Thermodynamic processes

Push the plunger of a bicycle pump shut with your thumb over the nozzle and you feel two things at once: the air inside resists you, and the barrel grows faintly warm under your palm. Nothing mysterious is happening — you are simply watching a gas change state, and the way it changes (fast, slow, insulated, in contact with the room) decides everything about how much you had to push and how warm it got. CBSE Class 11 gives this everyday act a precise vocabulary: a thermodynamic process is any path a gas's pressure, volume and temperature take from one state to another, and the whole of the Thermodynamics chapter is organised around naming and classifying these paths. A gas heated in a sealed steel can behaves nothing like the same gas expanding slowly against a piston, even though both start and end somewhere on a $P$ – $V$ graph. One keeps its volume locked and only pressure climbs — an isochoric process. The other trades volume for work at fixed pressure — isobaric. Squeeze a gas so gently that it stays at room temperature throughout and you get isothermal compression; squeeze it inside a thick, insulated cylinder so fast that no heat can escape and you get adiabatic compression — the very effect that warms your bicycle pump. Run a gas through a loop of such processes back to where it started and you get a cyclic process, the ancestor idea behind every engine, even though CBSE keeps engines themselves off this year's syllabus. This article builds the classification carefully: what stays constant in each named process, how each looks as a curve on a $P$ – $V$ diagram, why "slow" (quasi-static) is not the same promise as "reversible," and how the first law of thermodynamics, $\Delta Q = \Delta U + \Delta W$ , threads through every one of them — turning a page of Greek-rooted labels into a toolkit you can actually calculate with.

Key takeaways

A thermodynamic process is a path taken by a gas between an initial state $(P_1,V_1,T_1)$ and a final state $(P_2,V_2,T_2)$ ; named processes are classified by what is held fixed or by another defining condition.
Isothermal ( $T$ fixed): $PV = \text{constant}$ . Adiabatic (no heat exchange, $\Delta Q = 0$ ): $PV^\gamma = \text{constant}$ , a steeper curve than isothermal at the same point because $\gamma > 1$ .
Isochoric ( $V$ fixed): $\Delta W = 0$ , a vertical line on the $P$ – $V$ diagram. Isobaric ( $P$ fixed): $\Delta W = P\Delta V$ , a horizontal line.
Cyclic process: the system returns to its initial state, so $\Delta U = 0$ over the full loop and, by the first law, $\Delta Q_{\text{cycle}} = \Delta W_{\text{cycle}}$ ; the net work equals the area enclosed by the loop on a $P$ – $V$ diagram.
Quasi-static means infinitely slow, passing through well-defined equilibrium states throughout; reversible additionally requires no dissipation (friction, turbulence) so the path can be exactly undone; irreversible covers every real, naturally occurring process.
Heat engines, their efficiency, and refrigerators build directly on cyclic processes but are studied in Class 12 and beyond — CBSE Class 11 stops at classifying and computing work for the processes above.
Key idea: every thermodynamic process is named by what state variable it holds fixed (or by whether heat can enter at all), each has its own signature curve on a P-V diagram and its own work formula, and a cyclic process returns the system to its start so that the net heat absorbed exactly equals the net work done.

The core insight

Classify a gas process by what is held constant — temperature gives isothermal ( $PV=\text{constant}$ ), zero heat exchange gives adiabatic ( $PV^\gamma=\text{constant}$ ), fixed volume gives isochoric ( $\Delta W=0$ ), fixed pressure gives isobaric ( $\Delta W=P\Delta V$ ) — and a cyclic process closes the loop so that $\Delta U=0$ and $\Delta Q_{\text{cycle}}=\Delta W_{\text{cycle}}$ equals the area enclosed on the $P$ – $V$ diagram.

Where it fits in the syllabus

This article sits inside the CBSE Class 11 Physics Thermodynamics unit, right after the equation of state and the first law, and it is the organising layer that the rest of the chapter leans on: every later discussion of heat, work and internal energy in this course happens within one of these named processes. You are expected to classify a described change of state into isothermal, adiabatic, isochoric, isobaric, or cyclic, sketch it correctly on a $P$ – $V$ diagram, and compute the work done using $\Delta W = P\Delta V$ for the simple cases. Board-honest note: heat engines, the Carnot cycle, efficiency $\eta$ , and refrigerators/COP are explicitly out of the 2025-26 Class 11 prescription even though cyclic processes themselves are in scope — those quantitative engine ideas wait for Class 12 and beyond, and this article does not develop them beyond a one-line pointer.

The physical picture

Every process starts at some state $(P_1,V_1,T_1)$ and ends at another $(P_2,V_2,T_2)$ , connected by a path on which pressure and volume trade off according to the equation of state $PV = nRT$ . What makes a process "named" is a constraint on that path — something held fixed, or a condition on heat flow — and each constraint forces a distinctive shape onto the $P$ – $V$ curve.

PV = \text{constant} \qquad \text{(isothermal)}

where: $P$ is pressure, $V$ is volume, and the constant equals $nRT$ for the fixed temperature $T$ of that process; $n$ is the number of moles and $R = 8.314\ \text{J/mol}\,\cdot\,\text{K}$ . The physical reading is Boyle's law in disguise — squeeze the gas and its pressure rises exactly enough to keep $PV$ fixed, because temperature (and hence the average kinetic energy of the molecules) has not changed.

PV^\gamma = \text{constant} \qquad \text{(adiabatic)}

where: $\gamma = C_p/C_v$ is the ratio of specific heats (about $1.67$ for a monatomic gas, $1.4$ for a diatomic gas like air), a number greater than $1$ that is quoted from the kinetic theory of gases rather than derived here. The physical reading is that with no heat allowed in or out, any compression work done on the gas has nowhere to go except into raising its internal energy and hence its temperature — so pressure must rise faster with shrinking volume than the isothermal law alone would predict, which is exactly why the adiabatic curve is steeper.

\Delta W = P\,\Delta V \qquad \text{(isobaric, and the general definition of small-step gas work)}

where: $\Delta W$ is the work done by the gas as it expands (positive) or is compressed (negative), $P$ is the pressure at which the volume change happens, and $\Delta V = V_2 - V_1$ . For an isobaric process $P$ is literally constant throughout, so this formula applies directly over the whole change; for an isochoric process $\Delta V = 0$ so $\Delta W = 0$ regardless of how much the pressure swings.

Figure

Four curves on a pressure volume diagram showing an isochoric vertical line, an isobaric horizontal line, an isothermal curve and a steeper adiabatic curve through the same starting point, illustrating how each named process looks graphically

A cyclic process links several of these legs into a closed loop that returns exactly to the starting state. Because internal energy $U$ is a state variable — it depends only on the current state, not on the path taken to reach it — a full cycle has $\Delta U = 0$ . The first law $\Delta Q = \Delta U + \Delta W$ then collapses to $\Delta Q_{\text{cycle}} = \Delta W_{\text{cycle}}$ : whatever net heat the gas absorbed over the loop, it gave back out as net work, and that net work equals the area enclosed by the loop on the $P$ – $V$ diagram. Two further labels describe how any of these processes unfolds in time rather than what stays fixed. A quasi-static process happens infinitely slowly, so the gas is always essentially in equilibrium and $P$ , $V$ , $T$ are well-defined at every instant — this is the idealisation behind every smooth curve drawn on a $P$ – $V$ diagram. A reversible process is a quasi-static process with the further promise of zero dissipation, so it can be run backward and restore both the gas and its surroundings exactly. An irreversible process is everything else — friction, turbulence, a sudden burst, heat crossing a large temperature gap — and every real process that actually happens in nature is irreversible to some degree; reversibility is a limiting ideal, never a fact about the physical world.

Derivation ADeriving PV equals a constant for an isothermal change directly from the equation of state

The goal is to show, using nothing beyond the ideal-gas equation of state already established earlier in this chapter and ordinary algebra, that an isothermal process obeys $PV = \text{constant}$ — the relation quoted above — rather than simply asserting it. The idea is to write the equation of state once at the start of the process and once at the end, notice that temperature is the same both times by definition of "isothermal," and eliminate it.

Step 1. Write the equation of state at the initial state. For $n$ moles of gas at pressure $P_1$ , volume $V_1$ and temperature $T_1$ : $P_1V_1 = nRT_1$ .

Step 2. Write the equation of state at the final state, using the same fixed amount of gas $n$ : $P_2V_2 = nRT_2$ .

Step 3. Apply the definition of an isothermal process: temperature does not change along the path, so $T_1 = T_2 = T$ . Substituting this single condition into both equations of Step 1 and Step 2 gives $P_1V_1 = nRT$ and $P_2V_2 = nRT$ — the same right-hand side in both cases.

Step 4. Equate the two left-hand sides, since both equal the same quantity $nRT$ : $P_1V_1 = P_2V_2$ . This holds not just between the two labelled endpoints but at every intermediate point along a quasi-static isothermal path, because temperature is fixed throughout, not just at the ends — so the general statement is $PV = \text{constant}$ for the whole process, matching the boxed relation above. The curve this traces on a $P$ – $V$ diagram is a rectangular hyperbola, steepest at small $V$ and flattening out as $V$ grows, which is exactly the isothermal curve sketched in the figure. Verified: with $P_1=2\times10^5\ \text{Pa}$ , $V_1=1\times10^{-3}\ \text{m}^3$ , an isothermal halving of pressure to $P_2=1\times10^5\ \text{Pa}$ requires $V_2 = P_1V_1/P_2 = 2\times10^{-3}\ \text{m}^3$ , and indeed $P_1V_1 = P_2V_2 = 200\ \text{Pa}\,\cdot\,\text{m}^3$ .

Worked examples

Board level — work done across an isobaric and an isochoric leg

A gas undergoes a two-step process on a $P$ – $V$ diagram: A $\to$ B is isobaric expansion at $P = 2\times10^5$ Pa from $V_A = 1\times10^{-3}\ \text{m}^3$ to $V_B = 3\times10^{-3}\ \text{m}^3$ ; B $\to$ C is an isochoric pressure drop at constant $V_B$ down to $P = 1\times10^5$ Pa. Find the work done by the gas during A $\to$ B and during B $\to$ C, and state which state variable was held fixed in each leg.

Step 1. Identify the held-fixed quantity in A $\to$ B. Pressure stays at $P = 2\times10^5$ Pa throughout, so this leg is isobaric and $\Delta W = P\Delta V$ applies directly.

Step 2. Compute the isobaric work. $W_{AB} = P\Delta V = 2\times10^5 \times (3\times10^{-3} - 1\times10^{-3}) = 2\times10^5 \times 2\times10^{-3} = 400\ \text{J}$ .

Step 3. Identify the held-fixed quantity in B $\to$ C. Volume stays fixed at $V_B = 3\times10^{-3}\ \text{m}^3$ while pressure falls, so this leg is isochoric.

Step 4. Compute the isochoric work. Since $\Delta V = 0$ for the whole leg, $W_{BC} = P\Delta V = 0\ \text{J}$ , regardless of how far the pressure drops. Answer: the gas does $400\ \text{J}$ of work in the isobaric leg A $\to$ B and exactly $0\ \text{J}$ in the isochoric leg B $\to$ C.

Competitive level — closing the loop into a full cycle

Extend the previous process into a closed cycle: after B $\to$ C, the gas is compressed isobarically at $P=1\times10^5$ Pa from C back to a state D with $V_D = 1\times10^{-3}\ \text{m}^3$ , then finally D $\to$ A is an isochoric pressure rise at fixed $V_D$ back to $P=2\times10^5$ Pa, returning exactly to the starting state A. Find the work done on each new leg, the net work done by the gas over the full cycle A $\to$ B $\to$ C $\to$ D $\to$ A, and the net heat absorbed over that cycle.

Step 1. Work on leg C $\to$ D (isobaric compression at $P=1\times10^5$ Pa): $W_{CD} = P\Delta V = 1\times10^5 \times (1\times10^{-3} - 3\times10^{-3}) = 1\times10^5 \times (-2\times10^{-3}) = -200\ \text{J}$ . The negative sign means work is done on the gas as it is compressed.

Step 2. Work on leg D $\to$ A (isochoric, $\Delta V = 0$ ): $W_{DA} = 0\ \text{J}$ , exactly as in leg B $\to$ C, since volume never changes on this leg either.

Step 3. Sum all four legs for the net work over the cycle. From the board-level example, $W_{AB} = 400\ \text{J}$ and $W_{BC} = 0\ \text{J}$ ; adding this leg's $W_{CD} = -200\ \text{J}$ and $W_{DA} = 0\ \text{J}$ gives $W_{\text{net}} = 400 + 0 - 200 + 0 = 200\ \text{J}$ .

Step 4. Check this against the enclosed-area rule and apply the first law over the cycle. The loop A $\to$ B $\to$ C $\to$ D $\to$ A is a rectangle on the $P$ – $V$ diagram with height $\Delta P = 2\times10^5 - 1\times10^5 = 1\times10^5\ \text{Pa}$ and width $\Delta V = 3\times10^{-3} - 1\times10^{-3} = 2\times10^{-3}\ \text{m}^3$ , so its area is $1\times10^5 \times 2\times10^{-3} = 200\ \text{J}$ — matching the summed work exactly. Since the system returns to state A, $\Delta U_{\text{cycle}} = 0$ , so by the first law $\Delta Q_{\text{cycle}} = \Delta W_{\text{cycle}} = 200\ \text{J}$ . Answer: $W_{CD} = -200\ \text{J}$ , $W_{DA} = 0\ \text{J}$ , the net work done by the gas over the full cycle is $200\ \text{J}$ (equal to the rectangular area enclosed), and the net heat absorbed over the cycle is also $200\ \text{J}$ . Timing note: under three minutes. Once each leg is correctly labelled isobaric or isochoric, every work value is a one-line multiplication; the fast check is that a clockwise loop on a $P$ – $V$ diagram always gives positive net work equal to its enclosed area, so computing that area (here, a simple rectangle, $\Delta P\times\Delta V$ ) is a quick way to confirm the leg-by-leg sum without re-deriving the first law each time.

Common traps

Assuming every slow process is automatically reversible. Quasi-static (infinitely slow, well-defined states throughout) is necessary for reversibility but not sufficient — any friction or dissipation makes the process irreversible even if it is done gently.
Thinking a cyclic process means "nothing happened." Even though $\Delta U = 0$ over the whole loop, heat is absorbed and work is done on individual legs (as the worked example shows); only the net change in internal energy over the complete cycle is zero.
Confusing isochoric with isobaric, or reading the wrong axis as "the one held constant" on a $P$ – $V$ diagram — a vertical line is isochoric (volume fixed), a horizontal line is isobaric (pressure fixed).
Forgetting the negative sign when a gas is compressed. Work done by the gas is negative when volume decreases; losing this sign flips the answer in any multi-leg problem like the cycle above.
Believing an adiabatic process has no temperature change. No heat crosses the boundary, but temperature very much changes — compression work raises internal energy (and hence temperature) precisely because that energy has nowhere else to go.
Reaching for the Carnot engine or efficiency formulas. These are not part of the current CBSE Class 11 prescription; a cyclic process here is graded on work and heat via the first law, not on efficiency.

Common trapQuasi-static does not mean reversible

The single costliest mix-up in this topic is treating "done very slowly" and "reversible" as the same idea. A quasi-static process only promises that the system passes through a continuous sequence of equilibrium states, so that $P$ , $V$ and $T$ are well-defined at every instant — it says nothing about dissipation. A gas compressed extremely slowly but with a piston that has friction is still quasi-static, since the states along the way remain near-equilibrium, yet it is irreversible, because the frictional heat generated cannot be recovered by simply running the process backward. Reversibility demands the stronger, purely idealised condition of zero dissipation anywhere in the process, on top of being quasi-static. Every real process — however carefully controlled — has some friction, turbulence, or finite temperature gradient, so strictly reversible processes exist only as a limiting case used to bound what is physically achievable, never as something you can point to in a lab.

You should be able to…

Name each process type — isothermal, adiabatic, isochoric, isobaric, cyclic — and state exactly which quantity is held fixed, or which condition defines it, in each case.
Sketch all four named processes on the same $P$ – $V$ diagram starting from a common point, and explain why the adiabatic curve is steeper than the isothermal one.
Compute the work done by a gas using $\Delta W = P\Delta V$ for isobaric legs, recognise $\Delta W = 0$ for isochoric legs, and sum multiple legs to find the net work of a cyclic process.
Show, using only the equation of state and algebra, that $PV$ is constant along an isothermal path.
Distinguish quasi-static, reversible, and irreversible processes, and correctly judge a real-life scenario against all three labels.
Explain why $\Delta U = 0$ over a complete cycle and why this forces $\Delta Q_{\text{cycle}} = \Delta W_{\text{cycle}}$ , without invoking engines or efficiency.

Ask the tutor: Want to practise sketching a full multi-leg cycle on a $P$ – $V$ diagram and computing its net work and heat exchanged, or need help telling quasi-static, reversible and irreversible apart in tricky real-world scenarios? Ask the AI tutor for more worked cycles.

Why does the tyre-pump example count as adiabatic if I am not doing it in a vacuum flask?

Adiabatic only requires that heat has no time to cross the boundary, not that the container is a perfect thermos. A quick, hard push on a bicycle pump compresses the trapped air in a fraction of a second — far too fast for meaningful heat to conduct out through the metal barrel — so almost all the work you do goes straight into raising the internal energy of the air, and you feel that as warmth. Do the same compression over several minutes instead, letting the barrel stay at room temperature, and you would be much closer to an isothermal process. It is a comparison of timescales: how fast you compress versus how fast heat can escape.

Is a cyclic process the same thing as a heat engine?

Not in this course's sense, and this is a common exam trap. A cyclic process is only the geometric statement that a system, in a $P$ – $V$ diagram, returns exactly to its starting point — a shape on a graph. A heat engine is a device built around repeated cyclic processes, designed to convert some of the net heat absorbed into usable work, and judging how well it does that needs the idea of efficiency. CBSE Class 11 keeps cyclic processes (compute the work, apply $\Delta U=0$ ) but leaves engines, efficiency and Carnot's theorem for later study.

If PV equals a constant for isothermal and PV to the gamma equals a constant for adiabatic, why do the two curves cross at only one point on my diagram?

Both curves are drawn through the same starting state, so of course they share that one point — but away from it they separate because they obey different rules for how fast pressure must rise as volume shrinks. Since $\gamma$ is greater than $1$ , the adiabatic exponent makes pressure climb faster for the same drop in volume than the isothermal relation would predict, which is exactly why the adiabatic curve is steeper on both sides of the shared starting point. Two curves obeying genuinely different power laws in $V$ can only coincide at isolated points, not along a stretch, unless the laws are identical.

Do I need calculus to find the work done in an isothermal or adiabatic process for my board exam?

No — CBSE Class 11 keeps the numerical work you compute to the algebraic case $\Delta W = P\Delta V$ , which applies exactly when pressure is constant (isobaric) or trivially when volume is constant (isochoric, giving zero). For isothermal and adiabatic processes, where pressure changes continuously with volume, the exact work done needs integrating $P\,dV$ , beyond what this course expects you to reproduce. Board questions instead test classifying the process correctly, sketching its curve, and applying $PV=\text{constant}$ or $PV^\gamma=\text{constant}$ to relate states — not computing enclosed work numerically for those two.