Heat capacities Cp, Cv; relationship Cp − Cv = nR (ideal gas): Worked examples

This page is the workshop. On the parent note Heat capacities Cp, Cv you learned why $C_{p} > C_{v}$ and how Mayer's relation is born. Here we hammer it against every case the topic can throw at you — every gas type, the constant-volume and constant-pressure paths, the degenerate limits (solids, real gases), a word problem, and an exam twist.

Before we begin, one promise: every symbol used below was defined on the parent note. As a quick reminder in plain words:

Recall The four symbols we lean on

$C_{v}$ ::: heat to warm 1 mole by 1 K at fixed volume — units $J mol^{- 1} K^{- 1}$ .
$C_{p}$ ::: heat to warm 1 mole by 1 K at fixed pressure — always bigger than $C_{v}$ .
$R$ ::: the gas constant, $8.314 J mol^{- 1} K^{- 1}$ — the "expansion tax" per mole per kelvin.
$γ$ ::: the ratio $C_{p} / C_{v}$ , the adiabatic index.

The scenario matrix

Every question on this topic lives in exactly one of these cells. The examples below are labelled with the cell they cover.

#	Cell class	What makes it distinct	Example
A	Monatomic gas (He, Ar)	$C_{v} = \frac{3}{2} R$ , $γ = \frac{5}{3}$	Ex 1
B	Diatomic gas (N₂, O₂)	$C_{v} = \frac{5}{2} R$ , $γ = \frac{7}{5}$	Ex 2
C	Polyatomic gas (CO₂, CH₄)	more [[Degrees of Freedom	degrees of freedom]], $γ \to 1$
D	Two paths, same $Δ T$	compare $Q_{V}$ vs $Q_{P}$ , extract the work	Ex 4
E	Reverse problem	given $C_{p}$ or $γ$ , back out $C_{v}$	Ex 5
F	Degenerate: solid/liquid	$C_{p} - C_{v} \neq = R$ ; expansion work ≈ 0	Ex 6
G	Degenerate: real gas	van der Waals correction to Mayer	Ex 7
H	Word problem (real world)	heating air in a room / balloon	Ex 8
I	Exam twist	mixture of gases, or "which $γ$ ?" identity	Ex 9
J	Limiting behaviour	$γ$ as degrees of freedom $\to \infty$	Ex 10

The figure below shows the single geometric fact that unifies cells A–D: at constant pressure the gas pushes a piston, and that push is the entire difference between $C_{p}$ and $C_{v}$ .

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

Look at the red arrow — it is the piston's outward motion. The energy stored in that motion, per mole per kelvin, is exactly $R$ .

Example 1 — Cell A (monatomic)

Example 2 — Cell B (diatomic)

Worked example Nitrogen: heat at both paths

Given: 2 mol of N₂ ( $C_{v} = \frac{5}{2} R$ ), warmed from 300 K to 350 K. Find: $Q_{V}$ (constant volume) and $Q_{P}$ (constant pressure).

Forecast: Which is bigger, and by how much? Predict the gap in joules before computing.

$Δ T = 350 - 300 = 50 K$ . Why this step? Both formulas scale with the temperature change; fix it once.
Constant volume: $Q_{V} = n C_{v} Δ T = 2 \cdot \frac{5}{2} (8.314) \cdot 50 = 2078.5 J$ . Why this step? At fixed $V$ no work is done, so all heat becomes Internal Energy: $Q_{V} = Δ U$ .
Constant pressure: $C_{p} = C_{v} + R = \frac{7}{2} R$ , so $Q_{P} = 2 \cdot \frac{7}{2} (8.314) \cdot 50 = 2909.9 J$ . Why this step? At fixed $P$ the gas also does expansion work, needing extra heat.

Verify: The gap $Q_{P} - Q_{V} = 2909.9 - 2078.5 = 831.4 J$ must equal $n R Δ T = 2 (8.314) (50) = 831.4 J$ ✓ — the expansion tax, exactly as the first law predicts.

Example 3 — Cell C (polyatomic)

Example 4 — Cell D (two paths, extract the work)

Worked example Where did the extra heat go?

Given: 1 mol monatomic gas, $Δ T = 100 K$ , heated once at fixed $V$ , once at fixed $P$ . Find: the work done at constant pressure, and confirm it equals the heat difference.

Forecast: Guess the work in joules ( $R Δ T$ per mole) before reading on.

$Δ U = n C_{v} Δ T = 1 \cdot \frac{3}{2} (8.314) \cdot 100 = 1247.1 J$ . Why this step? $Δ U = n C_{v} Δ T$ holds for an ideal gas on any path — a fact worth over-using (see Mistake 3 on parent).
Constant-pressure work: $W = P Δ V = n R Δ T = 1 (8.314) (100) = 831.4 J$ . Why this step? Differentiating $P V = n R T$ at fixed $P$ gives $P d V = n R d T$ ; this is the piston push in the figure.
$Q_{P} = Δ U + W = 1247.1 + 831.4 = 2078.5 J$ . Why this step? Direct first-law bookkeeping: heat in = energy stored + work out.

Verify: $Q_{P}$ should also equal $n C_{p} Δ T = 1 \cdot \frac{5}{2} (8.314) (100) = 2078.5 J$ ✓ — two routes, one answer.

Example 5 — Cell E (reverse problem)

Example 6 — Cell F (degenerate: solid/liquid)

Worked example Why

$C_{p} - C_{v} \neq = R$ for water Given: liquid water, specific $c_{p} = 4.18 J g^{- 1} K^{- 1}$ , molar mass $M = 18 g mol^{- 1}$ . The real solid/liquid relation is $C_{p} - C_{v} = \frac{T V α ^{2}}{κ _{T}}$ . Find: molar $C_{p}$ , and explain why plugging into $C_{p} - C_{v} = R$ is wrong.

Forecast: For a liquid, will $C_{p} - C_{v}$ be near $R$ (8.3) or near zero?

Molar $C_{p} = c_{p} \cdot M = 4.18 \times 18 = 75.24 J mol^{- 1} K^{- 1}$ . Why this step? Mistake 2 on the parent — never mix per-gram and per-mole.
A liquid barely expands on heating, so $Δ V \approx 0$ , hence the expansion work $P d V \approx 0$ . Why this step? The whole $+ R$ came from ideal-gas expansion; kill the expansion, kill the $R$ .
Therefore $C_{p} \approx C_{v}$ for water, and $C_{p} - C_{v} = R = 8.314$ would be badly wrong here. Why this step? Marks the boundary of Mayer's validity — cell F is the "does not apply" cell.

Verify: Using measured constants for water at 298 K ( $α \approx 2.1 \times 1 0^{- 4} K^{- 1}$ , molar $V \approx 1.8 \times 1 0^{- 5} m^{3}$ , $κ_{T} \approx 4.6 \times 1 0^{- 10} Pa^{- 1}$ ): $\frac{T V α ^{2}}{κ _{T}} \approx 0.51 J mol^{- 1} K^{- 1}$ , which is $≪ 8.314$ ✓ — nowhere near $R$ .

Example 7 — Cell G (degenerate: real gas)

Example 8 — Cell H (word problem)

Worked example Heating the air in a room

Given: A sealed rigid room holds $n = 100 mol$ of air (treat as diatomic, $C_{v} = \frac{5}{2} R$ ). A heater raises the temperature by $Δ T = 5 K$ . Find: the heat supplied, and what changes if instead a window is open (constant pressure).

Forecast: Sealed vs open — which needs more heat for the same 5 K rise?

Sealed (rigid = constant volume): $Q_{V} = n C_{v} Δ T = 100 \cdot \frac{5}{2} (8.314) \cdot 5 = 10392.5 J$ . Why this step? A rigid room can't expand, so this is the pure $Δ U$ case.
Open window (air free to expand at atmospheric $P$ ): $Q_{P} = n C_{p} Δ T = 100 \cdot \frac{7}{2} (8.314) \cdot 5 = 14549.5 J$ . Why this step? Now warm air pushes out through the window — expansion work is paid.

Verify: Difference $= 14549.5 - 10392.5 = 4157.0 J$ , which must equal $n R Δ T = 100 (8.314) (5) = 4157.0 J$ ✓ — the open-window "energy leak" is precisely the expansion tax.

Example 9 — Cell I (exam twist: gas mixture)

Example 10 — Cell J (limiting behaviour)

Recall Rapid self-test

Extra heat for constant- $P$ vs constant- $V$ over $Δ T$ , $n$ moles ::: $n R Δ T$ $γ$ for a diatomic gas ::: $7/5 = 1.40$ Does $C_{p} - C_{v} = R$ hold for liquid water? ::: No — negligible expansion, $C_{p} \approx C_{v}$ $γ$ as degrees of freedom $\to \infty$ ::: approaches 1

Connections

Parent topic — the derivation these examples exercise.
First Law of Thermodynamics — the $Q = Δ U + W$ bookkeeping behind every example.
Ideal Gas Law — supplies $P d V = n R d T$ , the source of the $R$ .
Degrees of Freedom — sets $C_{v}$ , hence which cell (A/B/C/J) a gas falls in.
Real Gases (van der Waals) — cell G's correction beyond ideal Mayer.