PhysicsISC Class 11 · Thermodynamics

First Law of Thermodynamics: Work, Heat and Gas Processes

Updated 17 October 2026 · Fact-checked

The first law states ΔQ = ΔU + W: the heat supplied to a system equals the increase in its internal energy plus the work it does. For an ideal gas ΔU = nC_vΔT, and each process (isochoric, isobaric, isothermal, adiabatic) simplifies the law differently.

Understand First Law and Thermodynamic Processes

Sign convention. Q is positive when heat is supplied to the gas, W is positive when the gas does work (expands), and ΔU is positive when the internal energy rises. The internal energy of an ideal gas depends only on temperature, ΔU = nC_vΔT.

Work by a gas. W = ∫p dV, the area under the p-V curve. For constant pressure W = pΔV. In a cycle, the net work is the area enclosed: positive for a clockwise loop.

Processes of an ideal gas.
1. *Isochoric* (constant V): W = 0, so ΔQ = ΔU = nC_vΔT.
2. *Isobaric* (constant p): W = pΔV = nRΔT and ΔQ = nC_pΔT.
3. *Isothermal* (constant T): ΔU = 0, so ΔQ = W = nRT ln(V₂/V₁). The pV product is constant.
4. *Adiabatic* (no heat exchange): ΔQ = 0, so W = −ΔU. pV^γ = constant, TV^(γ−1) = constant, and W = (p₁V₁ − p₂V₂)/(γ − 1) = nR(T₁ − T₂)/(γ − 1).

Heat capacities. Mayer's relation: C_p − C_v = R, and γ = C_p/C_v. For a monatomic gas C_v = 3R/2 and γ = 5/3; for a diatomic gas C_v = 5R/2 and γ = 7/5 (at ordinary temperatures).

Reading problems. Write what is constant, list Q, W and ΔU, and check that ΔQ = ΔU + W holds. An adiabatic expansion cools the gas because the work is done at the expense of internal energy; an adiabatic compression heats it.

First Law and Thermodynamic Processes: formulas and key facts

First law: $Δ Q = Δ U + W$; W is work done by the gas.
Ideal gas: $Δ U = n C_{v} Δ T ∣ C_{p} - C_{v} = R$; γ = C_p/C_v.
Isothermal: $W = n R T l n (V_{2} / V_{1})$; ΔU = 0, so Q = W.
Adiabatic: $p V^{γ} = co n s t an t ∣ T V^{(} γ - 1) = co n s t an t ∣ W = n R (T_{1} - T_{2}) / (γ - 1)$; Q = 0.

First Law and Thermodynamic Processes 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.

Pumping a bicycle tyre quickly

Suppose air (taken as a monatomic ideal gas for this model, γ = 5/3, C_v = 3R/2 with R = 8.3 J mol⁻¹ K⁻¹) in a pump, 1 mol at 300 K, is compressed so rapidly that no heat escapes, and its volume falls to one-eighth.

Find the final temperature and the work done on the gas.

Try it first, then show the solutionSolution

T V^(γ−1) = constant: T₂ = 300 × 8^(2/3) = 300 × 4 = 1200 K.
W on gas = ΔU = nC_vΔT = 1 × 12.45 × 900 = 1.12 × 10⁴ J.
The pump gets hot because the work goes into internal energy.

Answer: T₂ = 1200 K; work done on the gas = 1.12 × 10⁴ J.

Memory hook: No heat flow means all the work becomes internal energy.

Tricks to remember First Law and Thermodynamic Processes

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

I-I-I-A
Isochoric: W = 0. Isobaric: W = pΔV. Isothermal: ΔU = 0. Adiabatic: Q = 0.
Use it when: The first step in every process question.
Clockwise loop, positive work
The enclosed area of a clockwise p-V cycle is the net work done by the gas.
Use it when: Engine cycles.

First Law and Thermodynamic Processes: solved examples

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

Example 1: Isothermal process · 1 mark · MCQBoard level

In an isothermal process on an ideal gas, the change in internal energy is:
(a) positive
(b) negative
(c) zero
(d) equal to the work done

Try it first, then show the solutionSolution

U depends on temperature only, and T is constant, so ΔU = 0.

Answer: (c) zero

Example 2: Applying the first law · 3 marks · Short answerBoard level

A gas absorbs 500 J of heat and does 200 J of work. In a second process 150 J of heat leaves the gas while 100 J of work is done on it. Find ΔU in each case.

Try it first, then show the solutionSolution

Case 1: ΔU = Q − W = 500 − 200. (1 mark)
= 300 J. (0.5 mark)
Case 2: Q = −150 J and W = −100 J (work done on the gas). (1 mark)
ΔU = −150 − (−100) = −50 J. (0.5 mark)

Answer: ΔU = +300 J; ΔU = −50 J.

Example 3: Isobaric heating · 4 marks · Short answerHarder board level

Two moles of a monatomic ideal gas are heated at constant pressure from 300 K to 400 K (R = 8.3 J mol⁻¹ K⁻¹). Find W, ΔU and Q.

Try it first, then show the solutionSolution

W = nRΔT = 2 × 8.3 × 100 = 1660 J. (1 mark)
ΔU = n(3R/2)ΔT = 2 × 12.45 × 100 = 2490 J. (1 mark)
Q = ΔU + W = 4150 J. (1 mark)
Check: nC_pΔT = 2 × 20.75 × 100 = 4150 J. (1 mark)

Answer: W = 1660 J; ΔU = 2490 J; Q = 4150 J.

Example 4: Isothermal expansion · 3 marks · Short answerHarder board level

A cylinder holds 0.5 mol of an ideal gas at 400 K. It expands isothermally to three times its volume (R = 8.3 J mol⁻¹ K⁻¹, ln 3 = 1.099). Find the work done and the heat absorbed.

Try it first, then show the solutionSolution

W = nRT ln(V₂/V₁) = 0.5 × 8.3 × 400 × ln 3. (1 mark)
= 1660 × 1.099 = 1824 J. (1 mark)
ΔU = 0, so Q = W = 1824 J absorbed. (1 mark)

Answer: W = Q = 1.82 × 10³ J.

Example 5: Adiabatic compression · 5 marks · Long answerHOTS / competency

A cylinder holds 1 mol of a monatomic ideal gas at 300 K. It is compressed adiabatically to one-quarter of its volume (γ = 5/3, R = 8.3 J mol⁻¹ K⁻¹, 4^(2/3) = 2.52). Find the final temperature, the pressure ratio p₂/p₁ and the work done on the gas.

Try it first, then show the solutionSolution

T₂ = T₁(V₁/V₂)^(γ−1) = 300 × 4^(2/3) = 300 × 2.52 = 756 K. (1 mark)
p₂/p₁ = (V₁/V₂)^γ = 4^(5/3) = 4 × 2.52 = 10.08. (1 mark)
Check with pV = nRT: p₂/p₁ = (T₂/T₁)(V₁/V₂) = 2.52 × 4 = 10.08. (1 mark)
ΔU = nC_vΔT = 1 × 12.45 × 456 = 5677 J. (1 mark)
Q = 0, so the work done on the gas = 5.68 × 10³ J. (1 mark)

Answer: T₂ = 756 K; p₂/p₁ = 10.1; work on gas = 5.68 × 10³ J.

Where students lose marks in First Law and Thermodynamic Processes

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

Mistake: Wrong sign for work
Treating work done on the gas as positive in ΔQ = ΔU + W.
Do this: W is work by the gas; use a negative W for compression.
Usually costs: 1 to 2 marks.
Mistake: Using C_p for ΔU
Writing ΔU = nC_pΔT in a constant-pressure process.
Do this: ΔU = nC_vΔT always for an ideal gas; C_p is for the heat at constant pressure.
Usually costs: 2 marks.
Mistake: Using the isothermal formula in an adiabatic process
Taking pV constant when Q = 0.
Do this: Adiabatic: pV^γ constant.
Usually costs: 2 marks.

Exam tips

List the signs of Q, W and ΔU, then check that the first law balances.
State which quantity is constant in the process.
Use absolute temperature (kelvin) for all gas laws.
Quote R, γ and ln 2 as used.

Practice questions on First Law and Thermodynamic Processes

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
In an adiabatic process:
(a) the temperature is constant
(b) no heat is exchanged
(c) the pressure is constant
(d) no work is done
Hint
Adiabatic means insulated from heat exchange.
Answer
(b) no heat is exchanged
Q2 · 2 marks · Short answerBoard level
Steam in a cylinder pushes a piston out at a steady pressure of 4 × 10⁵ Pa, its volume growing from 1.5 × 10⁻³ m³ to 4.5 × 10⁻³ m³. Find the work done by the steam.
Hint
W = pΔV.
Answer
W = 4 × 10⁵ × 3 × 10⁻³ = 1200 J.
Q3 · 4 marks · Short answerHarder board level
One mole of a diatomic gas (C_v = 5R/2, R = 8.3 J mol⁻¹ K⁻¹) is heated from 300 K to 340 K, once at constant volume and once at constant pressure. Find Q and W in each.
Hint
C_p = C_v + R = 7R/2.
Answer
Constant V: Q = 20.75 × 40 = 830 J, W = 0. Constant p: Q = 29.05 × 40 = 1162 J, W = R × 40 = 332 J (ΔU = 830 J in both).
Q4 · 3 marks · Short answerHarder board level
A gas goes round the rectangle A(10⁵ Pa, 2 L) → B(3 × 10⁵ Pa, 2 L) → C(3 × 10⁵ Pa, 4 L) → D(10⁵ Pa, 4 L) → A. Find the net work done by the gas and the net heat absorbed in a cycle.
Hint
Net work = enclosed area; ΔU = 0 over a cycle.
Answer
Area = (3 − 1) × 10⁵ × (4 − 2) × 10⁻³ = 400 J, clockwise, so W = +400 J. ΔU = 0, so Q = 400 J.
Q5 · 4 marks · Short answerHarder board level
A gas at 10⁵ Pa and volume 4 L is compressed adiabatically to 1 L (γ = 1.4). Find the final pressure and the work done on the gas. (4^1.4 = 6.96)
Hint
p V^γ is constant; W = (p₂V₂ − p₁V₁)/(γ − 1) for the work done on the gas.
Answer
p₂ = 10⁵ × 6.96 = 6.96 × 10⁵ Pa. Work on the gas = (6.96 × 10⁵ × 10⁻³ − 10⁵ × 4 × 10⁻³)/0.4 = (696 − 400)/0.4 = 740 J.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose 0.5 mol of a monatomic ideal gas at 300 K in a cylinder is compressed to half its volume, first slowly with the cylinder in a water bath, then (starting again) quickly in an insulated cylinder (R = 8.3 J mol⁻¹ K⁻¹, ln 2 = 0.693, γ = 5/3, 2^(2/3) = 1.587).
(i) Find the work done on the gas in the slow compression. (1 mark)
(ii) State the heat flow in that case. (1 mark)
(iii) Find the final temperature in the fast compression. (1 mark)
(iv) Find the work done on the gas in the fast compression. (1 mark)
Hint
Slow in a bath is isothermal; fast and insulated is adiabatic.
Answer
(i) W = nRT ln 2 = 0.5 × 8.3 × 300 × 0.693 = 863 J on the gas. (ii) ΔU = 0, so 863 J of heat leaves the gas to the bath. (iii) T₂ = 300 × 1.587 = 476 K. (iv) W = ΔU = 0.5 × 12.45 × 176 = 1.10 × 10³ J.

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.State the first law of thermodynamics.Show answer
ΔQ = ΔU + W.
2.Mayer's relation?Show answer
C_p − C_v = R.
3.Work in an isothermal process?Show answer
nRT ln(V₂/V₁).
4.What happens to the temperature in an adiabatic expansion?Show answer
It falls.

First Law and Thermodynamic Processes: frequently asked questions

Is the first law just conservation of energy?

Yes. It says the heat supplied is either stored as internal energy or used to do work.

Why is C_p greater than C_v?

At constant pressure the gas also does work in expanding, so extra heat is needed for the same temperature rise.

Does the internal energy of an ideal gas change in a cycle?

No. The gas returns to its initial state, so ΔU = 0 and the net heat equals the net work.