This MCQ module is based on: Degrees of Freedom Mean Free Path
TOPIC 22 OF 33
Degrees of Freedom Mean Free Path
🎓 Class 11
Physics
CBSE
Theory
Ch 12 – Kinetic Theory
⏱ ~14 min
🌐 Language: [gtranslate]
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Degrees of Freedom Mean Free Path
12.9 Degrees of Freedom
The number of independent ways a molecule can store kinetic energy is called its degrees of freedom (f). For a single point mass moving in 3D space, f = 3 (independent x, y, z translations). For more complex molecules:
| Molecule | Translational | Rotational | Vibrational (high T) | Total f (rigid) |
|---|---|---|---|---|
| Monatomic (He, Ne, Ar) | 3 | 0 | 0 | 3 |
| Diatomic (H₂, N₂, O₂) | 3 | 2 (about ⊥ axes) | +2 above ~1000 K | 5 |
| Triatomic linear (CO₂) | 3 | 2 | + several | 5 |
| Triatomic non-linear (H₂O, NH₃) | 3 | 3 | + several | 6 |
12.10 Law of Equipartition of Energy
Equipartition principle: In thermal equilibrium at temperature T, the average energy associated with each quadratic degree of freedom equals
\[\boxed{\,\overline{E}_{1\,\text{dof}} = \tfrac{1}{2} k_B T\,}\]
A molecule with f quadratic degrees of freedom therefore has mean energy (f/2) k_B T.
"Quadratic" means the kinetic-energy term ½ m v_x², or rotational term ½ I ω², or vibrational ½ k x² + ½ μ ẋ². Each such squared term contributes ½ k_B T per molecule on average. (A vibration has TWO quadratic terms — kinetic and potential — giving 2 × ½ k_B T = k_B T per vibrational mode.)
12.11 Specific Heat Capacities
Total internal energy of one mole of an ideal gas with f degrees of freedom:
U = (f/2) N_A k_B T = (f/2) R T
Molar heat capacity at constant volume:
C_v = dU/dT = (f/2) R
For an ideal gas Mayer's relation gives \(C_p - C_v = R\), so \(C_p = (f/2 + 1) R\). The ratio:
γ = C_p / C_v = (f + 2) / f
12.11.1 Predicted Specific Heats
| Gas type | f | C_v | C_p | γ | Examples |
|---|---|---|---|---|---|
| Monatomic | 3 | (3/2)R = 12.47 | (5/2)R = 20.78 | 5/3 = 1.67 | He, Ne, Ar |
| Rigid diatomic | 5 | (5/2)R = 20.78 | (7/2)R = 29.10 | 7/5 = 1.40 | H₂, N₂, O₂ at 300 K |
| Diatomic + vibration | 7 | (7/2)R = 29.10 | (9/2)R = 37.41 | 9/7 = 1.29 | diatomic at very high T |
| Triatomic (non-linear) | 6 | 3R = 24.94 | 4R = 33.26 | 4/3 = 1.33 | H₂O vapour |
All C values in J·mol⁻¹·K⁻¹.
12.11.2 Solids — Dulong-Petit Law
In a solid each atom can vibrate about its equilibrium position in three independent directions. Each vibration carries 2 quadratic terms ⇒ 6 × (½ k_B T) = 3 k_B T per atom; per mole:
U_solid = 3 R T ⇒ C = 3 R ≈ 25 J·mol⁻¹·K⁻¹ (Dulong-Petit)
This explains why most metals at room temperature have nearly the same molar heat capacity (~25 J/mol·K).
12.12 Mean Free Path
A molecule does not travel in a straight line for long — it collides with another molecule on average every mean free path λ. Treat each molecule as a hard sphere of diameter d. As the test molecule moves with mean speed v̄ relative to the others, in time t it sweeps a "collision tube" of cross-section π d². The number of molecules in that tube = n × π d² × v̄ × t (using the relative speed √2 v̄):
\[\boxed{\,\lambda = \dfrac{1}{\sqrt{2}\,\pi\, d^2\, n}\,}\]
Smaller molecules and lower density both increase λ (cleaner straight-line motion).
For air at STP (d ≈ 3 × 10⁻¹⁰ m, n ≈ 2.7 × 10²⁵ m⁻³): λ ≈ 1/(√2 × π × 9 × 10⁻²⁰ × 2.7 × 10²⁵) ≈ 9.3 × 10⁻⁸ m ≈ 100 nm — about 300× the molecular size. Each molecule collides ~10⁹ times per second.
Worked Examples
Example 12.10: γ from f
Predict γ for argon, oxygen and water vapour at 300 K assuming rigid molecules.
Ar: monatomic, f = 3 ⇒ γ = (3+2)/3 = 5/3 ≈ 1.67.
O₂: rigid diatomic, f = 5 ⇒ γ = 7/5 = 1.40.
H₂O: non-linear triatomic, f = 6 ⇒ γ = 8/6 = 1.33.
Measured values: 1.67, 1.40, 1.33 — equipartition holds beautifully.
O₂: rigid diatomic, f = 5 ⇒ γ = 7/5 = 1.40.
H₂O: non-linear triatomic, f = 6 ⇒ γ = 8/6 = 1.33.
Measured values: 1.67, 1.40, 1.33 — equipartition holds beautifully.
Example 12.11: Mean free path of N₂ at STP
Estimate λ for N₂ at STP given d ≈ 3.7 × 10⁻¹⁰ m, n = 2.69 × 10²⁵ m⁻³.
λ = 1/(√2 π d² n) = 1/(√2 × π × (3.7 × 10⁻¹⁰)² × 2.69 × 10²⁵)
= 1/(√2 × π × 1.37 × 10⁻¹⁹ × 2.69 × 10²⁵) = 1/(1.64 × 10⁷) ≈ 6.1 × 10⁻⁸ m ≈ 61 nm.
= 1/(√2 × π × 1.37 × 10⁻¹⁹ × 2.69 × 10²⁵) = 1/(1.64 × 10⁷) ≈ 6.1 × 10⁻⁸ m ≈ 61 nm.
Example 12.12: Collision frequency
If v_rms of N₂ at 300 K is 517 m/s, what is the collision frequency ν = v̄/λ given the value of λ above?
v̄ ≈ 0.92 × v_rms ≈ 476 m/s. ν = v̄/λ = 476 / 6.1 × 10⁻⁸ ≈ 7.8 × 10⁹ collisions/s. About 10 billion collisions per second per molecule.
Interactive: Mean Free Path Explorer L3 Apply
Vary number density n and molecular diameter d; observe λ and the collision frequency at 300 K.
Activity 12.4 — Predicting γ from Speed of Sound
L4 Analyse
Predict: The speed of sound in an ideal gas is v_s = √(γ RT/M). For air (M = 29 g/mol) at 300 K, v_s = 347 m/s. What value of γ does this imply, and what does that say about air's molecular structure?
- Solve v_s² = γRT/M for γ: γ = v_s² M / (RT).
- γ = (347)² × 0.029 / (8.314 × 300) = 120409 × 0.029 / 2494 = 1.40.
- This matches f = 5 — exactly what rigid diatomic gases (N₂, O₂) predict.
Conclusion: Speed of sound is a sensitive probe of γ and hence of molecular degrees of freedom. Air behaves as a rigid diatomic gas at room temperature — vibrations are "frozen out" by quantum mechanics. Around T ~ 1000 K, vibrational modes activate and γ falls towards 9/7 ≈ 1.29.
Competency-Based Questions
A 1-mole sample of nitrogen (N₂, rigid diatomic) is at 300 K. R = 8.314 J/mol·K, k_B = 1.38 × 10⁻²³ J/K, N_A = 6.022 × 10²³.
Q1. L1 Remember State the law of equipartition of energy.
In thermal equilibrium at temperature T, every quadratic degree of freedom of a molecule contributes (1/2) k_B T to the average energy. A molecule with f such degrees has mean energy (f/2)k_B T.
Q2. L2 Understand Why does a diatomic gas have only 2, not 3, rotational degrees of freedom?
Rotation about the bond axis has very small moment of inertia I, so the rotational kinetic energy ½ I ω² is essentially zero (or quantum-mechanically frozen). Only the two axes perpendicular to the bond contribute meaningful rotational energy.
Q3. L3 Apply Compute the internal energy of the 1-mole N₂ sample.
U = (f/2) RT = (5/2)(8.314)(300) = 6235.5 J ≈ 6.2 kJ.
Q4. L3 Apply Compute C_v, C_p, and γ for the gas above.
C_v = (5/2)R = 20.78 J/mol·K; C_p = C_v + R = (7/2)R = 29.10 J/mol·K; γ = 7/5 = 1.40.
Q5. L5 Evaluate The molar heat capacity of solid copper at 300 K is measured to be 24.5 J/mol·K, while at 50 K it drops to ~5 J/mol·K. Explain in terms of equipartition + quantum effects.
At 300 K, all 6 quadratic terms (3 KE + 3 PE) are excited ⇒ C = 3R = 24.9 J/mol·K (Dulong-Petit). At 50 K, the high-frequency vibrational modes are frozen out by quantum mechanics (k_B T < ℏω), so equipartition fails and C drops sharply (Einstein/Debye theory). Equipartition is a high-T limit.
Assertion-Reason Questions
Assertion (A): γ for monatomic gases is greater than for diatomic.
Reason (R): Diatomic molecules have more degrees of freedom, lowering γ = (f+2)/f.
Answer: A. γ_mono = 5/3, γ_dia = 7/5. As f grows γ decreases — extra degrees of freedom dilute the energy contribution.
Assertion (A): The mean free path increases when a gas is compressed.
Reason (R): λ ∝ 1/n.
Answer: D. Compression increases n, so λ decreases. R is correct as an algebraic statement, but A is false.
Assertion (A): Solids have molar specific heat about 3R at room temperature.
Reason (R): Each atom has 6 quadratic degrees of freedom (3 KE + 3 PE).
Answer: A. By equipartition, U = 6 × (½)k_BT × N_A = 3RT per mole, so C = dU/dT = 3R ≈ 25 J/mol·K — the Dulong-Petit law.
Frequently Asked Questions - Degrees of Freedom Mean Free Path
What is the main concept covered in Degrees of Freedom Mean Free Path?
In NCERT Class 11 Physics Chapter 12 (Kinetic Theory), "Degrees of Freedom Mean Free Path" covers core principles and equations needed for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Degrees of Freedom Mean Free Path useful in real-life applications?
Real-life applications of Degrees of Freedom Mean Free Path from NCERT Class 11 Physics Chapter 12 include engineering design, satellite mechanics, sports biomechanics, transportation safety, and electrical/electronic devices. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Degrees of Freedom Mean Free Path?
Key formulas in Degrees of Freedom Mean Free Path (NCERT Class 11 Physics Chapter 12 Kinetic Theory) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 12?
NCERT Class 11 Physics Chapter 12 (Kinetic Theory) is structured so each part builds on the previous one. Degrees of Freedom Mean Free Path connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Degrees of Freedom Mean Free Path?
CBSE board questions from Degrees of Freedom Mean Free Path typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the Degrees of Freedom Mean Free Path lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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