Adiabatic process

Pump a bicycle tyre hard and fast and the barrel gets noticeably warm in your hand — not because your hand is doing anything special, but because the air trapped inside is being squeezed too quickly for the heat to escape. No heat goes in or out; every joule of the work you do with your arm has nowhere to go except straight into the gas's internal energy, and that shows up as a temperature rise you can feel through the metal. Let the same gas expand quickly instead — puncture a sealed canister of compressed CO2 and the nozzle frosts over — and the opposite happens: the gas does work pushing its way out, that energy comes only from its own internal store since nothing is flowing in from outside, and the gas cools sharply enough to freeze water vapour in the air around it. Both are examples of an adiabatic process: a change of state for a gas in which $\Delta Q = 0$ throughout, either because the container is well insulated or because the change happens too fast for meaningful heat exchange with the surroundings. It sits right next to the isothermal process in the CBSE Class 11 syllabus as the other extreme case of gas behaviour, and the two make a natural contrast: isothermal holds temperature fixed by letting heat flow freely; adiabatic holds heat exchange at exactly zero and lets temperature swing instead. This article builds the governing law $PV^\gamma = \text{constant}$ from the first law of thermodynamics, works two contrasting numerical examples — one at board level, one competitive-exam pace — and pins down the traps that most reliably cost marks: mixing up adiabatic with isothermal, and getting the direction of temperature change backwards.

Key takeaways

An adiabatic process is one with no heat exchange with the surroundings at any stage: $\Delta Q = 0$ , always — this is the defining condition, not an approximation.
Applying the first law $\Delta Q = \Delta U + \Delta W$ with $\Delta Q = 0$ gives $\Delta U = -\Delta W$ : all work done by (or on) the gas comes directly out of (or into) its internal energy.
The governing relation for a quasi-static adiabatic change of an ideal gas is $PV^\gamma = \text{constant}$ , where $\gamma = C_P/C_V > 1$ ; equivalent forms are $TV^{\gamma-1} = \text{constant}$ and $T^\gamma P^{1-\gamma} = \text{constant}$ .
Adiabatic expansion always cools a gas; adiabatic compression always heats it — the sign is fixed by $\Delta U = -\Delta W$ and never reverses.
On a $P$ – $V$ diagram, an adiabatic curve through any point is always steeper than an isothermal curve through the same point, because $\gamma > 1$ .
The work done in a quasi-static adiabatic process between two states is $W = \dfrac{P_1V_1 - P_2V_2}{\gamma - 1} = \dfrac{nR(T_1 - T_2)}{\gamma - 1}$ .
Key idea: with heat exchange locked at zero, every joule of work a gas does adiabatically is paid for entirely out of its own internal energy, so $\Delta U = -\Delta W$ and the state variables obey $PV^\gamma = \text{constant}$ .

The core insight

An adiabatic process forbids heat exchange ( $\Delta Q = 0$ ) at every instant, so the first law collapses to $\Delta U = -\Delta W$ and the state of an ideal gas moves along $PV^\gamma = \text{constant}$ , cooling on expansion and heating on compression.

Where it fits in the syllabus

Adiabatic processes are named explicitly inside the CBSE Class 11 Physics Thermodynamics unit's prescribed scope, alongside isothermal, reversible, irreversible and cyclic processes, with the relation $PV^\gamma = \text{constant}$ given directly in the syllabus text. You are expected to define the process by $\Delta Q = 0$ , derive $\Delta U = -\Delta W$ from the first law, apply $PV^\gamma = \text{constant}$ and its $T$ – $V$ and $T$ – $P$ equivalents to connect two states, and reason physically about why compression heats and expansion cools a gas. The value of $\gamma = C_P/C_V$ itself is supplied from the Kinetic Theory unit (equipartition of energy), not derived here. Detailed applications inside heat engines — such as the adiabatic strokes of a Carnot cycle, or engine efficiency $\eta$ — are out of scope for the current 2025-26 prescription; a single forward pointer to Class 12/higher studies is as far as this section goes on that front.

The physical picture

An adiabatic process is defined purely by the absence of heat flow, at every instant of the change, not merely at the start and end:

\Delta Q = 0

This happens for one of two physical reasons: either the system is genuinely insulated (a thick-walled, lagged cylinder), or the change is fast enough that heat simply has no time to cross the boundary before the process is over — a rapid compression or expansion of a gas in an ordinary, unlagged container still counts as adiabatic if it happens quickly enough compared to the sluggish rate of heat conduction. Substituting $\Delta Q = 0$ into the first law $\Delta Q = \Delta U + \Delta W$ (with $\Delta W = P\,\Delta V$ , the work done by the gas) immediately gives:

\Delta U = -\Delta W

Read this literally: internal energy and work done are now locked to each other with a minus sign, and nothing else feeds the balance. If the gas expands and does positive work ( $\Delta W > 0$ ), its internal energy must fall by exactly that amount — there is no heat coming in from outside to make up the shortfall, so the temperature drops. If the gas is compressed and work is done on it ( $\Delta W < 0$ ), internal energy rises by exactly that amount, and the temperature climbs — with nowhere for the extra energy to escape as heat, it all stays inside as faster molecular motion.

Figure

Two cylinders side by side, one with a piston moving outward and an arrow showing internal energy falling and temperature dropping, the other with a piston moving inward and an arrow showing internal energy rising and temperature climbing, both cylinders drawn with insulated walls and no heat arrows crossing the boundary

Now bring in the gas's equation of state. For a quasi-static adiabatic change of an ideal gas, combining $\Delta U = -\Delta W$ with $PV = nRT$ and the definitions of $C_V$ and $C_P$ (a route worked step by step in the derivation below) leads to the relation quoted directly in the syllabus:

PV^\gamma = \text{constant}

where: $P$ and $V$ are the pressure and volume of the gas at any instant during the process, and $\gamma = C_P/C_V$ is the ratio of specific heats — a fixed number for a given gas ( $5/3$ for monatomic, $7/5$ for diatomic, always greater than $1$ ). Because $PV = nRT$ , this same law can be rewritten in terms of temperature instead of pressure, giving two equally useful equivalent forms:

TV^{\gamma - 1} = \text{constant}, \qquad T^\gamma P^{1-\gamma} = \text{constant}

The physical reading of $PV^\gamma = \text{constant}$ is that pressure falls faster with increasing volume than plain Boyle's law ( $PV = \text{constant}$ ) would predict, because in the adiabatic case temperature is falling too, and a cooler gas exerts less pressure for the same volume than a gas held at constant temperature. This is exactly why an adiabatic curve on a $P$ – $V$ diagram is steeper than an isothermal curve through the same starting point — the extra factor of $\gamma > 1$ in the exponent drags the curve down more sharply as $V$ grows.

Figure

A pressure versus volume graph showing two curves starting from the same point, an isothermal curve labelled PV equals constant that falls gently, and a steeper adiabatic curve labelled PV to the gamma equals constant that falls more sharply, with both curves ending lower for the adiabatic case at the same final volume

Finally, the work done during a quasi-static adiabatic process carrying the gas from state $1$ to state $2$ can be written directly in terms of the endpoint pressures and volumes, or equivalently the endpoint temperatures:

W = \frac{P_1V_1 - P_2V_2}{\gamma - 1} = \frac{nR(T_1 - T_2)}{\gamma - 1}

where: $n$ is the number of moles and $R = 8.314\ \text{J/mol}\,\cdot\,\text{K}$ is the universal gas constant. This is the quantity we use directly in the second worked example below, and it is simply $\Delta U = -\Delta W$ rearranged and expressed through $C_V = R/(\gamma - 1)$ .

Derivation ADeriving PV to the gamma equals constant from the first law

The goal is to start from nothing more than the first law of thermodynamics applied to a tiny adiabatic step, $\Delta Q = 0$ , and the ideal gas equation of state $PV = nRT$ , and arrive at the syllabus relation $PV^\gamma = \text{constant}$ for a quasi-static adiabatic process, using only the definitions $C_V = \left(\frac{dU}{dT}\right)_V$ and $\gamma = C_P/C_V$ — no calculus beyond recognising that small steps add up, and the integration itself is quoted rather than carried out symbol by symbol, since it lies just past the algebra-only ceiling of this course.

Step 1. Write the first law for one infinitesimal step of the process. For a small change, $dQ = dU + P\,dV$ . Since the process is adiabatic, $dQ = 0$ at every step, not just overall, so $dU = -P\,dV$ throughout the whole path — this is the local version of $\Delta U = -\Delta W$ used above, now applied instant by instant so it can be combined with the equation of state.

Step 2. Express $dU$ using the molar specific heat at constant volume. For an ideal gas, internal energy depends only on temperature, so $dU = nC_V\,dT$ regardless of what the volume is doing. Substituting into Step 1 gives $nC_V\,dT = -P\,dV$ , a relation purely between temperature and volume changes.

Step 3. Bring in the equation of state to eliminate $P$ . From $PV = nRT$ , $P = nRT/V$ . Substituting into Step 2: $nC_V\,dT = -\dfrac{nRT}{V}\,dV$ . The $n$ cancels from both sides, leaving $C_V\,\dfrac{dT}{T} = -R\,\dfrac{dV}{V}$ — a relation between the fractional change in temperature and the fractional change in volume, with all the extensive quantities gone.

Step 4. Integrate both sides along the process (quoted result, standard for this kind of relation) and use $R = C_P - C_V$ : the left side integrates to $C_V\ln T$ and the right side to $-R\ln V$ , so $C_V\ln T + R\ln V = \text{constant}$ . Dividing through by $C_V$ and writing $R/C_V = (C_P-C_V)/C_V = \gamma - 1$ gives $\ln T + (\gamma-1)\ln V = \text{constant}$ , which exponentiates to $TV^{\gamma-1} = \text{constant}$ — one of the syllabus forms already.

Step 5. Convert to the $P$ – $V$ form using $T \propto PV/n$ (from the equation of state, with $n$ fixed). Substituting $T = PV/(nR)$ into $TV^{\gamma-1} = \text{constant}$ gives $\dfrac{PV}{nR}\cdot V^{\gamma-1} = \text{constant}$ , and since $n$ and $R$ are themselves constants for a fixed amount of gas, this simplifies directly to $PV^\gamma = \text{constant}$ — the relation stated in the syllabus, now built from the first law and the equation of state rather than merely quoted.

Worked examples

Board level - final pressure after adiabatic expansion

1 mole of a diatomic ideal gas ( $\gamma = 1.4$ ) at initial pressure $P_1 = 2\times10^5$ Pa and volume $V_1 = 4\times10^{-3}\ \text{m}^3$ expands adiabatically to volume $V_2 = 8\times10^{-3}\ \text{m}^3$ . Find the final pressure.

Step 1. Identify the relation to use. Since the process is adiabatic, $P_1V_1^\gamma = P_2V_2^\gamma$ connects the two states directly — there is no need for the amount of gas or the temperature here, only pressures and volumes.

Step 2. Rearrange for the unknown and substitute. $P_2 = P_1\left(\dfrac{V_1}{V_2}\right)^\gamma = 2\times10^5 \times \left(\dfrac{4\times10^{-3}}{8\times10^{-3}}\right)^{1.4} = 2\times10^5 \times (0.5)^{1.4}$ .

Step 3. Evaluate the fractional power. $(0.5)^{1.4} = e^{1.4\ln 0.5} = e^{1.4\times(-0.6931)} = e^{-0.9704} = 0.3789$ .

Step 4. Combine to get the final pressure. $P_2 = 2\times10^5 \times 0.3789 = 7.58\times10^4\ \text{Pa}$ .

The final pressure is $P_2 \approx 7.58\times10^4$ Pa. This is noticeably lower than the $10^5$ Pa that Boyle's law (isothermal doubling of volume) would give for the same volume change — confirming, numerically, that the adiabatic curve falls faster than the isothermal one, because the gas is also cooling as it expands.

Competitive level - temperature and work in adiabatic compression

1 mole of an ideal monatomic gas ( $\gamma = 5/3$ , $C_V = \tfrac32 R$ ) initially at $T_1 = 400$ K is compressed adiabatically so that its volume is halved. Find the final temperature, and determine the work done on the gas. Take $R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}$ .

Step 1. Choose the temperature–volume form, since the problem gives temperature and a volume ratio directly. $T_1V_1^{\gamma-1} = T_2V_2^{\gamma-1}$ , so $T_2 = T_1\left(\dfrac{V_1}{V_2}\right)^{\gamma-1} = 400\times(2)^{5/3-1} = 400\times 2^{2/3}$ , since halving the volume means $V_1/V_2 = 2$ .

Step 2. Evaluate the fractional power. $2^{2/3} = e^{(2/3)\ln 2} = e^{0.4621} = 1.5874$ .

Step 3. Combine to get the final temperature. $T_2 = 400\times1.5874 = 634.96\ \text{K} \approx 635\ \text{K}$ — a rise of about $235$ K from compressing the gas to half its volume, with no heat added at all.

Step 4. Find the work done by the gas using $W = \dfrac{nR(T_1-T_2)}{\gamma-1} = \dfrac{1\times8.31\times(400-635)}{5/3-1} = \dfrac{8.31\times(-235)}{2/3} = \dfrac{-1952.85}{0.6667} = -2929.2\ \text{J}$ .

Step 5. Interpret the sign. $W = -2929.2$ J means the gas does negative work, i.e. about $2929.2$ J of work is done on the gas by whatever compresses it. Checking consistency with Step 1's temperature rise: $\Delta U = -\Delta W = -(-2929.2) = +2929.2\ \text{J}$ , a rise in internal energy — exactly matching a temperature that climbed rather than fell.

Final answers: $T_2 \approx 635$ K, and about 2929 J of work is done on the gas (equivalently $W = -2929.2$ J done by the gas). Timing note: in a competitive-exam setting, budget roughly 90 seconds for the $T_2$ calculation (recognise the $TV^{\gamma-1}$ form immediately rather than re-deriving it) and another 60-90 seconds for the work calculation — the sign check in Step 5 is the fastest way to catch an arithmetic slip before it costs the mark, since a positive $W$ alongside a temperature rise would immediately flag an error.

Common traps

Confusing adiabatic ( $\Delta Q = 0$ ) with isothermal ( $\Delta T = 0$ ). These describe opposite special cases; an adiabatic process generally does change temperature substantially, while an isothermal process by definition does not change it at all.
Using $PV = \text{constant}$ instead of $PV^\gamma = \text{constant}$ for adiabatic problems — dropping the exponent $\gamma$ and applying Boyle's law by reflex is one of the single most common numerical slip-ups in this topic.
Getting the sign of the temperature change backwards. Adiabatic expansion always cools a gas; adiabatic compression always heats it. Students under time pressure sometimes reverse this and get an internally inconsistent answer (as the sign check in the second worked example shows how to catch).
Forgetting that $\gamma$ depends on the type of gas — monatomic ( $5/3$ ), diatomic ( $7/5$ ), or polyatomic — and plugging in the wrong value without checking which gas the question specifies.
Treating a fast, uninsulated process as automatically isothermal. A sudden compression or expansion, even in an ordinary unlagged cylinder, can still be adiabatic if it happens faster than heat can conduct away — "adiabatic" is about the rate and boundary conditions of heat transfer, not merely about whether the container looks insulated.

Common trapDo not assume an adiabatic process keeps temperature constant

The single biggest misconception in this topic is bleeding the definition of isothermal into adiabatic because both sound like "special, simple" processes. They are not the same, and they are not even similar — they sit at opposite ends of how a gas can change state. Isothermal processes hold $\Delta T = 0$ by letting heat flow freely in or out to compensate for any work done, which is only possible if the process happens slowly in good thermal contact with a reservoir. Adiabatic processes hold $\Delta Q = 0$ by cutting off heat exchange entirely, which is only possible if the process happens fast or the system is insulated — and as a direct consequence, temperature is exactly what is free to change, often by a large amount, as the two worked examples above show (pressure falling faster than Boyle's law, temperature rising by 235 K under compression). If a question says "adiabatic," never write $\Delta U = 0$ or assume $T_1 = T_2$ ; if it says "isothermal," never reach for $PV^\gamma = \text{constant}$ .

You should be able to…

State the defining condition of an adiabatic process ( $\Delta Q = 0$ at every instant) and distinguish it clearly from an isothermal process ( $\Delta T = 0$ ).
Derive $\Delta U = -\Delta W$ from the first law for an adiabatic process, and explain in words why expansion cools a gas and compression heats it.
Use $PV^\gamma = \text{constant}$ and its equivalent $TV^{\gamma-1} = \text{constant}$ and $T^\gamma P^{1-\gamma} = \text{constant}$ forms to relate an initial and a final state.
Follow the first-law-to- $PV^\gamma$ derivation step by step, including why $C_V\,dT/T = -R\,dV/V$ appears along the way.
Compute the work done during a quasi-static adiabatic process using $W = \dfrac{nR(T_1-T_2)}{\gamma-1}$ , and sanity-check the sign against the direction of the temperature change.
Compare the steepness of adiabatic and isothermal curves on the same $P$ – $V$ diagram and explain the physical reason for the difference.

Ask the tutor: Want to practise switching between the $PV^\gamma$ , $TV^{\gamma-1}$ , and $T^\gamma P^{1-\gamma}$ forms, or work through the first-law derivation one more time in your own words? Ask the AI tutor for more adiabatic process problems.

Why does compressing a gas quickly make it hot, when compressing it slowly with the container touching ice would not?

Both compressions do positive work on the gas, but what happens to that energy is completely different. Compress it quickly (adiabatically) and there is no time for heat to escape, so by $\Delta U = -\Delta W$ every joule of work done on the gas stays inside as internal energy, and temperature rises — exactly like the bicycle pump. Compress it slowly while it touches ice, and heat has plenty of time to leave through the walls as fast as the compression generates it, keeping $\Delta T \approx 0$ (an isothermal, not adiabatic, process) — the work done still equals the heat rejected to the ice, so the ice slowly melts instead of the gas heating up. The physical difference is entirely about how much time the process gives heat to escape, not about the amount of work done.

Is a truly adiabatic process even possible, since nothing is perfectly insulated?

In the real world, no process is perfectly adiabatic in the strict sense of zero heat leakage over infinite time — every real material conducts at least a little heat. But "adiabatic" in this syllabus is a useful idealisation for two very common real situations: genuinely well-insulated systems over the timescale of interest (a vacuum flask, a lagged cylinder), and, more commonly in exam problems, processes that happen so fast that negligible heat can cross the boundary before the process finishes — sound waves compressing and rarefying air, or a sudden puncture releasing compressed gas. In both cases the approximation $\Delta Q \approx 0$ is excellent for the duration that matters, which is all physics needs from the model.

How is gamma different for monatomic and diatomic gases, and why does it matter here?

$\gamma = C_P/C_V$ comes from how many ways a gas molecule can store energy (its degrees of freedom), which is covered in the Kinetic Theory unit, not derived fresh here. A monatomic gas (like helium or argon) can only store energy in translational motion, giving $\gamma = 5/3 \approx 1.67$ . A diatomic gas (like the oxygen and nitrogen that make up most of air) can also store energy in rotation, giving a lower value $\gamma = 7/5 = 1.4$ . It matters directly in this topic because $\gamma$ sits as the exponent in $PV^\gamma = \text{constant}$ and in the denominator of the work formula — using the wrong $\gamma$ for the gas in a question (as the two worked examples above deliberately use different values, $1.4$ and $5/3$ ) changes every numerical answer that follows.

Can an adiabatic process also be a cyclic process, and does it fit into a heat engine?

A single adiabatic process is just one leg of a change of state — it takes a gas from one state to another without heat exchange. Real heat engines (like the Carnot cycle) do use adiabatic legs, alternating them with isothermal legs, to form a complete closed loop back to the starting state. However, the detailed working of heat engines, their efficiency, and named cycles like the Carnot cycle are explicitly outside the current CBSE Class 11 prescription for 2025-26 — that territory is studied in Class 12 and beyond. For this syllabus, treat the adiabatic process as a single, self-contained change of state governed by $PV^\gamma = \text{constant}$ , not as a piece of engine design.