Kinetic Theory of Gases (KTG) — NEET High-Yield Notes
Kinetic theory of gases connects microscopic molecular behavior (speed, collisions, momentum) to macroscopic measurable properties (Pressure, Volume, Temperature).
1. Postulates of Ideal Gas & Pressure Derivation
- Gases consist of large numbers of tiny, identical, spherical particles in continuous random motion.
- Volume of molecules is negligible compared to the total container volume.
- No intermolecular forces of attraction or repulsion exist between molecules.
- All collisions (molecule-molecule and molecule-wall) are perfectly elastic, conserving momentum and kinetic energy.
Fig 1: Molecular Collisions against Container Walls and Pressure Derivation
Kinetic Pressure Equation
P = 13 ρ vrms2 = 13 MV vrms2
where ρ = density, M = total mass, V = volume, vrms = root mean square speed.
2. Molecular Speeds & Maxwell-Boltzmann Distribution
Fig 2: Maxwell-Boltzmann Distribution of Molecular Speeds at T1 and T2
For an ideal gas of molar mass M at absolute temperature T:
- Most Probable Speed (vmp): Speed possessed by maximum fraction of molecules:vmp = √2 R TM ≈ 1.414 √RT/M
- Average (Mean) Speed (vavg): Arithmetic average of all molecular speeds:vavg = √8 R Tπ M ≈ 1.596 √RT/M
- Root-Mean-Square Speed (vrms):vrms = √3 R TM = √3 kB Tm ≈ 1.732 √RT/M
NEET Invariant Golden Order: vmp < vavg < vrms (RAM ratio: √2 : √8/π : √3 = 1 : 1.128 : 1.224).
3. Law of Equipartition of Energy & Degrees of Freedom
Each independent quadratic coordinate contributing to the total energy of a molecule represents 1 degree of freedom (f). In thermal equilibrium at temperature T, the average kinetic energy associated with each degree of freedom per molecule is:
Eper DOF = 12 kB T
Total translational kinetic energy for 1 mole of any ideal gas (translational f = 3) is strictly E = 32 RT.
Fig 3: Degrees of Freedom (f), Specific Heat Capacities, and Specific Heat Ratio (γ)
Specific Heat Relations
Cv = f2 R | Cp = Cv + R = (f2 + 1) R | γ = CpCv = 1 + 2f
4. Mean Free Path (λ)
Fig 4: Collision Cylinder of Diameter d and Mean Free Path Dependences
The mean free path is the average distance traveled by a gas molecule between successive elastic collisions:
λ = 1√2 n π d2 = kB T√2 π d2 P
where n = number of molecules per unit volume (N/V = P / kBT), and d = molecular diameter.
Kinetic Theory of Gases — Formula Matrix & 20 Golden Facts
Master Formula Matrix
| Quantity | Formula | NEET Key Concept |
|---|---|---|
| Ideal Gas Pressure | P = 13 ρ vrms2 | P = (2/3) Etrans, vol |
| Translational KE (1 mole) | E = 32 R T | Depends ONLY on T, independent of gas type! |
| Translational KE (1 molecule) | E = 32 kB T | kB = R/NA = 1.38 × 10-23 J/K |
| vrms | √3RT/M = √3P/ρ | Inversely proportional to √M |
| vavg | √8RT/(πM) | vavg ≈ 0.92 vrms |
| vmp | √2RT/M | Peak of Maxwell curve |
| Mean Free Path λ | 1√2 n π d2 | λ ∝ T / P |
| Specific Heat Ratio γ | 1 + 2f | Mono: 5/3, Dia: 7/5, Poly: 4/3 |
20 Golden NEET Must-Know Facts & Traps
1. Translational kinetic energy per mole is (3/2)RT for ALL gases (He, O2, CO2 all have identical translational KE at same T).
2. Internal energy of an ideal gas depends strictly on Temperature (U = U(T)), not on volume or pressure (Joule's Law).
3. At absolute zero (T = 0 K), all molecular translation ceases and vrms = 0 according to classical KTG.
4. Ratio of speeds: vmp : vavg : vrms = √2 : √8/π : √3 ≈ 1 : 1.128 : 1.224.
5. As temperature increases, Maxwell speed distribution curve shifts right and flattens, but total area under curve remains 1.
6. Hydrogen (H2) molecules have the highest vrms among all gases at a given temperature because M = 2 g/mol is minimum.
7. Earth has an atmosphere because vrms of O2 and N2 (≈ 0.5 km/s) is much less than escape velocity (11.2 km/s). Moon lacks atmosphere because its escape velocity (2.38 km/s) is smaller than vrms of warm gases.
8. Each vibrational mode contributes 2 degrees of freedom (1 kinetic + 1 potential) = kBT of energy per molecule.
9. For rigid diatomic gas: f = 5 (3 trans + 2 rot), Cv = (5/2)R, Cp = (7/2)R, γ = 7/5 = 1.40.
10. For vibrating diatomic gas at high temperature: f = 7, Cv = (7/2)R, Cp = (9/2)R, γ = 9/7 = 1.28.
11. Mean free path is inversely proportional to the square of molecular diameter: λ ∝ 1/d2.
12. At constant volume (rigid sealed container), as temperature T increases, mean free path λ remains constant because number density n = N/V does not change!
13. Pressure P = (2/3) Ev, where Ev is the translational kinetic energy per unit volume.
14. Dalton's Law of Partial Pressures: Total pressure P = P1 + P2 + ... for non-reacting ideal gases.
15. Molar mass of a gas mixture: Mmix = (Σ ni Mi) / (Σ ni).
16. Equivalent Cv of mixture: Cv,mix = (n1 Cv1 + n2 Cv2) / (n1 + n2).
17. Equivalent γ of mixture: γmix = (n1 Cp1 + n2 Cp2) / (n1 Cv1 + n2 Cv2).
18. Real gases behave like ideal gases at High Temperature and Low Pressure (intermolecular attraction negligible, volume of molecules negligible).
19. Pressure of a gas does NOT depend on container shape; it depends only on number density n and temperature T.
20. Collision frequency ν = vavg / λ ∝ P / √T.
Kinetic Theory of Gases — 8 Solved NEET Numericals (3-Box Layout)
Problem 1: Temperature at which RMS Speed is Doubled
Given: RMS speed of oxygen molecules at 27°C (T1 = 300 K) is v. At what temperature will the RMS speed become 2v?
Step-by-Step Calculation:
vrms = √3RT/M ∝ √T.
v2v1 = √T2T1 → 2 = √T2300.
Squaring both sides: 4 = T2300 → T2 = 4 × 300 = 1200 K = 1200 - 273 = 927°C.
vrms = √3RT/M ∝ √T.
v2v1 = √T2T1 → 2 = √T2300.
Squaring both sides: 4 = T2300 → T2 = 4 × 300 = 1200 K = 1200 - 273 = 927°C.
NEET Trap Alert: Always convert Celsius to Kelvin first! Doubling speed requires 4 times the absolute temperature, NOT 2 × 27 = 54°C.
Problem 2: Ratio of RMS Speeds for Different Gases
Given: Find the ratio of RMS speed of Hydrogen (H2, M = 2) at 300 K to Oxygen (O2, M = 32) at 1200 K.
Step-by-Step Calculation:
vrms ∝ √T/M.
vH2vO2 = √TH2TO2 × MO2MH2.
= √3001200 × 322 = √14 × 16 = √4 = 2 : 1.
vrms ∝ √T/M.
vH2vO2 = √TH2TO2 × MO2MH2.
= √3001200 × 322 = √14 × 16 = √4 = 2 : 1.
NEET Shortcut: Under the square root, (1/4) × 16 = 4, sqrt(4) = 2. Ratio is 2 : 1.
Problem 3: Translatory Kinetic Energy of 2 moles of Gas
Given: 2 moles of nitrogen gas at 27°C (300 K). Gas constant R = 8.314 J/mol⋅K. Find total translational kinetic energy.
Step-by-Step Calculation:
Translational kinetic energy for n moles: Etrans = n (32 R T).
Etrans = 2 × 32 × 8.314 × 300 = 3 × 2494.2 = 7482.6 J ≈ 7.48 kJ.
Translational kinetic energy for n moles: Etrans = n (32 R T).
Etrans = 2 × 32 × 8.314 × 300 = 3 × 2494.2 = 7482.6 J ≈ 7.48 kJ.
NEET Trap Alert: Even though nitrogen is diatomic (total f = 5), translational kinetic energy uses strictly f = 3! Total internal energy would be (5/2)nRT = 12.47 kJ.
Problem 4: Mean Free Path Variation with Pressure and Temp
Given: A gas in a container has mean free path λ = 100 nm at 300 K and 1 atm. If pressure is doubled (2 atm) and temperature is raised to 600 K, find new mean free path λ'.
Step-by-Step Calculation:
λ = kB T√2 π d2 P ∝ TP.
λ'λ = T'T × PP' = 600300 × 12 = 2 × 12 = 1.
λ' = λ = 100 nm.
λ = kB T√2 π d2 P ∝ TP.
λ'λ = T'T × PP' = 600300 × 12 = 2 × 12 = 1.
λ' = λ = 100 nm.
NEET Shortcut: T doubled and P doubled cancels out completely; λ remains identical!
Problem 5: Specific Heat Ratio (γ) of a Gas Mixture
Given: A mixture consists of 1 mole of Helium (monatomic, Cv1 = 1.5R) and 2 moles of Oxygen (diatomic, Cv2 = 2.5R). Find γ of mixture.
Step-by-Step Calculation:
Cv,mix = n1 Cv1 + n2 Cv2n1 + n2 = 1(1.5R) + 2(2.5R)1 + 2 = 1.5R + 5.0R3 = 6.53 R = 136 R.
Cp,mix = Cv,mix + R = 136 R + R = 196 R.
γmix = Cp,mixCv,mix = 19/613/6 = 1913 ≈ 1.46.
Cv,mix = n1 Cv1 + n2 Cv2n1 + n2 = 1(1.5R) + 2(2.5R)1 + 2 = 1.5R + 5.0R3 = 6.53 R = 136 R.
Cp,mix = Cv,mix + R = 136 R + R = 196 R.
γmix = Cp,mixCv,mix = 19/613/6 = 1913 ≈ 1.46.
NEET Shortcut: γmix must lie strictly between γdia (1.40) and γmono (1.67). Since oxygen has 2 moles and helium 1, γ is closer to 1.40. 19/13 = 1.46.
Problem 6: Pressure and Volume Relation with Kinetic Energy
Given: If pressure of an ideal gas is 1.5 × 105 N/m2 and volume is 2 m3, find total translational kinetic energy.
Step-by-Step Calculation:
From kinetic pressure equation: P = 23 (EV) → E = 32 P V.
E = 32 × (1.5 × 105) × 2 = 3 × 1.5 × 105 = 4.5 × 105 J = 450 kJ.
From kinetic pressure equation: P = 23 (EV) → E = 32 P V.
E = 32 × (1.5 × 105) × 2 = 3 × 1.5 × 105 = 4.5 × 105 J = 450 kJ.
NEET Shortcut: Etrans = 1.5 × (P × V) directly!
Problem 7: Change in Internal Energy upon Heating at Constant Volume
Given: 4 moles of a rigid diatomic gas is heated from 30°C to 80°C (ΔT = 50 K) at constant volume. Find ΔU. (R = 8.3 J/mol⋅K)
Step-by-Step Calculation:
For rigid diatomic gas, f = 5, Cv = 52 R.
ΔU = n Cv ΔT = 4 × (52 R) × 50 = 10 × R × 50 = 500 R.
ΔU = 500 × 8.3 = 4150 J = 4.15 kJ.
For rigid diatomic gas, f = 5, Cv = 52 R.
ΔU = n Cv ΔT = 4 × (52 R) × 50 = 10 × R × 50 = 500 R.
ΔU = 500 × 8.3 = 4150 J = 4.15 kJ.
NEET Shortcut: At constant volume, heat given Q = ΔU = n Cv ΔT.
Problem 8: Molecular Collision Frequency Estimate
Given: A gas molecule has average speed vavg = 400 m/s and mean free path λ = 2 × 10-7 m. Find the collision frequency (collisions per second).
Step-by-Step Calculation:
Collision frequency ν = vavgλ.
ν = 4002 × 10-7 = 200 × 107 = 2 × 109 collisions/s (2 GHz).
Collision frequency ν = vavgλ.
ν = 4002 × 10-7 = 200 × 107 = 2 × 109 collisions/s (2 GHz).
NEET Shortcut: Frequency = Speed / Distance. Gas molecules collide billions of times per second at STP!
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