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Turn P^a V^b = Constant into a Heat Capacity

A polytropic process follows P V^x = constant. Isobaric, isochoric, isothermal and adiabatic processes are all special cases, and for any x the molar heat capacity, the work done and the way temperature changes follow from one formula.

NSEJS · Thermal Physics · 5 linked questions · 100% coverage

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Polytropic Processes is the central idea in this question pattern. Identify the condition in the stem before applying the method below.

Study deeper: NCERT Science · Thermal Physics

Molar heat capacity in a polytropic process

For P V^

x = constant, the molar heat capacity is C = Cv + R/(1 - x)

It reduces correctly in the special cases: x = 0 (isobaric) gives Cv + R = Cp, x = γ (adiabatic) gives zero, x = 1 (isothermal) gives infinity, and very large x (isochoric) gives Cv. For one mole of a monatomic gas with PV⁵ = constant, C = 3R/2 + R/(1 - 5) = 3R/2 - R/4 = 5R/4.

When the process is described as P²

V = constant, rewrite it as P V^(1/2) = constant, so x = 1/2, and when it is P V^(2/3) = constant, x = 2/3

Which way does the temperature go?

Because T is proportional to PV, a process P V^

x = constant has T proportional to V^(1 - x)

If x is less than 1, temperature rises as the gas expands, so internal energy rises with it; if x is greater than 1, the temperature falls on expansion.

Both P² V = constant and P V^(2/3) = constant have x less than 1, so the gas heats up as it expands and its internal energy increases

Work in a polytropic or power-law process

For P = aV² (that is, P/V² = a), work is the integral of P dV, W = a(V2³ - V1³)/3

PV = RT for one mole, aV³ = RT, so W = R(T2 - T1)/3

The same trick (replace the power of V by RT/P or RT/a) converts any power-law work into a function of the temperature change.

Heat capacity from the heat-to-work ratio

If a diatomic gas does work Q/4 when heat Q is supplied, then ΔU = Q - Q/4 = 3Q/4 and the heat capacity is C = Q/(n ΔT) = (4/3) Cv = (4/3)(5R/2) = 10R/3. Always find ΔU from the first law, relate it to n Cv ΔT, and then divide the supplied heat by n ΔT.

OlympIQ exam lens

✓For P V^x = constant: C = Cv + R/(1 - x); x = 0, 1, γ and infinity give Cp, infinity, 0 and Cv
✓Rewrite P^a V^b = constant as P V^(b/a) = constant to find x
✓T is proportional to V^(1 - x); for x < 1 expansion raises temperature and internal energy
✓For P proportional to V², W = R(T2 - T1)/3 for one mole
✓With a given heat-to-work ratio, get ΔU from the first law and then C = Q/(n ΔT)

OlympIQ trap

Do not read the exponent on P as the polytropic index. In P²V = constant the index of V is 1/2 only after you take the square root of the whole relation; applying C = Cv + R/(1 - x) with x = 2 or x = 1 gives a wrong heat capacity.

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