Exercises — γ = Cp - Cv — for monatomic, diatomic, polyatomic

WHAT: A monatomic gas is a single ball-like atom. It can only move (translate) in 3 directions: $x,y,z$. It cannot be felt to spin (a point has no measurable moment of inertia). WHY: Each independent way of holding energy is one degree of freedom. So $f=3$ (translation only). $\gamma =1+\frac{2}{f}=1+\frac{2}{3}=\frac{5}{3}\approx \mathrm{1.67.}$ This is the left-most (pale-yellow) dot in the figure — smallest $f$, largest γ.

WHY these DOF: a diatomic molecule is a dumbbell — it translates in 3 directions and tumbles about 2 perpendicular axes, so $f=5$ at room temperature (bond-axis spin and vibration are quantum-frozen, so they store no energy). WHY $C_p=C_v+R$: the extra $R$ is the expansion work per mole per kelvin the gas must do at constant pressure (Mayer's Relation). $C_v=\frac{f}{2}R=\frac{5}{2}R,C_p=C_v+R=\frac{5}{2}R+R=\frac{7}{2}R.$

False. Since $C_p=C_v+R$ and both $C_v>0$ and $R>0$, we get $C_p>C_v$, so $\gamma =C_p/C_v>1$ always. The reason: at constant pressure the gas also does expansion work, which always needs extra heat (see First Law of Thermodynamics).

WHY $f=5$: nitrogen is a diatomic dumbbell, so 3 translation + 2 rotation give $f=5$ (vibration frozen at room temperature). This is the chalk-blue dot in the figure. WHY equipartition sets $C_v$: each active degree of freedom stores $\frac{1}{2}RT$ per mole, so $U=\frac{f}{2}RT$ and $C_v=dU/dT=\frac{f}{2}R$ (Degrees of Freedom & Equipartition Theorem). $C_v=\frac{5}{2}R=2.5\times 8.314=20.785{\text{J mol}}^{-1}{\text{K}}^{-1}.$ WHY add $R$ for $C_p$: Mayer's relation says the constant-pressure capacity is the constant-volume one plus the expansion-work cost $R$. $C_p=C_v+R=20.785+8.314=29.099{\text{J mol}}^{-1}{\text{K}}^{-1}.$ Sanity check: $\gamma =29.099/20.785=1.40$ ✓ — matches $7/5$.

WHY invert: γ is a one-to-one function of $f$ (the curve in the figure never repeats a height), so a measured γ pins down exactly one $f$. Solve $\gamma =1+2/f$ for $f$: $f=\frac{2}{\gamma -1}=\frac{2}{1.40-1}=\frac{2}{0.40}=5.$ WHY this identifies the gas: $f=5$ means 3 translation + 2 rotation, the signature of a dumbbell → diatomic (e.g. ${\text{O}}_2$, ${\text{N}}_2$). On the figure this reads as "which dot sits at height 1.40?" — the chalk-blue one at $f=5$.

WHAT: At constant volume, $Q=n{C}_v\Delta T$ (no work, all heat → internal energy). WHY $C_v$ and not $C_p$: volume is fixed, so no expansion work — First Law of Thermodynamics gives $dQ=dU$. Argon monatomic: $C_v=\frac{3}{2}R=1.5\times 8.314=12.471$. $Q=n{C}_v\Delta T=2\times 12.471\times (400-300)=2494.2\text{J}.$

At constant pressure: $Q=n{C}_p\Delta T$ with $C_p=\frac{7}{2}R=3.5\times 8.314=29.099$. $\Delta T=\frac{Q}{n{C}_p}=\frac{5000}{3\times 29.099}=57.27\text{K}.$ Internal energy part: $\Delta U=n{C}_v\Delta T=3\times 20.785\times 57.27=3571.4\text{J}$. Work part — WHY $W=nR\Delta T$: work done by an expanding gas is $W=P\Delta V$ (First Law of Thermodynamics). At constant pressure the ideal-gas law $PV=nRT$ gives $P\Delta V=nR\Delta T$ (multiply the temperature change by $nR$, since $P$ and $n$ are held fixed). So the expansion work is $W=P\Delta V=nR\Delta T=3\times 8.314\times 57.27=1428.6\text{J}.$ Check: $3571.4+1428.6=5000$ J ✓. The work fraction is $R/C_p=8.314/29.099=0.286$, i.e. $28.6\%$ of the heat went to pushing the piston.

WHY 2, not 1: a vibrating bond stores energy in both motion (kinetic) and stretch (potential); equipartition gives $\frac{1}{2}RT$ to each quadratic term, so one vibration = 2 degrees of freedom. New $f=5+2=7$. $\gamma =1+\frac{2}{7}=\frac{9}{7}\approx \mathrm{1.286.}$ Room-T value was $7/5=1.40$. So heating it up (more active DOF) slides the gas rightward and downward along the curve in the figure — γ drops toward 1.

Start from the ratio and divide top and bottom by $C_v$: $\frac{C_p-C_v}{C_p}=\frac{\frac{C_p}{C_v}-1}{\frac{C_p}{C_v}}=\frac{\gamma -1}{\gamma }=1-\frac{1}{\gamma }.$ For monatomic $\gamma =5/3$: $1-\frac{1}{5/3}=1-\frac{3}{5}=\frac{2}{5}=\mathrm{0.4.}$ Meaning: $40\%$ of constant-pressure heat becomes expansion work for a monatomic gas.

WHY add energies: internal energy is extensive — total $U$ is the sum of each gas's $U$ (see Internal Energy of Ideal Gas). $C_{v,\text{mix}}=\frac{n_1\left(\frac{3}{2}R\right)+n_2\left(\frac{5}{2}R\right)}{n_1+n_2}=\frac{2(1.5)+3(2.5)}{5}R=\frac{3+7.5}{5}R=2.1R.$ Numerically $C_{v,\text{mix}}=2.1\times 8.314=17.459{\text{J mol}}^{-1}{\text{K}}^{-1}.$ $C_{p,\text{mix}}=C_{v,\text{mix}}+R=3.1R=25.773{\text{J mol}}^{-1}{\text{K}}^{-1}.$ ${\gamma }_{\text{mix}}=\frac{C_{p,\text{mix}}}{C_{v,\text{mix}}}=\frac{3.1}{2.1}=\mathrm{1.476.}$ Notice it lands between $5/3=1.67$ (pure He) and $7/5=1.40$ (pure O₂), as it must.

WHY only γ matters here: in $v=\sqrt{\gamma RT/M}$ the quantities $R$, $T$, $M$ are identical for both gases, so they cancel in a ratio and only γ survives (see Speed of Sound in Gas). $\frac{v_{\text{mono}}}{v_{\text{dia}}}=\sqrt{\frac{{\gamma }_{\text{mono}}}{{\gamma }_{\text{dia}}}}=\sqrt{\frac{5/3}{7/5}}=\sqrt{\frac{25}{21}}=\mathrm{1.091.}$ Sound travels about $9.1\%$ faster in the monatomic gas — a stiffer, springier medium because more of its heat goes into pressure (less into hidden rotation).

WHY this law: in an Adiabatic Process no heat leaves, and $P{V}^{\gamma }$ stays constant — γ appears as the exponent. $P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }\Rightarrow \frac{P_2}{P_1}={\left(\frac{V_1}{V_2}\right)}^{\gamma }={\left(\frac{1}{2}\right)}^{5/3}.$ $\frac{P_2}{P_1}=2^{-5/3}=\mathrm{0.3150.}$ Pressure drops to about $31.5\%$ of its start — steeper than the isothermal $\frac{1}{2}$, because the gas also cools.

As $f$ grows, $2/f\to 0$, so $\gamma \to 1^{+}$ (approaching 1 from above, never reaching it) — this is the flat right-hand tail of the curve in the figure. Physical meaning: with enormously many storage boxes, almost all incoming heat gets shared among internal motions, leaving very little to change temperature. The extra $R$ of expansion work becomes negligible relative to the huge $C_v$, so $C_p/C_v\to 1$. A gas can never actually hit γ = 1 because $R$ is a fixed, non-zero cost of expansion.

Use $f=2/(\gamma -1)$:

• At $300$ K: $f=2/0.400=5.00$.
• At $2000$ K: $f=2/0.350=5.714$. The effective $f$ rose from $5$ toward $7$. The non-integer $5.71$ means the vibrational mode is partially unfrozen — quantum mechanics lets it activate gradually with temperature, so at $2000$ K it contributes a fraction of its full 2 DOF. Full activation ($f=7$) would give $\gamma =9/7=1.286$.

From $\gamma =C_p/C_v$ we get $C_p=\gamma C_v$. Put this into Mayer's $C_p-C_v=R$: $\gamma C_v-C_v=R\Rightarrow C_v(\gamma -1)=R\Rightarrow \boxed{C_v=\frac{R}{\gamma -1}}.$ Then $C_p=\gamma C_v=\frac{\gamma R}{\gamma -1}.$ Verify (diatomic, $\gamma =7/5$): $C_v=\frac{R}{7/5-1}=\frac{R}{2/5}=\frac{5}{2}R✓,C_p=\frac{(7/5)R}{2/5}=\frac{7}{2}R✓.$ These match the parent table exactly — the formulas are just Mayer + definition rearranged, no equipartition needed.

L1: read γ = 1+2/f. L2: plug and invert. L3: split heat into $\Delta U$ and $W$. L4: mixtures add capacities, γ feeds sound & adiabatics. L5: prove the general $C_v=R/(\gamma -1)$ and read fractional $f$ as partial quantum activation.