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Thermodynamics · Part 6 of 10

Adiabatic Process

Pump a tyre fast and the pump gets hot with no heat supplied. That is the adiabatic process, Q = 0. This part derives $P V^{γ}$ = constant, shows why the adiabat is γ times steeper than the isotherm, finds the adiabatic work, and ends with the isothermal and adiabatic bulk moduli.

Builds on: Part 4 · Molar Specific Heats: Cv, Cp and γ, Part 5 · Isochoric, Isobaric and Isothermal Processes.

Video coming soon

Derivation

$P V^{γ}$ = constant

$d Q = 0$ : $n C_{v} d T + P d V = 0$ . With $d T = (P d V + V d P) / n R$ and $C_{v} + R = C_{p}$ : $γ d V / V + d P / P = 0$ , so $P V^{γ}$ = constant, $T V^{γ - 1}$ = constant and $T^{γ} P^{1 - γ}$ = constant.

Board with the adiabatic derivation. — Full derivation from the first law.

An isotherm and a steeper adiabat crossing at one point. — Isotherm and adiabat through the same state.

Slopes

Adiabat γ times steeper

Isotherm: $d P / d V = - P / V$ . Adiabat: $d P / d V = - γ P / V$ . Through any point the adiabat falls γ times faster, because an expanding adiabatic gas also cools.

Adiabatic work

$W = - Δ U$

$W = \frac{P _{1} V _{1} - P _{2} V _{2}}{γ - 1} = \frac{n R ( T _{1} - T _{2} )}{γ - 1}$ . Expansion: T falls, W > 0. Compression: T rises, W < 0.

Adiabatic expansion with the area under it shaded. — Work = area under the adiabat.

Bulk modulus

$B_{i so} = P$ , $B_{a d i} = γ P$

$B = - V d P / d V$ . Isothermal: $B = P$ . Adiabatic: $B = γ P$ . Sound is adiabatic, so $v = γ P / ρ$ .

Summary

Key formulas

adiabatic: P V^{γ} = const

Adiabatic Process

T V^{γ - 1} = const

Adiabatic Process

W = \frac{P _{1} V _{1} - P _{2} V _{2}}{γ - 1} = \frac{n R ( T _{1} - T _{2} )}{γ - 1}

Adiabatic Process

Worked examples

One for every idea

Adiabatic Process

1. A gas with γ = 1.5 is compressed adiabatically from 16 litres to 12 litres. Find the ratio of final to initial pressure and of final to initial temperature.

$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$ : $\frac{P _{2}}{P _{1}} = (\frac{16}{12})^{1.5} \approx 1.54$ .
$T V^{γ - 1}$ constant: $\frac{T _{2}}{T _{1}} = (\frac{16}{12})^{0.5} \approx 1.15$ .
Both > 1: adiabatic compression heats the gas.

$P_{2} / P_{1} \approx 1.54$ , $T_{2} / T_{1} \approx 1.15$ . (The JSON calls this gas diatomic; a diatomic gas has γ = 1.4, so the video treats it simply as a gas with γ = 1.5.)

JEE-style question

Your turn

A monatomic ideal gas at 300 K expands adiabatically to 8 times its volume. Its final temperature is:

(a)75 K

(b)37.5 K

(c)9.4 K

(d)300 K

Show the answer and the traps

T₂ = T₁(V₁/V₂)^(γ − 1) = 300 × (1/8)^(2/3) = 300/4 = 75 K: option a.

37.5 K drops the power. 9.4 K uses γ instead of γ − 1.

And 300 K forgets that the gas cools as it does work with no heat coming in.

Watch out

Common mistakes

Swapping γ and γ − 1

P V^{γ}

for pressure,

T V^{γ - 1}

for temperature.

Thinking Q = 0 means no temperature changeAdiabatic compression always heats the gas; expansion cools it.

Using B = P for soundSound compressions are adiabatic:

B = γ P

Practice

Try these

1. Air (γ = 1.4) at 27 °C is compressed adiabatically to one-eighth of its volume. Find the final temperature.

$T_{2} = 300 \times 8^{0.4} \approx 689$ K (about 416 °C).

2. 2 mol of a monatomic gas expands adiabatically, cooling from 500 K to 300 K. Find the work done by the gas.

$W = \frac{n R ( T _{1} - T _{2} )}{γ - 1} = \frac{2 \times 8.314 \times 200}{2/3} \approx 4988$ J.

3. A gas at

1 0^{5}

Pa has an adiabatic bulk modulus of

1.4 \times 1 0^{5}

Pa. Find γ and its isothermal bulk modulus.

γ = 1.4; $B_{i so} = P = 1 0^{5}$ Pa.

Adiabatic Process

PVγ = constant

Adiabat γ times steeper

W=−ΔU

Biso​=P, Badi​=γP

Key formulas

One for every idea

Your turn

Common mistakes

Try these

$P V^{γ}$ = constant

$W = - Δ U$

$B_{i so} = P$ , $B_{a d i} = γ P$