PhysicsCBSE Class 11 · Thermodynamics

First Law, Work and P-V Diagrams Class 11

Updated 17 October 2026 · Fact-checked

The first law says that the heat Q supplied to a system equals the increase in its internal energy plus the work W done by it: Q = ΔU + W. The work done by a gas is W = ∫P dV, the area under the process on a P-V diagram; for an ideal gas, ΔU = nC_vΔT depends only on temperature change, so it is the same for every path between two states, while Q and W depend on the path.

Both paths join A to B so ΔU is the same, but path 1 has more area under it, so the gas does more work W and needs more heat Q.

Illustration drawn for QuestionBuddy.

Understand First Law, Work and P-V Diagrams

First law. For a system, $Δ Q = Δ U + Δ W$ , where $Δ Q$ is the heat supplied *to* the system, $Δ U$ the increase in internal energy and $Δ W$ the work done *by* the system (conventions: heat in positive, work done by the gas positive). The internal energy $U$ is a state function; $Q$ and $W$ are not (they depend on the path).

Work by a gas. For a small volume change, $d W = P d V$ , so $W = \int_{V_{1}}^{V_{2}} P d V$ = area under the curve on the P-V diagram. At constant pressure, $W = P Δ V$ ; at constant volume, $W = 0$ . Work is positive in expansion and negative in compression. For a closed cycle, the net work equals the area enclosed (positive if the cycle runs clockwise on the P-V diagram), and $Δ U = 0$ over a cycle so $Q_{n e t} = W_{n e t}$ .

Ideal gas. $P V = n R T$ and $U$ depends only on $T$ : $Δ U = n C_{v} Δ T$ for every process, with $C_{v} = \frac{3}{2} R$ (monatomic), $\frac{5}{2} R$ (diatomic) and $C_{p} = C_{v} + R$ . Also $Δ U = \frac{f}{2} Δ (P V)$ with $f = 3$ or $5$ .

Special processes.

*Isochoric* (constant $V$ ): $W = 0$ , $Q = Δ U = n C_{v} Δ T$ .
*Isobaric* (constant $P$ ): $W = P Δ V = n R Δ T$ , $Q = n C_{p} Δ T$ .
*Cyclic*: $Δ U = 0$ , $Q = W$ . The efficiency of a cycle used as an engine is $η = W_{n e t} / Q_{ab sor b e d}$ .

Path dependence. For two paths from the same initial to the same final state, $Δ U$ is the same; the path under which the gas does more work needs more heat.

First Law, Work and P-V Diagrams: formulas and key facts

First law: $Q = Δ U + W$; Heat in positive, work by the gas positive.
Work done by a gas: $W = \int_{V_{1}}^{V_{2}} P d V$; Area under the P-V curve.
Isobaric work: $W = P Δ V = n R Δ T$; Constant pressure.
Internal energy of an ideal gas: $Δ U = n C_{v} Δ T$; Any process.
Heat capacities: $C_{p} = C_{v} + R$; Mayer's relation.
Cycle efficiency: $η = \frac{W _{n e t}}{Q _{in}}$; Less than 1.

How to solve First Law, Work and P-V Diagrams questions

Find ΔU from the temperature (or PV) change first, then the work from the area, then the heat from the first law.

1Write the initial and final states (P, V, T) from the question or from PV = nRT.
2Compute ΔU = nC_vΔT (or (f/2)Δ(PV)); it depends only on the end states.
3Compute W as the area under the path on the P-V diagram (rectangle, triangle or trapezoid).
4Find Q = ΔU + W for the path, keeping signs.
5For a cycle, sum the heat absorbed (positive Q) for the efficiency η = W_net/Q_in.
6Check: over a closed cycle ΔU = 0 and Q_net = W_net.

First Law, Work and P-V Diagrams 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

Suppose a pump does 120 J of work compressing air, and during the compression the air loses 30 J of heat to the pump walls.

Find the change in the air's internal energy.

Try it first, then show the solutionSolution

W (by the gas) = −120 J; Q = −30 J.
ΔU = Q − W = −30 − (−120) = +90 J.

Answer: +90 J.

Memory hook: Work done on the gas raises its energy.

A gas in a cylinder

Suppose a gas at constant pressure 1.0 × 10⁵ Pa expands from 2.0 L to 5.0 L.

Find the work done by the gas.

Try it first, then show the solutionSolution

W = PΔV = 1.0 × 10⁵ × 3.0 × 10⁻³ = 300 J.

Answer: 300 J.

Memory hook: Convert litres to m³ (1 L = 10⁻³ m³).

Tricks to remember First Law, Work and P-V Diagrams

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

Q = U + W, sign check
Heat in and work out are positive.
Use it when: Sign conventions.
Area under the curve
Work is the area to the V-axis; closed loop gives the net work.
Use it when: P-V diagram questions.
U depends on T only
For an ideal gas, ΔU = nCvΔT in every process.
Use it when: ΔU for any path.
Isochoric means no work
Q = ΔU at constant volume.
Use it when: Quick Q.

First Law, Work and P-V Diagrams: solved examples

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

Example 1: Compression with heat loss · 2 marks · Very short answerBoard level

150 J of work is done on a gas while it loses 40 J of heat to the surroundings. Find the change in its internal energy.

Try it first, then show the solutionSolution

Work done by the gas W = −150 J; heat supplied Q = −40 J. (1)
ΔU = Q − W = −40 + 150 = +110 J. (1)

Answer: +110 J.

Example 2: Isobaric heating of a monatomic gas · 3 marks · Short answerHarder board level

Three moles of a monatomic ideal gas are heated at constant pressure from 250 K to 350 K (R = 8.314 J/mol K). Find the work done by the gas, the change in internal energy and the heat absorbed.

Try it first, then show the solutionSolution

W = nRΔT = 3 × 8.314 × 100 = 2494 J. (1)
ΔU = nC_vΔT = 3 × 1.5 × 8.314 × 100 = 3741 J. (1)
Q = ΔU + W = 6235 J (= nC_pΔT with C_p = 2.5R). (1)

Answer: W = 2494 J; ΔU = 3741 J; Q = 6235 J.

Example 3: A straight-line process · 3 marks · Short answerHarder board level

A monatomic ideal gas is taken from A (2 L, 4 × 10⁵ Pa) to B (6 L, 2 × 10⁵ Pa) along a straight line on a P-V diagram. Find the work done, the change in internal energy and the heat absorbed.

Try it first, then show the solutionSolution

W = area of the trapezoid = ½(4 + 2) × 10⁵ × 4 × 10⁻³ = 1200 J. (1)
P_AV_A = 800 J, P_BV_B = 1200 J; ΔU = (3/2)(1200 − 800) = 600 J. (1)
Q = ΔU + W = 1800 J. (1)

Answer: W = 1200 J; ΔU = 600 J; Q = 1800 J.

Example 4: A rectangular cycle · 4 marks · Long answerHarder board level

One mole of a monatomic ideal gas follows the cycle A → B → C → D → A, where A(2 L, 1 × 10⁵ Pa) → B(2 L, 3 × 10⁵ Pa) is isochoric, B → C(4 L, 3 × 10⁵ Pa) isobaric, C → D(4 L, 1 × 10⁵ Pa) isochoric and D → A isobaric. Find the net work, the heat absorbed and the efficiency.

Try it first, then show the solutionSolution

W_net = area = (3 − 1) × 10⁵ × 2 × 10⁻³ = 400 J. (1)
Heat absorbed in AB = ΔU = 1.5 × V ΔP = 1.5 × 2 × 10⁻³ × 2 × 10⁵ = 600 J. (1)
Heat absorbed in BC = (5/2)PΔV = 2.5 × 3 × 10⁵ × 2 × 10⁻³ = 1500 J, so Q_in = 2100 J. (1)
η = 400/2100 = 19%. (Heat rejected in CD and DA: 1200 + 500 = 1700 J; 2100 − 1700 = 400 J.) (1)

Answer: W = 400 J; Q_in = 2100 J; η ≈ 19%.

Example 5: Two paths between the same states · 4 marks · Long answerHOTS / competency

A monatomic ideal gas goes from A(1 × 10⁵ Pa, 2 L) to B(3 × 10⁵ Pa, 6 L) by (I) an isochoric rise to 3 × 10⁵ Pa followed by an isobaric expansion, or (II) an isobaric expansion to 6 L followed by an isochoric rise. Find W and Q for each path, and say what the results show.

Try it first, then show the solutionSolution

ΔU = 1.5(3 × 10⁵ × 6 × 10⁻³ − 10⁵ × 2 × 10⁻³) = 1.5(1800 − 200) = 2400 J, same for both. (1)
Path I: W = 3 × 10⁵ × 4 × 10⁻³ = 1200 J, so Q = 3600 J. (1)
Path II: W = 10⁵ × 4 × 10⁻³ = 400 J, so Q = 2800 J. (1)
ΔU is a state function; W and Q depend on the path. (1)

Answer: Path I: W = 1200 J, Q = 3600 J; path II: W = 400 J, Q = 2800 J; ΔU = 2400 J for both.

Where students lose marks in First Law, Work and P-V Diagrams

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

Mistake: Using the wrong sign for work in compression.
Treating all work as positive.
Do this: Work done by the gas is negative in compression.
Usually costs: The ΔU mark.
Mistake: Forgetting to convert litres to m³.
1 L = 1 dm³.
Do this: 1 L = 10⁻³ m³.
Usually costs: The answer mark.
Mistake: Calculating ΔU as Q − PΔV for a cyclic path incorrectly.
Forgetting ΔU = 0 over a cycle.
Do this: Over a cycle, Q_net = W_net.
Usually costs: 1 mark.
Mistake: Using C_p for ΔU.
Mixing heat capacities.
Do this: ΔU = nC_vΔT for any ideal-gas process.
Usually costs: Method mark.

Exam tips

Write the first law with your sign convention at the start.
Draw the P-V diagram and shade the area for work.
Compute ΔU from the end states before the path-dependent terms.
For cycles tabulate Q, W and ΔU for each leg and check the sums.
Give the efficiency as a percentage.

Practice questions on First Law, Work and P-V Diagrams

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
The work done by a gas in a closed cycle on a P-V diagram is
(a) zero (b) the area enclosed by the loop (c) equal to ΔU (d) equal to the heat rejected
Hint
ΔU = 0 over the cycle.
Answer
(b).
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The internal energy of an ideal gas depends only on its temperature.
Reason (R): The molecules of an ideal gas exert no forces on each other, so there is no potential energy of interaction.
(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
Only kinetic energy remains.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
A gas expands at a constant pressure of 1.5 × 10⁵ Pa from 2 L to 5 L. Find the work done by the gas.
Hint
W = PΔV.
Answer
W = 1.5 × 10⁵ × 3 × 10⁻³ = 450 J.
Q4 · 3 marks · Short answerHarder board level
One mole of a gas absorbs 20.8 J of heat per kelvin at constant volume (= 5R/2). It is heated by 20 K at constant pressure. Find C_p, γ, the heat absorbed, ΔU and the work done.
Hint
C_p = C_v + R.
Answer
C_v = 20.8, C_p = 29.1 J/mol K, γ = 1.4. Q = 29.1 × 20 = 582 J; ΔU = 20.8 × 20 = 416 J; W = R ΔT = 166 J.
Q5 · 4 marks · Long answerHOTS / competency
When 1.00 g of water at 100 °C and 1 atm (1.013 × 10⁵ Pa) is boiled to steam, its volume changes from 1.00 cm³ to 1671 cm³. The heat supplied is 2256 J. Find the work done against the atmosphere, the increase in internal energy and the fraction of the heat used for expansion, and explain where the rest goes.
Hint
W = PΔV with ΔV = 1670 cm³.
Answer
W = 1.013 × 10⁵ × 1.67 × 10⁻³ ≈ 169 J. ΔU = 2256 − 169 ≈ 2087 J. Fraction = 169/2256 ≈ 7.5%. The rest increases the internal (potential) energy of the molecules as they are separated against their attractions.

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.First law?Show answer
Q = ΔU + W.
2.Work in isobaric process?Show answer
PΔV.
3.ΔU of an ideal gas?Show answer
nC_vΔT.
4.C_p − C_v?Show answer
R.
5.Net work of a cycle?Show answer
Area enclosed.
6.Heat at constant volume?Show answer
ΔU.

First Law, Work and P-V Diagrams: frequently asked questions

Why is ΔU = 0 for a cycle?

U is a state function, and the gas returns to its initial state.

Does heat equal work in a cycle?

Yes: Q_net = W_net because ΔU = 0.

How does Q differ from ΔU?

Q also includes the work done by the system; at constant volume they are equal.

What is Mayer's relation?

C_p − C_v = R for an ideal gas, because at constant pressure the extra heat does expansion work.

What is the sign of work in a clockwise cycle?

Positive net work done by the gas.