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Thermodynamics · Part 8 of 10

Polytropic Process

Every standard process is one member of a family, $P V^{n}$ = constant. This stretch part (JEE Advanced, NSEP, INPhO) finds the molar heat capacity of any polytropic process, $C = \frac{R}{γ - 1} + \frac{R}{1 - n}$ , recovers the four special cases, and compares slopes on the P–V diagram.

Builds on: Part 5 · Isochoric, Isobaric and Isothermal Processes, Part 6 · Adiabatic Process.

Video coming soon

One family

$P V^{n}$ = constant

n = 0 isobaric, n = 1 isothermal, n = γ adiabatic, n → ∞ isochoric. Through a common point the slope $d P / d V = - n P / V$ grows with n, from flat to vertical.

Polytropic curves for n = 0, 1, gamma and infinity through one point. — Four processes through one point.

Heat capacity

$C = \frac{R}{γ - 1} + \frac{R}{1 - n}$

For one mole $W = \frac{R ( T _{1} - T _{2} )}{n - 1}$ , so $P d V = - \frac{R d T}{n - 1}$ and the first law gives C. Here n is the polytropic index, not the number of moles.

Special cases

n = 0, 1, γ, ∞

n = 0: $C = C_{p}$ . n = 1: C = ∞. n = γ: C = 0. n → ∞: $C = C_{v}$ . For 1 < n < γ, C is negative: the gas cools while absorbing heat.

Table of n, process and C. — C for each special process.

Summary

Key formulas

P V^{n} = const

Standard Processes as a Special Case of Polytropic Process

C = \frac{R}{γ - 1} + \frac{R}{1 - n}

Standard Processes as a Special Case of Polytropic Process

\frac{d P}{d V} = - n \frac{P}{V}

Indicator Diagram for a Polytropic Process

Worked examples

One for every idea

Standard Processes as a Special Case of Polytropic Process

1. A monatomic gas ( $γ = \frac{5}{3}$ ) undergoes a polytropic process with n = 2. Find its molar heat capacity.

$C = \frac{R}{γ - 1} + \frac{R}{1 - n} = \frac{R}{2/3} + \frac{R}{1 - 2}$ .
$= 1.5 R - R$ .

$C = 0.5 R \approx 4.16$ J/mol·K.

Indicator Diagram for a Polytropic Process

2. Two polytropic curves pass through the same point on a P–V diagram, with n = 1.2 and n = 1.5. Which is steeper there, and why?

$\frac{d P}{d V} = n \frac{P}{V}$ , and P and V are the same for both at that point.
So the slope is proportional to n: ratio 1.5/1.2 = 1.25.

The n = 1.5 curve is steeper (1.25 times).

JEE-style question

Your turn

One mole of a diatomic ideal gas (γ = 1.4) undergoes the process PV³ = constant. Its molar heat capacity in this process is:

(a)2R

(b)3R

(c)2.5R

(d)0

Show the answer and the traps

C = R/0.4 + R/(1 − 3) = 2.5R − 0.5R = 2R: option a.

3R flips the sign of R/(1 − n). 2.5R is just C_v.

And zero is the adiabatic case, n = γ, not n = 3.

Watch out

Common mistakes

Confusing the polytropic index n with the number of molesIn

P V^{n}

, n is a pure number; the moles are a separate quantity.

Flipping the sign of R/(1 − n)

C = \frac{R}{γ - 1} + \frac{R}{1 - n}

; for n > 1 the second term is negative.

Practice

Try these

1. Find the molar heat capacity of a diatomic gas (γ = 7/5) in the process

P V^{1/2}

= constant.

$C = 2.5 R + \frac{R}{1 - 0.5} = 4.5 R$ .

2. A monatomic gas follows

P V^{1.3}

= constant. Find C and say what is unusual about it.

$C = 1.5 R - \frac{R}{0.3} \approx - 1.83 R \approx - 15.2$ J/mol·K: negative, so the gas cools while absorbing heat.

3. One mole of ideal gas expands along

P V^{2}

= constant from V to 2V, starting at 400 K. Find the final temperature.

$T \propto P V \propto V^{- 1}$ : 200 K.

Polytropic Process

PVn = constant

C=γ−1R​+1−nR​

n = 0, 1, γ, ∞

Key formulas

One for every idea

Your turn

Common mistakes

Try these

$P V^{n}$ = constant

$C = \frac{R}{γ - 1} + \frac{R}{1 - n}$