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  3. PV Diagrams and Cyclic Processes

Thermodynamics · JEE & NEET Physics

PV Diagrams and Cyclic Processes: notes and previous year questions

Reading states, work as a signed area, path dependence, cycles, and engines versus refrigerators.

PV Diagrams and Cyclic Processes in short

  • Each point on a P–V diagram is an equilibrium state; T = PV/nR, so up and to the right is hotter.
  • Only slow (quasi-static) processes can be drawn as lines.
  • Work is the area under the path: positive moving right, negative moving left, zero straight up or down.
  • 1 atm × 1 L ≈ 100 J.

1Reading a P–V diagram

A P–V diagram has volume along the bottom and pressure up the side. Each point is one equilibrium state of the gas, and its temperature follows from T=PV/nRT = PV/nR. The isotherms (curves of constant T) lie further out for hotter states, so up and to the right is hotter. A line joining states is a process.

A line can be drawn only for a slow (quasi-static) process, in which the gas passes through equilibrium states. A sudden change, such as a free expansion, has only its two end points.

2Work is area

W=∫V1V2P dVW = \int_{V_1}^{V_2} P\,dVThe area under the path on the P–V diagram.
  • Moving right (expansion): W>0W > 0.
  • Moving left (compression): W<0W < 0.
  • Vertical (constant volume): W=0W = 0.
  • Straight-line paths: split the area into rectangles, triangles and trapeziums.
  • 1 atm×1 L≈105 Pa×10−3 m3=1001\ \text{atm} \times 1\ \text{L} \approx 10^5\ \text{Pa} \times 10^{-3}\ \text{m}^3 = 100 J.

3Work depends on the path

From A (1 L, 4 atm) to B (4 L, 1 atm): straight, the work is 750 J; down first then across at 1 atm, only 300 J; across first at 4 atm then down, 1200 J. The end states are the same, so ΔU\Delta U is the same on every path, and the heat Q=ΔU+WQ = \Delta U + W differs by the same amounts as the work.

4Cycles

In a cyclic process the gas returns to its starting state, so ΔU=0\Delta U = 0 and the net heat taken in equals the net work done:

Qnet=Wnet=∮P dVQ_{\text{net}} = W_{\text{net}} = \oint P\,dVThe area enclosed by the loop.

Going right along the top adds the area under the top path; coming back left along the bottom subtracts the area under the bottom path; what is left is the area inside. Find it from the shape, or add the work leg by leg.

5Which way round?

DirectionNet workMachine
ClockwiseW > 0: the gas does net workheat engine (car engines)
Counter-clockwiseW < 0: net work is done on the gasrefrigerator, air conditioner, heat pump

Clockwise loops expand at high pressure and compress at low pressure, so more work comes out than goes in. Counter-clockwise loops do the opposite. The heart's pressure–volume loop (a muscle-driven pump) also has an area equal to the work of one beat, about 1 J.

6A worked cycle

A (1 L, 10 atm) → B (4 L, 10 atm) at constant pressure → C (4 L, 2.5 atm) at constant volume → back to A in a straight line.

  1. AB: W=10×3=30W = 10 \times 3 = 30 atm L = +3000 J.
  2. BC: vertical, W=0W = 0.
  3. CA: moving left under a trapezium, 12(10+2.5)×3=18.75\tfrac{1}{2}(10 + 2.5) \times 3 = 18.75 atm L, so W=−1875W = -1875 J.
  4. Net: 3000−1875=11253000 - 1875 = 1125 J = the triangle's area; this is also the net heat absorbed.

Summary

Key ideas

  • Each point on a P–V diagram is an equilibrium state; T = PV/nR, so up and to the right is hotter.
  • Only slow (quasi-static) processes can be drawn as lines.
  • Work is the area under the path: positive moving right, negative moving left, zero straight up or down.
  • 1 atm × 1 L ≈ 100 J.
  • W and Q depend on the path; ΔU depends only on the end states.
  • Over a cycle ΔU = 0, so net Q = net W = the area inside the loop.
  • Clockwise loops are heat engines (W > 0); counter-clockwise loops are refrigerators or heat pumps (W < 0).

Every equation

State
T=PVnRT = \frac{PV}{nR}
Work
W=∫P dVW = \int P\,dV
Rectangle
W=P ΔVW = P\,\Delta V
Trapezium
W=12(P1+P2)(V2−V1)W = \tfrac{1}{2}(P_1 + P_2)(V_2 - V_1)
Cycle
ΔU=0, Qnet=Wnet\Delta U = 0,\ Q_{\text{net}} = W_{\text{net}}
Loop
Wnet=∮P dVW_{\text{net}} = \oint P\,dV
Units
1 atm L≈100 J1\ \text{atm L} \approx 100\ \text{J}

Previous year questions with solutions

Real JEE and NEET questions on pv diagrams and cyclic processes. Try each one before you open the solution.

Q1NEET 2026One correct option

An electric heater supplies heat to a system at a rate of 100 W . If the system performs work at a rate of 75 J/s75\ \mathrm{J/s}, then the rate at which internal energy increases will be:

  1. A75 W
  2. B100 W
  3. C125 W
  4. D25 W
Show answer and solution

Answer: Option D

The first law holds for every second of the process, so it can be written as rates: heat in per second == rise of internal energy per second ++ work done per second.

100=dUdt+75100 = \dfrac{dU}{dt} + 75, so dUdt=25 W\dfrac{dU}{dt} = 25\ \mathrm{W}.

The trap is A, 75 W75\ \mathrm{W}, the rate of doing work, and B, 100 W100\ \mathrm{W}, the whole heating rate — as if no work were done. C, 125 W125\ \mathrm{W}, adds the work to the heat, getting the sign of WW backwards: work done BY the system is energy leaving it.

Q2JEE Main 2026One correct option

Heat is supplied to a diatomic gas at constant pressure. Then the ratio of ΔQ:ΔU:ΔW\Delta Q:\Delta U:\Delta W is ______.

  1. A2 : 3 : 5
  2. B5 : 3 : 2
  3. C2 : 5 : 7
  4. D7 : 5 : 2
Show answer and solution

Answer: Option D

At constant pressure, per mole and per kelvin: ΔQ=CP ΔT\Delta Q = C_{P}\,\Delta T, ΔU=CV ΔT\Delta U = C_{V}\,\Delta T and ΔW=R ΔT\Delta W = R\,\Delta T. For a diatomic gas CV=52RC_{V} = \dfrac{5}{2}R and CP=72RC_{P} = \dfrac{7}{2}R, so

ΔQ:ΔU:ΔW=72:52:1=7:5:2\Delta Q : \Delta U : \Delta W = \dfrac{7}{2} : \dfrac{5}{2} : 1 = 7 : 5 : 2

Check with the first law: 7=5+27 = 5 + 2.

The trap is B, 5:3:25 : 3 : 2, which is the monatomic gas (52:32:1\dfrac{5}{2} : \dfrac{3}{2} : 1). A and C list the same numbers backwards, in the order W:U:QW : U : Q — and neither passes the check that QQ is the sum of the other two.

Q3JEE Main 2026Numerical answer

A diatomic gas (γ=1.4)(\gamma =1.4) does 100 J of work when it is expanded isobarically. Then the heat given to the gas ____\_\_\_\_ J.

Show answer and solution

Answer: 350

At constant pressure W=nR ΔTW = nR\,\Delta T and Q=nCP ΔTQ = nC_{P}\,\Delta T, so QW=CPR\dfrac{Q}{W} = \dfrac{C_{P}}{R}. With γ=1.4=75\gamma = 1.4 = \dfrac{7}{5} the gas is diatomic, CP=72RC_{P} = \dfrac{7}{2}R, and QW=72\dfrac{Q}{W} = \dfrac{7}{2} (the same as γγ−1=1.40.4=3.5\dfrac{\gamma}{\gamma - 1} = \dfrac{1.4}{0.4} = 3.5).

Q=3.5×100=350 JQ = 3.5 \times 100 = 350\ \mathrm{J}: 100 J100\ \mathrm{J} goes into work and 250 J250\ \mathrm{J} into internal energy.

The trap is 250 J250\ \mathrm{J}, which is ΔU\Delta U and forgets the work itself. Another is 140 J140\ \mathrm{J}, γ×100\gamma \times 100, which treats γ\gamma as the ratio of heat to work — that ratio is CPR\dfrac{C_{P}}{R}, not CPCV\dfrac{C_{P}}{C_{V}}.

Practice questions, easy to hard

Three questions from the pv diagrams and cyclic processes practice ladder: one easy, one medium, one hard.

Q4One correct option

Two simple processes follow straight from the first law.

Constant volume (isochoric): the gas pushes nothing back, so W=0W = 0 and all the heat goes into internal energy, Q=ΔU=nCV ΔTQ = \Delta U = nC_{V}\,\Delta T.

Constant pressure (isobaric): W=p ΔV=nR ΔTW = p\,\Delta V = nR\,\Delta T (from pV=nRTpV = nRT), and Q=ΔU+W=n(CV+R) ΔT=nCP ΔTQ = \Delta U + W = n(C_{V} + R)\,\Delta T = nC_{P}\,\Delta T. So the fraction of the heat that becomes work is

WQ=RCP=1−1γ\dfrac{W}{Q} = \dfrac{R}{C_{P}} = 1 - \dfrac{1}{\gamma}

and the fraction that becomes internal energy is CVCP=1γ\dfrac{C_{V}}{C_{P}} = \dfrac{1}{\gamma}.

A monatomic ideal gas is given 500 J500\ \mathrm{J} of heat at constant pressure. How much work does it do?

  1. A300 J300\ \mathrm{J}
  2. B500 J500\ \mathrm{J}
  3. C200 J200\ \mathrm{J}
  4. D143 J143\ \mathrm{J}
Show answer and solution

Answer: Option C

For a monatomic gas CV=32RC_{V} = \dfrac{3}{2}R and CP=52RC_{P} = \dfrac{5}{2}R, so WQ=RCP=25\dfrac{W}{Q} = \dfrac{R}{C_{P}} = \dfrac{2}{5} and W=25×500=200 JW = \dfrac{2}{5} \times 500 = 200\ \mathrm{J}. The other 300 J300\ \mathrm{J} raises the internal energy.

The trap is A, 300 J300\ \mathrm{J}, which is ΔU\Delta U, not WW. B, 500 J500\ \mathrm{J}, sends all the heat into work — but the gas warms up as it expands at constant pressure. D, 143 J143\ \mathrm{J}, uses the diatomic fraction 27\dfrac{2}{7}.

Q5Numerical answer

Many processes follow pVx=constantpV^{x} = \text{constant} for some number xx — a polytropic process. x=0x = 0 is constant pressure, x=1x = 1 is isothermal, x=γx = \gamma is adiabatic. Integrating p dVp\,dV gives the work for any x≠1x \neq 1:

W=p1V1−p2V2x−1=nR(T1−T2)x−1W = \dfrac{p_{1}V_{1} - p_{2}V_{2}}{x - 1} = \dfrac{nR(T_{1} - T_{2})}{x - 1}

(With x=γx = \gamma this is the adiabatic formula you already know.)

One mole of an ideal gas is heated so that its pressure stays proportional to its volume, p=kVp = kV — that is, pV−1pV^{-1} constant, x=−1x = -1. Its temperature rises from 300 K300\ \mathrm{K} to 400 K400\ \mathrm{K}. How much work does it do, in joules? Take R=8.3 J mol−1 K−1R = 8.3\ \mathrm{J\,mol^{-1}\,K^{-1}}.

Show answer and solution

Answer: 415 J

W=nR(T1−T2)x−1=8.3×(300−400)−1−1=−830−2=415 JW = \dfrac{nR(T_{1} - T_{2})}{x - 1} = \dfrac{8.3 \times (300 - 400)}{-1 - 1} = \dfrac{-830}{-2} = 415\ \mathrm{J}.

Check by area: p=kVp = kV is a straight line through the origin, and the area under it is k(V22−V12)2=p2V2−p1V12=R ΔT2\dfrac{k(V_{2}^{2} - V_{1}^{2})}{2} = \dfrac{p_{2}V_{2} - p_{1}V_{1}}{2} = \dfrac{R\,\Delta T}{2}.

The trap is reading "pp proportional to VV" as x=+1x = +1: written as pVxpV^{x} constant, p=kVp = kV is pV−1=kpV^{-1} = k, so x=−1x = -1 (and x=+1x = +1 would divide by zero). Another is nR ΔT=830 JnR\,\Delta T = 830\ \mathrm{J}, the constant-pressure answer, twice too big.

Q6One correct option

Take logs of pVx=constantpV^{x} = \text{constant}: log⁡p=−xlog⁡V+constant\log p = -x\log V + \text{constant}. So on a graph of log⁡p\log p against log⁡V\log V, every polytropic process is a straight line, and its slope is −x-x. Reading the slope gives xx, and xx gives the heat capacity C=CV+R1−xC = C_{V} + \dfrac{R}{1 - x}.

On a graph of log⁡p\log p against log⁡V\log V, a process is a straight line of slope −γ-\gamma. Which process is it?

  1. Aisothermal
  2. Bisobaric
  3. Cadiabatic
  4. Disochoric
Show answer and solution

Answer: Option C

Slope −γ-\gamma means x=γx = \gamma: pVγpV^{\gamma} is constant, the adiabatic, with C=0C = 0.

The trap is A, isothermal, which is x=1x = 1 and slope −1-1. B, isobaric, is x=0x = 0, a horizontal line. D, isochoric, keeps log⁡V\log V fixed — a vertical line, with no finite slope at all.