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Isobaric, isochoric, isothermal and adiabatic processes look like four topics with four formulas. They are one family. Any process that follows PVⁿ = constant is called polytropic, and the four NCERT processes are special values of n. This post derives the work and the molar heat capacity for any n, shows why the heat capacity can be negative, and works two JEE-style examples.
Is this in the syllabus?
Neither syllabus uses the word "polytropic". The Syllabus for JEE (Main) 2026 lists "the first law of thermodynamics, isothermal and adiabatic processes", and the JEE (Advanced) 2026 Information Brochure lists "First law of thermodynamics and its applications (only for ideal gases)". Processes such as P ∝ V or PV² = constant appear in problems as applications of the first law to an ideal gas. Check the current year's syllabus before your exam.
What do n = 0, 1, γ and infinity mean?
n = 0: P is constant, an isobaric process
n = 1: PV is constant, isothermal for an ideal gas
n = γ: adiabatic, where γ = Cp/Cv
n very large: V is constant, isochoric
On a graph of log P against log V, PVⁿ = constant is a straight line of slope −n, a quick way to read n from data.
How much work does the gas do?
Integrate P dV with P = K/Vⁿ. For n ≠ 1:
W = (P₁V₁ − P₂V₂)/(n − 1) = μR(T₁ − T₂)/(n − 1)
Here μ is the number of moles and W is the work done by the gas. Put n = γ and this is NCERT's adiabatic result, equation 11.16 in the chapter Thermodynamics. For n = 1, use the isothermal result μRT ln(V₂/V₁).
What is the molar heat capacity?
NCERT writes the first law as ΔQ = ΔU + ΔW (equation 11.1) and defines the molar heat capacity as C = ΔQ/(μΔT) (equation 11.6). For an ideal gas ΔU = μCvΔT, and the work formula gives ΔW = −μRΔT/(n − 1). Divide by μΔT:
C = Cv − R/(n − 1)
Check the special cases. n = 0 gives Cv + R = Cp. n = γ gives zero, as an adiabatic process should, because Cv = R/(γ − 1). A very large n gives Cv. At n = 1, C is infinite: heat flows but the temperature does not change.
Worked example 1: the process P ∝ V
Two moles of a monatomic ideal gas are heated along P = kV from 300 K to 400 K. NCERT's chapter Kinetic Theory gives Cv = 3R/2 for a monatomic gas. Take R = 8.314 J/(mol K), the NIST CODATA value rounded.
Here n = −1, so C = 3R/2 − R/(−2) = 3R/2 + R/2 = 2R.
ΔU = μCvΔT = 2 × 1.5 × 8.314 × 100 = 2494.2 J.
W = μR(T₁ − T₂)/(n − 1) = 2 × 8.314 × (300 − 400)/(−2) = 831.4 J.
Q = μCΔT = 2 × 2 × 8.314 × 100 = 3325.6 J.
Check: ΔU + W = 2494.2 + 831.4 = 3325.6 J. It matches.
Worked example 2: heat in, temperature down
One mole of a diatomic ideal gas (Cv = 5R/2, γ = 1.4) at 400 K expands along PV^1.2 = constant until its volume doubles.
From PV = μRT, TV^(n − 1) stays constant, so T₂ = 400 × 2^(−0.2) ≈ 400 × 0.8706 ≈ 348.2 K, and ΔT ≈ −51.78 K.
W = R(T₁ − T₂)/(n − 1) = 8.314 × 51.78/0.2 ≈ 2152 J.
C = 2.5R − R/0.2 = 2.5R − 5R = −2.5R.
Q = μCΔT = (−2.5 × 8.314) × (−51.78) ≈ +1076 J.
ΔU = μCvΔT = 2.5 × 8.314 × (−51.78) ≈ −1076 J. Check: ΔU + W = −1076 + 2152 = 1076 J = Q.
The gas absorbed about 1076 J of heat, yet its temperature fell by about 52 K.
Why do students find this confusing?
Loverude, Kautz and Heron (2002) studied university students reasoning about the adiabatic compression of an ideal gas. Although the students had studied the first law, few recognised its relevance, and many did not tell apart heat, temperature, work and internal energy. Example 2 tests exactly that: heat in does not mean temperature up. Write ΔU, W and Q as three separate lines every time.
Which mistakes cost marks?
Using the adiabatic formula with γ when the process has a different n.
Mixing sign conventions. The formula above gives work done by the gas, as in NCERT physics. Our post on thermodynamics sign conventions explains the chemistry version.
Forgetting that n can be negative, as in P ∝ V.
Using Cv = 3R/2 for a diatomic gas. NCERT gives 5R/2 for a diatomic gas at ordinary temperatures.
A practice routine
Day 1: derive C = Cv − R/(n − 1) from memory, then check the four special cases.
Day 2: three problems with n = −1, 2 and 1/2, finding W, ΔU and Q with a check line.
Day 3: read n from two given states using logarithms, then solve the energy balance.
Day 4: a gas held by a spring, from our post on gas under a spring-loaded piston, where one set-up gives P ∝ V.
If the special cases blur together, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, each with its own lesson. The Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer.
Take one problem from your thermodynamics exercise today and solve it twice: with its own formula, then with the polytropic formula for the right n.
