Heat capacities Cp, Cv; relationship Cp − Cv = nR (ideal gas): Question bank

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The picture below is the whole story: at constant $V$ the heat arrow feeds only the $U$ -tank; at constant $P$ the same heat splits, and the extra sliver is the expansion tax $R$ .

Figure — Heat capacities Cp, Cv; relationship Cp − Cv = nR (ideal gas)

True or false — justify

The relation

C_{p} - C_{v} = R

holds for a single mole of any gas at any pressure

False — it is exact only for an ideal gas; real gases follow

C_{p} - C_{v} = \frac{T V α ^{2}}{κ _{T}}

(symbols defined in the Edge cases section) and only approach

R

at low pressure. See van der Waals gases.

For liquids and solids,

C_{p}

and

C_{v}

are essentially equal

True — they barely expand when heated, so the

P d V

"expansion tax" is tiny and

C_{p} \approx C_{v}

C_{p}

is always greater than

C_{v}

for a substance that expands on heating

True — heating at constant pressure must also pay for expansion work against the surroundings, so it needs extra heat, making

C_{p} > C_{v}

For an ideal gas,

Δ U = n C_{v} Δ T

only when the volume is held constant

False — for an ideal gas

U

depends only on

T

, so

Δ U = n C_{v} Δ T

holds in every process (isobaric, isothermal, adiabatic). See Internal Energy.

The heat capacity ratio

γ

increases as a molecule gains more degrees of freedom

False — more modes raise

C_{v}

, and since

γ = 1 + R / C_{v}

, a larger

C_{v}

lowers

γ

. See Degrees of Freedom.

Along an isothermal ideal-gas process,

Δ U = 0

even though heat flows in

True — constant

T

means

Δ U = n C_{v} Δ T = 0

; all the heat leaves as work. See Isothermal vs Adiabatic Processes.

A monatomic ideal gas has

γ = 5/3

regardless of which element it is

True — with only 3 translational modes,

C_{v} = \frac{3}{2} R

and

C_{p} = \frac{5}{2} R

for He, Ar, Ne alike, giving

γ = 5/3

Molar heat capacity and specific heat capacity are numerically the same for a given gas

False — molar

C

is per mole, specific

c

is per gram; they differ by the molar mass,

C = c \times M

Spot the error

"Since the derivation of

C_{p} - C_{v} = R

just used the first law and enthalpy, it must apply to all matter."

The hidden ideal-gas steps are

d U = n C_{v} d T

(no intermolecular forces) and

P d V = n R d T

(ideal gas law); both fail for real gases, so the generality is false.

"At constant volume no work is done, therefore

Q_{v} = n C_{p} Δ T

Wrong capacity — at constant volume

δ Q = d U

, so

Q_{v} = n C_{v} Δ T

;

C_{p}

belongs to the constant-pressure path.

C_{p}

uses enthalpy only as a mathematical trick with no physical meaning."

Enthalpy

H = U + P V

captures the internal energy plus the

P V

work; at constant pressure

δ Q = d H

, so

C_{p}

genuinely counts the expansion energy. See Enthalpy.

"For 3 moles of gas, Mayer's relation gives

C_{p} - C_{v} = 3 R

C_{p}, C_{v}

are molar (per mole) it is still just

R

; the

n R

form applies only when they are the total (extensive) heat capacities.

"Since an adiabatic process transfers no heat, we cannot use

C_{v}

C_{p}

there."

We still can —

Δ U = n C_{v} Δ T

holds in any ideal-gas process, and

γ = C_{p} / C_{v}

actually governs the adiabatic curve

P V^{γ} = const

. See Adiabatic Process.

"Water's heat capacity is

4.18

, so its molar value is also

4.18 J mol^{- 1} K^{- 1}

That

4.18

is per gram; multiplying by the molar mass

18

gives

\approx 75.2 J mol^{- 1} K^{- 1}

per mole.

"The extra heat in the constant-pressure case became internal energy that we can recover later."

No — that extra

n R Δ T

left the system as expansion work pushing the piston; it is not stored in the gas.

Why questions

Why does

C_{p}

involve enthalpy

H

while

C_{v}

involves internal energy

U

At constant

V

δ Q = d U

; at constant

P

δ Q = d H = d (U + P V)

, so enthalpy automatically bundles in the

P V

work. See First Law of Thermodynamics.

Why is the difference exactly

R

per mole and not some other constant?

Because

P d V = n R d T

for an ideal gas at constant pressure, so the per-mole expansion cost per kelvin is precisely

R

. See Ideal Gas Law.

Why does adding degrees of freedom leave

C_{p} - C_{v}

unchanged but change

γ

New modes add equally to both

C_{p}

and

C_{v}

, so their difference stays

R

, but their ratio shrinks toward 1.

Why can we treat

Δ U = n C_{v} Δ T

as path-independent for an ideal gas?

U

depends only on

T

(no intermolecular potential energy), so any two states at the same temperature have the same

U

no matter the route.

Why does the real-gas correction take the form

C_{p} - C_{v} = \frac{T V α ^{2}}{κ _{T}}

The general (exact, any-substance) result is

C_{p} - C_{v} = T V \frac{α ^{2}}{κ _{T}}

, where

α = \frac{1}{V} (\frac{\partial V}{\partial T})_{P}

is how eagerly the volume swells per kelvin, and

κ_{T} = - \frac{1}{V} (\frac{\partial V}{\partial P})_{T}

is how easily it is squeezed by pressure. Heating at constant

P

expands the material by

α

; that expansion does work, but a stiffer material (small

κ_{T}

) resists more, so the "tax"

T V α^{2} / κ_{T}

grows with how much it expands (

α^{2}

) and shrinks with how squishy it is — the picture below traces this chain.

$C_{p} - C_{v} = R$ is ideal-gas-only and per-mole; the exact any-substance version $C_{p} - C_{v} = T V α^{2} / κ_{T}$ is $\geq 0$ (because of $α^{2}$ ) so $C_{p} \geq C_{v}$ always, with equality where $α = 0$ ; $Δ U = n C_{v} Δ T$ is always true for an ideal gas; more degrees of freedom lower $γ$ but never touch the difference $R$ ; and the extra constant-pressure heat is spent, not stored.

Connections

First Law of Thermodynamics — the source of $δ Q = d U$ and $δ Q = d H$ .
Enthalpy — why $C_{p}$ carries the $P V$ work.
Ideal Gas Law — supplies $P d V = n R d T$ .
Degrees of Freedom — controls $C_{v}$ and hence $γ$ .
Internal Energy — why $Δ U = n C_{v} Δ T$ is path-independent.
Adiabatic Process · Isothermal vs Adiabatic Processes — where $γ$ and $Δ U$ reappear.
Real Gases (van der Waals) — where $C_{p} - C_{v} = R$ breaks.