Heat capacities Cp, Cv; relationship Cp − Cv = nR (ideal gas): Visual walkthrough

Before we begin, three words in plain English, each anchored to a picture we will draw:

Heat — energy that flows into the gas because we hold a flame under it. We write a tiny sip of it as $δ Q$ . The little $δ$ just means "a very small amount".
Internal energy $U$ — the total wiggling/whizzing energy of all the molecules inside. Faster molecules ⇒ hotter gas ⇒ bigger $U$ . (See Internal Energy.)
Work $W$ — energy the gas spends by pushing something (a piston). A small sip is $δ W$ .

We also use $T$ (temperature, the "hotness"), $P$ (pressure, the sideways push per unit area on the walls), $V$ (volume, the space the gas fills), $n$ (number of moles = how much gas), and $R$ (a fixed number, $8.314$ , called the gas constant).

Step 1 — Two heaters, one target temperature

WHAT. We set up the whole story: same gas, same one-degree temperature rise, two containers.

WHY. Every symbol below refers to these two experiments. If you can see them, the algebra later is just bookkeeping.

PICTURE.

On the left, a rigid steel box — walls cannot move, so $V$ is frozen. On the right, a cylinder with a free piston that carries a weight; the weight fixes the push, so $P$ is frozen. In both we add heat until the thermometer climbs by the same $Δ T$ (read "change in $T$ ").

Our goal: prove the right-hand experiment always needs more heat, and find exactly how much more.

Step 2 — The First Law: heat splits into two jobs

WHAT. State the accounting rule energy obeys.

WHY. Heat can do only two things when it enters a gas: bump up the internal energy, or get spent as pushing work. We need a law that says "energy in = energy stored + energy spent". That is the First Law of Thermodynamics.

PICTURE.

The yellow arrow $δ Q$ pours in. It splits: the blue part $d U$ stays as faster molecules; the red part $δ W$ leaves as piston-pushing.

When the piston moves out by a sliver, the work is force × distance = (pressure × area) × distance = $P \times$ (volume swept). So:

$δ W = P d V$

Here $d V$ is the tiny volume increase. Substitute:

$d U = δ Q - P d V$

Step 3 — The rigid box pins down $C_{v}$

WHAT. Apply the First Law to the left (constant-volume) experiment.

WHY. With walls locked, the piston cannot move, so no work escapes. This isolates a clean meaning for $C_{v}$ .

PICTURE.

The red "work" arrow is crossed out — $d V = 0$ , so $P d V = 0$ . All of $δ Q$ becomes $d U$ .

Set $d V = 0$ in Step 2:

$d U = δ Q (constant volume)$

Every sip of heat becomes internal energy. Since $C_{v}$ is the heat per degree per mole, for $n$ moles:

$δ Q = n C_{v} d T \Rightarrow d U = n C_{v} d T$

Term by term: $n$ = moles of gas; $C_{v}$ = heat per mole per degree at fixed volume; $d T$ = tiny temperature rise.

Step 4 — Why $C_{p}$ needs a new quantity: Enthalpy

WHAT. Build the tool that makes the constant-pressure case as clean as Step 3.

WHY THIS TOOL. In the constant- $P$ box, heat is not simply $d U$ (some leaks as work) and not simply the work (some stays as $d U$ ). We want one bookkeeping quantity whose change equals the heat added at constant pressure. That quantity is Enthalpy, defined to bundle "internal energy + the room it pushes into".

PICTURE.

The blue block is $U$ ; the green block stacked on top is $P V$ . Together they form $H$ . Heating at constant pressure grows the whole tower.

Take a tiny change of $H = U + P V$ . When both $P$ and $V$ can nudge, the product $P V$ changes by $P d V + V d P$ (product rule: each factor gets its turn):

$d H = d U + P d V + V d P$

Now plug in Step 3's gift $d U = n C_{v} d T$ :

$d H = n C_{v} d T + P d V + V d P$

Step 5 — The piston box pins down $C_{p}$

WHAT. Freeze the pressure and read off what $C_{p}$ must equal.

WHY. Constant pressure means $d P = 0$ , which kills the last term and leaves a clean equation — the mirror image of Step 3.

PICTURE.

The weight on the piston is fixed, so $P$ never changes: $d P = 0$ . The term $V d P$ vanishes (crossed out in the figure).

Set $d P = 0$ in Step 4:

$d H = n C_{v} d T + P d V (constant pressure)$

But what is $d H$ at constant pressure? It is precisely the heat we add — that is exactly why we invented $H$ . And heat per degree per mole at constant $P$ is $C_{p}$ , so:

$d H = n C_{p} d T$

Equate the two expressions for $d H$ :

$n C_{p} d T = n C_{v} d T + P d V$

Read it aloud: the constant-pressure heat ( $n C_{p} d T$ ) equals the internal-energy part ( $n C_{v} d T$ ) plus the pushing work ( $P d V$ ). The gap between the two heat capacities is that leftover $P d V$ . Next step evaluates it.

Step 6 — The Ideal Gas Law turns $P d V$ into $n R d T$

WHAT. Compute the pushing work $P d V$ at constant pressure.

WHY THIS TOOL. We need to trade "volume change" for "temperature change" so both sides speak the same language ( $d T$ ). The link between $P, V, T$ for an ideal gas is $P V = n R T$ — that is the Ideal Gas Law.

PICTURE.

The plot shows $V$ climbing straight-line with $T$ at fixed $P$ (Charles's law). The slope of that line is $\frac{n R}{P}$ , so a step $d T$ makes $V$ step by $d V = \frac{n R}{P} d T$ .

Start from $P V = n R T$ . Hold $P$ fixed and take a tiny change of both sides. The left side changes by $P d V$ (since $P$ is constant, only $V$ moves); the right by $n R d T$ :

$P d V = n R d T$

Term by term: $R = 8.314$ = the gas constant; multiplying by $n$ and $d T$ gives the exact pushing work per temperature step. This is the "expansion tax".

Step 7 — Substitute and collapse to Mayer's relation

WHAT. Put Step 6 into Step 5 and simplify.

WHY. Everything is now in terms of $d T$ ; the $d T$ 's cancel and a pure relation between $C_{p}$ and $C_{v}$ survives.

Replace $P d V$ with $n R d T$ in Step 5's boxed equation:

$n C_{p} d T = n C_{v} d T + n R d T$

Divide every term by $n d T$ (allowed: $n > 0$ and $d T \neq = 0$ ):

Step 8 — Edge & degenerate cases (never hit a scenario we didn't show)

WHAT. Check the corners where the tidy formula bends.

WHY. A derivation you can trust must survive its own limits.

Case A — $Δ T = 0$ (isothermal). No temperature change ⇒ $d U = 0$ and $δ Q = P d V$ : all heat becomes work. Mayer's relation still holds (it's about $C_{p}, C_{v}$ , not about a specific heating), but here neither heat capacity does the tracking — see Isothermal vs Adiabatic Processes.

Case B — no heat, $δ Q = 0$ (adiabatic). Then $d U = - P d V$ : the gas cools as it expands, powering the piston from its own internal energy. This is where $γ = C_{p} / C_{v}$ takes over — see Adiabatic Process.

Case C — liquids & solids. Their volume barely grows on heating, so the pushing work $P d V \approx 0$ , giving $C_{p} \approx C_{v}$ . The right-hand column of the figure shows a nearly-immovable "piston".

Case D — real gases. Molecules do attract each other (see Real Gases (van der Waals)), so $U$ depends slightly on $V$ too, and Step 3's gift leaks. The exact replacement is $C_{p} - C_{v} = \frac{T V α ^{2}}{κ _{T}}$ . Only the ideal gas gives the clean $R$ .

Case E — more Degrees of Freedom. Adding rotation/vibration raises $C_{v}$ , but the difference $C_{p} - C_{v}$ stays $R$ — the tax is fixed. Only the ratio $γ = 1 + R / C_{v}$ shrinks.

The one-picture summary

One diagram, the whole chain: heat $δ Q$ enters → splits into stored $d U = n C_{v} d T$ (blue) and pushing $P d V = n R d T$ (red) → the constant-pressure heat $n C_{p} d T$ is their sum (yellow) → cancel $n d T$ ⇒ $C_{p} - C_{v} = R$ .

Recall Feynman retelling — explain the walkthrough to a friend

Picture two identical jars of air. Jar A is welded shut; Jar B has a lid that can slide up freely with a fixed weight on top. I heat both until the thermometer rises by the same amount.

In Jar A the lid can't move, so all my heat goes into making molecules zoom faster — that's the internal energy. The heat per degree here is $C_{v}$ .

In Jar B, as the air warms it pushes the lid up. Some of my heat is "wasted" lifting the lid instead of speeding molecules. So Jar B needs extra heat to reach the same temperature. The heat per degree here is $C_{p}$ , and it's bigger.

How much extra? Exactly the lifting work. For an ideal gas that work per degree per mole is a fixed number $R$ . So $C_{p}$ beats $C_{v}$ by precisely $R$ : that's $C_{p} - C_{v} = R$ , and for $n$ jars-worth of gas, $C_{p} - C_{v} = n R$ . The whole derivation is just this story written in symbols — with a special quantity $H$ invented so Jar B's heat is as easy to write as Jar A's.

Recall

Why is $C_{p} > C_{v}$ always? ::: Constant-pressure heating also does expansion work $P d V$ , so it needs extra heat beyond the internal-energy rise. What single equation replaces $P d V$ with $n R d T$ at constant $P$ ? ::: Differentiate $P V = n R T$ at constant $P$ : $P d V = n R d T$ . For an ideal gas, is $d U = n C_{v} d T$ only at constant volume? ::: No — it holds for every process, because $U$ depends on $T$ alone. Why does $γ$ shrink for polyatomic gases while $C_{p} - C_{v}$ stays $R$ ? ::: Extra degrees of freedom raise $C_{v}$ ; the difference is fixed at $R$ , so the ratio $1 + R / C_{v}$ falls.