Cp − Cv = R
The extra heat at constant pressure pays for the work of expansion.
Lesson at a glanceShow the key points
- Constant volume: (dQ)_V = dU. Constant pressure: (dQ)_p = dU + p dV = dU + nR dT.
- So Cp − Cv = R (27.6); Cp > Cv.
- Cv = (1/n) dU/dT (27.7); U = nCvT (27.8); dU = nCvdT for an ideal gas in any process.
- γ = Cp/Cv; at constant pressure ΔU/ΔQ = 1/γ.
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.
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