A student argues that since heating at constant volume involves no work, Cv captures 'pure' heat input, making Cv larger than Cp. What is wrong with this reasoning?
AThe student is correct — Cv > Cp because no energy is wasted on work
BCp > Cv because at constant pressure, extra heat must supply both the temperature rise and the expansion work against the surroundings
CCv = Cp for all ideal gases because internal energy depends only on temperature
DCp > Cv only for monatomic gases; for diatomic gases they are equal
The student has the causal direction backwards. At constant pressure, the gas expands and does PdV work on its surroundings. That work energy leaves the system without raising temperature, so you must supply more heat to achieve the same ΔT as you would at constant volume. This is why Cp is always larger than Cv — the 'extra' heat goes into expansion work, quantified exactly as R per mole for ideal gases.
Question 2 Multiple Choice
A monatomic ideal gas has Cv = (3/2)R. What is its Cp, and what is the physical origin of the difference?
ACp = (3/2)R — the same as Cv because monatomic gases have no rotational modes
BCp = (5/2)R — the extra R accounts for the PdV expansion work done at constant pressure
CCp = (7/2)R — the extra 2R accounts for both rotational and vibrational modes
DCp = 2R — the extra R/2 accounts for kinetic energy of translation in the pressure direction
By Mayer's relation, Cp = Cv + R = (3/2)R + R = (5/2)R. The extra R is the same for all ideal gases regardless of molecular complexity — it comes from the P(∂V/∂T)_P = R work term, not from any internal degrees of freedom. For a diatomic ideal gas, Cv = (5/2)R and Cp = (7/2)R — still differing by exactly R.
Question 3 True / False
The difference Cp − Cv = R holds exactly for all ideal gases, regardless of whether they are monatomic, diatomic, or polyatomic.
TTrue
FFalse
Answer: True
Yes — the derivation of Mayer's relation uses only two properties of ideal gases: (1) PV = nRT, which gives P(∂V/∂T)_P = nR, and (2) internal energy depends only on temperature, so (∂U/∂T)_P = (∂U/∂T)_V = Cv. Neither assumption depends on molecular structure. The extra R always comes from expansion work, making Cp − Cv = R universal for ideal gases regardless of how many degrees of freedom the molecule has.
Question 4 True / False
For a real gas at high pressure near its condensation point, the difference Cp − Cv will be approximately R, just as for an ideal gas.
TTrue
FFalse
Answer: False
For real gases with significant intermolecular interactions, Cp − Cv deviates from R. The exact relation is Cp − Cv = −T(∂P/∂V)_T(∂V/∂T)²_P, which reduces to R only when (∂U/∂V)_T = 0 — i.e., when molecules don't interact and compressing them doesn't change their potential energy. Near condensation, intermolecular attractions are strong and (∂U/∂V)_T ≠ 0, so the departure from R is a measurable diagnostic of intermolecular interactions.
Question 5 Short Answer
Why does heating an ideal gas at constant pressure require more energy than heating it to the same final temperature at constant volume?
Think about your answer, then reveal below.
Model answer: At constant pressure, the gas expands as it heats, doing work on its surroundings (W = PΔV = nRΔT for an ideal gas). That work energy leaves the system without contributing to the temperature rise. To reach the same final temperature, you must supply extra heat equal to nRΔT to compensate for the energy lost to expansion work. At constant volume, no expansion occurs and all heat goes directly into raising internal energy (and thus temperature). The difference in heat required per degree — Cp − Cv = R — equals exactly the work done per mole per kelvin of temperature rise.
The key is that heat and work are both forms of energy transfer. At constant pressure, the total heat input must cover two things: the internal energy increase (same as at constant volume) plus the expansion work done against the surroundings. For ideal gases, the work term PdV = nRdT per degree, so Cp = Cv + R. This is why Cₚ > Cᵥ for any substance, and why the difference equals exactly R for ideal gases — it is a direct measure of the expansion work term in the first law.