Lesson 3: Kinetic Theory of Gases and Radiation

Link to earlier classes : The heat, temperature and radiation of classes 9-10 are explained here through the molecules of a gas.

In the figure : Molecules in the vessel move in all directions; their impacts on the walls create pressure.

Rule 1: Gas laws — Boyle (T constant) : P V = constant. Charles (P constant) : V ÷ T = constant. Gay-Lussac (V constant) : P ÷ T = constant. Avogadro : equal volumes at equal P, T hold equal numbers of molecules. T always in kelvin.

• Easy : 2 atm, 3 L gas expands to 6 L at constant T : P₂ = 2 × 3 ÷ 6 = 1 atm.
• Medium : 2 L of gas at 300 K heated to 450 K at constant pressure : V₂ = 2 × 450 ÷ 300 = 3 L.
• Exam (CET) : A closed vessel at 27 °C has pressure P; at 327 °C the pressure is 2P (300 K → 600 K). (Options : 1) P 2) 2P 3) 12P 4) P ÷ 2 → answer 2)

Rule 2: Ideal gas equation — P V = n R T = N k_B T. R = 8.314 J mol⁻¹ K⁻¹; k_B = 1.38 × 10⁻²³ J/K; R = N_A k_B; N_A = 6.022 × 10²³.

• Easy : n = 2 mol, T = 300 K, V = 0.05 m³ : P = 2 × 8.314 × 300 ÷ 0.05 = 9.98 × 10⁴ Pa.
• Medium : P = 10⁵ Pa, V = 1 L = 10⁻³ m³, T = 300 K : N = P V ÷ (k_B T) = 2.42 × 10²² molecules.
• Exam (CET) : At equal P, V, T two gases hold the same number of molecules, even if their masses differ. (Options : 1) more in H₂ 2) more in O₂ 3) equal 4) cannot say → answer 3)

Rule 3: Assumptions of the kinetic model — Molecules are tiny hard spheres; their volume is negligible; collisions are perfectly elastic; no forces except in collisions; motion is random; paths between collisions are straight; wall collisions are elastic.

• Easy : At high pressure and low T a real gas departs from ideal behaviour.
• Medium : At low pressure and high T a real gas is close to ideal.
• Exam (CET) : With no intermolecular force an ideal gas's internal energy is purely kinetic. (Options : 1) kinetic + potential 2) potential only 3) kinetic only 4) zero → answer 3)

Derivation : combining the laws into P V = n R T

• Step 1 : Boyle : V ∝ 1/P (T, n fixed). Charles : V ∝ T (P, n fixed). Avogadro : V ∝ n.
• Step 2 : All together : V ∝ n T ÷ P.
• Step 3 : Replace the proportionality by a constant R : P V = n R T. With N = n N_A and R = N_A k_B : P V = N k_B T.

Spot the difference

In the figure : A molecule moves straight until it collides and then changes direction; the gap between collisions is λ.

Rule 1: Mean free path — The average distance a molecule travels between two successive collisions : λ = 1 ÷ (√2 π d² n) = k_B T ÷ (√2 π d² P) (n = number density, d = molecular diameter).

• Easy : O₂, d = 360 pm, T = 300 K, P = 10⁵ Pa : λ = 7.19 × 10⁻⁸ m ≈ 72 nm.
• Medium : Pressure reduced 100 times (T constant) : λ increases 100 times.
• Exam (CET) : P and T both doubled ⇒ λ ∝ T ÷ P is unchanged. (Options : 1) doubles 2) halves 3) same 4) four times → answer 3)

Rule 2: Dependence of λ — λ ∝ T (P constant), λ ∝ 1 ÷ P (T constant), λ ∝ 1 ÷ d²; so larger molecules and denser gas ⇒ smaller λ.

• Easy : d doubled ⇒ λ becomes one quarter.
• Medium : T from 300 K to 600 K (P constant) ⇒ λ doubles.
• Exam (CET) : In a vacuum tube λ becomes larger than the tube so molecules hit only the walls. (Options : 1) λ small 2) λ large 3) λ zero 4) λ constant → answer 2)

Derivation : derivation of λ = 1 ÷ (√2 π d² n)

• Step 1 : Treat the other molecules as at rest and the moving one as having diameter d. In time t it sweeps a cylinder of volume π d² v t.
• Step 2 : The number of molecules in it, n π d² v t, equals the number of collisions.
• Step 3 : λ = distance ÷ collisions = v t ÷ (n π d² v t) = 1 ÷ (n π d²).
• Step 4 : All molecules move, so the average relative speed is √2 times larger ⇒ λ = 1 ÷ (√2 π d² n); with n = P ÷ (k_B T), λ = k_B T ÷ (√2 π d² P).

Spot the difference

In the figure : A molecule strikes the x-wall of the cubical box; the impacts of all molecules together give the pressure.

Rule 1: Pressure of a gas — P = ⅓ ρ v_rms² = N m ⟨v²⟩ ÷ (3 V) (ρ = density). Pressure = momentum delivered per second per unit area by the molecules.

• Easy : ρ = 1.2 kg/m³, v_rms = 500 m/s : P = ⅓ × 1.2 × 250000 = 1.0 × 10⁵ Pa.
• Medium : ρ and v_rms both doubled : P becomes 8 times (ρ doubles, v² is 4 times).
• Exam (CET) : Volume halved (T constant) : n doubles ⇒ P doubles. (Options : 1) half 2) same 3) double 4) four times → answer 3)

Rule 2: rms speed — v_rms = √⟨v²⟩ = √(3 k_B T ÷ m) = √(3 R T ÷ M) (M = molar mass in kg/mol). v_rms ∝ √T and ∝ 1 ÷ √M.

• Easy : O₂ (M = 0.032), 300 K : v_rms = 484 m/s.
• Medium : He (M = 0.004), 300 K : v_rms = 1368 m/s.
• Exam (CET) : At the same T the v_rms ratio of H₂ to O₂ = √(32 ÷ 2) = 4. (Options : 1) 16 2) 4 3) 2 4) 8 → answer 2)

Rule 3: Temperature and kinetic energy — The average kinetic energy per molecule ⟨KE⟩ = ½ m v_rms² = (3/2) k_B T depends only on T, not on the type of gas. Temperature measures the mean kinetic energy of the molecules.

• Easy : 300 K : ⟨KE⟩ = 1.5 × 1.38 × 10⁻²³ × 300 = 6.21 × 10⁻²¹ J.
• Medium : T doubled ⇒ ⟨KE⟩ doubles but v_rms rises only √2 times.
• Exam (CET) : At the same T, ⟨KE⟩ per molecule of H₂ and O₂ is equal; v_rms differs. (Options : 1) H₂ larger 2) O₂ larger 3) equal 4) zero → answer 3)

Derivation : derivation of P = N m ⟨v²⟩ ÷ 3V

• Step 1 : N molecules in a cube of side L. One molecule hits the wall with velocity vₓ and rebounds; its momentum change is 2 m vₓ.
• Step 2 : It returns to the same wall after 2L ÷ vₓ ⇒ impacts per second = vₓ ÷ 2L.
• Step 3 : Force from one molecule = 2 m vₓ × vₓ ÷ 2L = m vₓ² ÷ L. Total force F = m Σvₓ² ÷ L = N m ⟨vₓ²⟩ ÷ L.
• Step 4 : Motion is random ⇒ ⟨vₓ²⟩ = ⟨v²⟩ ÷ 3; pressure P = F ÷ L² = N m ⟨v²⟩ ÷ (3 L³) = N m ⟨v²⟩ ÷ 3V.
• Step 5 : Compare with P V = N k_B T : ⅓ N m ⟨v²⟩ = N k_B T ⇒ ½ m ⟨v²⟩ = (3/2) k_B T.

Spot the difference

In the figure : A monatomic molecule only translates in 3 directions; a diatomic molecule can also rotate about 2 axes.

Rule 1: Degrees of freedom — The independent variables needed to describe a molecule (f). Monatomic f = 3; diatomic (moderate T) f = 5 (3 translation + 2 rotation); non-linear polyatomic f = 6. Vibration counts only at very high T.

• Easy : He, Ne, Ar : f = 3.
• Medium : O₂, N₂, H₂ : f = 5.
• Exam (CET) : Linear CO₂ is taken as f = 5 at ordinary T; H₂O, CH₄ : f = 6. (Options : 1) 3 2) 5 3) 6 4) 7 → answer 3 for H₂O)

Rule 2: Equipartition of energy — In equilibrium every quadratic term (degree of freedom) carries an average ½ k_B T. So per molecule f × ½ k_B T and for n moles U = (f ÷ 2) n R T.

• Easy : 1 mol He, 300 K : U = (3/2) × 8.314 × 300 = 3741 J.
• Medium : 2 mol diatomic, 300 K : U = (5/2) × 2 × 8.314 × 300 = 12471 J.
• Exam (CET) : The U of an ideal gas depends only on T, not on P or V. (Options : 1) T only 2) P only 3) P and V 4) V → answer 1)

Rule 3: Specific heat capacities — C_V = (f ÷ 2) R; Mayer's relation C_P − C_V = R; γ = C_P ÷ C_V = 1 + 2 ÷ f. Monatomic γ = 5/3; diatomic γ = 7/5.

• Easy : Monatomic : C_V = 1.5 × 8.314 = 12.47; C_P = 20.79 J mol⁻¹ K⁻¹.
• Medium : 1 mol monatomic gas heated by 20 K at constant volume : Q = C_V ΔT = 249.4 J.
• Exam (CET) : For a diatomic gas C_P ÷ C_V = 1.4. (Options : 1) 1.67 2) 1.4 3) 1.33 4) 1.2 → answer 2)

Derivation : Mayer's relation C_P − C_V = R

• Step 1 : One mole of ideal gas heated by dT at constant volume : dQ₁ = C_V dT (no work ⇒ only U rises).
• Step 2 : At constant pressure for the same dT : dQ₂ = C_P dT = dU + P dV. U depends only on T ⇒ dU = C_V dT.
• Step 3 : From P V = R T at constant P : P dV = R dT.
• Step 4 : So C_P dT = C_V dT + R dT ⇒ C_P − C_V = R.

Spot the difference

In the figure : Incident radiation divides into three parts : reflected, absorbed, transmitted.

Rule 1: Thermal radiation — Every body above 0 K emits and absorbs energy as electromagnetic waves; no medium is needed. Parts of the incident radiation : a + r + t = 1 (absorption, reflection, transmission coefficients).

• Easy : For an opaque body t = 0 ⇒ a + r = 1.
• Medium : A body absorbs 70 % and reflects 20 % : t = 1 − 0.7 − 0.2 = 0.1.
• Exam (CET) : Black clothes warm up more in the sun (large a), white less (large r). (Options : 1) black less 2) black more 3) equal 4) white more → answer 2)

Rule 2: Blackbody — A perfect blackbody absorbs all radiation of all wavelengths : a = 1, r = t = 0. It also emits the maximum possible radiation at its temperature. Ferry's blackbody = a hollow sphere blackened inside with a small hole.

• Easy : Radiation entering the hole is reflected repeatedly inside and absorbed ⇒ the hole acts as black.
• Medium : The Sun and a hot furnace behave almost like blackbodies.
• Exam (CET) : Lamp black a ≈ 0.95–0.99; polished metal a ≈ 0.05. (Options : 1) polished metal is a good blackbody 2) lamp black is 3) both 4) neither → answer 2)

Rule 3: Kirchhoff's law — At the same temperature and wavelength, (emissive power ÷ absorptive power) is the same for all bodies (= emissive power of the blackbody). A good absorber is a good emitter. Emissivity e = R of the body ÷ R of the blackbody (0 ≤ e ≤ 1; e = a).

• Easy : A blackened pan heats up quickly and cools quickly.
• Medium : A shiny kettle keeps tea warm longer (small e).
• Exam (CET) : A body with a = 0.6 has e = 0.6 (same T, λ). (Options : 1) 0.4 2) 0.6 3) 1 4) 0 → answer 2)

Derivation : Kirchhoff's law from a thought experiment

• Step 1 : A closed enclosure at temperature T holds a given body and a blackbody; the system is in thermal equilibrium.
• Step 2 : In equilibrium a body emits exactly what it absorbs (else its temperature would change) : e_body = a × (enclosure radiation E).
• Step 3 : For the blackbody a = 1 ⇒ E_b = E.
• Step 4 : Hence e_body ÷ a = E_b — same for every body; so relative emissivity = a.

Spot the difference

In the figure : Curves for three temperatures; as the temperature rises the peak moves left and up.

Rule 1: Spectral distribution — A blackbody emits all wavelengths but not equally. Each T has a curve with one peak. As T rises (1) the peak shifts to shorter λ, (2) the peak height and (3) the area under the curve (total radiation) increase.

• Easy : A furnace at 1500 K looks red; a star at 6000 K looks white-blue.
• Medium : A heated iron rod goes red → orange → yellowish white as T rises.
• Exam (CET) : The area under the curve ∝ T⁴; doubling T makes it 16 times. (Options : 1) 2× 2) 4× 3) 8× 4) 16× → answer 4)

Rule 2: Wien's displacement law — λ_max T = b, b = 2.897 × 10⁻³ m K. The peak wavelength is inversely proportional to T; a hotter body radiates mostly at shorter wavelengths.

• Easy : Sun, T = 5800 K : λ_max = 2.897 × 10⁻³ ÷ 5800 = 499 nm (green-yellow).
• Medium : Human body at 310 K : λ_max = 9.35 μm (infrared).
• Exam (CET) : A star with λ_max = 580 nm ⇒ T = b ÷ λ_max = 4995 K. (Options : 1) 2500 K 2) 5000 K 3) 7500 K 4) 10000 K → answer 2)

Rule 3: Using the law — The surface temperature of stars and distant objects follows from their λ_max. For two bodies λ₁ T₁ = λ₂ T₂.

• Easy : λ_max doubled ⇒ T halved.
• Medium : A has λ_max = 400 nm, B has 800 nm : T_A ÷ T_B = 2. (T_B = 3621 K, T_A = 7242 K).
• Exam (CET) : A red star is cooler, a blue star is hotter. (Options : 1) blue cooler 2) red cooler 3) same 4) cannot say → answer 2)

Spot the difference

In the figure : The R–T curve rises as T⁴; doubling T makes R sixteen times.

Rule 1: Stefan–Boltzmann law — Radiation emitted per unit area per second by a blackbody : R = σ T⁴; σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴; [M T⁻³ K⁻⁴]. For an ordinary body R = e σ T⁴. Total power P = e σ A T⁴.

• Easy : Blackbody at 300 K : R = 5.67 × 10⁻⁸ × 300⁴ = 459.3 W/m².
• Medium : Blackbody A = 0.01 m², T = 1000 K : P = σ A T⁴ = 567 W.
• Exam (CET) : T doubled ⇒ R × 16; T tripled ⇒ × 81. (Options : 1) 4 2) 8 3) 16 4) 81 → answer 3 for doubling)

Rule 2: Net radiation — A body (T) in surroundings (T₀) : net power P_net = e σ A (T⁴ − T₀⁴). T > T₀ ⇒ it loses heat; T = T₀ ⇒ P_net = 0 (equilibrium).

• Easy : e = 0.8, A = 0.5 m², T = 600 K, T₀ = 300 K : P_net = 0.8 × 5.67 × 10⁻⁸ × 0.5 × (600⁴ − 300⁴) = 2756 W.
• Medium : Taking T₀ = 0 K gives P = e σ A T⁴ (emission only).
• Exam (CET) : At T = T₀ the body still radiates but the net is zero; temperature stays constant. (Options : 1) radiation stops 2) net zero 3) T rises 4) T falls → answer 2)

Rule 3: Applications — The law is used for the Sun's surface temperature, stars, a lamp filament and body heat loss. A filament's steady temperature : electric power supplied = e σ A T⁴.

• Easy : Filament : e = 0.5, A = 5 × 10⁻⁵ m², power 60 W : T = [60 ÷ (e σ A)]^¼ = 2551 K.
• Medium : A star of radius R and temperature T : total power L = 4π R² σ T⁴.
• Exam (CET) : Two spherical blackbodies : r₁ ÷ r₂ = 2, T₁ ÷ T₂ = 1 ÷ 2 ⇒ P₁ ÷ P₂ = 4 × (1/16) = ¼. (Options : 1) ¼ 2) 4 3) 1 4) 16 → answer 1)

Derivation : the net-emission formula

• Step 1 : At temperature T the body emits power e σ A T⁴.
• Step 2 : Surroundings at T₀ send power on the body and by Kirchhoff (a = e) it absorbs e σ A T₀⁴.
• Step 3 : Net = emitted − absorbed.
• Step 4 : P_net = e σ A (T⁴ − T₀⁴).

Spot the difference