Polytropic Processes
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Direct answer
A polytropic process is any quasistatic path obeying PV^x = constant, and the familiar processes are just particular x values: x = 0 is isobaric, x = 1 is isothermal, x = γ is adiabatic and x → ∞ is isochoric. The work done by the gas is W = (P1V1 − P2V2)/(x − 1) = nR(T1 − T2)/(x − 1), which stays finite for every x except 1, where the isothermal limit gives nRT ln(V2/V1) instead. The molar heat capacity of the path is C = Cv + R/(1 − x): zero for the adiabat, infinite for the isotherm, and — the JEE favourite — negative whenever 1 < x < γ, meaning the gas warms while losing heat. One exponent therefore runs the entire thermodynamic family.
What you must remember
- The exponent map: x = 0 isobaric, x = 1 isothermal, x = γ adiabatic, x → ∞ isochoric; on a P–V diagram the slope steepens as x grows, so an adiabat rises more sharply than an isotherm through the same point.
- Work formula: W = nR(T1 − T2)/(x − 1); for x = 1 use W = nRT ln(V2/V1); the isochoric work is zero, recovered as x → ∞.
- Molar heat capacity of the path: C = Cv + R/(1 − x); substituting x = γ gives zero (adiabatic), x = 1 gives infinity (isothermal, heat flows with no temperature change).
- Negative heat capacity: for 1 < x < γ, C < 0 — the gas heats up while rejecting heat, or cools while absorbing; processes faster than isothermal but slower than adiabatic behave this way.
- Temperature–volume link: T V^(x−1) = constant, so for x > 1 an expansion always cools the gas; the magnitude steepens with x.
- Finding x from data: if P ∝ 1/V^(3/2) then x = 3/2; if the graph through (P0, V0) and (P0/8, 2V0) fits PV^x, then (1/8) = 2^(−x) gives x = 3.
- Monoatomic benchmarks: Cv = 3R/2 and γ = 5/3; diatomic Cv = 5R/2, γ = 7/5 — needed before C = Cv + R/(1 − x) yields numbers.
One gas, three exponents
Take one mole of a monoatomic gas (Cv = 3R/2) at 300 K expanding from 10 litres to 20 litres along PV^(3/2) = constant. Since T V^(1/2) = constant, doubling V multiplies T by 1/√2, so T2 = 212 K. The work done is W = R(T1 − T2)/(x − 1) = 8.31 × 88/0.5 ≈ 1463 J. The molar heat capacity of the path is C = 3R/2 + R/(1 − 1.5) = 3R/2 − 2R = −R/2 — negative, because 1 < 1.5 < 5/3 lies between isothermal and adiabatic. Consistency check: heat exchanged Q = C ΔT = (−R/2)(−88) ≈ +366 J absorbed, and indeed ΔU = Cv ΔT = (3R/2)(−88) ≈ −1097 J satisfies Q = ΔU + W. The three bookkeeping lines close perfectly, which is the fastest self-check in thermodynamics.
The negative-C result deserves a moment because it is the most quoted outcome of polytropic thinking. Along PV^(3/2) = constant the gas expands, does positive work, and cools — yet the path is slower than an adiabat, so heat trickles in while the temperature falls. Nothing paradoxical is happening: the work term dominates the energy ledger, and C merely reports the sign of the mismatch.
How the exam frames it
JEE Main tends to hand you x directly — "a gas undergoes PV² = constant" — and ask for work or final temperature, so the marks live in clean substitution and remembering the (x − 1) rather than (1 − x) in the denominator; check your sign against physics (expansion with x > 1 must cool the gas, so W must come out positive). JEE Advanced prefers to hide the exponent: a graph whose two labelled points pin down x, or "P is proportional to 1/V^(3/2)", or asking which of five paths has negative molar heat capacity. The multi-correct trap pairs a true statement (negative C exists) with a false companion (C is negative for all x > 1 — false, only up to γ). And keep the gas's identity consistent across Cv, γ and R; mixing a diatomic Cv with a monoatomic γ is a quiet frequency-seen error.
Frequently asked questions
Which value of x corresponds to each standard process?
x = 0 gives constant pressure, x = 1 the isotherm, x = γ the adiabat and x → ∞ the isochore; every other x defines a genuine polytropic path between these archetypes.
What is the work done in a polytropic process?
W = (P1V1 − P2V2)/(x − 1) = nR(T1 − T2)/(x − 1) for x ≠ 1, with the isothermal case x = 1 requiring W = nRT ln(V2/V1).
When is the molar heat capacity of a process negative?
Whenever 1 < x < γ, since C = Cv + R/(1 − x) then dips below zero — the gas can absorb heat while cooling, or release heat while warming.
Why is C infinite for an isothermal process?
Because any finite heat transfer produces zero temperature change, so C = Q/(nΔT) divides by zero — the definition, not a paradox.
How do you extract x from a P–V graph?
Use the labelled points in PV^x = constant: the ratio of pressures equals the ratio of volumes raised to −x, and a couple of powers of two usually make x obvious by inspection.