University Physics — Year 1 · Bachelor Year 1

20Kinetic Theory and the Perfect Gas

The air in a classroom weighs as much as a grown man, and contains more molecules than there are grains of sand on Earth, each flying at the speed of a rifle bullet and colliding ten billion times a second with its neighbors. Nobody can follow one of them; nobody needs to. A handful of averages — pressure, temperature, volume, amount — describe the air completely for every purpose of engineering, and a simple picture of molecules bouncing off walls explains where those averages come from and how they are related. This chapter opens the thermodynamics part of the volume by looking at matter from both ends: the microscopic chaos and the macroscopic calm, and the kinetic theory that connects them in the case of the perfect gas.

20.1 Two scales of description

Definition 20.1 (Microscopic and macroscopic scales)

At the microscopic scale matter is made of molecules: $1\,\mathrm{mol}$ contains $N_A = 6.02 \times 10^{23}\,$ of them, each about $0.3\,\mathrm{nm}$ across; in a gas at ordinary conditions they are $3\,\mathrm{nm}$ apart, move at hundreds of meters per second and collide every $70\,\mathrm{nm}$ or so. At the macroscopic scale a system is described by a few state variables — pressure $P$ , volume $V$ , temperature $T$ , amount of substance $n$ — that are averages over enormous numbers of molecules. A variable is extensive if it doubles when the system is doubled ( $V$ , $n$ , mass, energy), intensive if it does not ( $P$ , $T$ , density).

Definition 20.2 (Thermodynamic equilibrium)

A system is in thermodynamic equilibrium when its state variables are uniform and constant in time and no macroscopic flow of matter or energy crosses it. Only then are $P$ and $T$ defined for the system as a whole; the equation of state links them to $V$ and $n$ .

Definition 20.3 (Pressure)

A fluid at rest pushes on every surface element $\dd S$ of a wall with a force $\dd\vect F = P\,\dd S\,\vect n$ normal to it, toward the wall; $P$ is the pressure, in pascals ( $1\,\mathrm{Pa} = 1\,\mathrm{N}/\mathrm{m}^{2}$ ); $1\,\mathrm{bar} = 1 \times 10^{5}\,\mathrm{Pa}$ , $1\,\mathrm{atm} = 1.013 \times 10^{5}\,\mathrm{Pa}$ . At a given point it does not depend on the orientation of the surface (Chapter 21).

Definition 20.4 (Temperature)

Two systems in contact through a wall that lets energy pass reach, in time, a common state: they are then at the same temperature, and two bodies each in equilibrium with a third are in equilibrium with each other (the zeroth law). The kelvin scale is fixed by $k_B = 1.380\,649 \times 10^{-23}\,\mathrm{J}/\mathrm{K}$ exactly; Celsius: $\theta = T - 273.15$ . The kinetic theory below gives $T$ its microscopic meaning.

20.2 The kinetic model of the perfect gas

Definition 20.5 (Perfect gas, microscopic model)

A perfect gas (ideal gas) is a collection of $N$ molecules, treated as points of mass $m$ , moving freely and at random — no interaction except brief collisions — in a container of volume $V$ . At equilibrium the velocity distribution is isotropic and stationary; $n^* = N/V$ is the number density and $\langle v^2\rangle$ the mean square speed.

Theorem 20.6 (Kinetic pressure)

The pressure exerted by the gas on the walls is

P = \tfrac13\,n^*m\langle v^2\rangle = \tfrac23\,n^*\langle\epsilon_k\rangle ,

where $\langle\epsilon_k\rangle = \tfrac12 m\langle v^2\rangle$ is the mean kinetic energy of a molecule.

Partial proof. Take the simplified model in which the molecules all have speed $u$ and move along the three axes, one sixth in each direction. A molecule hitting the wall $x = 0$ head-on bounces back elastically and delivers the momentum $2mu$ to it. In time $\dd t$ , the molecules moving toward the wall that reach an area $S$ are those within the slab $u\,\dd t$ thick, one sixth of the $n^*Su\,\dd t$ molecules there. Force $=$ momentum per time $= (n^*Su\,\dd t/6)(2mu)/\dd t = \tfrac13 n^*mu^2S$ , so $P = \tfrac13 n^*mu^2$ . Averaging over the actual distribution of speeds and directions replaces $u^2$ by $\langle v^2\rangle$ (the isotropy gives $\langle v_x^2\rangle = \tfrac13\langle v^2\rangle$ ); the Year 2 volume does it in full. ∎

Left: the kinetic origin of pressure — molecules striking a wall and bouncing back deliver momentum to it; their number per second times 2mu is the force. Right: the actual speeds are spread (Maxwell’s distribution, here for nitrogen at two temperatures); the model’s single u is replaced by the root-mean-square speed. — Left: the kinetic origin of pressure — molecules striking a wall and bouncing back deliver momentum to it; their number per second times $2mu$ is the force. Right: the actual speeds are spread (Maxwell’s distribution, here for nitrogen at two temperatures); the model’s single $u$ is replaced by the root-mean-square speed.

Definition 20.7 (Kinetic temperature; equation of state)

The temperature of a perfect gas is defined by the mean kinetic energy of its molecules,

\langle\epsilon_k\rangle = \tfrac12 m\langle v^2\rangle = \tfrac32 k_BT ,

so that, combining with Theorem 20.6,

PV = Nk_BT = nRT, \qquad R = N_Ak_B = 8.314\,\mathrm{J}/(\mathrm{mol}\,\mathrm{K}) ,

the equation of state of the perfect gas. The root-mean-square speed is $u = \sqrt{\langle v^2\rangle} = \sqrt{3k_BT/m} = \sqrt{3RT/M}$ , $M$ the molar mass.

Example 20.8 (Numbers for air)

Nitrogen at $300\,\mathrm{K}$ : $u = \sqrt{3 \times 8.314 \times 300/0.028} = 517\,\mathrm{m}/\mathrm{s}$ ; hydrogen, $1930\,\mathrm{m}/\mathrm{s}$ ; the heavier the slower, as $1/\sqrt M$ . At $1\,\mathrm{bar}$ and $300\,\mathrm{K}$ , $n^* = P/k_BT = 2.4 \times 10^{25}\,\mathrm{m}^{-3}$ : $24\,\mathrm{L}$ per mole, $2.7 \times 10^{22}\,$ molecules in a liter. A classroom of $200\,\mathrm{m}^{3}$ : $8300\,\mathrm{mol}$ , $240\,\mathrm{kg}$ of air.

Remark 20.9 (Why the model works, and when it fails)

Between collisions a molecule of air flies a mean free path $\lambda \approx 1/(\sqrt2\pi d^2n^*) \approx 70\,\mathrm{nm}$ — two hundred times its size: the gas is mostly empty, interactions are rare, and $PV = nRT$ holds to better than a percent at ordinary pressures. At high pressure or low temperature the molecules’ own volume and their attraction matter: van der Waals’s equation $(P + an^2/V^2)(V - nb) = nRT$ corrects for both, and below a critical temperature the gas liquefies (Chapter 25).

The Clapeyron diagram of a perfect gas: at each temperature the isotherm is a hyperbola P = nRT/V, higher for higher T. A transformation is a path in this plane; a cycle, a closed loop (). — The Clapeyron diagram of a perfect gas: at each temperature the isotherm is a hyperbola $P = nRT/V$ , higher for higher $T$ . A transformation is a path in this plane; a cycle, a closed loop (Chapter 24).

20.3 Internal energy of the perfect gas

Definition 20.10 (Internal energy)

The internal energy $U$ of a system is the sum of the kinetic energies of its molecules in the frame where the system is at rest and of their mutual potential energies; it is an extensive state function.

Proposition 20.11 (Internal energy of perfect gases)

For a perfect gas $U$ depends on $T$ alone (first Joule law):

\begin{gather*} U = \tfrac32 nRT \quad (\text{monatomic: He, Ne, Ar}), \\ U = \tfrac52 nRT \quad (\text{diatomic at ordinary temperatures: N}_2, \text{O}_2, \text{H}_2), \end{gather*}

so the molar heat capacity at constant volume $C_{V,m} = \dd U_m/\dd T$ is $\tfrac32 R = 12.5\,\mathrm{J}/(\mathrm{mol}\,\mathrm{K})$ and $\tfrac52 R = 20.8\,\mathrm{J}/(\mathrm{mol}\,\mathrm{K})$ respectively.

Partial proof. No mutual potential energy by hypothesis, so $U = N\langle\epsilon_k\rangle = \tfrac32 Nk_BT$ for point molecules. A diatomic molecule also rotates: each of its two rotational degrees of freedom carries, at equilibrium, the same $\tfrac12 k_BT$ as each translational one (the equipartition theorem, admitted; the Year 3 volume derives it and explains why the vibration joins in only above a thousand kelvins). ∎

Example 20.12 (A room’s worth)

The $8300\,\mathrm{mol}$ of air in the classroom at $300\,\mathrm{K}$ : $U = \tfrac52 \times 8300 \times 8.314 \times 300 = 5.2 \times 10^{7}\,\mathrm{J}$ — the energy of a liter and a half of gasoline, held in molecular motion. Warming the room by $1$ K costs $\tfrac52 nR = 170\,\mathrm{kJ}$ (at constant volume).

20.4 Condensed phases

Proposition 20.13 (Model of the incompressible, indilatable phase)

Liquids and solids are modeled, at this level, as phases of fixed volume — neither compressible nor dilatable — whose internal energy depends on $T$ alone: $\dd U = C\,\dd T$ with $C = mc$ , $c$ the specific heat capacity ( $4180\,\mathrm{J}/(\mathrm{kg}\,\mathrm{K})$ for water, $450\,\mathrm{J}/(\mathrm{kg}\,\mathrm{K})$ for steel). The real, small deviations are measured by the thermal expansion coefficient $\alpha = (1/V)(\partial V/\partial T)_P$ ( $\sim1 \times 10^{-5}\,\mathrm{K}^{-1}$ for metals, $2 \times 10^{-4}\,\mathrm{K}^{-1}$ for liquids) and the isothermal compressibility $\chi_T = -(1/V)(\partial V/\partial P)_T$ ( $\sim5 \times 10^{-10}\,\mathrm{Pa}^{-1}$ for water, against $1/P \approx 1 \times 10^{-5}\,\mathrm{Pa}^{-1}$ for a gas).

Proof. Admitted at this level. ∎

Example 20.14 (How incompressible)

Water at the bottom of the Mariana trench ( $1100\,\mathrm{bar}$ ) is compressed by $\chi_T\Delta P \approx 5\times10^{-10} \times 1.1\times10^8 = 5\%$ ; a $30\,\mathrm{m}$ steel rail warmed by $50$ K lengthens by $\alpha L\Delta T = 18\,\mathrm{mm}$ — the reason for expansion joints. Against a gas, whose volume halves under a doubled pressure, the model of a fixed volume is good to a few percent in most situations of this volume.

20.5 Exercises

Exercise 20.1 ★

Number of molecules in $1.0\,\mathrm{L}$ of gas at $0{}^{\circ}\mathrm{C}$ and $1\,\mathrm{atm}$ ; mean distance between neighbors; compare with the molecular size $0.3\,\mathrm{nm}$ .

Solution

Solution of Exercise 20.1.

$N = PV/k_BT = 1.013\times10^5 \times 10^{-3}/(1.38\times10^{-23} \times 273) = 2.7 \times 10^{22}$ ; $n^{*-1/3} = (3.7\times10^{-26})^{1/3} = 3.3\,\mathrm{nm}$ , ten times the molecular size.

Exercise 20.2 ★

Root-mean-square speeds of H $_2$ , N $_2$ , O $_2$ and CO $_2$ at $300\,\mathrm{K}$ , and of N $_2$ at $77\,\mathrm{K}$ (boiling nitrogen).

Solution

Solution of Exercise 20.2.

$u = \sqrt{3RT/M}$ : H $_2$ $1930\,\mathrm{m}/\mathrm{s}$ , N $_2$ $517\,\mathrm{m}/\mathrm{s}$ , O $_2$ $484\,\mathrm{m}/\mathrm{s}$ , CO $_2$ $412\,\mathrm{m}/\mathrm{s}$ ; N $_2$ at $77\,\mathrm{K}$ : $262\,\mathrm{m}/\mathrm{s}$ .

Exercise 20.3 ★

A car tire of volume $30\,\mathrm{L}$ is inflated to a gauge pressure of $2.5\,\mathrm{bar}$ at $20{}^{\circ}\mathrm{C}$ . Amount and mass of air inside; absolute pressure after a drive that heats it to $50{}^{\circ}\mathrm{C}$ .

Solution

Solution of Exercise 20.3.

$P = 3.5\,\mathrm{bar}$ absolute: $n = PV/RT = 3.5\times10^5 \times 0.030/(8.314 \times 293) = 4.3\,\mathrm{mol}$ , $125\,\mathrm{g}$ . At $323\,\mathrm{K}$ , $P \propto T$ : $3.86\,\mathrm{bar}$ (gauge $2.85\,\mathrm{bar}$ ).

Exercise 20.4 ★

Estimate the mean free path of air molecules at $1\,\mathrm{bar}$ , $300\,\mathrm{K}$ ( $d = 0.37\,\mathrm{nm}$ ), and the collision frequency of one molecule. At what pressure does the mean free path reach $1\,\mathrm{m}$ (a “vacuum”)?

Solution

Solution of Exercise 20.4.

$n^* = P/k_BT = 2.4 \times 10^{25}\,\mathrm{m}^{-3}$ ; $\lambda = 1/(\sqrt2\pi \times 1.37\times 10^{-19} \times 2.4\times10^{25}) = 6.8 \times 10^{-8}\,\mathrm{m}$ ; frequency $u/\lambda \approx 500/7\times10^{-8} = 7 \times 10^{9}\,\mathrm{s}^{-1}$ . $\lambda \propto 1/P$ : $1\,\mathrm{m}$ at $P \approx 10^5 \times 7\times10^{-8} = 7 \times 10^{-3}\,\mathrm{Pa}$ .

Exercise 20.5 ★★

Density of air at $1.013\,\mathrm{bar}$ and $20{}^{\circ}\mathrm{C}$ ( $M = 29\,\mathrm{g}/\mathrm{mol}$ ), then at $100{}^{\circ}\mathrm{C}$ . Lift of a $2800\,\mathrm{m}^{3}$ hot-air balloon (difference of the two weights of air).

Solution

Solution of Exercise 20.5.

$\rho = PM/RT$ : $1.20\,\mathrm{kg}/\mathrm{m}^{3}$ at $293\,\mathrm{K}$ , $0.946\,\mathrm{kg}/\mathrm{m}^{3}$ at $373\,\mathrm{K}$ . Lift $(1.20 - 0.95) \times 2800 \times 9.81 = 7.0\,\mathrm{kN}$ : about $720\,\mathrm{kg}$ .

Exercise 20.6 ★★

Air is $21\%$ O $_2$ by molecules. Partial pressure of oxygen at sea level; at $5000\,\mathrm{m}$ where $P = 0.54\,\mathrm{atm}$ ; at the summit of Everest ( $0.33\,\mathrm{atm}$ ). Why does the body struggle there?

Solution

Solution of Exercise 20.6.

$P_{\mathrm{O}_2} = 0.21P$ : $0.21\,\mathrm{atm}$ ; $0.11\,\mathrm{atm}$ at $5000\,\mathrm{m}$ ; $0.07\,\mathrm{atm}$ on Everest — a third of sea level. Oxygen enters the blood in proportion to its partial pressure; at a third, the lungs cannot saturate the hemoglobin.

Exercise 20.7 ★★

One mole of CO $_2$ at $300\,\mathrm{K}$ in $1.0\,\mathrm{L}$ : pressure from the perfect-gas law, then from van der Waals ( $a = 0.364\,\mathrm{Pa}\,\mathrm{m}^{6}/\mathrm{mol}^{2}$ , $b = 4.27 \times 10^{-5}\,\mathrm{m}^{3}/\mathrm{mol}$ ). Which correction dominates here?

Solution

Solution of Exercise 20.7.

Perfect gas: $P = RT/V = 2494/10^{-3} = 24.9\,\mathrm{bar}$ . Van der Waals: $RT/(V - b) - a/V^2 = 2494/9.57\times10^{-4} - 0.364/10^{-6} = 26.1 - 3.6 = 22.4\,\mathrm{bar}$ : the attraction ( $-3.6$ bar) outweighs the excluded volume ( $+1.2$ bar); real CO $_2$ is $10\%$ below the ideal value.

Exercise 20.8 ★★

Internal energy of the air in a $50\,\mathrm{m}^{3}$ room at $1\,\mathrm{bar}$ , $300\,\mathrm{K}$ ; energy to warm it by $5\,\mathrm{K}$ at constant volume; time with a $2\,\mathrm{kW}$ heater (no losses).

Solution

Solution of Exercise 20.8.

$n = PV/RT = 10^5 \times 50/(8.314 \times 300) = 2000\,\mathrm{mol}$ ; $U = \tfrac52 nRT = 1.25 \times 10^{7}\,\mathrm{J}$ ; $\Delta U = \tfrac52 nR \times 5 = 208\,\mathrm{kJ}$ ; $104\,\mathrm{s}$ at $2\,\mathrm{kW}$ .

Exercise 20.9 ★★

A $30\,\mathrm{m}$ steel rail ( $\alpha = 1.2 \times 10^{-5}\,\mathrm{K}^{-1}$ ) between $-10{}^{\circ}\mathrm{C}$ and $40{}^{\circ}\mathrm{C}$ : length change. A brass ring ( $\alpha = 1.9 \times 10^{-5}\,\mathrm{K}^{-1}$ ) of inner diameter $49.95\,\mathrm{mm}$ must slip over a $50.00\,\mathrm{mm}$ shaft: by how much must it be heated?

Solution

Solution of Exercise 20.9.

$\Delta L = \alpha L\Delta T = 1.2\times10^{-5} \times 30 \times 50 = 18\,\mathrm{mm}$ . Ring: $\Delta d/d = \alpha\Delta T = 0.05/49.95 = 10^{-3}$ : $\Delta T = 10^{-3}/ 1.9\times10^{-5} = 53\,\mathrm{K}$ (the hole expands like the metal around it).

Exercise 20.10 ★★★

Energy per molecule at $300\,\mathrm{K}$ ( $\tfrac32 k_BT$ ) in joules and eV; molar heat capacities $C_{V,m}$ of argon and of nitrogen; why is that of nitrogen larger though its molecules are lighter?

Solution

Solution of Exercise 20.10.

$\tfrac32 k_BT = 6.2 \times 10^{-21}\,\mathrm{J} = 0.039\,\mathrm{eV}$ . Argon $\tfrac32 R = 12.5\,\mathrm{J}/(\mathrm{mol}\,\mathrm{K})$ ; nitrogen $\tfrac52 R = 20.8\,\mathrm{J}/(\mathrm{mol}\,\mathrm{K})$ : the diatomic molecule also stores energy in rotation (two extra degrees of freedom), whatever its mass.

Exercise 20.11 ★★★

Redo the derivation of the kinetic pressure in the six-direction model, then show that for an isotropic distribution of velocities $\langle v_x^2\rangle = \tfrac13\langle v^2\rangle$ and that the pressure formula is unchanged.

Solution

Solution of Exercise 20.11.

Six-direction model: as in the text, $P = \tfrac13 n^*mu^2$ . Isotropy: $\langle v_x^2\rangle = \langle v_y^2\rangle = \langle v_z^2\rangle$ and their sum is $\langle v^2\rangle$ , so each is $\tfrac13\langle v^2\rangle$ . The flux argument with a distribution gives $P = n^*m\langle v_x^2\rangle$ (molecules with $v_x > 0$ hitting the wall, momentum $2mv_x$ each, flux $n^*v_x/2$ per speed class, averaged), i.e. $\tfrac13 n^*m\langle v^2\rangle$ — the same.

Exercise 20.12 ★★★

Root-mean-square speeds of H $_2$ and O $_2$ at $300\,\mathrm{K}$ compared with the Earth’s escape velocity, and with the Moon’s ( $2.4\,\mathrm{km}/\mathrm{s}$ ). Knowing that the fraction of molecules faster than $3u$ is small but not zero, explain why the Earth has lost its hydrogen and the Moon its whole atmosphere.

Solution

Solution of Exercise 20.12.

H $_2$ : $1.9\,\mathrm{km}/\mathrm{s}$ , O $_2$ : $0.48\,\mathrm{km}/\mathrm{s}$ ; Earth $11.2\,\mathrm{km}/\mathrm{s}$ , Moon $2.4\,\mathrm{km}/\mathrm{s}$ . For O $_2$ on Earth, $v_e/u = 23$ : the tail of the distribution beyond that is utterly negligible; for H $_2$ , $v_e/u \approx 6$ : a minute but steady fraction escapes each year, and over billions of years the hydrogen is gone. On the Moon, $v_e/u$ is $5$ for O $_2$ and $1.2$ for H $_2$ : everything leaked away.

20.6 Problem: The air in the room

Problem 20.1

Weekend problem — a classroom of air: counting its molecules, weighing it, timing their flights, and recovering from the same model the lift of a balloon, the thinness of mountain air and the speed of sound

A classroom measures $10 \times 8 \times 2.5\,\mathrm{m}$ ; its air is at $P = 1.013 \times 10^{5}\,\mathrm{Pa}$ , $T = 293\,\mathrm{K}$ , molar mass $M = 29.0\,\mathrm{g}/\mathrm{mol}$ , $79\%$ N $_2$ and $21\%$ O $_2$ by molecules. $R = 8.314\,\mathrm{J}/(\mathrm{mol}\,\mathrm{K})$ , $k_B = 1.38 \times 10^{-23}\,\mathrm{J}/\mathrm{K}$ , $N_A = 6.02 \times 10^{23}$ , $g = 9.81\,\mathrm{m}/\mathrm{s}^{2}$ .

Part I — Counting and weighing.

Amount of air (moles) and number of molecules in the room.
Mass of the air; compare with a person.
Number density $n^*$ and mean distance between molecules ( $n^{*-1/3}$ ); compare with the molecular diameter $0.37\,\mathrm{nm}$ .
Root-mean-square speed of N $_2$ and of O $_2$ ; why does the heavier gas move more slowly at the same temperature?
Mean kinetic energy of one molecule; total translational kinetic energy of the room’s air; compare with the kinetic energy of a car at $100\,\mathrm{km}/\mathrm{h}$ .
Mean free path $\lambda \approx 1/(\sqrt2\pi d^2n^*)$ and the mean time between collisions of a molecule.

Part II — Pressure from impacts.

Using the six-direction model with $u = 500\,\mathrm{m}/\mathrm{s}$ , compute the number of impacts per second on $1\,\mathrm{cm}^{2}$ of wall.
Compute the momentum delivered per second, and check that it reproduces the atmospheric pressure.
Total force of the air on one wall of $8\,\mathrm{m}$ by $2.5\,\mathrm{m}$ ; why does the wall not move?
The room is warmed to $303\,\mathrm{K}$ with its windows closed and sealed: new pressure, and the net force on the wall now.
Same warming with a window open: what leaves, and how much?

Part III — Buoyancy.

Density of the room’s air, and of air at $373\,\mathrm{K}$ at the same pressure.
A hot-air balloon of volume $2800\,\mathrm{m}^{3}$ is filled with air at $373\,\mathrm{K}$ : weight of the hot air, weight of the displaced cold air, and the available lift (Archimedes, Chapter 21).
The envelope, basket, burner and gas weigh $300\,\mathrm{kg}$ : how many $75\,\mathrm{kg}$ passengers can it lift?
To what temperature should the air be heated to double the lift, and why is that not done (the envelope is nylon)?
A helium balloon of the same volume at $293\,\mathrm{K}$ ( $M = 4\,\mathrm{g}/\mathrm{mol}$ ): lift? Why do tourist balloons use hot air nonetheless?

Part IV — Thin air and the speed of sound.

Partial pressure of oxygen in the room.
At the summit of Everest $P = 0.33\,\mathrm{atm}$ and $T \approx 250\,\mathrm{K}$ : number density of oxygen molecules compared with the room’s.
The atmosphere’s pressure falls with height roughly as $P(z) = P_0\eu^{-z/H}$ with $H = RT/Mg$ (Chapter 21): compute $H$ for $260\,\mathrm{K}$ and the height at which $P = P_0/2$ .
The speed of sound in a perfect gas is $c_s = \sqrt{\gamma RT/M}$ with $\gamma = 1.4$ for air (Chapter 22): compute it at $293\,\mathrm{K}$ and compare with the rms speed; interpret the closeness.
How does $c_s$ vary with $T$ ? Speed of sound in the room at $303\,\mathrm{K}$ , and in helium at $293\,\mathrm{K}$ ( $\gamma = 5/3$ ): why does a helium voice sound high?
At the top of Everest, is sound faster or slower than in the room? By how much?
The room’s air is replaced by argon at the same $P$ and $T$ : what changes in the count, the mass, the speeds, the internal energy?
Internal energy of the room’s air ( $U = \tfrac52 nRT$ ) and the energy to warm it by $10\,\mathrm{K}$ at constant volume.
Summarize in three lines what the kinetic model delivered: which macroscopic quantities it explained, and from which single microscopic average.

Solution

Solution of Problem 20.1.

1. $V = 200\,\mathrm{m}^{3}$ : $n = PV/RT = 1.013\times10^5 \times 200/(8.314 \times 293) = 8320\,\mathrm{mol}$ ; $N = nN_A = 5.0 \times 10^{27}$ .

2. $m = nM = 241\,\mathrm{kg}$ : three people.

3. $n^* = N/V = 2.5 \times 10^{25}\,\mathrm{m}^{-3}$ ; distance $n^{*-1/3} = 3.4\,\mathrm{nm}$ , nine molecular diameters.

4. N $_2$ : $\sqrt{3 \times 8.314 \times 293/0.028} = 511\,\mathrm{m}/\mathrm{s}$ ; O $_2$ : $478\,\mathrm{m}/\mathrm{s}$ . Same $\tfrac32 k_BT$ per molecule: $\tfrac12 mu^2$ fixed, so $u \propto 1/\sqrt m$ .

5. $\tfrac32 k_BT = 6.1 \times 10^{-21}\,\mathrm{J}$ ; total $N \times 6.1\times10^{-21} = 3.0 \times 10^{7}\,\mathrm{J}$ ; a $1000\,\mathrm{kg}$ car at $100\,\mathrm{km}/\mathrm{h}$ : $3.9 \times 10^{5}\,\mathrm{J}$ — the room’s molecular motion is eighty cars’ worth.

6. $\lambda = 1/(\sqrt2\pi \times 1.37\times10^{-19} \times 2.5\times10^{25}) = 66\,\mathrm{nm}$ ; time $\lambda/u = 66\times10^{-9}/500 = 1.3 \times 10^{-10}\,\mathrm{s}$ .

7. Impacts per second on $S$ : $\tfrac16 n^*Su = \tfrac16 \times 2.5\times 10^{25} \times 10^{-4} \times 500 = 2.1 \times 10^{23}\,\mathrm{s}^{-1}$ .

8. Each gives $2mu = 2 \times 4.8\times10^{-26} \times 500 = 4.8 \times 10^{-23}\,\mathrm{kg}\,\mathrm{m}/\mathrm{s}$ ( $m = M/N_A = 4.8 \times 10^{-26}\,\mathrm{kg}$ ): force $2.1\times10^{23} \times 4.8\times10^{-23} = 10\,\mathrm{N}$ on $1\,\mathrm{cm}^{2}$ : $1.0 \times 10^{5}\,\mathrm{Pa}$ . The model reproduces atmospheric pressure.

9. $F = PS = 1.013\times10^5 \times 20 = 2.0 \times 10^{6}\,\mathrm{N}$ — two hundred tonnes; the same pressure acts on the other side.

10. $P' = P \times 303/293 = 1.048 \times 10^{5}\,\mathrm{Pa}$ ; net $\Delta P\,S = 3.5 \times10^3 \times 20 = 70\,\mathrm{kN}$ outward: seven tonnes on the wall.

11. Air flows out until the pressure equalizes: $n \propto 1/T$ at fixed $P$ , $V$ : $8320 \times (1 - 293/303) = 275\,\mathrm{mol}$ , $8\,\mathrm{kg}$ leave.

12. $\rho = PM/RT$ : $1.205\,\mathrm{kg}/\mathrm{m}^{3}$ at $293\,\mathrm{K}$ ; $0.946\,\mathrm{kg}/\mathrm{m}^{3}$ at $373\,\mathrm{K}$ .

13. Hot air $0.946 \times 2800 \times 9.81 = 26.0\,\mathrm{kN}$ ; displaced cold air $1.205 \times 2800 \times 9.81 = 33.1\,\mathrm{kN}$ ; lift $7.1\,\mathrm{kN}$ ( $725\,\mathrm{kg}$ ).

14. $725 - 300 = 425\,\mathrm{kg}$ : five passengers.

15. Lift $\propto \rho_c - \rho_h = \rho_c(1 - 293/T)$ : doubling $1 - 293/373 = 0.215$ needs $T = 293/0.57 = 514\,\mathrm{K}$ ( $240{}^{\circ}\mathrm{C}$ ), well above what nylon tolerates (about $120{}^{\circ}\mathrm{C}$ ).

16. Helium $\rho = 1.205 \times 4/29 = 0.166\,\mathrm{kg}/\mathrm{m}^{3}$ : lift $(1.205 - 0.166) \times 2800 \times 9.81 = 28.5\,\mathrm{kN}$ ( $2.9\,\mathrm{t}$ ), four times more — but helium is expensive and lost at every landing, while hot air is free and its lift can be adjusted with the burner.

17. $0.21 \times 1.013\times10^5 = 2.1 \times 10^{4}\,\mathrm{Pa}$ .

18. $n^*_{\mathrm{O}_2} = P_{\mathrm{O}_2}/k_BT$ : room $2.1\times10^4/ (1.38\times10^{-23} \times 293) = 5.3 \times 10^{24}\,\mathrm{m}^{-3}$ ; Everest $0.21 \times 0.33 \times 1.013\times10^5/(1.38\times10^{-23} \times 250) = 2.0 \times 10^{24}\,\mathrm{m}^{-3}$ : $38\%$ of the room’s.

19. $H = RT/Mg = 8.314 \times 260/(0.029 \times 9.81) = 7.6\,\mathrm{km}$ ; $P_0/2$ at $H\ln 2 = 5.3\,\mathrm{km}$ .

20. $c_s = \sqrt{1.4 \times 8.314 \times 293/0.029} = 343\,\mathrm{m}/\mathrm{s}$ , against $u = 503\,\mathrm{m}/\mathrm{s}$ for the mixture: $c_s/u = \sqrt{\gamma/3} = 0.68$ . Sound is a disturbance carried by the molecules themselves, so it travels at a speed of the order of theirs — a little less, since only the component along the propagation matters.

21. $c_s \propto \sqrt T$ : $343\sqrt{303/293} = 349\,\mathrm{m}/\mathrm{s}$ . Helium: $\sqrt{(5/3) \times 8.314 \times 293/0.004} = 1010\,\mathrm{m}/\mathrm{s}$ : the resonances of the vocal tract scale with $c_s$ , so every formant is three times higher.

22. Colder: $343\sqrt{250/293} = 317\,\mathrm{m}/\mathrm{s}$ , $8\%$ slower; the pressure does not enter.

23. Same $N$ (same $P$ , $V$ , $T$ ); mass $\times 40/29 = 333\,\mathrm{kg}$ ; speeds $\times\sqrt{29/40}$ : $u = 428\,\mathrm{m}/\mathrm{s}$ ; $U = \tfrac32 nRT$ instead of $\tfrac52$ : $60\%$ .

24. $U = \tfrac52 \times 8320 \times 8.314 \times 293 = 5.1 \times 10^{7}\,\mathrm{J}$ ; $\Delta U = \tfrac52 nR \times 10 = 1.7 \times 10^{6}\,\mathrm{J}$ .

25. One average, $\langle v^2\rangle$ , fixes the pressure ( $\tfrac13 n^*m\langle v^2\rangle$ ), defines the temperature ( $\tfrac12 m\langle v^2\rangle = \tfrac32 k_BT$ ), gives the equation of state and the internal energy, and sets the scale of the speed of sound.

Terms defined in this chapter

See all 393 terms in the glossary