On this page
A question gives you 4 g of helium and 16 g of oxygen in one cylinder and asks for γ, or asks how hot the mixture gets after an adiabatic squeeze. Most students know not to average the two gammas. Fewer can do the reverse problem, or spot when the question has hidden the mole numbers inside masses. This post covers those cases with one shortcut and a check you can do in your head.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus both list degrees of freedom, the law of equipartition of energy and its "applications to specific heat capacities of gases" under Kinetic Theory. The JEE (Advanced) 2026 brochure lists "Specific heats (Cv and Cp for monoatomic and diatomic gases)". The NCERT Class 11 chapter Kinetic Theory (2026-27 reprint) derives Cv = (3/2)R and γ = 5/3 for monatomic gases, Cv = (5/2)R and γ = 7/5 for rigid diatomic gases, and Cp − Cv = R for any ideal gas. It does not work a mixture example, but the method follows directly from adding internal energies. Check the syllabus for your year.
Why do you add Cv and not γ?
Internal energy adds. If n₁ moles of gas 1 and n₂ moles of gas 2 are heated by ΔT at constant volume, the heat needed is n₁Cv₁ΔT + n₂Cv₂ΔT. So the mixture's molar Cv is
Cv(mix) = (n₁Cv₁ + n₂Cv₂) / (n₁ + n₂)
Cp − Cv = R still holds for the mixture, because it is still an ideal gas. Then γ = Cp/Cv. The gammas themselves are ratios, and ratios do not average.
A faster form: since Cv = R/(γ − 1) for each ideal gas,
n/(γmix − 1) = n₁/(γ₁ − 1) + n₂/(γ₂ − 1)
For a monatomic gas 1/(γ − 1) = 3/2, for rigid diatomic 5/2. That is just f/2, half the degrees of freedom.
Worked example 1: a mixture given in grams
4 g of helium and 16 g of oxygen. Find γ.
Convert to moles: helium 4/4 = 1 mol; oxygen 16/32 = 0.5 mol. Total 1.5 mol.
Cv(mix) = (1 × 1.5R + 0.5 × 2.5R) / 1.5 = 2.75R / 1.5 = (11/6)R ≈ 1.833R.
Cp(mix) = Cv + R = (17/6)R.
γ = 17/11 ≈ 1.545.
Two tempting wrong answers: averaging the gammas by moles gives about 1.578, and averaging them by mass gives about 1.453. Both often appear as options.
Worked example 2: working backwards from γ
A mixture of helium and nitrogen has γ = 1.5. What fraction of the molecules are helium?
Let the helium mole fraction be x. Use the 1/(γ − 1) form: 1/(1.5 − 1) = 2, so
2 = 1.5x + 2.5(1 − x), which gives x = 0.5.
Equal moles. Check by the long route: Cv = 0.5 × 1.5R + 0.5 × 2.5R = 2R, Cp = 3R, γ = 1.5.
Worked example 3: an adiabatic compression
The mixture from example 1 starts at 300 K and is compressed adiabatically to half its volume. Find the final temperature.
For an adiabatic process TV^(γ−1) is constant, so T₂ = T₁ × 2^(γ−1) = 300 × 2^(6/11) ≈ 300 × 1.459 ≈ 438 K. Using the wrong averaged γ of 1.578 would give about 448 K, a difference large enough to change an answer option. For more on adiabatic and spring-loaded piston set-ups, see our post on gas, piston and spring problems.
Worked example 4: a non-linear molecule
Equal moles of methane and helium. Treat methane as a rigid non-linear molecule with 3 translational and 3 rotational degrees of freedom and no vibration, which is NCERT's polyatomic formula with f = 0: Cv = 3R. Then Cv(mix) = (3R + 1.5R)/2 = 2.25R and γ = 3.25/2.25 = 13/9 ≈ 1.44.
Be careful with carbon dioxide. CO₂ is linear, so it has only 2 rotational degrees of freedom; as a rigid molecule it behaves like a diatomic gas, with Cv = 2.5R. Read what the question tells you to assume.
Which mistakes cost marks?
Averaging γ directly, by moles or by mass.
Using grams as weights. Always convert to moles.
Including vibration by default. NCERT treats diatomic molecules as rigid unless told otherwise.
Forgetting that Cp = Cv + R per mole. Adding R to the total heat capacity of 1.5 mol, instead of the molar value, gives the wrong γ.
Applying the result to reacting gases. The method assumes the gases do not react.
The idea that each degree of freedom carries the same share of energy is easier to trust once you have seen molecules moving. The Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read. Our earlier guide to kinetic theory of gases covers rms speed and the basic mixture example.
A practice routine
Day 1: Make a table of f, Cv/R and 1/(γ − 1) for monatomic, rigid diatomic and non-linear rigid molecules. Learn it.
Day 2: Four mixture questions given in grams. For each, also compute the wrong averaged-γ answer.
Day 3: Two backwards questions: find the mole ratio from γ.
Day 4: Two adiabatic questions on mixtures, one compression and one expansion.
Tonight, find γ for 2 g of hydrogen mixed with 4 g of helium, then check it with the 1/(γ − 1) shortcut.
