Cyclic process

Picture a gas trapped in a cylinder, pushed through a whole sequence of expansions and compressions, heatings and coolings — and then, after all that drama, arriving back at exactly the pressure, volume and temperature it started with. Nothing about the gas itself has changed. And yet something very real has happened around it: heat has flowed in, work has been done, energy has been shuffled from one form to another and handed out to the surroundings. This loop-and-return is a cyclic process, and it is the single idea that turns the first law of thermodynamics from a bookkeeping equation into the working principle behind every engine ever built. The trick that makes a cyclic process tractable is a fact you already half-know: internal energy $U$ depends only on the state of the gas, not on the road taken to reach it. So if the gas comes back to where it began, $U$ must come back too — its net change over the full loop is exactly zero, no matter how tangled the path. Drop that single fact into the first law and a remarkably clean result falls out: the net heat absorbed over a cycle equals the net work delivered by the cycle. Nothing is left over, nothing is hidden inside the gas. This article grounds that idea in the exact CBSE Class 11 syllabus — the first law $\Delta Q = \Delta U + \Delta W$ , the reading of work as the area under a $P$ – $V$ curve, and the closed-loop version of both. You will see why area enclosed by a loop (not area under a curve) is the number that matters, why the direction you trace the loop — clockwise or counter-clockwise — flips the sign of that work, and how to break any cyclic problem into its isobaric and isochoric legs and add them up with confidence.

Key takeaways

A cyclic process returns the gas to its exact starting state (same $P$ , $V$ , $T$ ) after one or more intermediate steps.
Because internal energy $U$ is a state function, $\Delta U_{cycle} = 0$ for any cyclic process — however complicated the path.
The first law $\Delta Q = \Delta U + \Delta W$ then collapses, over a full cycle, to $\Delta Q_{cycle} = \Delta W_{cycle}$ : net heat in equals net work out.
On a $P$ – $V$ diagram, work for any step is the area under that step's curve; for a closed loop, the net work equals the area enclosed by the loop.
Tracing the loop clockwise gives net positive work done by the gas (heat-engine sense); tracing it counter-clockwise gives net negative work — work is done on the gas instead.
Solve any cyclic problem leg by leg: compute $W = P\,\Delta V$ on each isobaric leg, note $W=0$ on each isochoric leg, sum them, then apply $\Delta Q_{cycle} = \Delta W_{cycle}$ .
Key idea: because internal energy is a state function, a cyclic process always has ΔU = 0, so the net heat absorbed equals the net work done, and that net work is simply the signed area enclosed by the loop on a P-V diagram.

The core insight

A cyclic process returns a gas to its starting state, so $\Delta U_{cycle}=0$ , which forces $\Delta Q_{cycle}=\Delta W_{cycle}$ — the net heat absorbed equals the net work done, and that work is the area enclosed by the loop on a $P$ – $V$ diagram, positive if traced clockwise and negative if traced counter-clockwise.

Where it fits in the syllabus

This is part of the CBSE Class 11 Physics unit on Thermodynamics, sitting right after you have studied thermal equilibrium, the zeroth law, the first law, and the named processes — isothermal, adiabatic, isochoric, isobaric. The syllabus explicitly lists cyclic processes among the changes of state of a gas you must be able to analyse using the first law, $\Delta Q = \Delta U + \Delta W$ with $\Delta W = P\,\Delta V$ . Everything below stays inside that scope: no heat-engine efficiency, no Carnot cycle, no entropy — those belong to the second law's qualitative statements only at this level, or to Class 12 and beyond, and are flagged as such rather than developed here.

The physical picture

A gas is described by the state variables pressure $P$ , volume $V$ and temperature $T$ , linked by an equation of state. Internal energy $U$ is a state function: two gas samples in the same state have the same $U$ , full stop, regardless of the sequence of heatings, coolings, expansions or compressions that got them there. A cyclic process is any sequence of processes that starts at some state A and, after any number of intermediate steps, returns exactly to state A.

\Delta U_{cycle} = U_{final} - U_{initial} = 0 \qquad (1)

where: $U_{final}$ and $U_{initial}$ are the internal energy at the end and start of the loop. Because the final state is identical to the initial one — same $P$ , $V$ , $T$ — their difference is exactly zero. The physical reading is that this is true regardless of what happened inside the loop: the gas could have been heated then cooled, expanded then compressed, in a straight rectangle or a wandering curve — the only thing that matters for equation (1) is that it ends where it began. Feed this into the first law of thermodynamics for the whole cycle:

\Delta Q_{cycle} = \Delta U_{cycle} + \Delta W_{cycle} = \Delta W_{cycle} \qquad (2)

where: $\Delta Q_{cycle}$ is the net heat absorbed by the gas over the complete loop, and $\Delta W_{cycle} = \oint P\,dV$ is the net work done by the gas over the loop (the small circle on the integral sign is the standard way of writing "add this up all the way around a closed path"). The physical reading: whatever heat goes in over the cycle comes back out as mechanical work, with nothing left stored in the gas, because the gas has nowhere left to store it — it is back at its starting $U$ . Work for a small step is $dW = P\,dV$ , so the work for any one leg of a process is the area under that leg's curve on a $P$ – $V$ diagram. Add the legs of a closed loop together and the parts of the area that are shared by an outward leg and a returning leg cancel, leaving only the area actually enclosed by the loop.

Trace the loop clockwise (expand at the higher pressure, compress at the lower pressure): the enclosed area comes out positive, and the gas does net positive work on the surroundings — this is how a heat engine converts absorbed heat into usable work.
Trace the loop counter-clockwise (expand at the lower pressure, compress at the higher pressure): the enclosed area comes out negative, and the surroundings do net work on the gas — heat is expelled rather than delivered.

Figure

A pressure volume diagram showing a rectangular cycle ABCDA traced clockwise, with an isobaric expansion from A to B at the higher pressure, an isochoric drop from B to C, an isobaric compression from C to D at the lower pressure, and an isochoric rise from D back to A, with the enclosed rectangle shaded to show that its area equals the net work done by the gas over one cycle

Derivation

Derivation AThe net work over a cyclic process equals the enclosed area

The goal is to show, starting only from the definition of work done by an expanding gas and the fact that internal energy is a state function, that the net work delivered over one complete cyclic process equals the signed area enclosed by that cycle's loop on a $P$ – $V$ diagram, with a clockwise trace giving positive net work and a counter-clockwise trace giving negative net work — the exact tool needed to solve every rectangular-cycle problem in this chapter.

Step 1. Start from the definition of work in a small step. For a gas at pressure $P$ undergoing a tiny volume change $dV$ , the work done by the gas is $dW = P\,dV$ . Summing (integrating) this along any path from an initial state to a final state gives the area under that path on a $P$ – $V$ diagram — positive while the gas expands ( $dV>0$ ) and negative while it is compressed ( $dV<0$ ).

Step 2. Split a cyclic process into an outward leg and a return leg. Any closed loop on a $P$ – $V$ diagram can be thought of as an "upper" path taking the gas from state A to state B and a "lower" path bringing it back from B to A (or vice versa for a more general shape, but the rectangular loops of this chapter make the split literal: two isobaric legs at the top and bottom pressures, two isochoric legs at the side volumes). The work on the outward leg is the area under that leg; the work on the return leg is minus the area under that leg, because the volume change has the opposite sign.

Step 3. Subtract the two areas. Net work over the full loop, $W_{cycle} = W_{outward} + W_{return}$ , is the area under the outward leg minus the area under the return leg. Geometrically, this difference is exactly the area enclosed between the two legs — the region bounded by the closed curve — because the part of the area common to both legs cancels out. This is the key geometric fact: only the enclosed area survives, not the full area down to the volume axis.

Step 4. Fix the sign with the direction of tracing. If the upper (higher-pressure) leg is traversed while volume is increasing — the loop goes clockwise — the outward leg's area exceeds the return leg's area, so $W_{cycle}$ is positive: the gas does net work on the surroundings. If instead the higher-pressure leg is traversed while volume is decreasing — the loop goes counter-clockwise — the return leg's area exceeds the outward leg's, so $W_{cycle}$ is negative: the surroundings do net work on the gas.

Step 5. Connect to heat via the first law. Because $U$ is a state function, $\Delta U_{cycle}=0$ for any closed loop, so the first law $\Delta Q = \Delta U + \Delta W$ gives $\Delta Q_{cycle} = \Delta W_{cycle}$ exactly, for every cyclic process regardless of shape. The enclosed area is therefore not just the net work — it is also the net heat exchanged over the cycle, with the same sign. Verified: for the rectangular cycle of Worked example 1 below (P1 = 2×10^5 Pa, P2 = 1×10^5 Pa, V1 = 1×10^-3 m^3, V2 = 3×10^-3 m^3, clockwise), the leg-by-leg sum W_AB + W_BC + W_CD + W_DA = 400 + 0 - 200 + 0 = 200 J matches the enclosed-area formula delta P times delta V = (2-1)×10^5 × (3-1)×10^-3 = 200 J.

Worked examples

Board level - work and heat over a clockwise rectangular cycle

One mole of an ideal gas is taken through a rectangular cycle ABCDA on a $P$ – $V$ diagram, traced clockwise. A to B: isobaric expansion at $P_1 = 2\times10^5\ \text{Pa}$ , volume rises from $V_1 = 1\times10^{-3}\ \text{m}^3$ to $V_2 = 3\times10^{-3}\ \text{m}^3$ . B to C: isochoric cooling at constant $V_2$ , pressure falls from $P_1$ to $P_2 = 1\times10^5\ \text{Pa}$ . C to D: isobaric compression at $P_2$ , volume falls from $V_2$ back to $V_1$ . D to A: isochoric heating at constant $V_1$ , pressure rises from $P_2$ back to $P_1$ , closing the loop. Find the net work and net heat exchanged over one cycle.

Step 1. Work on the isobaric legs. A to B, expansion at constant $P_1$ : $W_{AB} = P_1(V_2-V_1) = 2\times10^5 \times (3-1)\times10^{-3} = 400\ \text{J}$ . C to D, compression at constant $P_2$ : $W_{CD} = P_2(V_1-V_2) = 1\times10^5\times(1-3)\times10^{-3} = -200\ \text{J}$ .

Step 2. Work on the isochoric legs. B to C and D to A both have $dV=0$ , so $W_{BC}=W_{DA}=0$ — pressure changes at fixed volume do no work.

Step 3. Add all four legs. $W_{cycle} = 400 + 0 - 200 + 0 = 200\ \text{J}$ . Cross-check with the enclosed-area shortcut: $\Delta P\times\Delta V = (2-1)\times10^5\times(3-1)\times10^{-3} = 200\ \text{J}$ — matches.

Step 4. Apply the first law over the cycle. Since $\Delta U_{cycle}=0$ , $\Delta Q_{cycle} = \Delta W_{cycle}$ . Answer: the gas absorbs a net 200 J of heat and delivers exactly 200 J of net work over one cycle.

Competitive level - counter-clockwise cycle and heat rejected

A gas undergoes a cyclic process consisting of two isochoric and two isobaric steps forming a rectangle on the $P$ – $V$ diagram, traced counter-clockwise: pressure rises from $1\times10^5\ \text{Pa}$ to $3\times10^5\ \text{Pa}$ at constant volume $2\times10^{-3}\ \text{m}^3$ ; volume then rises from $2\times10^{-3}\ \text{m}^3$ to $5\times10^{-3}\ \text{m}^3$ at constant pressure $3\times10^5\ \text{Pa}$ ; pressure then drops back to $1\times10^5\ \text{Pa}$ at constant volume $5\times10^{-3}\ \text{m}^3$ ; and finally volume returns to $2\times10^{-3}\ \text{m}^3$ at constant pressure $1\times10^5\ \text{Pa}$ . Find the net work and net heat exchanged, and state whether the gas absorbs or rejects heat over the cycle.

Step 1. Compute the enclosed rectangle's dimensions. $\Delta P = (3-1)\times10^5 = 2\times10^5\ \text{Pa}$ , $\Delta V = (5-2)\times10^{-3} = 3\times10^{-3}\ \text{m}^3$ , so the enclosed area is $\Delta P\times\Delta V = 2\times10^5\times3\times10^{-3} = 600\ \text{J}$ .

Step 2. Fix the sign from the tracing direction. Here pressure is raised first, while volume is still at its smaller value, and only then does the gas expand at the higher pressure; the compression back happens at the lower pressure. That order is counter-clockwise on the $P$ – $V$ diagram (the mirror image of the Board-level example), so the net work done by the gas is negative: $W_{cycle} = -600\ \text{J}$ .

Step 3. Apply the first law. $\Delta U_{cycle}=0$ again, so $\Delta Q_{cycle} = \Delta W_{cycle} = -600\ \text{J}$ . Answer: the surroundings do 600 J of net work on the gas, and the gas rejects 600 J of heat to the surroundings over one cycle. Timing note: under two minutes once you recognise the rectangle. Compute $\Delta P\times\Delta V$ first for the magnitude — never re-derive four individual leg terms if the loop is a plain rectangle. Then settle the sign purely from the order of legs: pressure-up-then-volume-up-then-pressure-down-then-volume-down is counter-clockwise (negative work by the gas); volume-up-then-pressure-down-then-volume-down-then-pressure-up is clockwise (positive work by the gas). Do not try to guess the sign from the numbers alone — always trace the sequence of legs in your head or on the diagram.

Common traps

Thinking $\Delta U_{cycle}=0$ means no heat or work happens during the cycle — it only means the net change in $U$ over the whole loop is zero; heat and work flow throughout each leg, they simply balance out to zero for $U$ specifically.
Forgetting that the sign of net work depends on the direction of tracing (clockwise vs counter-clockwise), not on the shape or size of the loop — the same rectangle can give $+200\ \text{J}$ or $-200\ \text{J}$ depending purely on which way it is traversed.
Confusing the area enclosed by a closed loop with the area under one open curve — an open path's area-under-the-curve is the work for that single process only; only a closed loop's enclosed area gives the net cyclic work.
Assuming every cyclic process must be built from nameable straight-line legs (isobaric, isochoric) — in general a cycle can be any closed curve, including smooth isothermal or adiabatic arcs; the enclosed-area rule still applies, only the geometry of the area gets harder.
Skipping the isochoric legs when adding up work "because nothing seems to happen" — pressure genuinely changes on those legs and heat is exchanged, it is only the work term that vanishes since $dV=0$ .

You should be able to…

State and justify why $\Delta U_{cycle}=0$ for any cyclic process, using the fact that internal energy is a state function.
Apply $\Delta Q_{cycle} = \Delta W_{cycle}$ to a given cycle once $\Delta U_{cycle}=0$ is established.
Compute the net work over a rectangular cycle both by summing leg-by-leg work and by the shortcut $\Delta P\times\Delta V$ , and check the two agree.
Determine the sign of net work (positive or negative) purely from the direction — clockwise or counter-clockwise — in which the cycle is traced.
Distinguish the area enclosed by a closed loop from the area under a single open path, and know which one applies to a cyclic problem.
Recognise that a cycle need not be built from named straight-line legs — smooth curved cycles obey the same $\Delta U_{cycle}=0$ and enclosed-area rules.

Ask the tutor: Stuck on tracing the direction of a cycle that isn't a neat rectangle, or want your leg-by-leg work sum double-checked against the enclosed-area shortcut? Ask the AI tutor to walk through a cyclic-process problem with you step by step.

If the internal energy comes back to the same value, doesn't that mean the gas ends up unchanged in every way, including no net heat or work

Not quite — "unchanged" applies only to the state of the gas (its $P$ , $V$ , $T$ , and hence $U$ ), not to what flowed through it. Over the cycle, the gas can absorb a net amount of heat and deliver an equal amount of net work, or vice versa, without its own internal energy budget shifting at all, because heat in and work out balance each other exactly. Think of it like a toll booth: cars (energy) pass through constantly, but the booth itself (the gas's state) is unchanged at the end of the day.

How do I quickly tell if a rectangular cycle is clockwise or counter-clockwise just from the description of the legs

Trace the sequence of legs in your head, starting from any corner, and watch whether pressure or volume changes first. If the gas expands (volume increases) while at the higher pressure, and is compressed (volume decreases) while at the lower pressure, the loop is clockwise and the net work by the gas is positive. If instead the gas expands at the lower pressure and is compressed at the higher pressure, the loop is counter-clockwise and the net work is negative. Sketching even a rough box with an arrow on each leg takes ten seconds and removes all doubt.

Why don't I just add up delta Q for each leg to check the answer, the way I add up delta W

You can, but it needs the process type on each leg, which usually is not given as plainly as pressure and volume. For an isochoric leg, $\Delta W = 0$ , so $\Delta Q = \Delta U$ for that leg, and $\Delta U$ needs the molar specific heat $C_v$ and the temperature change (from $PV=nRT$ ). For an isobaric leg, $\Delta Q = nC_p\Delta T$ . Once you know $n$ and the gas's degrees of freedom you can do this leg by leg and it must sum to the same $\Delta Q_{cycle}$ as $\Delta W_{cycle}$ — a genuinely useful check, but the direct area-enclosed route is faster whenever only $P$ and $V$ are given.

Is this the same thing as a heat engine, and will I need Carnot's cycle for this chapter

A cyclic process is the general geometric idea — any closed loop on a $P$ – $V$ diagram — and it is exactly what makes heat engines possible, since an engine must return to its starting state every stroke to run continuously. But heat-engine efficiency, the Carnot cycle, and refrigerator COP are not part of the current CBSE Class 11 prescription; they are studied in Class 12 and beyond. For this chapter, stick to $\Delta U_{cycle}=0$ , $\Delta Q_{cycle}=\Delta W_{cycle}$ , and the enclosed-area rule — that is everything the board expects here.