Specific Heats of Gases
Why gases have two heat capacities, and how molecular shape predicts them.
Two Heat Capacities
For a gas, 'heat per degree' depends on how you heat it.
Ask “how much heat warms this gas by one degree?” and the honest answer is “it depends how you do it”. Gases need a process attached to their heat capacity.
Heat capacity depends on the process
The specific heat capacity is the heat per unit mass per degree of temperature rise:
Take a gas from state a to state b. The temperature change is fixed by the end states, but by the first law ΔQ = ΔU + ΔW, and the work ΔW depends on the path. A path with more area under it needs more heat for the same ΔT. So for a gas, s depends on the process.
Two processes are singled out:
Per mole instead of per kilogram, the molar heat capacities:
0.56 g of nitrogen (28 g/mol) in a rigid flask is warmed from 20°C to 40°C. Cᵥ = 20.8 J K⁻¹ mol⁻¹. How much heat is needed?
A 0.50 m³ tank holds helium at 1.0 × 10⁵ Pa and 300 K. Cᵥ = 12.5 J K⁻¹ mol⁻¹. How much heat raises it to 350 K (ignore the tank’s expansion)?
- 1For a gas, the heat for a given ΔT depends on the process (through the work done).
- 2Specific heats s_V and s_p; molar heat capacities C_V and C_p.
- 3For solids and liquids the difference is small.
Check yourself
Cp − Cv = R
The extra heat at constant pressure pays for the work of expansion.
At constant pressure, part of the heat escapes as work, so Cₚ is bigger than Cᵥ. For an ideal gas the difference is exactly R.
Deriving Cₚ − Cᵥ = R
Take n moles of ideal gas at p, V, T.
Constant volume
No work is done, so all the heat becomes internal energy: (dQ)ᵥ = dU.
Constant pressure
Now let it push a light piston against a constant force pA. The same temperature rise dT needs heat (dQ)ₚ = dU′ + p dV. From pV = nRT at constant p, p dV = nR dT. For an ideal gas the internal energy depends only on temperature, so dU′ = dU. Therefore
Dividing by n dT:
Cᵥ and internal energy
Since (dQ)ᵥ = dU:
A gas has Cᵥ = 6.0 cal mol⁻¹ K⁻¹. Taking R = 2.0 cal mol⁻¹ K⁻¹, find γ = Cₚ/Cᵥ.
A gas with γ = 1.4 is heated at constant pressure with 210 J. How much goes into internal energy, and how much is work?
- 1At constant volume all heat raises U; at constant pressure some becomes work, so Cₚ > Cᵥ.
- 2Ideal gas: Cₚ − Cᵥ = R.
- 3Cᵥ = (1/n) dU/dT; U = nCᵥT.
- 4At constant pressure, ΔU/ΔQ = 1/γ.
Check yourself
Videos, simulations & further reading (3)
Measuring Cp and Cv
A long stream of gas for Cp; a delicate steam balance for Cv.
Gases are light, so they hold little heat. Measuring their heat capacities takes patience: either a long stream of gas, or a very sensitive balance.
Cₚ by Regnault’s method
Gas stored at high pressure in a large tank (kept at constant temperature in water) flows through a copper coil in a hot oil bath at θ₁, then through a second coil in a calorimeter of water. A screw valve is adjusted continuously so that a manometer after it shows constant pressure: the gas flows at constant pressure. The water warms from θ₂ to θ₃.
The gas enters the calorimeter coil at θ₁ and leaves at a temperature that rises from θ₂ to θ₃, on average (θ₂ + θ₃)/2. With W the water equivalent of calorimeter and coil, m the water and s its specific heat:
How many moles passed?
A manometer on the tank gives its pressure before (p₁) and after (p₂). For the tank of volume V at temperature T:
Cᵥ by Joly’s differential steam calorimeter
Two identical hollow copper spheres hang from the arms of a sensitive balance inside a steam chamber, each with a small pan below to catch condensed water. Both are evacuated and balanced; then gas (mass m₁) is put into one and the balance restored with weights.
Steam is let in. It condenses on both spheres, heating them from θ₁ to θ₂, but more condenses on the sphere with gas, because it must also warm the gas. The extra condensed mass m₂ is found by rebalancing. Heated wire coils stop drops forming on the suspensions, and small umbrella shields keep drips off. Since the volume is constant:
(M = molar mass; the spheres’ small expansion is neglected.)
0.80 g of air (29 g/mol) is warmed from 20°C to 100°C in a Joly calorimeter. The extra steam condensed is 22 mg; L = 2.26 × 10⁶ J/kg. Find Cᵥ.
- 1Regnault (Cₚ): gas at constant pressure carries heat from a hot bath into a calorimeter; nCₚ[θ₁ − (θ₂ + θ₃)/2] = (W + m)s(θ₃ − θ₂); n from the tank's pressure drop.
- 2Joly (Cᵥ): the extra steam condensing on a gas-filled sphere: Cᵥ = Mm₂L/[m₁(θ₂ − θ₁)].
Check yourself
Videos, simulations & further reading (1)
Isothermal & Adiabatic Processes
Slow and in contact, or fast and insulated: the two limits of a gas process.
Squeeze a gas slowly in a metal cylinder and it stays at room temperature. Squeeze it fast in a bicycle pump and the pump gets hot. These are the two limits: isothermal and adiabatic.
T constant. Ideal gas: ΔU = 0, so heat in = work done, W = nRT ln(V_f/V_i). Boyle’s law holds. Heat capacity ΔQ/(nΔT) is infinite. Achieved by working slowly in contact with a large reservoir (e.g. a metal cylinder in air).
No heat exchanged; heat capacity zero. Work done by the gas comes from its internal energy: expanding, it cools; compressed, it warms. Achieved with insulation or by acting quickly.
pV^γ is constant in a reversible adiabatic process
With dQ = 0 the first law gives 0 = dU + p dV = nCᵥdT + p dV. From pV = nRT, n dT = (p dV + V dp)/R. Substituting and using Cᵥ + R = Cₚ:
Integrating between the initial and final states:
Dry air at 20°C and 8.0 atm is suddenly released to 1.0 atm. Treating it as reversible and adiabatic with γ = 1.40, find the final temperature.
Air at 300 K (γ = 1.4) is compressed quickly to half its volume. Estimate its temperature.
Work in an adiabatic process
With pV^γ = K, W = ∫K V^(−γ) dV between V₁ and V₂:
- 1Isothermal: T fixed, W = nRT ln(V_f/V_i) = Q; heat capacity infinite.
- 2Adiabatic: Q = 0, W = −ΔU; heat capacity zero; expansion cools, compression heats.
- 3pV^γ, T^γ/p^(γ−1) and TV^(γ−1) are constant in a reversible adiabatic process.
- 4Adiabatic work W = (p₁V₁ − p₂V₂)/(γ − 1).
Check yourself
Videos, simulations & further reading (7)
Equipartition of Energy
½kT for every way a molecule can store energy.
Count the ways a molecule can move and you can predict the heat capacity of a gas, before measuring anything.
Degrees of freedom
A point-like molecule has kinetic energy ½mvₓ² + ½mv_y² + ½mv_z²: three independent terms. That suits monatomic gases.
A diatomic molecule is like a dumbbell. Besides moving, it can rotate about the two axes perpendicular to its length: ½I_xω_x² + ½I_yω_y². (Rotation about its own axis contributes nothing, since point atoms have no moment of inertia about it.) That makes 5 terms. If the atoms also vibrate along the bond, add ½μv² + ½kr²: 7 terms.
The equipartition theorem
For a monatomic gas this gives (3/2)kT per molecule, as kinetic theory found. For n moles with f degrees of freedom, U = n(f/2)RT, so
Measured at 15°C (Cᵥ in J K⁻¹ mol⁻¹, γ):
Red line: equipartition prediction fR/2.
| Gas type | f | Cᵥ (J K⁻¹ mol⁻¹) | Cₚ | γ |
|---|---|---|---|---|
| Monatomic | 3 | 12.5 | 20.8 | 1.67 |
| Diatomic, rotating | 5 | 20.8 | 29.1 | 1.40 |
| Diatomic, also vibrating | 7 | 29.1 | 37.4 | 1.29 |
Measured values agree well: He and Ar have Cᵥ ≈ 12.5 and γ ≈ 1.67; H₂, O₂, N₂ and CO have Cᵥ ≈ 20.8 and γ ≈ 1.40, so ordinary diatomic molecules don’t vibrate at room temperature. CO₂, H₂O and CH₄ behave as if f = 6.
Find the internal energy of 2.0 g of helium (4.0 g/mol) at 300 K.
Find the internal energy of 2.0 g of nitrogen (28 g/mol) at 273 K.
- 1Degrees of freedom: independent energy terms; 3 monatomic, 5 rigid diatomic, 7 vibrating diatomic.
- 2Equipartition: ½kT per degree of freedom.
- 3Cᵥ = fR/2, Cₚ = Cᵥ + R, γ = 1 + 2/f: 1.67 monatomic, 1.40 diatomic.
- 4Diatomic gases don't vibrate at room temperature; CO₂, H₂O, CH₄ act as f ≈ 6.