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Two formulas, nCvΔT and nCpΔT, cover most heat questions on ideal gases in JEE and NEET. The trouble is knowing which one a question wants. Students often use nCvΔT for the heat in a constant-pressure process, or nCpΔT for the change in internal energy, and lose the marks on a correct-looking line. This post sets out exactly when each formula gives the heat, works three examples, and gives you a check that catches the mix-up.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus puts the first law of thermodynamics in Physics Unit 8 and "applications to specific heat capacities of gases" in Unit 9, and lists heat capacity and molar heat capacity in Chemistry Unit 4. The NEET (UG) 2026 syllabus has the same three entries. The JEE (Advanced) 2026 brochure lists "Specific heats (Cv and Cp for monoatomic and diatomic gases)". NCERT covers the processes in Class 11 Physics, Chapter 11 Thermodynamics (sections 11.8.4 and 11.8.5), and the values of Cv and Cp in Chapter 12 Kinetic Theory. Both are in the 2026-27 reprint. Check your own exam year's syllabus.
What happens to heat at constant volume?
NCERT's isochoric section says that no work is done on or by the gas, so "the heat absorbed by the gas goes entirely to change its internal energy and its temperature". So
Q = ΔU = nCvΔT (constant volume)
Chemistry writes the same idea as ΔU = qV in Class 11 Chemistry, Unit 5 Thermodynamics, the heat a bomb calorimeter measures.
And at constant pressure?
In an isobaric process NCERT gives the work as W = P(V₂ − V₁) = nR(T₂ − T₁), and says the heat "goes partly to increase internal energy and partly to do work". Adding the two parts:
Q = nCvΔT + nRΔT = nCpΔT (constant pressure)
That is NCERT's Cp − Cv = R (Eq. 11.8) read backwards. Chemistry calls this heat qp = ΔH.
The values come from equipartition. NCERT's Kinetic Theory chapter gives Cv = (3/2)R for a monatomic gas and (5/2)R for a rigid diatomic gas, with Cp = Cv + R. Its Table 12.2 lists measured values of 12.5 and 20.8 J mol⁻¹ K⁻¹ for Cv, and 20.8 and 29.1 J mol⁻¹ K⁻¹ for Cp.
Worked example 1: the same temperature rise, two ways
Heat 2.0 mol of nitrogen, a diatomic gas, from 300 K to 350 K. Use R = 8.314 J mol⁻¹ K⁻¹, so Cv = 20.8 and Cp = 29.1 J mol⁻¹ K⁻¹.
At constant volume:
Q = nCvΔT = 2.0 × 20.8 × 50 = 2080 J.
W = 0, so ΔU = 2080 J.
At constant pressure:
Q = nCpΔT = 2.0 × 29.1 × 50 = 2910 J.
W = nRΔT = 2.0 × 8.314 × 50 = 831 J.
ΔU = Q − W = 2079 J, the same as before (the 1 J gap is rounding).
The constant-pressure heat is larger by the factor Cp/Cv = 7/5. For a monatomic gas the factor is 5/3.
Worked example 2: when you know P and V, not n or T
An ideal diatomic gas expands at a constant 2.0 × 10⁵ Pa from 1.5 L to 4.0 L. Since PV = nRT, nRΔT = PΔV, so you never need n.
W = PΔV = 2.0 × 10⁵ × 2.5 × 10⁻³ = 500 J.
ΔU = (Cv/R) × PΔV = (5/2) × 500 = 1250 J.
Q = (Cp/R) × PΔV = (7/2) × 500 = 1750 J.
Check: Q − W = 1250 J = ΔU. Only 2/7 of the heat became work; for a monatomic gas the share is R/Cp = 2/5.
Worked example 3: same start and end, different heat
A monatomic ideal gas goes from state A (1.0 × 10⁵ Pa, 10 L) to state C (2.0 × 10⁵ Pa, 20 L) by two routes.
Route 1, isochoric then isobaric:
Raise P at 10 L: Q = (3/2)VΔP = 1.5 × 0.010 × 1.0 × 10⁵ = 1500 J.
Expand at 2.0 × 10⁵ Pa: Q = (5/2)PΔV = 2.5 × 2.0 × 10⁵ × 0.010 = 5000 J. Total 6500 J, with W = 2000 J.
Route 2, isobaric then isochoric:
Expand at 1.0 × 10⁵ Pa: Q = 2.5 × 1.0 × 10⁵ × 0.010 = 2500 J.
Raise P at 20 L: Q = 1.5 × 0.020 × 1.0 × 10⁵ = 3000 J. Total 5500 J, with W = 1000 J.
Both give ΔU = (3/2)(P_C V_C − P_A V_A) = 1.5 × (4000 − 1000) = 4500 J. The heat and work differ by 1000 J between routes; the internal energy change does not.
Why do students mix these up?
Meltzer (2004) studied 653 students in three introductory university physics courses and interviewed 32 more. Many over-generalised the state-function idea to heat and work, and 20% or fewer could use the first law effectively even after instruction. Example 3 is the fix: work it once by hand and the path dependence of Q stops being an abstract statement. For the sign conventions that differ between your physics and chemistry books, see our post on thermodynamics sign conventions.
Which mistakes cost marks?
Writing ΔU = nCpΔT for an isobaric process. ΔU is always nCvΔT.
Using Q = nCvΔT for a process that is not at constant volume.
Using Cv = (3/2)R for air or oxygen. Diatomic gases near room temperature have (5/2)R.
Mixing litres and cubic metres. 1 L = 10⁻³ m³, so 1 Pa × 1 L = 10⁻³ J.
Forgetting that Q can be negative. Cooling at constant pressure gives Q = nCpΔT with ΔT negative.
A practice routine
Write Q, W and ΔU for each of the two processes in a three-line table, from memory.
Redo example 1 for 2.0 mol of argon.
Redo example 3 for a diatomic gas. Does ΔU still match between routes?
Take four past JEE Main thermodynamics questions, label each step "isochoric" or "isobaric" before writing any formula, then solve.
When a step-by-step check would help, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Today, solve example 3 again without looking, and confirm that both routes give 4500 J.
