Worked examples — γ = Cp - Cv — for monatomic, diatomic, polyatomic

This page is a workout. The parent note built the two master formulas; here we drill them against every kind of question an exam or a lab can throw at you. Before each example you must forecast the answer — guessing first is how the idea sticks.

Two tools we will use again and again (both proven in the parent):

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The scenario matrix

Every question about $\gamma$, $C_p$, $C_v$ falls into one of these cells. The examples below are labelled with the cell(s) they hit, and together they touch every one.

#	Cell class	What makes it tricky	Covered by
A	Forward: shape → γ	pick correct $f$ (rotation count)	Ex 1
B	Backward: γ → f → shape	invert the formula, read off molecule type	Ex 2
C	Numeric $C_p,C_v$ in J/mol·K	keep track of units, use Mayer not memory	Ex 3
D	Mixture of gases	energies add, then divide by total moles	Ex 4
E	Degenerate / limiting $f$	$f\to \infty$ and smallest possible $f=3$	Ex 5
F	High-T unfreezing (vibration)	$f$ jumps $5\to 7$; γ shifts	Ex 6
G	Real-world word problem (sound speed)	γ feeds another formula	Ex 7
H	Exam twist: ratio trap	per-kg vs per-mole; γ unaffected	Ex 8

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Example 1 — Cell A (forward: shape → γ)

Step 1. Neon is monatomic. A point atom can translate along $x,y,z$ → 3 translational boxes; it cannot be felt to spin (0 rotational) and has no bond to vibrate. So $f=3$. Why this step? $f$ is fixed by molecular shape, and a single atom is the simplest shape — only translation counts.

Step 2. $C_v=\frac{f}{2}R=\frac{3}{2}R$. Why this step? Equipartition gives $\frac{1}{2}R$ per box, times $f=3$ boxes.

Step 3. $C_p=C_v+R=\frac{3}{2}R+R=\frac{5}{2}R$. Why this step? Mayer's relation: constant-pressure heating also pays the expansion-work bill of $R$.

Step 4. $\gamma =\frac{C_p}{C_v}=\frac{5/2}{3/2}=\frac{5}{3}\approx 1.67$.

Verify: Cross-check with $\gamma =1+2/f=1+2/3=5/3$. ✓ Both routes agree.

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Example 2 — Cell B (backward: γ → f → shape)

Step 1. Invert the master formula: $f=\frac{2}{\gamma -1}$. Why this step? We are handed $\gamma$ and asked for $f$ — the inverse relation answers exactly "which $f$ produces this $\gamma$?", just as arctan answers "which angle has this tan?".

Step 2. $f=\frac{2}{1.40-1}=\frac{2}{0.40}=5$.

Step 3. $f=5=\underset{\text{translation}}{\underbrace{3}}+\underset{\text{rotation}}{\underbrace{2}}$. Two rotational boxes = a dumbbell = diatomic (e.g. ${\text{O}}_2$, ${\text{N}}_2$) at room temperature. Why this step? Split $f$ into the always-present 3 translations plus rotations; 2 rotations is the signature of a linear two-atom molecule (spin about its own bond axis is frozen).

Verify: Forward-check: $\gamma =1+2/5=7/5=1.40$. ✓ Matches the given value.

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Example 3 — Cell C (numeric heat capacities)

Step 1. Diatomic ⇒ $f=5$, so $C_v=\frac{5}{2}R=2.5\times 8.314=20.785\text{J/mol⋅K}$. Why this step? Plug the shape's $f$ into $C_v=\frac{f}{2}R$; keep numbers so units come out in $\text{J/mol⋅K}$.

Step 2. $C_p=C_v+R=20.785+8.314=29.099\text{J/mol⋅K}$. Why this step? Mayer again — avoid memorizing $C_p$ separately; it's always $C_v+R$.

Step 3. $\gamma =\frac{C_p}{C_v}=\frac{29.099}{20.785}=1.400$.

Verify: The gap $C_p-C_v=29.099-20.785=8.314=R$. ✓ Exactly one $R$, as Mayer demands. Units: J/mol·K throughout. ✓

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Example 4 — Cell D (mixture)

Step 0 (state the condition). Because both gases share one temperature $T$, we may write each gas's internal energy with the same $T$ and simply add them. If they were at different temperatures we would first have to let them reach thermal equilibrium. Why this step? Internal energy of an ideal gas is $U=\frac{f}{2}nRT$ — it only makes sense to sum two energies at a common $T$.

Step 1. Internal energy is extensive — it adds up (same $T$ for both): $U=n_1\left(\frac{3}{2}RT\right)+n_2\left(\frac{5}{2}RT\right).$ Why this step? Each gas stores its own energy independently; total energy is the sum, not an average.

Step 2. Divide by total moles to get the molar (per-mole) $C_v$: $C_{v,\text{mix}}=\frac{n_1\frac{3}{2}R+n_2\frac{5}{2}R}{n_1+n_2}=\frac{2(1.5)+3(2.5)}{5}R=\frac{3+7.5}{5}R=\frac{10.5}{5}R=2.1R.$ Why this step? Molar heat capacity is energy per mole, so we normalize by the $5$ total moles.

Step 3. $C_{p,\text{mix}}=C_{v,\text{mix}}+R=2.1R+R=3.1R$ (Mayer holds for the ideal mixture too).

Step 4. ${\gamma }_{\text{mix}}=\frac{3.1}{2.1}=1.476$.

Verify: ${\gamma }_{\text{mix}}=1.476$ lies between $1.40$ and $1.67$, as any weighted average must. Nearer to $1.40$ because oxygen dominates the mole count. ✓

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Example 5 — Cell E (degenerate & limiting $f$)

Figure: horizontal axis = degrees of freedom $f$; vertical axis = heat-capacity ratio $\gamma$. The red curve is $\gamma =1+2/f$. Three gases are marked as coloured dots: monatomic (He, Ne, Ar) at $(3,5/3)$, diatomic (O₂, N₂, H₂) at $(5,7/5)$, and non-linear polyatomic (H₂O, CH₄) at $(6,4/3)$. The dashed grey line is the asymptote $\gamma =1$, which the curve approaches but never touches.

Step 1 (largest γ). The smallest physically allowed $f$ is $f=3$ (a point atom can never have fewer than 3 translational boxes). Then $\gamma =1+\frac{2}{3}=\frac{5}{3}\approx 1.67$. Why this step? $\gamma =1+2/f$ decreases as $f$ grows, so it is maximized by the smallest $f$. There is no $f<3$ ideal gas, so $5/3$ is a hard ceiling.

Step 2 (limit). As $f\to \infty$: $\frac{2}{f}\to 0$, so $\gamma \to 1+0=1$. Why this step? A molecule with enormously many storage boxes shares each joule of heat among so many motions that constant-P and constant-V heating become nearly identical — $C_p\approx C_v$, so $\gamma \to 1^{+}$.

Step 3 (the shape of the curve). Look at the figure: the red curve starts high at the blue dot $(3,1.67)$, drops through the orange diatomic dot $(5,1.40)$ and the green polyatomic dot $(6,1.33)$ — concrete examples of that green point are H₂O and CH₄ (non-linear, 3 rotational boxes) — and then flattens toward the dashed asymptote $\gamma =1$ but never reaches it.

Verify: $\gamma (3)=5/3=1.6\overline{6}$ (max ✓); ${\lim }_{f\to \infty }(1+2/f)=1$ (approached from above, never equal ✓). Every real gas therefore obeys $1<\gamma \le 5/3$.

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Example 6 — Cell F (high-temperature unfreezing)

Step 1. Room-T diatomic: $f=5$ (3 translation + 2 rotation), $\gamma =7/5=1.40$. Why this step? This is our baseline before vibration wakes up.

Step 2. Vibration unfreezes ⇒ add 2 quadratic boxes: $f=5+2=7$. Why this step? At high $T$ the quantum "freeze" lifts; each vibrational mode contributes potential and kinetic energy — 2 quadratic DOF.

Step 3. ${\gamma }_{\text{hot}}=1+\frac{2}{7}=\frac{9}{7}\approx 1.286$.

Verify: $\gamma$ dropped from $1.40$ to $9/7\approx 1.286$. ✓ More boxes ⇒ γ closer to 1, exactly the trend of the curve in Ex 5. This is why $\gamma$ is not strictly constant — it shifts only when new DOF unfreeze.

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Example 7 — Cell G (real-world word problem: speed of sound)

Step 1 (air). $v=\sqrt{\frac{(1.4)(8.314)(300)}{0.029}}$. Numerator $=1.4\times 8.314\times 300=3491.88$. Divide by $M$: $3491.88/0.029=120409.7$. Root: $v\approx 347.0\text{m/s}$. Why this step? The Speed of Sound in Gas formula turns our thermodynamic $\gamma$ directly into a measurable everyday number — this is why γ matters outside a textbook.

Step 2 (argon). $v_{\text{Ar}}=\sqrt{\frac{(5/3)(8.314)(300)}{0.040}}$. Numerator $=1.6667\times 8.314\times 300=4157.0$. Divide: $4157.0/0.040=103925$. Root: $v_{\text{Ar}}\approx 322.4\text{m/s}$. Why this step? Same formula, new molecule — the heavier mass ($0.040$ vs $0.029$) and larger $\gamma$ ($1.67$ vs $1.40$) compete.

Step 3 (compare). $322.4<347.0$, so sound travels slower in argon. Why this step? The heavier molar mass wins: $v\propto \sqrt{\gamma /M}$, and $M$ increased by a bigger factor than $\gamma$.

Verify: Units: $\sqrt{\frac{\text{J/mol⋅K}\cdot \text{K}}{\text{kg/mol}}}=\sqrt{\frac{\text{J}}{\text{kg}}}=\sqrt{{\text{m}}^2/{\text{s}}^2}=\text{m/s}$. ✓ Air ≈ 347 m/s matches the textbook ~343 m/s (rounding of $M$ and $T$). ✓

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Example 8 — Cell H (exam twist: per-kg vs per-mole trap)

Step 1. Convert per-kg to per-mole: $C_v=c_v\times M=650\times 0.032=20.8\text{J/mol⋅K}$. Why this step? "Molar" means per mole; multiply the per-kg value by kg-per-mole so the kg cancels: $\frac{\text{J}}{\text{kg⋅K}}\cdot \frac{\text{kg}}{\text{mol}}=\frac{\text{J}}{\text{mol⋅K}}$.

Step 2. $C_p=C_v+R=20.8+8.314=29.114\text{J/mol⋅K}$.

Step 3. $\gamma =\frac{C_p}{C_v}=\frac{29.114}{20.8}=1.400$. Why this step? γ is a ratio of two heat capacities in the same units — the molar mass factor cancels top and bottom, so γ is identical whether you use per-kg or per-mole values.

Verify: $C_v\approx 20.8=\frac{5}{2}R$ ✓ (recovers the diatomic value), and $\gamma =1.40=7/5$ ✓. The trap: you did not need $M$ to get γ — only to get the molar numbers. γ is mass-independent.

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Recall Which cell is this? (self-test)

"$\gamma =1.33$, what molecule?" — which matrix cell? ::: Cell B (backward: invert to $f=6$, non-linear polyatomic). "Sound speed in helium?" — which cell? ::: Cell G (γ feeds the $v=\sqrt{\gamma RT/M}$ formula). "Given $c_v$ in J/kg·K, find γ." — which cell? ::: Cell H (per-kg trap; γ unaffected). "Largest γ possible?" — which cell and answer? ::: Cell E; $\gamma =5/3$ at $f=3$.

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Connections

• Parent: γ = Cp/Cv — formulas drilled here
• Degrees of Freedom & Equipartition Theorem — the source of $f$ and $C_v=\frac{f}{2}R$
• Mayer's Relation — used in every example ($C_p=C_v+R$)
• Speed of Sound in Gas — Example 7 word problem
• First Law of Thermodynamics — underlies Mayer's relation
• Internal Energy of Ideal Gas — $U=\frac{f}{2}RT$, used in the mixture example
• Adiabatic Process — where γ is used next