Questions: Heat Capacity at Constant Volume and Pressure

Question 1

You heat 1 mole of an ideal monatomic gas by 10 K at constant volume, then separately heat a second mole of the same gas by 10 K at constant pressure. Which process requires more heat input?

• A. They require the same heat — the temperature rise is identical, so the energy added must be equal
• B. Constant pressure requires more heat — additional energy must be supplied to do PdV expansion work
• C. Constant volume requires more heat — confining the gas causes more internal collisions, needing more energy
• D. Constant pressure requires more heat — higher pressure means more molecular resistance to heating

Answer: (B)

At constant volume, no expansion occurs (W = 0), so all heat goes into internal energy: Q_V = nCᵥΔT. At constant pressure, the gas expands as it warms, doing work W = PΔV = nRΔT on its surroundings. To achieve the same ΔT, you must supply that extra work on top of the internal energy increase: Q_P = nCᵥΔT + nRΔT = nCₚΔT. Since Cₚ = Cᵥ + R > Cᵥ, the constant-pressure process always requires more heat for the same temperature rise. The extra heat doesn't go into the gas — it goes into the surroundings as expansion work.

Question 2

The Mayer relation Cₚ = Cᵥ + R holds for ideal gases. What does the R term physically represent?

• A. The extra heat needed to overcome intermolecular attractions as the gas expands
• B. The work per mole done by the gas expanding against constant pressure as temperature rises by 1 K
• C. The additional rotational kinetic energy that activates specifically at constant pressure
• D. The energy lost to friction between gas molecules during the expansion process

Answer: (B)

For an ideal gas at constant pressure, when temperature rises by ΔT, the gas expands by ΔV = nRΔT/P. The work done by the gas is W = PΔV = nRΔT. Per mole per kelvin, this is R ≈ 8.314 J/(mol·K). This is exactly the difference Cₚ − Cᵥ = R. The gas molecules themselves have the same kinetic energy as in the constant-volume case (since ΔU = nCᵥΔT is the same), but extra energy equal to nRΔT was needed to push back the atmosphere. There are no intermolecular attractions in an ideal gas (option A is wrong).

Question 3

For all ideal gases, Cₚ > Cᵥ, and the ratio γ = Cₚ/Cᵥ approaches 1 as the number of thermally active degrees of freedom increases.

• T. True
• F. False

Answer: True

Both claims are correct. Cₚ = Cᵥ + R always, so Cₚ is always larger. The ratio γ = Cₚ/Cᵥ = (Cᵥ + R)/Cᵥ = 1 + R/Cᵥ. As more degrees of freedom activate (translation, rotation, vibration), Cᵥ grows while R stays constant, so R/Cᵥ shrinks and γ → 1. Monatomic: Cᵥ = 3R/2, γ = 5/3 ≈ 1.67. Diatomic at room temperature: Cᵥ = 5R/2, γ = 7/5 = 1.40. At high temperature with vibration active: Cᵥ = 7R/2, γ = 9/7 ≈ 1.29.

Question 4

For an ideal gas heated at constant volume, the heat added equals the change in enthalpy ΔH.

• T. True
• F. False

Answer: False

At constant volume, Q = ΔU = nCᵥΔT. It is constant pressure, not constant volume, where Q equals the change in enthalpy: Q_P = ΔH = nCₚΔT. This is precisely why enthalpy was defined as H = U + PV — it is the natural thermodynamic potential for constant-pressure processes because ΔH automatically accounts for both the internal energy change and the PV work. At constant volume, enthalpy changes by ΔH = ΔU + VΔP = nCᵥΔT + nRΔT = nCₚΔT, which is not equal to Q_V.

Question 5

Explain why Cₚ is always greater than Cᵥ for an ideal gas, and what the difference Cₚ − Cᵥ = R represents physically.

Model answer: At constant volume, all the heat supplied goes into increasing the internal energy (kinetic energy of the molecules), so Cᵥ = (1/n)(ΔU/ΔT). At constant pressure, the gas must not only increase its internal energy by the same amount but also expand and do work on the surroundings. By the first law, Q_P = ΔU + W = nCᵥΔT + PΔV. For an ideal gas, PΔV = nRΔT, so Q_P = n(Cᵥ + R)ΔT, giving Cₚ = Cᵥ + R. The R represents the mechanical work done by one mole of ideal gas expanding against constant external pressure when heated by 1 K — energy that leaves the gas as work rather than staying as internal energy.

This also explains why Cₚ is the relevant heat capacity for most laboratory and industrial processes, which occur at (approximately) constant atmospheric pressure rather than constant volume. Measuring Q at constant pressure and dividing by nΔT gives Cₚ directly — and Cᵥ can be recovered via the Mayer relation. The elegance of the Mayer relation is that it connects a purely mechanical quantity (R, from PV = nRT) to measured thermal properties (Cᵥ and Cₚ).