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Foundation5 past questions

Degrees of freedom and thermal properties

Degrees of freedom determine the distribution of energy among molecules, affecting internal energy, specific heats, and the ratio of specific heats (γ).

Why this shows up in the exam

NEET tests your ability to use the equipartition theorem, calculate internal energy, and relate γ to molecular structure.

How NEET tests this

Direct recall · 3 QsNumerical · 3 Qs

Learn the idea

Degrees of freedom (f) tell how many independent quadratic energy terms each molecule has; each such term carries (½)kBT of average thermal energy.

🧠 Memory hook: Mono‑atomic = 3 (just move), Linear di‑atomic = 5 (3 move + 2 turn), Non‑linear = 6 (add one more turn).

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • Equipartition theorem: each quadratic degree of freedom contributes (½)kBT per molecule (or (½)RT per mole).
  • For a mono‑atomic gas f=3 (only translation).
  • For a linear di‑atomic gas f=5 (3 translation + 2 rotation, vibration ignored at ordinary T).
  • For a non‑linear poly‑atomic gas f=6 (3 translation + 3 rotation).
  • Cv (molar) = (f/2) R and Cp = Cv + R.
  • γ = Cp/Cv = (f+2)/f.

How to approach it

  1. 11. Identify the molecular type (mono‑atomic, linear di‑atomic, non‑linear).
  2. 22. Write the appropriate f using the NCERT table (3,5,6).
  3. 33. Use equipartition to get Cv = (f/2)R and then Cp = Cv+R.
  4. 44. Form γ = Cp/Cv = (f+2)/f or use Cp‑Cv = R as needed.

Worked example — watch it click

The average thermal energy for a monoatomic gas is: (kB is Boltzmann constant and T absolute temperature)

  • ✅3/2 kBT
  • B)5/2 kBT
  • C)7/2 kBT
  • D)1/2 kBT

The concept behind this problem

The example asks for the average thermal energy of a mono‑atomic gas, which directly uses the equipartition rule with f=3 to give (3/2)kBT.

Step by step

  1. 1For a monoatomic gas, there are 3 translational degrees of freedom.
  2. 2By equipartition theorem, each degree of freedom contributes (1/2)kᵦT to the average energy.
  3. 3Therefore, average thermal (kinetic) energy per molecule = 3 × (1/2)kᵦT = (3/2)kᵦT.

Watch out

A common slip is to add rotational contributions for a mono‑atomic gas, leading to a wrong answer like 5/2 kBT.

Common slip-ups that cost marks

  • •Treating vibrational modes as active at room temperature – they are frozen for most NEET problems.
  • •Mixing per‑molecule (kBT) and per‑mole (RT) forms of the theorem.
  • •Confusing the number of atoms with the number of degrees of freedom.

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Practise it

These are real questions from past NEET papers that test this exact idea.

Question 1 of 5NEET 2015

The ratio of specific heats Cₚ/Cᵥ = γ in terms of degree of freedom (n) is given by:

Push further

More challenging

4 harder questions built from the past papers above — a step up in difficulty, with distractors designed so you can't get there by elimination. Written and checked by our reviewers, not from a real paper.

Question 1 of 4

If the difference between the molar specific heats at constant pressure and constant volume for nitrogen gas (N₂) is R, what would be the difference between the specific heats per unit mass (cₚ - cᵥ) for helium gas (He)? (Molar mass of N₂ = 28 g/mol, Molar mass of He = 4 g/mol)

More from Kinetic Theory of Gases