an ideal gas depends only on T.
Worked example 3 — Cv and Cp from γ alone
A gas has γ = 1.40. Find Cv and Cp.
Use Cp = γCv and Cp − Cv = R, so (γ − 1)Cv = R.
Cv = R/(γ − 1) = 8.314 ÷ 0.40 = 20.785 J mol⁻¹ K⁻¹
Cp = γ Cv = 1.40 × 20.785 = 29.099 J mol⁻¹ K⁻¹
Check: Cp − Cv = 29.099 − 20.785 = 8.314 ✓.
Worked example 4 — Degrees of freedom from γ
For 1 mole of an ideal gas, Cv = (f/2)R, where f is the number of active degrees of freedom. Then γ = Cp/Cv = (f/2 + 1)/(f/2) = 1 + 2/f.
If γ = 1.67: 2/f = 0.67, so f = 2 ÷ 0.67 = 2.99 ≈ 3, which means a monatomic gas (3 translational degrees).
If γ = 1.40: 2/f = 0.40, so f = 5, which means a diatomic gas (3 translational + 2 rotational).
Common mistakes that cost marks
- Using Cp − Cv = R for solids, liquids or real gases. The exact value R holds for ideal gases. For a real gas, Cp − Cv is larger than R; for solids and liquids it is very small.
- Mixing units. Adding 8.314 J to a Cv given in cal, or 0.0821 to a Cv given in J. Convert first.
- Forgetting n. Cp − Cv = R is per mole. For n moles the difference is nR.
- Saying Cp < Cv. Cp is always greater than Cv for a gas, so γ > 1 always.
- Using Cp for ΔU. ΔU = n Cv ΔT for an ideal gas in any process, not only at constant volume. Likewise ΔH = n Cp ΔT in any process.
- Calling the ratio "Cv/Cp". γ = Cp/Cv, the larger over the smaller.
Where this appears in exams
| Exam level | Typical use |
|---|---|
| CBSE/ICSE Class 11 | Derivation of Cp − Cv = R, heat at constant P and V, ΔU and ΔH of ideal gases |
| JEE / NEET | γ and degrees of freedom, adiabatic processes (PVᵞ = constant), work and heat in cycles |
| Class 12 / Boards | Short derivation questions and 2–3 mark numericals; check the current board sample paper for the format |
Always check the latest official syllabus and notification of your exam for the exact topics included.
Check your gas numbers. Use the free Ideal Gas Law (PV = nRT) calculator to find n, P, V or T. Then apply Cp, Cv and ΔT from this article to get q, ΔU and the work done.
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