MHT-CET Physics · Thermodynamics

The Four Processes and the Adiabatic Relations

Holding temperature, heat, pressure or volume fixed gives the isothermal, adiabatic, isobaric and isochoric processes; an adiabatic change of an ideal gas follows PV^γ = constant, equivalently TV^(γ−1) = constant, which is steeper on a P–V graph than an isothermal.

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Why this matters

36 PYQs, 11 of them HARD. Thirteen identify a process from a statement or a graph. Sixteen apply the adiabatic relations — the final pressure or temperature after a sudden compression, P against T, density against pressure. Seven chain two processes or compare them, including gases that obey PV² or VP² = constant. Three cards.

Concept 1 of 3: Recognising Each Process

Each process fixes one quantity. Isothermal: T constant, so ΔU = 0, PV = constant, a hyperbola on P–V, and all heat becomes work. Adiabatic: no heat exchanged, the system insulated, PV^γ = constant, and a steeper curve than the isothermal through the same point; the temperature changes. Isobaric: P constant, a horizontal line on P–V. Isochoric: V constant, a vertical line on P–V, no work, dQ = dU. A process PV^n = constant with zero specific heat must be adiabatic, so n = γ. On a p–T graph an isothermal is a vertical line.

Definition

Isothermal: T const, ΔU = 0, Q = W, hyperbola.
Adiabatic: Q = 0, $PV^\gamma$ const, steeper than isothermal; ΔT ≠ 0.
Isobaric: P const, horizontal on P–V. Isochoric: V const, vertical on P–V, dQ = dU.
$PV^n$ const with zero specific heat ⇒ $n = \gamma$ .
On a cycle's P–V graph, the two steeper sides are the adiabatics.

Process equations

PV = \text{const (isothermal)}, \qquad PV^\gamma = \text{const (adiabatic)}

Worked example

A cycle on a P–V graph has two curved sides of different steepness and two straight sides. Which curves are adiabatic?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 1 · Q21Moderate

Example 1 · Thermodynamics · Isothermal, Adiabatic, Isobaric, and Isochoric Processes

The P-V graph of an ideal gas cycle is shown. The adiabatic process is described by the region

Calling the steeper curve isothermal

Adiabatic curves are the steeper ones — γ times the isothermal slope at the same point.

Thinking an adiabatic keeps temperature constant

No heat enters, but work changes the internal energy, so the temperature changes. Constant temperature is the isothermal.

Drill 12 more on recognising each process

Concept 2 of 3: The Adiabatic Relations

Three forms of one law: PV^γ, TV^(γ−1) and P^(1−γ)T^γ are all constant. A 'sudden' compression is adiabatic. Compress to 1/k of the volume and the pressure rises by k^γ and the temperature by k^(γ−1); powers of 2 make the numbers clean — 32^(2/5) = 4 and 8^(4/3) = 16. Pressure follows temperature as P ∝ T^(γ/(γ−1)), 3.5 for a diatomic gas. Density is inversely proportional to volume, so P ∝ ρ^γ. The rms speed goes as √T, so cutting it 4 times needs T to fall 16 times.

Definition

$PV^\gamma$ , $TV^{\gamma - 1}$ and $P^{1-\gamma}T^\gamma$ are constant.
Compressed to 1/k: $P \times k^\gamma$ , $T \times k^{\gamma - 1}$ (monoatomic to 1/8 at 300 K ⇒ 1200 K; to 1/27 ⇒ 9T).
$P \propto T^{\gamma/(\gamma - 1)}$ : diatomic 3.5; $P \propto \rho^\gamma$ : ρ × 32 ⇒ P × 128.
Gas column in a cylinder: $\dfrac{T_2}{T_1} = \left(\dfrac{L_1}{L_2}\right)^{\gamma - 1}$ .
$v_{\text{rms}} \propto \sqrt{T}$ : ÷4 ⇒ T ÷ 16 ⇒ with γ = 1.5, V × 256.

Adiabatic

PV^{\gamma} = \text{const}, \qquad TV^{\gamma - 1} = \text{const}

Worked example

Air (γ = 1.4) at 300 K is suddenly compressed to 1/32 of its volume. Final temperature?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 20 April Shift I · Q30Moderate

Example 2 · Thermodynamics · Isothermal, Adiabatic, Isobaric, and Isochoric Processes

A diatomic gas

\left( \gamma=\frac{7}{5} \right)

is compressed adiabatically to volume

\frac{V_{0}}{32}

, where

V_{0}

is its initial volume. The initial temperature of the gas is

T_{i}

in kelvin and the final temperature is

xT_{i}

in kelvin. The value of

x

Using γ in the temperature relation

Pressure goes as V^(−γ) but temperature as V^(−(γ−1)). Using γ for the temperature turns 4 into 128.

Forgetting that 'sudden' means adiabatic

A gas compressed suddenly has no time to exchange heat. Treating it as isothermal gives 4P instead of 8P for a four-fold compression with γ = 1.5.

Drill 15 more on the adiabatic relations

Concept 3 of 3: Chaining and Comparing Processes

Work through a chain one leg at a time, using the rule for each leg: an isothermal leg keeps PV, an adiabatic leg keeps PV^γ, an isobaric leg keeps P. To compare processes from the same start to the same volume, remember that the adiabatic changes pressure most: compressed to half, the isothermal pressure doubles and the adiabatic rises 2^γ times. A gas forced along PV^n = constant has T ∝ V^(1−n): for PV² = constant the temperature falls as it expands, for VP² = constant it rises as √V.

Definition

Isothermal V → 4V then adiabatic back to V (γ = 3/2): $\dfrac{P}{4} \times 4^{3/2} = 2P$ .
Same compression to V/8 (γ = 5/3): isothermal 8P, adiabatic 32P, ratio 1 : 4.
Same final pressure after doubling V: $P_{\text{iso}} : P_{\text{adia}} : P_{\text{isobar}} = 2 : 2^\gamma : 1$ .
$PV^2$ const ⇒ $T \propto \dfrac{1}{V}$ ; $VP^2$ const ⇒ $T \propto \sqrt{V}$ (2V ⇒ $\sqrt{2}\,T$ ).

Chaining

\text{isothermal: } P_1V_1 = P_2V_2, \qquad \text{adiabatic: } P_2V_2^\gamma = P_3V_3^\gamma

Worked example

A gas (γ = 5/3) at P, V expands isothermally to 2V, then adiabatically to 16V. Final pressure?

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 10th May Shift 1 · Q37Hard

Example 3 · Thermodynamics · Isothermal, Adiabatic, Isobaric, and Isochoric Processes

Initial pressure and volume of a gas are 'P' and 'V' respectively. First its volume is expanded to '4V' by isothermal process and then again its volume is reduced to 'V' by adiabatic process; then its final pressure if

\gamma=\frac{3}{2}

Applying one rule to the whole chain

Each leg has its own rule. PV^γ through an isothermal leg, or PV through an adiabatic one, gives a wrong pressure at the join.

Drill 6 more on chaining and comparing processes

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Recognising Each Process
Process equations
$PV = \text{const (isothermal)}, \qquad PV^\gamma = \text{const (adiabatic)}$
The Adiabatic Relations
Adiabatic
$PV^{\gamma} = \text{const}, \qquad TV^{\gamma - 1} = \text{const}$
Chaining and Comparing Processes
Chaining
$\text{isothermal: } P_1V_1 = P_2V_2, \qquad \text{adiabatic: } P_2V_2^\gamma = P_3V_3^\gamma$

Watch out for (5)

Calling the steeper curve isothermal
→ Recognising Each Process
Thinking an adiabatic keeps temperature constant
→ Recognising Each Process
Using γ in the temperature relation
→ The Adiabatic Relations
Forgetting that 'sudden' means adiabatic
→ The Adiabatic Relations
Applying one rule to the whole chain
→ Chaining and Comparing Processes

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