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Each quadratic degree of freedom holds one half of $kT$: the thermal energy and heat capacity of gases and solids, the Dulong-Petit law, and the modes that freeze out.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to count the degrees of freedom of a gas molecule or an atom in a solid, find a thermal energy and a heat capacity from equipartition, predict a metal's specific heat from the Dulong-Petit law, and say when and why the classical count fails.
You know the ideal gas law $PV = nRT = NkT$, that the average translational kinetic energy of a gas molecule is $\tfrac{3}{2}kT$, and that heating a system raises its internal energy unless the system does work. This lesson extends the $\tfrac{3}{2}kT$ result to molecules that rotate and to atoms bound in a solid, and turns it into a prediction of how much energy it takes to warm anything.
| Term | What it means |
|---|---|
| Degree of freedom | An independent way a particle can store energy that enters the energy as a squared term, such as $\tfrac{1}{2}mv_x^2$ or $\tfrac{1}{2}k_sx^2$. |
| Equipartition theorem | In thermal equilibrium at temperature $T$, each quadratic degree of freedom has average energy $\tfrac{1}{2}kT$. |
| Thermal energy | The part of a system's internal energy that equipartition counts: $U = \tfrac{f}{2}NkT$. |
| Heat capacity | The energy needed per kelvin of temperature rise, $C = Q/\Delta T$, in J/K. |
| Heat capacity at constant volume | $C_V = (\partial U/\partial T)_V$: all the energy goes into the system, none into work. |
| Specific heat capacity | Heat capacity per kilogram, $c = C/m$, in J/(kg K). |
| Molar heat capacity | Heat capacity per mole, in J/(mol K). |
| Dulong-Petit law | A simple solid's molar heat capacity is about $3R = 24.9$ J/(mol K). |
| Frozen degree of freedom | A mode whose smallest quantum of energy is much larger than $kT$, so it holds almost no thermal energy. |
When a system is in thermal equilibrium, its energy is shared out among all the ways it can be stored. The equipartition theorem makes this exact for any term in the energy that is quadratic — a velocity squared or a displacement squared: at temperature $T$, the average energy in each such term is
$$\bar{E} = \tfrac{1}{2}kT,$$
whatever the mass, the spring constant or the moment of inertia. Each quadratic term is called a degree of freedom. If each of $N$ particles has $f$ of them, the thermal energy of the system is
$$U_{\text{thermal}} = N f \cdot \tfrac{1}{2}kT = \tfrac{f}{2}NkT = \tfrac{f}{2}nRT.$$
Counting $f$ is the whole of the work.
The heat capacity follows by differentiation. At constant volume no work is done, so all the energy supplied raises $U$ and
$$C_V = \frac{dU}{dT} = \tfrac{f}{2}Nk = \tfrac{f}{2}nR.$$
A monatomic gas has $C_V = \tfrac{3}{2}R = 12.5$ J/(mol K), a diatomic gas $\tfrac{5}{2}R = 20.8$ J/(mol K), and a solid $3R = 24.9$ J/(mol K).
Another way: picture
Think of the energy as money shared equally among bank accounts, one account per degree of freedom, with temperature setting how much each account holds. Helium opens three accounts, nitrogen five, and an atom in a crystal six. Warming by one kelvin puts the same deposit, one half of $k$, into every account, so the molecule with more accounts needs more energy for the same warming. A frozen account is one whose minimum deposit is bigger than anything the temperature can pay, so it stays empty.
Another way: steps
The theorem is a consequence of the Boltzmann distribution, which later in the course gives the probability of a state of energy $E$ as proportional to $e^{-E/kT}$. For a term $E = cq^2$, where $q$ is a velocity or a displacement and $c$ a positive constant, the average is
$$\bar{E} = \frac{\int cq^2 e^{-cq^2/kT}\,dq}{\int e^{-cq^2/kT}\,dq} = \tfrac{1}{2}kT.$$
The constant $c$ cancels: substituting $x = q\sqrt{c/kT}$ turns both integrals into pure numbers times powers of $kT/c$, and the ratio leaves $kT$ times $\int x^2e^{-x^2}dx \big/ \int e^{-x^2}dx = \tfrac{1}{2}$. That is why a heavy molecule and a light one, or a stiff spring and a soft one, all hold the same average energy per term. Only the form of the energy matters.
The same calculation shows what does not get $\tfrac{1}{2}kT$. A term that is not quadratic — gravitational energy $mgh$, linear in height — has average $kT$ instead. And the integral assumes $q$ can take any value continuously. When quantum mechanics restricts the energy to steps, and the first step is much larger than $kT$, the mode almost never gets excited and holds almost nothing.
Every problem in this lesson follows the same route.
Checks. A molar $C_V$ must be a half-integer multiple of $R$: $1.5R$, $2.5R$, $3R$, $3.5R$. A monatomic gas has the smallest heat capacity of all ideal gases. For a solid, the molar heat capacity near room temperature should be close to $25$ J/(mol K); if the specific heat per kilogram seems to vary wildly between metals, that is because their molar masses do, not their physics. Finally, the answer for $U$ is the thermal energy only: it leaves out chemical bonds and the rest energy of the particles, which do not change in a warming.
Measured heat capacities agree with equipartition for monatomic gases and for most metals at room temperature, and fail in revealing ways elsewhere.
Hydrogen gas. Below about $80$ K its molar $C_V$ is $\tfrac{3}{2}R$, as if it could not rotate. Between about $100$ K and $1000$ K it is $\tfrac{5}{2}R$. Above a few thousand kelvin it climbs toward $\tfrac{7}{2}R$ as the bond starts to vibrate. Each step is a mode unfreezing when $kT$ becomes comparable to its quantum of energy. Hydrogen's small moment of inertia makes its rotational quanta unusually large, which is why the effect is so clear in it.
Diamond. Carbon atoms are light and the bonds very stiff, so the lattice vibrations have high frequencies and large quanta. At room temperature diamond's specific heat is $509$ J/(kg K), against a Dulong-Petit prediction of about $2080$: three quarters of its vibrational degrees of freedom are frozen. Einstein's 1907 explanation of this, treating each atom as a quantum oscillator, was one of the first successes of quantum theory. The same reasoning, taken up in the lesson on the Einstein solid, predicts that every solid's heat capacity falls to zero as $T \to 0$.
So equipartition is a high-temperature law. It is reliable when $kT$ is much larger than the spacing of a mode's energy levels, and silent when it is not.
Put a $1.0$ kg aluminum pan holding $1.0$ kg of water on a $2000$ W burner and the pan is hot long before the water. Equipartition says why.
Aluminum's molar mass is $0.02698$ kg/mol, so a kilogram is $37$ moles of atoms and, by Dulong-Petit, needs $37 \times 24.9 = 924$ J per kelvin; the measured value is $897$. Water's molar mass is $0.018$ kg/mol, so a kilogram is $55.5$ moles of molecules, and each molecule is a flexible cluster of three atoms held by hydrogen bonds to its neighbors, with many more ways to store energy than an aluminum atom. Water's specific heat is $4186$ J/(kg K), about $75$ J/(mol K), or nine times $\tfrac{1}{2}R$.
Warming both from $20$ °C to $90$ °C takes $897 \times 70 = 62\,800$ J for the pan and $4186 \times 70 = 293\,000$ J for the water, so of every joule the burner delivers, less than a fifth goes into the pan. At $2000$ W, and ignoring losses, the whole job takes $(62\,800 + 293\,000)/2000 \approx 178$ s — three minutes — while the pan alone would have needed half a minute. Cooks exploit the same numbers the other way: a heavy cast-iron skillet stores enough energy that a cold steak barely dents its temperature.
Before molecules could be imaged, heat capacities were one of the few ways to learn their shapes. Measured at constant volume near room temperature, the molar heat capacities of gases fall into groups. Helium, neon and argon give $12.5$ J/(mol K), which is $\tfrac{3}{2}R$: single atoms. Nitrogen, oxygen, hydrogen and carbon monoxide give $20.7$ to $21.0$, which is $\tfrac{5}{2}R$: dumbbells. Water vapor and methane give about $25$ to $27$, near $3R$: nonlinear molecules, which can rotate about three axes rather than two.
Carbon dioxide gives $28.5$ J/(mol K), which fits no whole count. It is a linear molecule, $\tfrac{5}{2}R = 20.8$ from translation and rotation, plus a partly excited bending vibration: its bending mode has a low frequency, so at room temperature it is neither frozen nor fully active. That same bending vibration is how carbon dioxide absorbs infrared radiation from the ground, which makes it a greenhouse gas while nitrogen and oxygen are not. A number measured with a calorimeter and a thermometer carries information about the shape and stiffness of a molecule far too small to see.
Two samples at the same temperature need not hold the same energy, and one kelvin of warming need not cost the same. Temperature fixes the average energy of each degree of freedom. A mole of nitrogen at room temperature holds five thirds of the thermal energy of a mole of helium beside it, and a mole of copper holds twice as much as the helium, because its atoms store energy in their springs as well as in their motion.
The opposite error is to count atoms rather than degrees of freedom: to reason that a nitrogen molecule, being two atoms, should hold twice helium's energy, $f = 6$. Two atoms joined rigidly cannot move independently, so the molecule has three translations and two rotations. A rotation about the bond itself stores nothing, because the atoms are too small to have a moment of inertia about it worth exciting. The count is five, until the bond is hot enough to vibrate.
A party balloon holds $0.20$ mol of helium at $300$ K. Count helium's degrees of freedom.
$f = 3$
Helium is monatomic: three translations and nothing else.
Find the average thermal energy of one atom.
$\tfrac{3}{2}kT = 1.5 \times 1.38 \times 10^{-23} \times 300 = 6.21 \times 10^{-21}\ \text{J}$
Each of the three terms holds one half of $kT$.
Count the atoms.
$N = nN_A = 0.20 \times 6.02 \times 10^{23} = 1.20 \times 10^{23}$
Avogadro's number of atoms per mole.
Multiply the number of atoms by the energy of each.
$U = N \times \tfrac{3}{2}kT = 1.20 \times 10^{23} \times 6.21 \times 10^{-21} \approx 748\ \text{J}$
Thermal energy is the sum over all the atoms.
Check with the molar form.
$U = \tfrac{3}{2}nRT = 1.5 \times 0.20 \times 8.31 \times 300 = 747.9\ \text{J}$
Since $R = N_Ak$, both routes must agree, and they do.
A room holds $9960$ mol of air at $15$ °C. Count the degrees of freedom of an air molecule.
$f = 3_{\text{trans}} + 2_{\text{rot}} = 5$
Air is almost all nitrogen and oxygen, both diatomic, with their vibrations frozen at room temperature.
Write the molar heat capacity at constant volume.
$\dfrac{C_V}{n} = \tfrac{5}{2}R = 2.5 \times 8.31 = 20.775\ \text{J/(mol K)}$
Differentiating $\tfrac{5}{2}RT$ per mole with respect to $T$.
Multiply by the amount for the whole room.
$C_V = 9960 \times 20.775 \approx 2.07 \times 10^{5}\ \text{J/K}$
Heat capacity is extensive: twice the air, twice the capacity.
Find the temperature change for warming to $20$ °C.
$\Delta T = 20 - 15 = 5\ \text{K}$
A change of five Celsius degrees is a change of five kelvin.
Multiply the heat capacity by the temperature change.
$Q = C_V\Delta T = 2.07 \times 10^{5} \times 5 \approx 1.03 \times 10^{6}\ \text{J}$
At constant volume all the energy supplied becomes thermal energy.
Find how long a $1.5$ kW heater takes.
$t = \dfrac{Q}{P} = \dfrac{1.03 \times 10^{6}}{1500} \approx 690\ \text{s} \approx 11.5\ \text{min}$
The air alone is quick to warm. A real room takes much longer because the walls, floor and furniture, with far larger heat capacities, warm too.
Count the degrees of freedom of one copper atom in the crystal.
$f = 3_{\text{kinetic}} + 3_{\text{potential}} = 6$
Each atom oscillates about its site in three directions, and each oscillation has a kinetic and a potential term.
Write the thermal energy and heat capacity of one mole.
$U = \tfrac{6}{2}RT = 3RT \quad\Rightarrow\quad \dfrac{C}{n} = 3R$
In a solid the volume barely changes, so $C_V$ and $C_P$ nearly agree.
Evaluate the molar heat capacity.
$3R = 3 \times 8.31 = 24.93\ \text{J/(mol K)}$
The same for every simple solid: this is the Dulong-Petit law.
Put copper's molar mass in kilograms per mole.
$M = 63.55\ \text{g/mol} = 0.06355\ \text{kg/mol}$
A specific heat is per kilogram.
Divide to get the specific heat.
$c = \dfrac{24.93}{0.06355} \approx 392\ \text{J/(kg K)}$
One kilogram of copper is $1/0.06355 = 15.7$ moles of atoms.
Compare with the measured value.
$c_{\text{measured}} = 385\ \text{J/(kg K)}, \qquad \dfrac{392 - 385}{385} \approx 2\%$
Copper's lattice vibrations are fully excited at room temperature, so equipartition works.
Predict diamond the same way and compare.
$c = \dfrac{24.93}{0.01201} \approx 2080\ \text{J/(kg K)}, \qquad c_{\text{measured}} = 509\ \text{J/(kg K)}$
A factor of four too high: diamond's stiff, light lattice has most of its vibrations frozen at room temperature.
Count nitrogen's degrees of freedom.
$f = 5$
Three translations and two rotations; the vibration is frozen.
Write the energy in terms of the quantities given.
$U = \tfrac{5}{2}nRT = \tfrac{5}{2}PV$
The ideal gas law replaces $nRT$ by $PV$, so neither $n$ nor $T$ is needed.
Substitute in SI units.
Two rigid tanks hold $3$ mol of helium and $3$ mol of nitrogen, both at room temperature. Each is warmed by $13$ K. How many times as much energy does the nitrogen need as the helium?
Complete the worked solution: what is the thermal energy of $3$ mol of argon at $200$ K, and what does it become when the kelvin temperature doubles? Use $R = 8.31$ J/(mol K).
Count argon's degrees of freedom and write its thermal energy.
$f = 3 \quad\Rightarrow\quad U = \tfrac{3}{2}nRT$
A single atom can move in three directions and has no rotation that stores energy.
Multiply out $nRT$.
$nRT = 3 \times 8.31 \times 200 =$ a J
This is the energy scale every equipartition result is a multiple of.
Take three halves of it.
$U = 1.5 \times nRT =$ b J
Each of the three degrees of freedom holds one half of $RT$ per mole.
Double the kelvin temperature.
$U' = \tfrac{3}{2}nR(2T) = 2U =$ c J
The thermal energy is proportional to the absolute temperature.
Match each system to the number of quadratic degrees of freedom per particle that equipartition gives it, near $200$ K unless stated.
| 3 | 5 | 7 | 6 | |
|---|---|---|---|---|
| an atom of helium gas | ||||
| a molecule of nitrogen gas at room temperature | ||||
| a molecule of nitrogen gas hot enough that its bond vibrates | ||||
| an atom in a crystal of copper |
Three samples each contain $1$ mol and are at $400$ K. Fill in the degrees of freedom per particle and the thermal energy of each sample. Use $R = 8.31$ J/(mol K).
| degrees of freedom $f$ | thermal energy (J) | |
|---|---|---|
| helium gas | ||
| nitrogen gas | ||
| copper solid |
A rigid laboratory tank holds $2$ mol of nitrogen at room temperature. How much energy must be supplied to warm it by $5$ K? Use $R = 8.31$ J/(mol K).
Answer: unit: J / MJ / kJ
Use equipartition to predict the specific heat capacity of zinc, whose molar mass is $65.38$ g/mol. Give your answer in J/(kg K). Use $R = 8.31$ J/(mol K).
Answer: J/(kg K)
A small sealed room holds $8000$ mol of air, treated as a diatomic ideal gas. A $1500$ W space heater warms the air at constant volume by $3$ K. Ignoring the walls and furniture, how long does it take? Use $R = 8.31$ J/(mol K).
Answer: unit: h / s / min
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Three samples each contain $3$ mol and are at $400$ K. Fill in the degrees of freedom per particle and the thermal energy of each sample. Use $R = 8.31$ J/(mol K).
| degrees of freedom $f$ | thermal energy (J) | |
|---|---|---|
| helium gas | ||
| nitrogen gas | ||
| copper solid |
You can find thermal energies and heat capacities by counting degrees of freedom. Explain to someone why a mole of nitrogen needs more energy than a mole of helium to warm by one kelvin, and why diamond breaks the Dulong-Petit law.
14. Your turn: a $2.0$ L flask holds nitrogen at $100$ kPa and room temperature. What is its thermal energy?, step 3
$U = 2.5 \times (1.00 \times 10^{5}) \times (2.0 \times 10^{-3}) = 500\ \text{J}$
A litre is $10^{-3}$ m³, and a pascal cubic meter is a joule.