NCERT Solutions for Class 11 Physics Chapter 12: Kinetic Theory
Complete NCERT solutions for Class 11 Physics Chapter 12 with full revision notes on gas laws, the kinetic theory of an ideal gas, equipartition of energy, specific heats and mean free path, all 10 exercises solved step by step, and every figure linked to its NCERT PDF page.
Chapter at a glance — complete revision notes
12.1 Introduction
Boyle discovered his law in 1661. Boyle, Newton and several others tried to explain the behaviour of gases by considering that gases are made up of tiny atomic particles. The actual atomic theory got established more than 150 years later. Soon after, the kinetic theory of gases was developed in which the pressure and temperature of a gas are related to the motion of its molecules. The pioneers were Maxwell, Boltzmann and Clausius, and the theory explains, in microscopic terms, the macroscopic laws of thermodynamics.
Kinetic theory is a bridge between the microscopic (molecular) and the macroscopic (P, V, T) description. It explains why the pressure of a gas is proportional to the mean kinetic energy per molecule, gives T a molecular meaning, and predicts the specific heats of gases and solids. It also shows where the picture fails, which led to quantum ideas: the specific heats of H₂ and of metals at low temperatures could not be understood classically.
12.2 Molecular Nature of Matter
The idea that matter is made of atoms goes back to the ancient Indian (Kanada, 6th century BC) and Greek (Democritus) thinkers. The modern atomic theory is due to Dalton, who proposed that atoms of an element are identical and that compounds form when atoms combine in simple whole-number ratios.
The laws that led to the molecular picture
| Law | Statement |
|---|---|
| Dalton’s law of multiple proportions | When two elements combine to form more than one compound, the masses of one that combine with a fixed mass of the other are in simple ratios |
| Gay-Lussac’s law of combining volumes | When gases combine in a chemical reaction at the same temperature and pressure, the volumes of the reactants and (gaseous) products are in simple whole-number ratios. E.g. 2 volumes of hydrogen + 1 volume of oxygen → 2 volumes of water vapour |
| Avogadro’s law (hypothesis, 1811) | Equal volumes of all gases at the same temperature and pressure contain the same number of molecules |
Avogadro’s law explains Gay-Lussac’s law, and it requires that the elementary gases like hydrogen, oxygen and nitrogen exist as diatomic molecules (H₂, O₂, N₂), not as single atoms. The number of molecules in one mole of any substance is the Avogadro number, NA = 6.022 × 1023 mol−1. At STP one mole of any gas occupies 22.4 L.
How big are molecules, and how far apart?
- Atomic size is of the order of 1 Å = 10−10 m. (An oxygen molecule has a diameter of about 3 Å.)
- In solids and liquids the atoms are in contact: the interatomic spacing is about 2 Å.
- In a gas at STP the average separation is about 33 Å, since the volume per molecule is 22.4 × 10−3/(6.02 × 1023) = 3.7 × 10−26 m³, whose cube root is 3.3 × 10−9 m. This is about 10 times the molecular size, so the molecules occupy only about 0.04 % of the volume of the gas (see Exercise 12.1).
- The molecules of a gas therefore move freely most of the time and interact only during collisions; this is the idea behind the ideal gas.
12.3 Behaviour of Gases
Gases at low pressures and high temperatures obey simple relations among P, V, T and the amount of gas. These are called the gas laws.
| Law | Statement | Mathematical form |
|---|---|---|
| Boyle’s law (1661) | At constant temperature, the pressure of a fixed mass of gas is inversely proportional to its volume | PV = constant, i.e. P1V1 = P2V2 |
| Charles’ law | At constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature | V/T = constant |
| Gay-Lussac’s (pressure) law | At constant volume, the pressure of a fixed mass of gas is directly proportional to its absolute temperature | P/T = constant |
| Avogadro’s law | At the same T and P, equal volumes contain equal numbers of molecules | V ∝ μ (moles) |
Ideal gas equation
Combining Boyle’s and Charles’ laws and Avogadro’s law gives the ideal gas equation (equation of state):
PV = μRT = N kB T
Here μ is the number of moles, R = 8.314 J mol−1 K−1 is the universal gas constant, N is the number of molecules and kB = R/NA = 1.381 × 10−23 J K−1 is the Boltzmann constant. An ideal gas is one that obeys this equation exactly at all pressures and temperatures. Another useful form is P = (ρ/M)RT with ρ the density and M the molar mass. At STP the molar volume is V = RT/P = 8.314 × 273.15/1.013 × 105 = 22.4 × 10−3 m³.
Deviations from the ideal gas behaviour
Real gases obey the ideal gas equation only at low pressure and high temperature. Fig. 12.1 plots PV/T against P for a fixed mass of carbon dioxide at three temperatures: PV/T would be a horizontal line (= μR) for an ideal gas. The curves fall below the line at moderate pressures (intermolecular attraction makes the gas more compressible) and rise at high pressures (the finite size of molecules, repulsion). All curves approach the ideal-gas value μR as P → 0, and the departure is smaller at higher temperature.
Fig. 12.2 shows the experimental P–V isotherms of CO₂ at different temperatures compared with the ideal-gas isotherms (Boyle’s law). At high temperature the real curve is close to the hyperbola of Boyle’s law. At low temperature the gas liquefies on compression: the isotherm has a flat portion (liquid and vapour coexist at constant pressure) followed by a very steep rise (the liquid is nearly incompressible).
Fig. 12.3 shows the experimental T–V curve at constant pressure. For an ideal gas V is proportional to T all the way to 0 K, so the volume would vanish at absolute zero. The real gas liquefies at a temperature well above 0 K and the curve departs from the straight line: volume ∝ T holds only at high temperature.
Why do real gases deviate?
- Finite size of molecules: the volume available is less than V, so P is higher than ideal at high pressure.
- Intermolecular attraction: molecules near the wall are pulled inward by the others, reducing the pressure on the wall (PV/T falls below the ideal value at moderate P).
- At low P and high T the molecules are far apart and fast, so both effects are negligible and the gas is nearly ideal.
The van der Waals equation (P + a/V²)(V − b) = RT takes both effects into account (a for attraction, b for molecular volume).
12.4 Kinetic Theory of an Ideal Gas
Assumptions of the kinetic theory
- A gas consists of a very large number of identical molecules, each of mass m, in continuous random motion.
- The size of the molecules is negligible compared with the average distance between them (the molecules are treated as point masses).
- Molecules exert no force on each other (no attraction or repulsion) except during collisions; between collisions each moves in a straight line with constant velocity.
- Collisions of molecules with each other and with the walls are perfectly elastic; the collision time is negligible compared with the time between collisions.
- Gravity is neglected: molecules are equally likely to be found anywhere in the container, and their velocities are equally likely to point in any direction (isotropy).
Pressure of an ideal gas
Consider a cubical or any-shaped container with n molecules per unit volume. A molecule of velocity (vx, vy, vz) strikes a wall of area A perpendicular to the x-axis (Fig. 12.4). The collision is elastic, so its x-component reverses and the momentum transferred to the wall is
Δp = 2 m vx
In a time Δt, all molecules with velocity component vx that lie within a distance vxΔt of the wall (volume A vxΔt) can strike it. Only half of them move towards the wall. So the number striking in Δt is ½ n A vxΔt, and the total momentum transferred is
Δptotal = (2 m vx)(½ n A vx Δt) = n m vx² A Δt
Force F = Δp/Δt and pressure P = F/A. Taking the average over all molecules, ⟨vx²⟩ = ⟨vy²⟩ = ⟨vz²⟩ = ⅓⟨v²⟩ (isotropy):
P = n m ⟨vx²⟩ = (1/3) n m ⟨v²⟩
Since n = N/V and mass density ρ = n m: P = (1/3)ρ⟨v²⟩, or PV = (1/3)N m ⟨v²⟩. In terms of the average translational kinetic energy per molecule ε = ½m⟨v²⟩:
P = (2/3) n ⟨½ m v²⟩ = (2/3) n ε
Pressure is two-thirds of the translational kinetic energy per unit volume.
Kinetic interpretation of temperature
Compare PV = (1/3) N m ⟨v²⟩ with PV = N kB T. This gives
½ m ⟨v²⟩ = (3/2) kB T
- The average translational kinetic energy of a molecule is (3/2)kBT: it depends only on temperature and not on the mass or the nature of the gas. This is the molecular meaning of temperature.
- The internal energy of one mole of a monatomic ideal gas is U = (3/2)RT; for N molecules, U = (3/2)N kB T.
- At T = 300 K, ε = (3/2)(1.38 × 10−23)(300) = 6.2 × 10−21 J = 0.039 eV.
Root-mean-square (rms) speed
vrms = √⟨v²⟩ = √(3kBT/m) = √(3RT/M)
Here M is the molar mass (in kg). Lighter molecules move faster at the same temperature, vrms ∝ √(T/M). For nitrogen at 300 K, vrms ≈ 517 m s−1; for hydrogen at 300 K, about 1930 m s−1; this is why light gases escape from the atmosphere of the Moon (and why Earth has lost its hydrogen and helium).
| Gas | Molar mass M (g) | vrms at 300 K (m s−1) |
|---|---|---|
| Hydrogen (H₂) | 2 | 1930 |
| Helium (He) | 4 | 1370 |
| Nitrogen (N₂) | 28 | 517 |
| Oxygen (O₂) | 32 | 484 |
| Carbon dioxide (CO₂) | 44 | 412 |
Law of equipartition of speeds; Graham’s law of diffusion
At the same temperature, ½m₁⟨v₁²⟩ = ½m₂⟨v₂²⟩, so the ratio of the rms speeds of two gases is v1/v2 = √(m2/m1). The rate of diffusion (effusion) through a porous wall or small hole is proportional to the average speed (Fig. 12.5), so the rate of diffusion of a gas is inversely proportional to the square root of its molar mass (Graham’s law). This is used to separate the isotopes of uranium (²³⁵UF₆ and ²³⁸UF₆) by gaseous diffusion.
Dalton’s law of partial pressures
For a mixture of non-interacting ideal gases, each gas exerts its own pressure as though it alone occupied the container: P = P1 + P2 + … The proof follows from P = (1/3)n m⟨v²⟩ for each gas with n = n1 + n2 + … if the gases are in thermal equilibrium: the total pressure is the sum of the pressures that each gas would exert alone.
12.5 Law of Equipartition of Energy
The kinetic energy of a molecule is not only translational. Diatomic and polyatomic molecules also rotate and vibrate, and these motions store energy. How is the total energy shared? Maxwell and Boltzmann showed that the energy is divided equally among all the degrees of freedom.
Degrees of freedom
The number of degrees of freedom f is the number of independent coordinates (or velocity components) needed to specify the energy of the molecule, i.e. the number of independent ways in which it can store energy.
| Molecule | Translational | Rotational | Vibrational modes | f (at ordinary T, no vibration) |
|---|---|---|---|---|
| Monatomic (He, Ne, Ar) | 3 | 0 | 0 | 3 |
| Diatomic (H₂, N₂, O₂, CO) | 3 | 2 | 1 (if excited) | 5 |
| Triatomic linear (CO₂) | 3 | 2 | 4 (if excited) | 5 |
| Triatomic non-linear (H₂O) | 3 | 3 | 3 (if excited) | 6 |
The law of equipartition of energy
In thermal equilibrium at absolute temperature T, the average energy associated with each degree of freedom is ½kBT per molecule (½RT per mole). Each quadratic term in the expression for the energy of a molecule counts as one degree of freedom. For example:
- Translation: ½m vx² + ½m vy² + ½m vz² (3 quadratic terms, 3 × ½kBT).
- Rotation: ½I ω² for each axis of rotation that has appreciable moment of inertia.
- Vibration: one mode has both kinetic energy ½m(dx/dt)² and potential energy ½kx², i.e. two quadratic terms, so each vibrational mode contributes 2 × ½kBT = kBT.
A molecule with f degrees of freedom has average energy (f/2)kBT, and one mole has internal energy U = (f/2)RT.
12.6 Specific Heat Capacity
Monatomic gases
Only translation: f = 3. U = (3/2)RT per mole, so Cv = dU/dT = (3/2)R = 12.5 J mol−1 K−1, Cp = Cv + R = (5/2)R = 20.8 J mol−1 K−1, γ = Cp/Cv = 5/3 = 1.67.
Diatomic gases (rigid molecules)
Translation + 2 rotations: f = 5. U = (5/2)RT, Cv = (5/2)R = 20.8, Cp = (7/2)R = 29.1 J mol−1 K−1, γ = 7/5 = 1.40.
Diatomic gases (with vibration)
If the vibrational mode is also active: f = 7, U = (7/2)RT, Cv = (7/2)R = 29.1, Cp = (9/2)R = 37.4 J mol−1 K−1, γ = 9/7 = 1.29. At room temperature most diatomic gases (H₂, N₂, O₂) do not show vibrational contribution, since the vibrational energy quantum is much larger than kBT; Cl₂ does show partly (see Ex 10.15).
Polyatomic gases
For a non-linear triatomic molecule (3 translations + 3 rotations) f = 6: Cv = 3R = 24.9 J mol−1 K−1, Cp = 4R = 33.2 J mol−1 K−1, γ = 4/3 = 1.33. If vibrations are active, each mode adds R to Cv.
| Nature of gas | Cv (J mol−1 K−1) | Cp (J mol−1 K−1) | Cp − Cv | γ |
|---|---|---|---|---|
| Monatomic | 12.5 | 20.8 | 8.31 | 1.67 |
| Diatomic | 20.8 | 29.1 | 8.31 | 1.40 |
| Triatomic (non-linear) | 24.93 | 33.24 | 8.31 | 1.33 |
General formulae: Cv = (f/2)R, Cp = (f/2 + 1)R, γ = 1 + 2/f. The predicted values agree with the experimental values for most gases at room temperature (He, Ne, Ar, H₂, N₂, O₂, CO), while the measured Cv is larger for Cl₂ and polyatomic gases, which shows that vibrational modes contribute.
Worked example (NCERT Example 12.8). A cylinder of 44.8 L contains helium at STP. Amount: 2 mol (since 22.4 L per mole). He is monatomic: Cv = (3/2)R. Volume is fixed so heat is determined by Cv: Q = 2 × 1.5 × 8.31 × 15.0 = 374 J for a 15 °C rise.
Specific heat of solids
Each atom of a solid vibrates about its mean position in three dimensions. Each vibration mode has kinetic + potential energy, so the average energy per atom is 3kBT (6 quadratic terms × ½kBT). For one mole U = 3RT. At constant pressure ΔV is negligible for a solid, so C = ΔQ/ΔT = ΔU/ΔT = 3R = 24.9 J mol−1 K−1 (Dulong–Petit law). This agrees with experiment at room temperature for most solids (carbon is an exception, and the law fails at low temperatures).
| Substance | Specific heat (J kg−1 K−1) | Molar specific heat (J mol−1 K−1) |
|---|---|---|
| Aluminium | 900.0 | 24.4 |
| Carbon | 506.5 | 6.1 |
| Copper | 386.4 | 24.5 |
| Lead | 127.7 | 26.5 |
| Silver | 236.1 | 25.5 |
| Tungsten | 134.4 | 24.9 |
Specific heat of water (ice and liquid)
Treating water as a solid with the 3 atoms of each molecule (H, H, O) each having energy 3kBT: U = 9RT per mole, so C = 9R = 74.8 J mol−1 K−1; the measured value is about 75.4 J mol−1 K−1 (4.18 J g−1 K−1 × 18 g), in good agreement.
12.7 Mean Free Path
Molecules in a gas move with speeds of the order of the speed of sound (about 500 m s−1), yet a gas leaking from a cylinder in a kitchen takes a considerable time to spread across the room, and the top of a smoke cloud holds together for hours. This is because molecules have a finite size and undergo incessant collisions: their paths are deflected again and again and they diffuse rather than fly straight.
Suppose the molecules are spheres of diameter d, and focus on a single molecule moving with average speed ⟨v⟩ among the others, which are taken to be at rest. In time Δt it sweeps out a cylinder of cross-section πd² (any other molecule whose centre comes within a distance d of its centre will collide with it) and length ⟨v⟩Δt (Fig. 12.7).
The volume of the cylinder is πd²⟨v⟩Δt, so with n molecules per unit volume the number of collisions in Δt is nπd²⟨v⟩Δt. The collision rate (frequency) is
collision frequency = n π d² ⟨v⟩
The mean free path l is the average distance travelled between two successive collisions: l = ⟨v⟩Δt/(number of collisions) = 1/(nπd²). Allowing for the fact that all the molecules are moving, the relative speed is greater by √2:
l = 1/(√2 π n d²) and time between collisions τ = l/⟨v⟩ ≈ 1/(√2 π n d² ⟨v⟩)
Since n = P/(kBT), l = kBT/(√2 π d² P): the mean free path is inversely proportional to the pressure and to the square of the molecular diameter, and increases with temperature at constant pressure.
Typical numbers (NCERT example)
For air at STP, with d = 2 × 10−10 m, n = 2.7 × 1025 m−3: l = 1/(1.414 × 3.14 × 2.7 × 1025 × 4 × 10−20) = 2 × 10−7 m (about 1000 times the molecular size and 100 times the average molecular separation 3 × 10−9 m); the collision frequency is about 2.5 × 109 s−1 (taking ⟨v⟩ ≈ 500 m s−1). At a pressure of 10−4 atm the mean free path rises to about 2 mm, and in a highly evacuated tube (10−7 atm) to more than 1 m.
| Quantity | Typical value for air at STP |
|---|---|
| Molecular diameter d | 2 × 10−10 m |
| Average separation | 3 × 10−9 m |
| Mean free path | 2 × 10−7 m |
| Mean speed | about 500 m s−1 |
| Collision frequency | about 2.5 × 109 s−1 |
| Time between collisions | about 4 × 10−10 s |
Chapter Summary and Formula Sheet
- The ideal gas equation PV = μRT = N kBT; real gases obey it at low pressure and high temperature.
- Kinetic theory: P = (1/3) n m⟨v²⟩ = (2/3) n ε, where ε is the mean translational kinetic energy per molecule.
- Temperature: ½m⟨v²⟩ = (3/2)kBT, vrms = √(3RT/M).
- Law of equipartition: energy ½kBT per molecule per degree of freedom; a vibrational mode counts as two.
- Cv = (f/2)R, Cp = (f/2 + 1)R: monatomic Cv = 3R/2, diatomic 5R/2, solids 3R.
- Mean free path l = 1/(√2 π n d²); Avogadro number NA = 6.022 × 1023 mol−1; kB = 1.381 × 10−23 J K−1.
| Quantity | Formula |
|---|---|
| Ideal gas | PV = μRT = N kBT |
| Pressure | P = ⅓ρ⟨v²⟩ |
| Average KE | (3/2)kBT |
| rms speed | √(3kBT/m) = √(3RT/M) |
| Mean energy (f degrees) | (f/2)kBT |
| Specific heats | Cv = (f/2)R; Cp = Cv + R; γ = 1 + 2/f |
| Solids | C = 3R |
| Mean free path | l = 1/(√2 π n d²) |
Points to Ponder
- Pressure of a fluid is not only exerted on the wall; pressure exists everywhere in a fluid. Any layer of gas inside the container is in equilibrium because the pressure is the same on both sides of the layer.
- The intermolecular distance in a gas at ordinary pressures and temperatures is only about 10 times the interatomic distance in solids and liquids. What is different is the mean free path, which is 100 times the interatomic distance and 1000 times the size of the molecule.
- Each quadratic term in the total energy expression of a molecule is a degree of freedom, with energy ½kBT. Each vibrational mode gives 2 (not 1) degrees of freedom: kinetic and potential, corresponding to the energy 2 × ½kBT = kBT.
- Air molecules in a room do not all fall and settle on the ground (due to gravity) because of their high speeds and incessant collisions. In equilibrium there is a very slight increase in density at lower heights: mgh is much less than the average kinetic energy ½mv² for ordinary heights.
- ⟨v²⟩ is not always equal to (⟨v⟩)². The average of a squared quantity is not necessarily the square of the average. Example: for speeds 1, 2, 3 m s−1, ⟨v⟩ = 2 and (⟨v⟩)² = 4, but ⟨v²⟩ = (1 + 4 + 9)/3 = 4.67.
Ex 12.1 — Fraction of molecular volume to the actual volume of oxygen at STP
Concept. In a gas the molecules occupy only a tiny part of the space available to them. The fraction is (volume of the molecules in one mole)/(molar volume), where each molecule is treated as a sphere of diameter d.
Question. Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3 Å.
Given
- d = 3 Å = 3 × 10−10 m, so r = 1.5 × 10−10 m
- Avogadro number NA = 6.022 × 1023 mol−1
- Molar volume at STP = 22.4 L = 22.4 × 10−3 m³
Solution
Step 1: volume of one molecule (sphere):
v = (4/3)πr³ = (4/3)(3.14)(1.5 × 10−10)³ = 1.41 × 10−29 m³
Step 2: volume of all molecules in one mole:
Vmol = NA v = 6.022 × 1023 × 1.41 × 10−29 = 8.5 × 10−6 m³
Step 3: fraction of the molar volume occupied by the molecules:
fraction = 8.5 × 10−6/(22.4 × 10−3) = 3.8 × 10−4
The molecules occupy only about 3.8 × 10−4 (≈ 0.04 %) of the volume of the gas. The remaining 99.96 % is empty space.
Interpretation
A fraction of 4 × 10−4 means the average spacing is (1/3.8 × 10−4)1/3 = 14 times the molecular diameter in volume terms, since the cube-root of the inverse fraction is 13.8. So molecules are separated by roughly 10 to 14 diameters: very far apart compared with their size. This justifies treating them as point masses with negligible interaction, except in collisions, as in the kinetic theory of an ideal gas.
Comparison with a liquid. In a liquid the molecules touch, so the fraction is of the order of 0.5 – 0.7 (the packing fraction); a gas at STP is about a thousand times more rarefied than a liquid, in agreement with the density ratio of air (1.3 kg m−3) to liquid air (about 900 kg m−3).
Effect of pressure. At 100 atm the fraction rises 100-fold to about 4 %, and here the finite size of the molecules and the intermolecular forces begin to matter (non-ideal behaviour, Fig. 12.1).
Extra depth. The estimate shows why the ideal gas model works at normal pressures: the “excluded volume” b of the van der Waals equation is only about 4 × NA v (four times the actual molecular volume) = 3.4 × 10−5 m³ mol−1, which is negligible against 22.4 × 10−3 m³. Only when the gas is compressed to a few hundred atmospheres does b become comparable to the volume.
Conclusion: The molecular volume is about 3.8 × 10−4 of the gas volume at STP.
Ex 12.2 — Molar volume at STP is 22.4 litres
Concept. The ideal gas equation PV = μRT gives the volume for any amount of any ideal gas. With μ = 1 mol, T = 273.15 K and P = 1 atm = 1.013 × 105 Pa, the volume is the same for all gases because R and the standard conditions are universal.
Question. Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, 0 °C). Show that it is 22.4 litres.
Given
- μ = 1 mol; T = 0 °C = 273.15 K
- P = 1 atm = 1.013 × 105 Pa (N m−2)
- R = 8.314 J mol−1 K−1
Solution
PV = μRT ⇒ V = μRT/P
V = (1 × 8.314 × 273.15)/(1.013 × 105) = 2271/(1.013 × 105) m³
V = 2.24 × 10−2 m³ = 22.4 × 10−3 m³ = 22.4 L
Since 1 m³ = 1000 L, 2.24 × 10−2 m³ = 22.4 L. Hence the molar volume of any ideal gas at STP is 22.4 litres, as required.
Why every gas has the same molar volume
The result does not contain the mass or nature of the gas: only μ, R, T and P. This is Avogadro’s law: equal volumes of all gases at the same T and P contain the same number of molecules, so one mole (NA = 6.022 × 1023 molecules) occupies the same volume, whatever the gas.
Useful consequences
| Quantity | Value at STP |
|---|---|
| Molar volume | 22.4 L = 22.4 × 10−3 m³ |
| Number density | NA/V = 6.022 × 1023/22.4 × 10−3 = 2.69 × 1025 m−3 (Loschmidt number) |
| Volume per molecule | 3.7 × 10−26 m³ |
| Mean separation | 3.3 × 10−9 m = 33 Å |
Other temperatures. At room temperature (27 °C) and 1 atm the molar volume is 22.4 × 300/273 = 24.6 L. At 25 °C it is 24.5 L, often used in chemistry. The value 22.4 L applies only at STP (0 °C); the IUPAC STP (1 bar, 273.15 K) gives 22.71 L, but the NCERT definition is 1 atm.
Check with units. J mol−1 K−1 × mol × K/(N m−2) = N m/(N m−2) = m³. Good.
Extra depth. Since P = nkBT, the number density at STP is n = P/kBT = 1.013 × 105/(1.381 × 10−23 × 273.15) = 2.69 × 1025 m−3, called the Loschmidt number. Dividing NA by this number again gives 22.4 L, a consistency check between the macroscopic constant R and the microscopic constant kB = R/NA.
Conclusion: Molar volume = RT/P = 22.4 L at STP.
Ex 12.3 — PV/T versus P graph for oxygen at two temperatures
Concept. For an ideal gas PV/T = μR = NkB is a constant (a horizontal line). Real gases deviate from it, particularly at high pressure and low temperature, and approach the ideal value as P → 0. The curve at the lower temperature deviates more.
Question. Figure 12.8 shows a plot of PV/T versus P for 1.00 × 10−3 kg of oxygen gas at two different temperatures. (a) What does the dotted plot signify? (b) Which is true: T1 > T2 or T1 < T_2? (c) What is the value of PV/T where the curves meet on the y-axis? (d) If we obtained similar plots for 1.00 × 10^-3 kg of hydrogen, would we get the same value of PV/T at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value (for the low-pressure, high-temperature region)? (MH2 = 2.02 u, MO2 = 32.0 u, R = 8.31 J mol−1 K−1.)
(a) The dotted plot
The dotted horizontal line is the behaviour of an ideal gas: PV/T = μR is independent of P. It shows that the ideal gas law holds for all pressures. The real-gas curves approach this line as P → 0.
(b) Which temperature is higher?
The curve labelled T1 lies closer to the dotted line (it deviates less) than the T2 curve. Real gases are closer to the ideal behaviour at a higher temperature, because the molecules have more kinetic energy and the intermolecular attraction becomes unimportant. Hence T1 > T2.
(c) Value of PV/T on the y-axis
At the point where both curves meet the y-axis, P → 0 and the gas is ideal, so PV/T = μR:
μ = 1.00 × 10−3 kg/(32.0 × 10−3 kg mol−1) = 3.125 × 10−2 mol
PV/T = μR = 3.125 × 10−2 × 8.31 = 0.26 J K−1
(d) Hydrogen
No. For the same mass of hydrogen the number of moles is different (M is 2.02 g mol−1, much smaller than 32 g mol−1), so μR would be different: μ = 1.00 × 10−3/(2.02 × 10−3) = 0.495 mol and PV/T = 0.495 × 8.31 = 4.1 J K−1, about 16 times the oxygen value. To get the same value 0.26 J K−1, the number of moles must be 3.125 × 10−2 mol, so
mass of H₂ = 3.125 × 10−2 × 2.02 g = 6.3 × 10−2 g = 6.3 × 10−5 kg
| Part | Answer |
|---|---|
| (a) | ideal-gas behaviour: PV/T = μR independent of P |
| (b) | T1 > T2 |
| (c) | 0.26 J K−1 |
| (d) | No; 6.3 × 10−5 kg of hydrogen |
Reasoning for (d). PV/T = μR depends only on the number of moles, not on the kind of gas. Hence equal numbers of moles (not equal masses) give equal values of PV/T. 3.125 × 10−2 mol of O₂ weighs 1.0 g while the same number of moles of H₂ weighs only 0.063 g.
Extra depth. The deviation of the curves below the line at moderate pressures is due to the attractive forces between molecules, which lower the pressure on the walls; the rising part at high pressure is due to the finite volume of the molecules. The temperature at which the initial slope of the PV/T–P curve becomes zero is called the Boyle temperature; above it the curve rises from the beginning.
Conclusion: (a) ideal gas; (b) T1 > T2; (c) 0.26 J K−1; (d) no; 6.3 × 10−5 kg of H₂.
Ex 12.4 — Mass of oxygen taken out of a cylinder
Concept. The gauge pressure is the excess over atmospheric pressure, so the absolute pressure is P = gauge + 1 atm. Using PV = μRT before and after, the number of moles in the cylinder is μ = PV/(RT); the mass removed is (μ1 − μ2)M.
Question. An oxygen cylinder of volume 30 litre has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol−1 K−1, molecular mass of O₂ = 32 u).
Given
- V = 30 L = 30 × 10−3 m³ (constant)
- Initial: P1 = 15 + 1 = 16 atm = 16 × 1.013 × 105 = 1.62 × 106 Pa; T1 = 300 K
- Final: P2 = 11 + 1 = 12 atm = 1.216 × 106 Pa; T2 = 290 K
- M = 32 g mol−1
Solution
Initial amount:
μ1 = P1V/(RT1) = (1.62 × 106 × 30 × 10−3)/(8.31 × 300) = 4.862 × 104/2493 = 19.5 mol
Mass m1 = 19.5 × 32 = 624 g.
Final amount:
μ2 = P2V/(RT2) = (1.216 × 106 × 30 × 10−3)/(8.31 × 290) = 3.647 × 104/2410 = 15.1 mol
Mass m2 = 15.1 × 32 = 484 g.
Mass withdrawn:
Δm = m1 − m2 = 624 − 484 ≈ 140 g = 0.14 kg
Common mistakes
- Using the gauge pressure directly instead of absolute pressure (gauge + 1 atm). With gauge pressures the answer would be wrong: 15 and 11 atm give 18.3 and 14.1 mol and a mass of 0.13 kg.
- Forgetting to convert temperatures to kelvin or litres to m³.
- Assuming the number of moles stays fixed; here the cylinder has lost gas, so the number of moles changes.
| State | P (atm, absolute) | T (K) | μ (mol) | Mass (g) |
|---|---|---|---|---|
| Initial | 16 | 300 | 19.5 | 624 |
| Final | 12 | 290 | 15.1 | 484 |
| Removed | 4.4 | 140 |
Sanity check. Pressure falls by 25 % while the temperature falls by 3.3 %, so the quantity of gas falls by about 1 − (12/290)/(16/300) = 1 − 0.776 = 22 %, i.e. 0.22 × 624 = 138 g, consistent with 140 g.
Extra depth. Note the large role of atmospheric pressure: a gauge pressure of zero does not mean the cylinder is empty; it still holds oxygen at 1 atm. A full oxygen cylinder at 150 atm and 30 litres contains about 180 moles = 5.8 kg of oxygen. The method is the “equation of state at two states” approach and is used in all cylinder-withdrawal problems.
Conclusion: Mass of oxygen taken out ≈ 0.14 kg.
Ex 12.5 — Growth of an air bubble rising from the bottom of a lake
Concept. The number of moles of air in the bubble is constant. Using the combined gas law P1V1/T1 = P2V2/T2 with absolute pressure (atmosphere + water pressure at depth) gives the final volume.
Question. An air bubble of volume 1.0 cm³ rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C?
Given
- Bottom: V1 = 1.0 cm³ = 1.0 × 10−6 m³; T1 = 12 °C = 285 K; depth h = 40 m
- P1 = P0 + ρgh = 1.013 × 105 + 1000 × 9.8 × 40 = 1.013 × 105 + 3.92 × 105 = 4.93 × 105 Pa
- Surface: P2 = P0 = 1.013 × 105 Pa; T2 = 35 °C = 308 K
Solution
P1V1/T1 = P2V2/T2 ⇒ V2 = V1(P1/P2)(T2/T1)
V2 = 1.0 × (4.93 × 105/1.013 × 105) × (308/285)
V2 = 1.0 × 4.87 × 1.081 = 5.26 cm³ ≈ 5.3 cm³
The bubble grows to about 5.3 cm³, more than five times its original volume.
Effect of each factor
| Factor | Change | Effect on volume |
|---|---|---|
| Pressure falls from 4.93 × 105 Pa to 1.013 × 105 Pa | ÷ 4.87 | × 4.87 |
| Temperature rises from 285 K to 308 K | × 1.081 | × 1.081 |
| Net | × 5.26 |
Most of the growth (a factor of 4.9) is due to the fall of pressure (Boyle’s law); the rise of temperature contributes only about 8 %.
Notes
- The pressure at depth 40 m is about 4.9 atm: each 10 m of water adds about 1 atm.
- Pressure inside the bubble equals the pressure of the surrounding water (surface tension is negligible for a bubble of this size).
- Divers must exhale while ascending: air in the lungs expands in the same way and can rupture the lungs.
Common mistake. Taking P1 = ρgh only and forgetting atmospheric pressure would give P1/P2 = 3.9 and V2 = 4.2 cm³, which is wrong.
Extra depth. Bubbles in carbonated drinks and in lakes grow as they rise for the same reason. The result also shows that the buoyant force on the bubble increases as it rises, because it displaces more water, so it accelerates upwards. In the deep sea at 100 m the pressure is about 11 atm and a 1 cm³ bubble would be about 1/11 as large as at the surface.
Conclusion: The bubble grows to about 5.3 cm³.
Ex 12.6 — Number of air molecules in a room
Concept. For an ideal gas PV = N kB T, so the number of molecules is N = PV/(kBT). The composition of the air (oxygen, nitrogen, water vapour, etc.) does not matter, because the equation counts all molecules together.
Question. Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m³ at a temperature of 27 °C and 1 atm pressure.
Given
- V = 25.0 m³; T = 27 °C = 300 K; P = 1 atm = 1.013 × 105 Pa
- Boltzmann constant kB = 1.381 × 10−23 J K−1
Solution
N = PV/(kBT) = (1.013 × 105 × 25.0)/(1.381 × 10−23 × 300)
N = 2.5325 × 106/(4.143 × 10−21) = 6.1 × 1026
The room contains about 6.1 × 1026 molecules.
Alternative route through moles
Number of moles μ = PV/(RT) = (1.013 × 105 × 25)/(8.314 × 300) = 1015 mol. Then N = μNA = 1015 × 6.022 × 1023 = 6.1 × 1026, the same result. (The room holds about 1 kmol of air; at STP it would be 25/22.4 = 1.12 kmol, and 6.1 × 1026 is 8 % fewer because the room is warmer than 0 °C.)
Mass of this air
With the average molar mass of air 29 g mol−1: m = 1015 × 29 g = 29.4 kg. So the air in an ordinary room weighs about 30 kg, around the mass of a child; it is easy to forget that air has mass.
| Quantity | Value |
|---|---|
| Moles | 1.0 × 103 mol |
| Molecules | 6.1 × 1026 |
| Mass of air | about 29 kg |
| Number density n = N/V | 2.4 × 1025 m−3 |
Estimate versus exact. The problem says “estimate”: the answer may be given as about 6 × 1026. A hundred-fold change in the answer would signal an error in powers of ten.
Extra depth. Dividing the molecular count by the room volume gives n = 2.4 × 1025 m−3, so each molecule has a volume of 4 × 10−26 m³ or, in a cubical cell, a side of 3.5 nm = 35 Å. A cubic micrometre of air still contains 2.4 × 107 molecules, which is why air behaves as a continuous fluid even on a microscopic scale.
More to know. The number of molecules in a room is astronomically large but the average spacing is still about 35 Å. Each molecule in the room undergoes about 5 × 109 collisions per second with its neighbours, which is why the air in the room is in perfect thermal equilibrium within a fraction of a second after any disturbance. The same approach gives the number of molecules in a sealed bulb of known volume at low pressure: for a vacuum of 10−7 atm in the same room, N would still be 6 × 1019, showing how hard it is to get an “empty” space.
Conclusion: Number of molecules ≈ 6.1 × 1026.
Ex 12.7 — Average thermal energy of a helium atom at three temperatures
Concept. The average translational kinetic energy of a molecule of any ideal gas is (3/2)kBT. For a monatomic gas like helium this is the whole thermal energy, since there are no rotational or vibrational modes.
Question. Estimate the average thermal energy of a helium atom at (i) room temperature (27 °C), (ii) the temperature on the surface of the Sun (6000 K), (iii) the temperature of 10 million kelvin (the typical core temperature in the case of a star).
Given
- Monatomic gas: average thermal energy E = (3/2)kBT
- kB = 1.38 × 10−23 J K−1; 1 eV = 1.6 × 10−19 J
Solution
(i) T = 27 °C = 300 K:
E = 1.5 × 1.38 × 10−23 × 300 = 6.2 × 10−21 J = 0.039 eV
(ii) T = 6000 K:
E = 1.5 × 1.38 × 10−23 × 6000 = 1.24 × 10−19 J = 0.78 eV
(iii) T = 107 K:
E = 1.5 × 1.38 × 10−23 × 107 = 2.07 × 10−16 J = 1.3 × 103 eV = 1.3 keV
| Temperature | Average thermal energy (J) | Average thermal energy (eV) |
|---|---|---|
| 300 K (room) | 6.2 × 10−21 | 0.039 |
| 6000 K (Sun’s surface) | 1.24 × 10−19 | 0.78 |
| 107 K (core of a star) | 2.07 × 10−16 | 1.3 × 103 (1.3 keV) |
Physical meaning
- At room temperature the thermal energy (0.04 eV) is far below the energy needed to excite or ionise a helium atom (first ionisation energy 24.6 eV), so helium stays neutral.
- At 6000 K the energy is 0.78 eV: still far below the ionisation energy, but many atoms in the high-energy tail of the distribution have enough energy to excite or ionise (partial ionisation in the Sun’s atmosphere).
- At 107 K the average energy (1.3 keV) greatly exceeds 24.6 eV, so the helium (and hydrogen) are completely ionised: the matter is a plasma of nuclei and free electrons. The energy of collisions is large enough that hydrogen nuclei can approach closely enough for nuclear fusion.
Rule of thumb. kBT = 1/40 eV at 300 K; kBT = 1 eV at 11 600 K. So (3/2)kBT is 0.039 eV at room temperature.
Extra depth. The average energy depends only on temperature, not on mass: a helium atom, a nitrogen molecule (its translational energy) and a dust particle in the same gas have the same average translational kinetic energy at the same temperature. For a diatomic gas the total thermal energy including rotation would be (5/2)kBT.
Conclusion: (i) 6.2 × 10−21 J (0.039 eV); (ii) 1.24 × 10−19 J (0.78 eV); (iii) 2.07 × 10−16 J (1.3 keV).
Ex 12.8 — Equal volumes of neon, chlorine and uranium hexafluoride at the same T and P
Concept. By Avogadro’s law (a consequence of PV = NkBT), equal volumes of all ideal gases at the same pressure and temperature contain the same number of molecules. At the same temperature all the molecules have the same average translational kinetic energy (3/2)kBT, so the lighter molecules have the greater rms speed.
Question. Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic) and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is vrms the largest?
Number of molecules
Yes. From PV = N kB T, N = PV/(kBT). P, V and T are the same, so N is the same for all three vessels: Avogadro’s law (equal volumes of gases at the same temperature and pressure contain equal numbers of molecules, independent of the gas, and whether it is monatomic, diatomic or polyatomic).
Root-mean-square speed
No. At the same temperature the average translational kinetic energy per molecule is the same, ½m⟨v²⟩ = (3/2)kBT, so
vrms = √(3kBT/m) = √(3RT/M) ∝ 1/√M
Molar masses: neon 20.2 g mol−1, chlorine (Cl₂) 70.9 g mol−1, uranium hexafluoride (UF₆) 352 g mol−1 (U = 238 + 6 × 19). The lightest molecule, neon, moves fastest, so vrms is largest for neon.
| Gas | Molar mass M (g mol−1) | vrms at 300 K (m s−1) | Relative to Ne |
|---|---|---|---|
| Neon | 20.2 | 609 | 1 |
| Chlorine | 70.9 | 324 | 0.53 |
| UF₆ | 352 | 146 | 0.24 |
Ratios: vNe/vCl₂ = √(70.9/20.2) = 1.87; vNe/vUF₆ = √(352/20.2) = 4.17. (Values at 300 K from vrms = √(3RT/M).)
Significance
The large difference between the speeds of ²³⁵UF₆ and ²³⁸UF₆ is only about 0.43 % (v ∝ 1/√M, with M = 349 and 352), but repeated gaseous diffusion through porous barriers in many stages accumulates this tiny difference and is used to enrich uranium for reactors (Graham’s law of diffusion).
Common misconception. It is wrong to say “a heavier molecule has more kinetic energy”. At the same T the average translational kinetic energy is equal; the heavier molecules are simply slower.
Extra depth. The rms speeds at 300 K: Ne 609 m s−1, Cl₂ 324 m s−1, UF₆ 146 m s−1; all are comparable to or lower than the speed of sound in the gas (e.g. 1/3 less than the speed of sound in neon). For the same T, the total translational energy of a vessel is the same, (3/2)NkBT, for the three vessels, but a vessel of the diatomic or polyatomic gas also stores rotational and vibrational energy, so its total internal energy is larger.
Conclusion: (i) Yes, equal numbers; (ii) no; vrms is largest for neon.
Ex 12.9 — Temperature at which argon has the same rms speed as helium at −20 °C
Concept. The rms speed of a gas is vrms = √(3RT/M). Equating the rms speeds of two different gases gives T/M equal for the two: T1/M1 = T2/M2.
Question. At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at −20 °C? (atomic mass of Ar = 39.9 u, of He = 4.0 u).
Given
- Helium: THe = −20 °C = 253 K, MHe = 4.0 g mol−1
- Argon: MAr = 39.9 g mol−1; TAr = ?
Solution
Equal rms speeds:
√(3RTAr/MAr) = √(3RTHe/MHe) ⇒ TAr/MAr = THe/MHe
TAr = THe × (MAr/MHe) = 253 × (39.9/4.0) = 253 × 9.975 = 2.52 × 103 K
The argon gas must be at about 2.5 × 103 K (≈ 2250 °C) for its atoms to move as fast as helium atoms at −20 °C.
Check by finding the common speed
vrms(He, 253 K) = √(3 × 8.31 × 253/4.0 × 10−3) = √(1.577 × 106) = 1256 m s−1. For argon at 2520 K: √(3 × 8.31 × 2520/39.9 × 10−3) = √(1.574 × 106) = 1255 m s−1. They agree.
Interpretation. Argon atoms are 10 times as heavy as helium atoms, so to have the same speed they need ten times the energy; since the average kinetic energy is proportional to T, the temperature must be 10 times larger.
| Gas | M (g mol−1) | T (K) | vrms (m s−1) |
|---|---|---|---|
| Helium | 4.0 | 253 | 1256 |
| Argon | 39.9 | 2520 | 1255 |
Common mistake. Using −20 °C directly in the formula without converting to 253 K gives a wrong answer; a ratio of temperatures must be in kelvin.
Extra depth. This is why the light gases hydrogen and helium escape from the Earth’s atmosphere (the escape speed is 11.2 km s−1) while heavier gases stay: the tail of the speed distribution for light gases at the high-temperature upper atmosphere exceeds the escape speed, while for heavy gases it does not. The Moon, with a smaller escape speed (2.4 km s−1), has lost almost all its atmosphere.
Conclusion: TAr ≈ 2.5 × 103 K.
Ex 12.10 — Mean free path and collision frequency of nitrogen
Concept. In a gas of molecular diameter d and number density n, the mean free path is l = 1/(√2 π n d²), with n = P/(kBT). The collision frequency is v/l, the collision time is d/v and the mean time between collisions is l/v.
Question. Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 °C. Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (MN2 = 28.0 u).
Given
- P = 2.0 atm = 2.026 × 105 Pa; T = 17 °C = 290 K
- r = 1.0 Å, so d = 2.0 × 10−10 m
- m = 28 × 1.66 × 10−27 = 4.65 × 10−26 kg; kB = 1.38 × 10−23 J K−1
Step 1: number density
n = P/(kBT) = 2.026 × 105/(1.38 × 10−23 × 290) = 5.06 × 1025 m−3
Step 2: mean free path
l = 1/(√2 π n d²) = 1/(1.414 × 3.14 × 5.06 × 1025 × 4.0 × 10−20) = 1/(8.99 × 106) = 1.1 × 10−7 m
Step 3: speed and collision frequency
vrms = √(3kBT/m) = √(3 × 1.38 × 10−23 × 290/4.65 × 10−26) = 5.1 × 102 m s−1
collision frequency = vrms/l = 5.1 × 102/1.1 × 10−7 = 4.6 × 109 s−1
The time between collisions: τ = l/vrms = 1.1 × 10−7/5.1 × 102 = 2.2 × 10−10 s.
Step 4: collision time
The collision time is the time during which a molecule is in contact with another: it moves a distance of the order of d at speed v: tc = d/vrms = 2.0 × 10−10/5.1 × 102 = 3.9 × 10−13 s.
τ/tc = 2.2 × 10−10/3.9 × 10−13 ≈ 5.6 × 102
The molecule moves freely for a time about 500 times longer than the duration of a collision. Collisions are therefore almost instantaneous events, which justifies the kinetic-theory assumption that molecules interact only briefly.
| Quantity | Value |
|---|---|
| Number density n | 5.06 × 1025 m−3 |
| Mean free path l | 1.1 × 10−7 m (about 550 d) |
| vrms | 5.1 × 102 m s−1 |
| Collision frequency | 4.6 × 109 s−1 |
| Free time τ | 2.2 × 10−10 s |
| Collision time tc | 3.9 × 10−13 s |
| τ/tc | about 5.6 × 102 |
Extra depth. The same calculation with P = 1 atm gives double the mean free path (2.2 × 10−7 m) and half the collision frequency. Strictly, the collision frequency should use the mean speed ⟨v⟩ = √(8kBT/πm) = 4.7 × 102 m s−1 (rather than vrms), which gives 4.3 × 109 s−1; the textbook uses vrms and obtains 5 × 109 s−1 – the exact value is unimportant for an estimate.
Conclusion: l ≈ 1.1 × 10−7 m; collision frequency ≈ 4.6 × 109 s−1; the free time is about 5.6 × 102 times the collision time.