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PhysicsMediumClass 11

Kinetic Theory — Speed + Degrees of Freedom

Thermodynamics & KTG

38

JEE Qs

10%

Hard

75

min

Master the formulas for all three types of speeds, understand the concept of degrees of freedom and their temperature dependence, and know the equipartition principle inside out for various molecular structures.

🧮 Key Formulas

v_rms = sqrt(3RT/M) = sqrt(3kT/m)
v_avg = sqrt(8RT/(pi*M)) = sqrt(8kT/(pi*m))
v_p = sqrt(2RT/M) = sqrt(2kT/m)
Order of speeds: v_p < v_avg < v_rms
Average kinetic energy per molecule per degree of freedom = (1/2)kT
Internal energy of an ideal gas U = (f/2)nRT
Degrees of freedom (f) for monoatomic gas = 3 (translational)
Degrees of freedom (f) for diatomic gas = 5 (3 translational + 2 rotational) at moderate temperatures
Degrees of freedom (f) for non-linear polyatomic gas = 6 (3 translational + 3 rotational) at moderate temperatures

✅ Key Points for JEE

  • 1The Maxwell-Boltzmann speed distribution curve illustrates that gas molecules possess a range of speeds, with the peak shifting towards higher speeds and broadening as temperature increases.
  • 2Degrees of freedom represent the number of independent ways a molecule can possess energy (translational, rotational, vibrational); their value depends on molecular structure and temperature.
  • 3The Equipartition of Energy Theorem states that for any system in thermal equilibrium, the total energy is equally distributed among its active degrees of freedom, with each degree of freedom contributing (1/2)kT per molecule or (1/2)RT per mole.
  • 4Vibrational degrees of freedom are 'activated' only at high temperatures due to their higher energy quanta, leading to an increase in total degrees of freedom and internal energy at elevated temperatures.
  • 5Root Mean Square speed (v_rms) is used when dealing with kinetic energy, Average speed (v_avg) is a simple arithmetic mean, and Most Probable speed (v_p) corresponds to the peak of the Maxwell-Boltzmann distribution curve.

⚠️ Common Mistakes

  • Confusing the formulas or physical significance of v_p, v_avg, and v_rms, or failing to recall their relative magnitudes.
  • Incorrectly counting degrees of freedom for diatomic or polyatomic molecules, especially neglecting or overcounting vibrational modes which depend on temperature.
  • Not using absolute temperature (in Kelvin) for all calculations involving gas speeds or internal energy.
  • Misapplying the Equipartition Theorem by incorrectly associating energy contribution with non-active degrees of freedom.

📝 Practice Questions

See all

Q29.Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : With the increase in the pressure of an ideal gas, the volume falls off more rapidly in an isothermal process in comparison to the adiabatic process. Reason (R) : In isothermal process, PV = constant, while in adiabatic process PVγ = constant. Here γ is the ratio of specific heats, P is the pressure and V is the volume of the ideal gas. In the light of the above statements, choose the correct answer from the options given below : (1) Both (A) and (R) are true and (R) is the correct (2) Both (A) and (R) are true but (R) is NOT the explanation of (A) correct explanation of (A) (3) (A) is true but (R) is false (4) (A) is false but (R) is true 2025 (29 Jan Shift 2) JEE Main Previous Year Paper

2025·Assertion ReasoningMedium

Q40.A cup of coffee cools from 90∘C to 80∘C in t minutes when the room temperature is 20∘C. The time taken by the similar cup of coffee to cool from 80∘C to 60∘C at the same room temperature is : (1) 13 10 t (2) 1013 t (3) 13 5 t (4) 135 t

2025·NumericalMedium

Q42. A poly-atomic molecule ( CV = 3R, CP = 4R, where R is gas constant) goes from phase space point A (PA = 105 Pa, VA = 4 × 10−6 m3) to point B (PB = 5 × 104 Pa, VB = 6 × 10−6 m3) to point 2025 (29 Jan Shift 2) JEE Main Previous Year Paper C (PC = 104 Pa, VC = 8 × 10−6 m3). A to B is an adiabatic path and B to C is an isothermal path. The net heat absorbed per unit mole by the system is : (1) 500R(ln 3 + ln 4) (2) 450R(ln 4 −ln 3) (3) 500R ln 2 (4) 400R ln 4

2025·MCQHard

Q29.An amount of ice of mass 10−3 kg and temperature −10∘C is transformed to vapour of temperature 110∘C by applying heat. The total amount of work required for this conversion is, (Take, specific heat of ice = 2100Jkg−1 K−1 , specific heat of water = 4180Jkg−1 K−1 , specific heat of steam = 1920Jkg−1 K−1 , Latent heat of ice = 3.35 × 105Jkg−1 and Latent heat of steam = 2.25 × 106 Jkg−1 ) (1) 3043 J (2) 3024 J (3) 3003 J (4) 3022 J

2025·MCQMedium

Q41.Two spherical bodies of same materials having radii 0.2 m and 0.8 m are placed in same atmosphere. The temperature of the smaller body is 800 K and temperature of the bigger body is 400 K . If the energy radiated from the smaller body is E, the energy radiated from the bigger body is (assume, effect of the surrounding temperature to be negligible), (1) 16 E (2) E (3) 64 E (4) 256 E

2025·MCQMedium

Q49.Three conductors of same length having thermal conductivity k1, k2 and k3 are connected as shown in figure. Area of cross sections of 1st and 2nd conductor are same and for 3rd conductor it is double of the 1st conductor. The temperatures are given in the figure. In steady state condition, the value of θ is _______ ∘C. (Given : k1 = 60Js−1 m−1 K−1, k2 = 120Js−1 m−1 K−1, k3 = 135Js−1 m−1 K−1 ) ∣ ∣∣ 2025 (22 Jan Shift 1) JEE Main Previous Year Paper

2025·NumericalMedium

NCERT Chapters

  • Class 11 Physics Ch 13: Kinetic Theory