PhysicsCBSE Class 11 · Thermodynamics

Isothermal and Adiabatic Processes of an Ideal Gas Class 11

Updated 17 October 2026 · Fact-checked

In a quasi-static isothermal process of an ideal gas, T is constant, PV = constant, ΔU = 0 and W = Q = nRT ln(V₂/V₁). In an adiabatic process Q = 0, so PV^γ = constant, TV^(γ−1) = constant and W = nC_v(T₁ − T₂) = (P₁V₁ − P₂V₂)/(γ − 1). On a P-V diagram the adiabat is steeper than the isotherm because γ > 1.

From the same start A, the adiabat falls more steeply than the isotherm, so at V₂ it ends lower (C below B) and the gas does less work.

Illustration drawn for QuestionBuddy.

Understand Isothermal and Adiabatic Processes of an Ideal Gas

Quasi-static processes. A process carried out so slowly that the system is always in equilibrium can be drawn as a curve on a P-V diagram and is *reversible* if no dissipative effects (friction, free expansion) occur.

Isothermal process ( $T$ constant, slow contact with a heat reservoir).

$P V = constant$ , $Δ U = 0$ for an ideal gas, so $Q = W$ .
$W = \int_{V_{1}}^{V_{2}} P d V = n R T ln \frac{V _{2}}{V _{1}} = n R T ln \frac{P _{1}}{P _{2}}$ .
The gas absorbs heat when it expands and gives out heat when compressed.

Adiabatic process ( $Q = 0$ , thermal insulation or a very fast process).

$P V^{γ} = constant$ , $T V^{γ - 1} = constant$ and $T^{γ} P^{1 - γ} = constant$ , with $γ = C_{p} / C_{v}$ ( $5/3$ monatomic, $7/5$ diatomic).
$W = \frac{P _{1} V _{1} - P _{2} V _{2}}{γ - 1} = n C_{v} (T_{1} - T_{2})$ , from the first law with $Q = 0$ so $Δ U = - W$ .
A gas cools in an adiabatic expansion and heats in an adiabatic compression.

Comparison. At any point of the P-V diagram, the slope of an isotherm is $- P / V$ and of an adiabat $- γ P / V$ , so the adiabat is steeper. Starting from the same state and expanding to the same volume, the gas does more work isothermally (the heat inflow keeps $P$ higher) than adiabatically.

Irreversibility. Free expansion of a gas into a vacuum has $W = 0$ , $Q = 0$ and $Δ U = 0$ : for an ideal gas the temperature does not change, unlike a quasi-static adiabatic expansion. It cannot reverse itself; real processes with friction, heat flow across a finite temperature difference, or uncontrolled expansion are irreversible.

Isothermal and Adiabatic Processes of an Ideal Gas: formulas and key facts

Isothermal law: $P V = constant$; Ideal gas, constant T.
Isothermal work: $W = n R T ln \frac{V _{2}}{V _{1}}$; $Q = W$ , $Δ U = 0$ .
Adiabatic law: $P V^{γ} = constant, T V^{γ - 1} = constant$; Q = 0.
Adiabatic work: $W = \frac{P _{1} V _{1} - P _{2} V _{2}}{γ - 1} = n C_{v} (T_{1} - T_{2})$; Reversible adiabatic.
Ratio of heat capacities: $γ = \frac{C _{p}}{C _{v}}$; 5/3, 7/5.
Slope comparison: $(\frac{d P}{d V})_{a d} = γ (\frac{d P}{d V})_{i so}$; Adiabat steeper.

How to solve Isothermal and Adiabatic Processes of an Ideal Gas questions

Decide the process type first; then use PV (isothermal) or PV^γ and TV^(γ−1) (adiabatic) and the first law for energy.

1Identify the process from the wording: constant temperature, insulated, rapid, or slow in a conducting vessel.
2Isothermal: use P₁V₁ = P₂V₂ and W = nRT ln(V₂/V₁); ΔU = 0.
3Adiabatic: use P₁V₁^γ = P₂V₂^γ and T₁V₁^(γ−1) = T₂V₂^(γ−1).
4Get W from ΔU = −W (adiabatic) with ΔU = nC_vΔT.
5Compare the processes using the first law and the P-V curves.
6State whether the gas heats or cools.

Isothermal and Adiabatic Processes of an Ideal Gas in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A bicycle pump heating up

Suppose air (γ = 1.4) at 300 K is compressed very quickly to one-fifth of its volume.

Estimate the temperature of the compressed air.

Try it first, then show the solutionSolution

Fast compression is adiabatic: T₂ = T₁(V₁/V₂)^(γ−1) = 300 × 5^0.4.
5^0.4 = 1.904, so T₂ ≈ 571 K.

Answer: About 571 K.

Memory hook: Sudden compression heats the gas.

Gas expanding slowly in a cylinder

Suppose 1 mol of an ideal gas at 300 K is allowed to expand slowly in a conducting cylinder to twice its volume.

Find the work done by the gas.

Try it first, then show the solutionSolution

Isothermal: W = RT ln 2 = 8.314 × 300 × 0.693 = 1729 J.

Answer: About 1.73 kJ.

Memory hook: The heat absorbed equals the work.

Tricks to remember Isothermal and Adiabatic Processes of an Ideal Gas

Short memory aids. Say each one aloud once, then use it on a question the same day.

Isothermal: ΔU = 0
Heat in equals work out.
Use it when: Energy bookkeeping.
Adiabatic: Q = 0
ΔU = −W; expansion cools, compression heats.
Use it when: Sign of ΔT.
Steeper adiabat
γ > 1 makes the adiabat steeper than the isotherm.
Use it when: P-V sketches.
Free expansion
W = Q = ΔU = 0, but irreversible.
Use it when: Conceptual traps.

Isothermal and Adiabatic Processes of an Ideal Gas: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Adiabatic expansion of a monatomic gas · 2 marks · Very short answerBoard level

A monatomic ideal gas at a pressure of 1.0 × 10⁵ Pa expands adiabatically to eight times its volume. Find the final pressure (γ = 5/3).

Try it first, then show the solutionSolution

P₂ = P₁(V₁/V₂)^γ = 10⁵ × (1/8)^(5/3). (1)
8^(5/3) = 32, so P₂ = 10⁵/32 = 3125 Pa. (1)

Answer: 3125 Pa.

Example 2: Isothermal expansion · 3 marks · Short answerHarder board level

One mole of an ideal gas at 300 K doubles its volume in an isothermal quasi-static expansion (R = 8.314 J/mol K, ln 2 = 0.693). Find the work done by the gas, the change in its internal energy and the heat absorbed.

Try it first, then show the solutionSolution

W = RT ln(V₂/V₁) = 8.314 × 300 × 0.693 ≈ 1729 J. (1)
ΔU = 0 for an isothermal process of an ideal gas. (1)
Q = ΔU + W = 1729 J absorbed from the surroundings. (1)

Answer: W = 1729 J; ΔU = 0; Q = 1729 J.

Example 3: Adiabatic work and volume change · 3 marks · Short answerHarder board level

Two moles of a monatomic ideal gas at 400 K expand adiabatically and cool to 300 K. Find the work done by the gas and the ratio of the final to the initial volume (R = 8.314 J/mol K).

Try it first, then show the solutionSolution

W = nC_v(T₁ − T₂) = 2 × 1.5 × 8.314 × 100 = 2494 J. (1)
T V^(2/3) = constant (γ − 1 = 2/3). (1)
V₂/V₁ = (T₁/T₂)^(3/2) = (4/3)^1.5 ≈ 1.54. (1)

Answer: 2494 J; 1.54.

Example 4: Compressing air isothermally and adiabatically · 4 marks · Long answerHarder board level

Air (γ = 1.4) at 300 K and 1 atm is compressed to one-tenth of its volume (a) slowly in a conducting cylinder and (b) quickly in an insulated cylinder. Find the final pressure in each and the final temperature in (b). Which process needs more work?

Try it first, then show the solutionSolution

(a) Isothermal: P₂ = P₁V₁/V₂ = 10 atm. (1)
(b) Adiabatic: P₂ = 10^1.4 atm = 25.1 atm. (1)
T₂ = 300 × 10^0.4 = 300 × 2.512 ≈ 754 K. (1)
The adiabat is steeper, so the area under it is larger: the adiabatic compression needs more work. (1)

Answer: (a) 10 atm; (b) 25.1 atm and 754 K; adiabatic needs more work.

Example 5: Free expansion versus adiabatic expansion · 4 marks · Long answerHOTS / competency

One mole of a monatomic ideal gas at 300 K occupies a volume V. (a) It expands freely into a vacuum to 2V. (b) Starting from the same state, it expands reversibly and adiabatically to 2V. Find the final temperature and the work done by the gas in each case, and explain why they differ.

Try it first, then show the solutionSolution

(a) Free expansion: W = 0, Q = 0, so ΔU = 0 and T stays at 300 K. (1)
(b) T V^(2/3) = constant: T₂ = 300/2^(2/3) = 300/1.587 ≈ 189 K. (1)
W = nC_vΔT = 1.5 × 8.314 × (300 − 189) ≈ 1.38 × 10³ J. (1)
In (a) the gas pushes against nothing and does no work, so its internal energy is unchanged; in (b) it does work on the piston at the expense of its internal energy. (1)

Answer: (a) 300 K, W = 0; (b) 189 K, W ≈ 1.38 kJ.

Where students lose marks in Isothermal and Adiabatic Processes of an Ideal Gas

The slips examiners see most often, and what to write instead.

Mistake: Using PV = constant for an adiabatic process.
Confusing the two laws.
Do this: Adiabatic: PV^γ = constant.
Usually costs: The whole numerical.
Mistake: Applying Q = 0 to an isothermal process.
Mixing the conditions.
Do this: Isothermal means ΔT = 0, not Q = 0.
Usually costs: Concept marks.
Mistake: Writing the work done as PΔV for a changing pressure.
Treating P as constant.
Do this: Use W = nRT ln(V₂/V₁) or (P₁V₁ − P₂V₂)/(γ − 1).
Usually costs: The answer mark.
Mistake: Saying free expansion cools an ideal gas.
Confusing with adiabatic expansion.
Do this: In free expansion W = 0 and T is unchanged.
Usually costs: Reasoning mark.

Exam tips

State the type of process and the reason in the first line.
Use R = 8.314 J/mol K and show ln 2 = 0.693 where needed.
Show T V^(γ−1) = constant before substituting.
State the sign of Q, W and ΔU for each case.
Draw both curves on one P-V diagram to compare.

Practice questions on Isothermal and Adiabatic Processes of an Ideal Gas

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
When a gas is compressed adiabatically, its temperature
(a) rises (b) falls (c) stays constant (d) first rises then falls
Hint
Work done on the gas raises U.
Answer
(a).
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): On a P-V diagram the adiabatic curve through a point is steeper than the isothermal curve.
Reason (R): For a gas γ = C_p/C_v is greater than 1.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Slopes are −P/V and −γP/V.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
A gas at 1.5 × 10⁵ Pa has its volume halved isothermally. Find its new pressure.
Hint
P₁V₁ = P₂V₂.
Answer
P₂ = 3.0 × 10⁵ Pa.
Q4 · 3 marks · Short answerHarder board level
A diatomic gas (γ = 1.4) at 300 K is compressed adiabatically to one-quarter of its volume. Find its final temperature (4^0.4 = 1.741).
Hint
TV^(γ−1) = constant.
Answer
T₂ = 300 × 4^0.4 = 300 × 1.741 ≈ 522 K.
Q5 · 4 marks · Long answerHOTS / competency
A gas (γ = 1.4) at 5 × 10⁵ Pa and 2 L expands to 4 L (a) adiabatically and (b) isothermally. Find the final pressure and the work done in each case (2^1.4 = 2.639, ln 2 = 0.693), and say which is larger and why.
Hint
Adiabatic: W = (P₁V₁ − P₂V₂)/(γ − 1).
Answer
(a) P₂ = 5 × 10⁵/2.639 = 1.89 × 10⁵ Pa; W = (1000 − 758)/0.4 ≈ 605 J. (b) P₂ = 2.5 × 10⁵ Pa; W = P₁V₁ ln 2 = 693 J. The isothermal work is larger because the heat absorbed keeps the pressure higher during the expansion.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Isothermal work?Show answer
nRT ln(V₂/V₁).
2.Adiabatic law?Show answer
PV^γ = constant.
3.γ for monatomic gas?Show answer
5/3.
4.ΔU in an isothermal process of an ideal gas?Show answer
Zero.
5.Temperature in free expansion of an ideal gas?Show answer
Unchanged.
6.Which curve is steeper?Show answer
The adiabat.

Isothermal and Adiabatic Processes of an Ideal Gas: frequently asked questions

Why does a gas cool on adiabatic expansion?

It does work at the expense of its internal energy, and with no heat supplied the temperature falls.

What is a quasi-static process?

A process carried out so slowly that the system is at every stage in equilibrium.

What does reversible mean?

A process that can be retraced exactly by an infinitesimal change in conditions, with no dissipation.

Why is the free expansion of a gas irreversible?

The gas never spontaneously returns to the original half of the container.

Do real gases follow PV^γ = constant?

Approximately, when the process is fast enough to be adiabatic and the gas is nearly ideal.