NEET UG Physics · Chapter 10

Kinetic Theory of Gases

Ideal gas equation, pressure of a gas, rms speed, degrees of freedom, equipartition and mean free path.

NCERT Class 11NEET: 1 Q per paper 10 min read

1. Chapter at a Glance

The kinetic theory of gases explains the behaviour of a gas — its pressure, temperature and heat capacity — from the motion of its tiny molecules. It shows that pressure comes from molecules hitting the walls and that temperature measures their average kinetic energy. NEET usually asks one question: on rms speed, the law of equipartition, degrees of freedom or mean free path.

What NEET asks most
  • rms speed and how it depends on T and molar mass.
  • Average kinetic energy per molecule, (3/2)kT.
  • Degrees of freedom and γ for mono-, di- and polyatomic gases.
  • Ideal gas equation in its different forms.
  • Mean free path and its dependence on pressure and size.

2. Ideal Gas and Its Assumptions

  • Ideal gas equation: PV = nRT = NkBT, where R = 8.314 J mol⁻¹ K⁻¹, kB = 1.38 × 10⁻²³ J K⁻¹ and NA = 6.02 × 10²³ mol⁻¹.
  • It combines Boyle's law (PV constant at fixed T), Charles' law (V ∝ T at fixed P) and Avogadro's law (equal volumes of gases at the same T and P contain equal numbers of molecules).
  • Kinetic theory assumptions: molecules are tiny compared with the space between them; they move randomly and collide elastically; there are no forces between them except during collisions; collisions take negligible time.
  • Real gases behave ideally at low pressure and high temperature.
  • Dalton's law: total pressure of a gas mixture = sum of partial pressures.

3. Pressure and Temperature

P = ⅓ n m vrms² = ⅓ ρ vrms²    PV = ⅔ Etrans
  • Average translational kinetic energy of a molecule = (3/2)kBT — the same for all gases at the same temperature.
  • Total translational KE of one mole = (3/2)RT.
  • rms speed vrms = √(3RT/M) = √(3kBT/m) = √(3P/ρ). It depends on T and M, not on pressure (at fixed T).
  • Other speeds: most probable √(2RT/M) and average √(8RT/πM). Order: vmp < vavg < vrms, in the ratio √2 : √(8/π) : √3.
lower Thigher T numberspeed
Molecular speed distribution: heating shifts the peak (most probable speed) to the right and flattens the curve; the total area stays the same.

4. Degrees of Freedom and Equipartition

The law of equipartition of energy states that, in equilibrium, each degree of freedom (each independent way a molecule can store energy) has an average energy of ½kBT.

Gas typeDegrees of freedom fCvCpγ = 1 + 2/f
Monoatomic (He, Ne, Ar)3 (translational)3R/25R/21.67
Diatomic, rigid (O₂, N₂, H₂)5 (3 trans + 2 rot)5R/27R/21.40
Diatomic, with vibration77R/29R/21.29
Polyatomic, rigid non-linear (CH₄)63R4R1.33

Internal energy of n moles U = (f/2)nRT. For a mixture: Cv,mix = (n₁Cv1 + n₂Cv2)/(n₁ + n₂).

5. Mean Free Path

λ = 1/(√2 π d² n) = kBT/(√2 π d² P)

It is the average distance a molecule travels between two collisions (n = molecules per unit volume, d = molecular diameter). It is larger at low pressure and for smaller molecules. At room conditions it is about 10⁻⁷ m, a few hundred times the molecular size.

6. Formula Master Table

QuantityFormula
Ideal gasPV = nRT = NkBT
Kinetic pressureP = ⅓ρvrms²
KE per molecule(3/2)kBT
rms speed√(3RT/M)
Energy per degree of freedom½kBT
Internal energyU = (f/2)nRT
γ from fγ = 1 + 2/f
Mean free pathλ = kBT/(√2πd²P)

7. Master Summary Table

ConceptKey pointNEET Note
PressureFrom molecular collisionsP = ⅓ρv²
TemperatureMeasures average KESame KE for all gases at same T
rms speed∝ √(T/M)Independent of P at fixed T
Equipartition½kT per degree of freedomDiatomic f = 5
Specific heatsCp − Cv = Rγ = 1 + 2/f
Mean free path∝ T/(d²P)Halving P doubles λ

8. Mind Map

Kinetic Theory
Ideal gas
  • PV = nRT
  • Assumptions
  • Gas laws
Pressure & T
  • P = ⅓ρv²
  • KE = (3/2)kT
Speeds
  • vrms, vavg, vmp
  • Speed distribution
Energy
  • Degrees of freedom
  • Equipartition
  • Mean free path

9. Tips and Tricks

Shortcuts
  • To make vrms n times, make T n² times.
  • At the same T: vrms(H₂) : vrms(O₂) = √(32/2) = 4 : 1.
  • Always convert M to kg mol⁻¹ (O₂: 0.032, N₂: 0.028, He: 0.004).
  • For mixtures, compute Cv first, then Cp = Cv + R, then γ.
Common NEET traps
  • Saying heavier molecules have more KE at the same T — KE is the same; they just move slower.
  • Using °C in (3/2)kT.
  • Giving a diatomic molecule 3 or 6 degrees of freedom at room temperature (it is 5).

10. Important One-liners

  • Temperature of a gas is a measure of the average kinetic energy of its molecules.
  • At absolute zero the kinetic energy of an ideal gas would be zero.
  • rms speed does not change with pressure at constant temperature.
  • Each degree of freedom contributes ½kT of energy per molecule.
  • γ is largest for monoatomic gases.
  • Mean free path increases when pressure decreases.
  • Gases deviate from ideal behaviour at high pressure and low temperature.

11. Solved Numericals: Average to Advanced

Five average and five advanced numericals. Take R = 8.314 J mol⁻¹ K⁻¹, kB = 1.38 × 10⁻²³ J K⁻¹. Click to reveal each answer and working.

Average level

N1Average

Find the rms speed of nitrogen molecules (M = 28 g mol⁻¹) at 27 °C.

Show answer & solution

Concept: vrms = √(3RT/M), T in K and M in kg mol⁻¹.

  1. T = 300 K; M = 0.028 kg mol⁻¹.
  2. vrms = √(3 × 8.314 × 300/0.028) = √(267 236).
  3. ≈ 517 m s⁻¹.

Answer: about 517 m s⁻¹

N2Average

Find the average translational kinetic energy of a gas molecule at 127 °C.

Show answer & solution

Concept: KE = (3/2)kBT.

  1. T = 400 K.
  2. KE = 1.5 × 1.38 × 10⁻²³ × 400 = 8.28 × 10⁻²¹ J.

Answer: 8.28 × 10⁻²¹ J

N3Average

At what temperature will the rms speed of oxygen molecules be half their value at 27 °C?

Show answer & solution

Concept: vrms ∝ √T.

  1. Half the speed needs one-fourth the temperature.
  2. T = 300/4 = 75 K (= −198 °C).

Answer: 75 K

N4Average

A gas of density 1.2 kg m⁻³ has an rms molecular speed of 500 m s⁻¹. Find its pressure.

Show answer & solution

Concept: P = ⅓ρvrms².

  1. P = ⅓ × 1.2 × 250 000.
  2. P = 1 × 10⁵ Pa (about 1 atmosphere).

Answer: 1 × 10⁵ Pa

N5Average

How many moles of gas are in a 10 L cylinder at 2 × 10⁵ Pa and 300 K?

Show answer & solution

Concept: n = PV/RT.

  1. V = 10 L = 0.01 m³.
  2. n = 2 × 10⁵ × 0.01/(8.314 × 300) = 2000/2494.
  3. n ≈ 0.80 mol.

Answer: about 0.80 mol

Advanced level

N6Advanced

One mole of helium is mixed with one mole of oxygen (rigid). Find Cv, Cp and γ of the mixture.

Show answer & solution

Concept: mole-weighted average of Cv.

  1. Cv,mix = (1 × 1.5R + 1 × 2.5R)/2 = 2R.
  2. Cp,mix = 2R + R = 3R.
  3. γ = 3R/2R = 1.5.

Answer: 2R; 3R; 1.5

N7Advanced

Find the internal energy of 5 g of helium at 300 K (M = 4 g mol⁻¹).

Show answer & solution

Concept: U = (f/2)nRT with f = 3 for a monoatomic gas.

  1. n = 5/4 = 1.25 mol.
  2. U = 1.5 × 1.25 × 8.314 × 300.
  3. U ≈ 4677 J.

Answer: about 4.68 kJ

N8Advanced

The absolute temperature of a gas is doubled and it is replaced by a gas of half the molar mass. By what factor does the rms speed change?

Show answer & solution

Concept: vrms ∝ √(T/M).

  1. T/M becomes (2T)/(M/2) = 4(T/M).
  2. vrms becomes √4 = 2 times.

Answer: doubles

N9Advanced

The pressure of a gas is halved and its absolute temperature is doubled. By what factor does its mean free path change?

Show answer & solution

Concept: λ = kBT/(√2πd²P) ∝ T/P.

  1. T/P becomes (2T)/(P/2) = 4(T/P).
  2. λ becomes 4 times.

Answer: 4 times

N10Advanced

Find vmp, vavg and vrms for helium (M = 4 g mol⁻¹) at 400 K.

Show answer & solution

Concept: RT/M = 8.314 × 400/0.004 ≈ 8.31 × 10⁵ m² s⁻².

  1. vmp = √(2 × 8.31 × 10⁵) ≈ 1290 m s⁻¹.
  2. vavg = √(8 × 8.31 × 10⁵/π) ≈ 1455 m s⁻¹.
  3. vrms = √(3 × 8.31 × 10⁵) ≈ 1579 m s⁻¹.
  4. Order vmp < vavg < vrms ✓.

Answer: about 1290, 1455 and 1579 m s⁻¹

12. NEET-pattern Questions with Detailed Answers

These questions are original and modelled on concepts repeatedly asked in NEET UG.

Q1Easy

The average kinetic energy of a gas molecule is directly proportional to:

  1. √T
  2. T
  3. T²
  4. 1/T
Show solution

KE = (3/2)kBT ∝ T (absolute temperature). The rms speed is ∝ √T.

Answer: (B)

Q2Easy

The number of degrees of freedom of a rigid diatomic molecule is:

  1. 3
  2. 5
  3. 6
  4. 7
Show solution

Three translational plus two rotational = 5 (rotation about the bond axis does not count).

Answer: (B)

Q3Medium

At the same temperature, the ratio of rms speeds of hydrogen and oxygen molecules is:

  1. 1 : 4
  2. 4 : 1
  3. 1 : 16
  4. 16 : 1
Show solution

v ∝ 1/√M: √(32/2) = 4, so 4 : 1.

Answer: (B)

Q4Medium

The internal energy of one mole of a monoatomic ideal gas at temperature T is:

  1. RT/2
  2. RT
  3. 3RT/2
  4. 5RT/2
Show solution

U = (f/2)RT with f = 3 ⇒ 3RT/2. 5RT/2 would be for a rigid diatomic gas.

Answer: (C)

Q5Medium

If the pressure of a gas is increased at constant temperature, the rms speed of its molecules:

  1. Increases
  2. Decreases
  3. Remains unchanged
  4. Becomes zero
Show solution

vrms = √(3RT/M) depends only on T and M; raising P at fixed T just raises the density.

Answer: (C)

Q6Medium

For a gas whose molecules have f degrees of freedom, γ equals:

  1. 1 + f/2
  2. 1 + 2/f
  3. 2/f
  4. f/(f + 2)
Show solution

Cv = fR/2, Cp = (f/2 + 1)R ⇒ γ = (f + 2)/f = 1 + 2/f.

Answer: (B)

Q7Hard

The rms speed of a gas at 27 °C is v. Its rms speed at 927 °C is:

  1. v√2
  2. 2v
  3. 4v
  4. v/2
Show solution

T changes from 300 K to 1200 K (4 times), so vrms becomes √4 = 2 times. Using °C (27 → 927) would wrongly suggest a much larger change.

Answer: (B)

Q8Hard

The mean free path of gas molecules depends on the molecular diameter d as:

  1. d
  2. 1/d
  3. 1/d²
  4. d²
Show solution

λ = 1/(√2πd²n): larger molecules present a bigger target (∝ d²) and collide more often.

Answer: (C)

13. Quick Revision Summary

Remember before the exam
  • PV = nRT = NkBT; P = ⅓ρvrms².
  • KE per molecule (3/2)kBT — same for all gases at the same T.
  • vrms = √(3RT/M); vmp < vavg < vrms.
  • ½kBT per degree of freedom; f = 3, 5, 6; γ = 1 + 2/f.
  • λ = kBT/(√2πd²P).
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