In short

A thermodynamic process is any path a gas takes from one (P, V, T) state to another. The two most-used paths carry names.

  • Isothermal process — temperature held constant; T = \text{const}. For an ideal gas this collapses to Boyle's law: PV = \text{const}. The work done by the gas in expanding from V_1 to V_2 is
W_\text{iso} = n R T \ln\!\left(\frac{V_2}{V_1}\right).

Because T does not change, the internal energy does not change either (\Delta U = 0 for an ideal gas), so all of the work comes from heat absorbed: Q = W.

  • Adiabatic process — no heat exchange with the surroundings; Q = 0. For an ideal gas this implies PV^\gamma = \text{const} and TV^{\gamma - 1} = \text{const}, where \gamma = C_p / C_v is the ratio of specific heats (7/5 for diatomic air, 5/3 for monatomic helium). The work done by the gas is
W_\text{adi} = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1} = \frac{n R (T_1 - T_2)}{\gamma - 1}.

Because no heat enters, the work comes from the gas's own internal energy, so the gas cools on adiabatic expansion and heats on adiabatic compression.

Why the adiabat is steeper than the isotherm on a P–V diagram. On an isotherm, (dP/dV)_T = -P/V (Boyle's law differentiated). On an adiabat, (dP/dV)_\text{adi} = -\gamma P/V. Since \gamma > 1, the adiabat slope is steeper by the factor \gamma — for air, the adiabat is 1.4 times steeper than the isotherm at any common point. The physical picture: on an adiabat, compression heats the gas, which adds pressure on top of the ordinary Boyle compression. The slope steepness is directly the strength of that extra kick.

Real-world mapping. Slow process with good thermal contact → isothermal (the ice-water-bath bicycle pump, the slow stroke of a refrigeration compressor). Fast process with poor thermal contact → adiabatic (the diesel engine's compression stroke, an air-gun's shot, the pressure wave of a firecracker, the compression stage of a home AC compressor). The line between them is how fast heat can escape compared to how fast the volume changes.

Hold a deflated football bladder in your hand. Pump it up with a bicycle pump while another person holds the barrel. Two things happen during each stroke: the air inside the pump's barrel gets compressed, and the barrel gets noticeably warm under the other person's fingers. Slow the stroke down and hold the piston at each level for a while. The barrel stays cool — heat has time to leak away through its steel wall. Speed the stroke up and push hard. The barrel gets so hot you can barely touch it.

Same gas. Same initial and final volumes. Two different amounts of temperature rise, two different amounts of heat flow, two different pressures for the same final volume. The difference is entirely in the path from state 1 to state 2 — how quickly you pushed, and how much heat escaped along the way. The thermodynamic variables (P, V, T) tell you where the gas is. The process — the path it takes — is an independent physical fact about how it got there.

Most real thermodynamic processes are messy mixtures of heat flow, work, and non-uniform temperatures. But two idealised limits bracket all real behaviour and are worth understanding with care:

Both are extreme. A diesel piston compresses the air in 4 ms — not quite adiabatic, but close enough that the adiabatic equation gets the answer right to a few per cent. A nitrogen cylinder used in a gym pulley-compressor opens its valve slowly enough that the gas inside re-equilibrates with room temperature — close enough to isothermal. Between the two extremes lie the real processes, and engineers quantify how close to one limit or the other they are.

This article derives the two ideal-process laws from PV = nRT and the first law of thermodynamics \Delta U = Q - W. It computes the work done along each and shows why the adiabat is steeper than the isotherm. It closes with the JEE-advanced treatment of mixed processes, the speed of sound derivation, and why rapid expansion of compressed air is a surprisingly effective way to make ice.

Sign conventions you must commit to memory

Before any derivation, fix the signs.

These are the conventions used in the Indian NCERT textbook, in JEE problem sets, and throughout this article. Chemistry textbooks sometimes use the opposite convention for W (positive when done on the system, yielding \Delta U = Q + W). Pick one and stick with it; flipping midway is how mistakes creep in.

For an ideal gas, internal energy depends only on temperature: U = U(T), specifically U = n C_v T plus a constant (for n moles at temperature T). See degrees of freedom and equipartition for the origin. The key takeaway: if T does not change, neither does U; \Delta U = 0. If T changes by \Delta T, \Delta U = n C_v \Delta T — always, for any process, because U depends only on T and not on how you got there.

The isothermal process — T held constant

Lock the gas inside a cylinder with a movable piston. Put the cylinder in a very large heat reservoir (a water bath, or simply a large room that cannot change temperature noticeably) at temperature T. Move the piston slowly enough that at every instant the gas is in thermal equilibrium with the reservoir. As the gas expands or contracts, its pressure and volume change — but its temperature stays pinned at T.

Isothermal setupA cylindrical gas vessel immersed in a large water bath at temperature T. The vessel has a movable piston on the right. An arrow on the piston indicates slow motion. Heat Q flows across the thin metal wall between the gas and the bath. Labels indicate that because motion is slow, the gas temperature equals T at every instant, and PV equals nRT remains true throughout.heat reservoir at temperature Tgas (T, P, V)slowQPV = nRTT = constΔU = 0Q = WW = nRT ln(V₂/V₁)
The isothermal setup. The gas exchanges heat freely with a large reservoir through a thin conducting wall; the piston moves slowly enough that $T_\text{gas} = T_\text{reservoir}$ at every instant. Because $T$ is fixed, $\Delta U = 0$ and every joule of work the gas does on the piston is replenished by a joule of heat from the reservoir.

Deriving PV = \text{const} for the isothermal path

Step 1. Start from the ideal gas law:

PV = nRT.

Step 2. Hold T constant. The right-hand side is then a constant:

P V = (nR) T = \text{constant along the path}. \tag{1}

Why: in an isothermal process, only P and V can change; T is fixed. So PV — the product of the two that change — must equal a fixed number (nRT) throughout. This is Boyle's law used as a path equation rather than a statement about a single experiment.

That is the isothermal path equation. On a P–V diagram, isotherms are rectangular hyperbolas of the form P = (nRT)/V. Higher-temperature isotherms sit further from the origin.

Work done in an isothermal process

For a gas expanding from V_1 to V_2 against its external pressure, the work done by the gas is

W = \int_{V_1}^{V_2} P\, dV.

Why: when the piston moves outward by an infinitesimal dV against pressure P, the gas does work F \, dx = P A\, dx = P\, dV. Integrating over the path gives the total work. See the first law of thermodynamics for a full derivation.

Step 3. For the isothermal path, substitute P = nRT/V:

W_\text{iso} = \int_{V_1}^{V_2} \frac{nRT}{V}\, dV = nRT \int_{V_1}^{V_2} \frac{dV}{V}.

Why: pull nRT out of the integral because it is constant. The remaining integrand is 1/V — a standard integral.

Step 4. Evaluate the integral:

W_\text{iso} = nRT\,[\ln V]_{V_1}^{V_2} = nRT (\ln V_2 - \ln V_1)
\boxed{\;W_\text{iso} = nRT \ln\!\left(\frac{V_2}{V_1}\right).\;} \tag{2}

Why: the integral of 1/V is \ln V. Subtracting gives \ln V_2 - \ln V_1 = \ln(V_2/V_1). For expansion (V_2 > V_1) the logarithm is positive, so W > 0 — the gas does positive work on its surroundings. For compression (V_2 < V_1) the logarithm is negative, so W < 0 — work is done on the gas.

Equivalent forms: since P_1 V_1 = P_2 V_2 on an isotherm, V_2/V_1 = P_1/P_2, so

W_\text{iso} = nRT\,\ln(P_1/P_2) = P_1 V_1 \ln(V_2/V_1).

These are the same number written three ways.

Heat absorbed in an isothermal process

Step 5. For an ideal gas, U depends only on T. Since T is constant,

\Delta U_\text{iso} = 0.

Step 6. The first law \Delta U = Q - W then forces

Q_\text{iso} = W_\text{iso} = nRT \ln(V_2/V_1). \tag{3}

Why: every joule of work the gas does comes from the heat reservoir, because the gas's own internal energy cannot change at constant T. This makes the isothermal process a perfect heat-to-work converter for a given volume range — a property that is central to the Carnot cycle.

For expansion, Q > 0 (heat flows into the gas). For compression, Q < 0 (heat flows out of the gas to the reservoir). Either way, all heat becomes work (or all work becomes heat), and the gas's temperature never wavers.

The adiabatic process — no heat crosses the boundary

Now insulate the cylinder perfectly — wrap it in thermos walls — or push the piston fast enough that the process is over before appreciable heat can cross the wall. In either limit, Q = 0. The first law reduces to

\Delta U = -W.

Why: setting Q = 0 in \Delta U = Q - W leaves \Delta U = -W. If the gas does positive work (expansion), its internal energy drops — the gas cools. If work is done on the gas (compression), its internal energy rises — the gas heats up.

This is the physical origin of the warm bicycle-pump barrel, the cold aerosol can after sustained use, the blazing hot compression stroke of a diesel engine, and the chill that leaves the nozzle of a compressed-air canister. In every one of these, mechanical work is being traded for internal energy with no thermal middleman.

Adiabatic setupA cylindrical gas vessel with thick insulating walls, marked as thermally insulated. The piston on the right moves rapidly, shown by a fast arrow. No heat Q crosses the boundary. Labels indicate Q equals zero, so the first law reduces to delta U equals minus W. Work done on the gas raises its temperature; work done by the gas lowers it. The path equation P V to the gamma equals constant holds only when gamma equals Cp over Cv, the adiabatic index.thermally insulated walls (Q = 0)gasfastQ = 0ΔU = −WPV^γ = constTV^(γ−1) = constW = (P₁V₁ − P₂V₂)/(γ−1)
The adiabatic setup. No heat crosses the boundary, either because the walls are perfectly insulating or because the process is so fast that heat has no time to flow. All of the work trades one-for-one against the gas's own internal energy, so the gas cools on expansion and heats on compression.

Deriving PV^\gamma = \text{const} — the path equation

Start from the first law in differential form,

dU = dQ - dW = -P\, dV \qquad (\text{adiabatic: } dQ = 0).

Step 1. For an ideal gas, dU = n C_v\, dT:

n C_v\, dT = -P\, dV. \tag{4}

Why: the internal energy of an ideal gas is U = n C_v T + const; its differential is dU = n C_v\, dT. This substitution links temperature change to volume change for any ideal-gas process — here specialised to adiabatic through dQ = 0.

Step 2. Use the ideal gas law PV = nRT to eliminate P:

P = \frac{nRT}{V}.

Substitute into (4):

n C_v\, dT = -\frac{n R T}{V} dV.

Why: the goal is to get a differential equation in T and V alone, so that you can separate variables and integrate. Replacing P by nRT/V achieves exactly that.

Step 3. Separate variables:

\frac{dT}{T} = -\frac{R}{C_v}\,\frac{dV}{V}. \tag{5}

Why: divide both sides by n C_v and by T. The n on the right cancels the n on the left; the temperature goes on the left with its differential, the volume on the right with its differential.

Step 4. Simplify R/C_v using Mayer's relation C_p - C_v = R (derived in isobaric and isochoric processes):

\frac{R}{C_v} = \frac{C_p - C_v}{C_v} = \frac{C_p}{C_v} - 1 = \gamma - 1.

Why: the ratio \gamma \equiv C_p/C_v is called the adiabatic index. For a monatomic gas, \gamma = 5/3; for a diatomic gas at room temperature, \gamma = 7/5; for a polyatomic rigid gas, \gamma = 4/3. Substituting R/C_v = \gamma - 1 will make the final answer depend on \gamma alone.

Equation (5) becomes

\frac{dT}{T} = -(\gamma - 1)\,\frac{dV}{V}. \tag{6}

Step 5. Integrate both sides:

\ln T = -(\gamma - 1)\ln V + \text{const}.
\ln T + (\gamma - 1)\ln V = \text{const}.
\ln(T V^{\gamma - 1}) = \text{const}.
\boxed{\;T V^{\gamma - 1} = \text{const}\;}. \tag{7}

Why: integrate term by term. \ln terms combine: a \ln T + b \ln V = \ln(T^a V^b). Exponentiating both sides converts the additive constant into a multiplicative one and gives the T-V form of the adiabatic relation.

Step 6. Convert (7) to a P-V relation using PV = nRT. Solve for T:

T = \frac{PV}{nR}.

Substitute into (7):

\frac{PV}{nR} \cdot V^{\gamma - 1} = \text{const}
P V^\gamma = \text{const} \cdot nR = \text{const}'
\boxed{\;P V^\gamma = \text{const}\;}. \tag{8}

Why: multiplying the two volume factors gives V^{1 + (\gamma - 1)} = V^\gamma. The constant nR absorbs into the overall constant — equation (8) is the P-V form of the adiabatic relation, and is the version you will see most often on a PV diagram.

Step 7. The third form, in T and P, follows by combining (7) and (8):

T^\gamma P^{1 - \gamma} = \text{const} \quad\Longleftrightarrow\quad T P^{(1 - \gamma)/\gamma} = \text{const}.

All three forms are equivalent; pick the pair of variables (T and V, or P and V, or T and P) that matches the quantities you know.

Work done in an adiabatic process

Adiabatic work is most cleanly computed from the first law:

Step 8. Because Q = 0, \Delta U = -W, so W = -\Delta U = -n C_v (T_2 - T_1) = n C_v (T_1 - T_2).

Why: if you know the starting and ending temperatures, the work is immediately n C_v \Delta T (with the right sign). No integral needed, because the path is already implicitly integrated into the ideal-gas U(T).

Step 9. Rewrite using R = C_v (\gamma - 1), so C_v = R/(\gamma - 1):

W_\text{adi} = \frac{nR (T_1 - T_2)}{\gamma - 1}.

Step 10. Convert to a P–V form using nR T = PV:

\boxed{\;W_\text{adi} = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}\;}. \tag{9}

Why: nRT_1 = P_1 V_1 and nRT_2 = P_2 V_2, so nR(T_1 - T_2) = P_1 V_1 - P_2 V_2. Divide by \gamma - 1 to get the P–V form. The sign is correct: for expansion (V_2 > V_1, hence P_2 < P_1, and specifically P_2 V_2 < P_1 V_1 along an adiabat with \gamma > 1) the numerator is positive, so W > 0 — the gas does positive work on its surroundings, at the cost of its own internal energy.

Sanity check: direct integration. On an adiabat, PV^\gamma = C, so P = C/V^\gamma and

W = \int_{V_1}^{V_2} \frac{C}{V^\gamma}\, dV = \frac{C}{1 - \gamma}\left[V^{1-\gamma}\right]_{V_1}^{V_2} = \frac{C\,(V_2^{1-\gamma} - V_1^{1-\gamma})}{1-\gamma}.

Using C = P_1 V_1^\gamma = P_2 V_2^\gamma, this reduces to (P_1 V_1 - P_2 V_2)/(\gamma - 1) after a couple of lines of algebra. The direct-integration answer matches equation (9). The first-law route is faster; they give the same number.

Isothermal vs adiabatic — why the adiabat is steeper

Plot both processes on a P–V diagram starting from the same (P_1, V_1).

$P$–$V$ paths starting from the same state $(V=1, P=2)$. Dark (ink) solid: isotherm $PV = 2$. Red solid: adiabat for a diatomic gas ($\gamma = 1.4$). Red dashed: adiabat for a monatomic gas ($\gamma = 5/3$). Both adiabats fall faster than the isotherm — the monatomic curve steepest of all — because on an adiabat, expansion cools the gas and the reduced temperature cuts the pressure below the isothermal value at every $V$.

At any given state (P, V) the slopes are:

The ratio of slopes is

\frac{(dP/dV)_Q}{(dP/dV)_T} = \gamma.

Why: both slopes are negative (pressure falls as volume rises), but the adiabatic slope is \gamma times as negative — i.e. steeper. For diatomic air (\gamma = 1.4) the adiabat is 1.4 times steeper than the isotherm at any common point. For monatomic helium (\gamma = 5/3 \approx 1.67) it is 1.67 times steeper. This steepness difference is not cosmetic — it is the visible trace of the temperature change that distinguishes the two processes.

Physical reading of the slope ratio. On an adiabatic expansion the gas cools, and the cooler gas naturally exerts lower pressure at a given volume than a warmer gas would. So when you expand along an adiabat, P drops below what the isotherm at the starting temperature would predict at the same volume. The drop is steeper by exactly the factor \gamma — which is why \gamma appears as the power of V in the adiabatic equation.

Explore both processes yourself

Drag the red dot to change the adiabatic index \gamma. The solid curve always shows the isotherm PV = 2; the red curve shows the adiabat PV^\gamma = 2 through the same starting point. As you crank \gamma up past 1, the adiabat visibly steepens. At \gamma = 1 the two curves coincide (an isotherm is the \gamma \to 1 limit of the adiabatic equation — which is really the limit where C_v \to \infty, i.e. the gas never warms no matter how much work is done on it).

Interactive: isotherm vs adiabat for variable γ A P versus V plot. A fixed dark curve shows the isotherm PV = 2. A red curve shows the adiabat P = 2 divided by V to the gamma, with gamma set by a draggable red dot on the right side of the plot area. Gamma varies from 1.0 to 1.8. As gamma increases the red curve falls off more steeply relative to the dark isotherm; at gamma 1 the red curve coincides with the isotherm. V P 1 2 3 4 5 0 1 2 3 4 γ (1, 2) drag red dot vertically isotherm PV = 2 adiabat PV^γ = 2
The isotherm (dark) is fixed at $PV = 2$. The adiabat (red) obeys $PV^\gamma = 2$ through the same $(V=1, P=2)$ point, with $\gamma$ adjustable from 1.0 to 1.8 (covering monatomic through polyatomic gases). Drag the red dot vertically along the right edge to change $\gamma$. Note how the adiabat steepens as $\gamma$ increases.

Worked examples

Example 1: Slow compression of air in a bicycle pump (isothermal)

You pump up a deflated football slowly enough for the pump barrel to stay at room temperature (T = 300\,\text{K}, ambient air). One full stroke compresses 2.0 L of air at 1.0\times 10^5\,\text{Pa} (atmospheric) into 0.5 L. Compute the final pressure, the work done on the gas, and the heat exchanged with the surroundings.

Slow bicycle pump compressionA bicycle pump barrel shown twice. Left: initial state with volume 2.0 litres at 1 atmosphere, label T = 300 K. Right: final state with volume 0.5 litres at 4 atmospheres, same T = 300 K because the process is slow and the barrel loses heat to the room. An arrow indicates slow compression. A small heat arrow points from the barrel outwards, showing that heat leaves the gas to the room during the compression.V₁ = 2.0 L, P₁ = 1 atmT = 300 Kslow0.5 L, 4 atmT = 300 KQ outn = P₁V₁/RT= 8.02×10⁻² mol
Bicycle pump compression, slow enough for the gas to stay at room temperature throughout. Heat flows out of the gas to keep $T$ fixed at 300 K.

Step 1. Identify the process.

Slow → isothermal, T = 300\,\text{K}. So PV = const along the path.

Why: "slow" in the problem statement is the clue — a slow process with a thin barrel wall stays isothermal with the room. Also, "pump barrel stays at room temperature" explicitly fixes T.

Step 2. Apply Boyle's law to find P_2.

P_1 V_1 = P_2 V_2 \quad\Rightarrow\quad P_2 = P_1 \cdot \frac{V_1}{V_2} = 1.0\times 10^5 \cdot \frac{2.0}{0.5} = 4.0\times 10^5\,\text{Pa}.

Why: the isothermal path is PV = const. With V_1 \to V_2 = 2.0 → 0.5 L, the volume reduces by 4×, so the pressure multiplies by 4×. Final pressure is 4 atm.

Step 3. Compute the number of moles.

n = \frac{P_1 V_1}{RT} = \frac{1.0\times 10^5 \times 2.0\times 10^{-3}}{8.314 \times 300} = \frac{200}{2494} \approx 0.0802\,\text{mol}.

Why: use the ideal-gas law at the initial state. Volume in SI (m³), pressure in pascals, temperature in kelvin; n comes out in moles.

Step 4. Compute the work done by the gas.

W_\text{iso} = nRT\,\ln(V_2/V_1) = 0.0802 \times 8.314 \times 300 \times \ln(0.5/2.0).
\ln(0.25) \approx -1.386.
W_\text{iso} = 0.0802 \times 8.314 \times 300 \times (-1.386) \approx -277\,\text{J}.

Why: for compression (V_2 < V_1), the logarithm is negative, so W is negative — the gas does negative work, which is another way of saying the pump does positive work on the gas. The magnitude is 277 joules.

Step 5. Compute the heat exchanged.

Q_\text{iso} = W_\text{iso} \approx -277\,\text{J}.

Why: for an isothermal process, \Delta U = 0 and the first law gives Q = W. Since W = -277 J, Q = -277 J as well — 277 J of heat flows out of the gas into the surroundings. That is exactly the heat the bicycle-pump barrel transfers to the room during slow compression.

Result: The final pressure is 4 atm. The pump does 277 J of work on the gas. The gas, in turn, dumps 277 J of heat into the room. Nothing has gained or lost internal energy — all of it has flowed through the gas as a "heat-to-work" conveyor belt in reverse.

What this shows: An isothermal compression is a perfect transducer from mechanical work to heat flow. All of the work done on the gas is transferred to the surroundings as heat. This is the idealised limit of the compression stage in a refrigerator's condenser — and why that coil gets hot.

Example 2: Diesel-engine compression stroke (adiabatic)

A diesel engine compresses air initially at P_1 = 1.0\times 10^5\,\text{Pa} and T_1 = 310\,\text{K} from V_1 = 600\,\text{cm}^3 to V_2 = 30\,\text{cm}^3. The compression ratio is 20:1 (characteristic of diesel). Treat the process as adiabatic (\gamma = 1.4 for air). Find the final pressure, the final temperature, and the work done on the gas. Explain why the air gets hot enough to ignite the fuel spray.

Diesel compression strokeA cylinder with a piston moving in fast. Initial state on the left: V = 600 cubic centimetres, P = 1 atmosphere, T = 310 kelvin. Final state on the right: V = 30 cubic centimetres, P = about 66 atmospheres, T = about 1027 kelvin. A fast arrow marks the rapid motion of the piston. Label notes the compression is adiabatic, Q = 0, so all work goes into heating the gas. A small flame icon near the final state indicates auto-ignition of the fuel spray.V₁ = 600 cm³P₁ = 1 atmT₁ = 310 Kfast, Q=030cm³~66 atm~1027 Kfuel ignites
Diesel compression. At 20:1 compression ratio, adiabatic compression drives the air well above the auto-ignition temperature of diesel fuel (about 480 K), so the injected fuel spray combusts without any spark.

Step 1. Identify the process.

Fast compression, no time for heat to escape → adiabatic. \gamma = 1.4 for air. Use PV^\gamma = const and TV^{\gamma - 1} = const.

Why: the compression stroke of a diesel engine takes about 4–5 milliseconds, which is too fast for appreciable heat to leave through the cylinder wall. Adiabatic is an excellent approximation; a full heat-transfer analysis corrects it by at most 5–10 per cent.

Step 2. Compute the final pressure from PV^\gamma = const.

P_2 = P_1 \left(\frac{V_1}{V_2}\right)^\gamma = 1.0\times 10^5 \times (20)^{1.4}.

20^{1.4} = ? Use 20^{1.4} = 20 \cdot 20^{0.4}. \log_{10}(20^{0.4}) = 0.4 \times 1.301 = 0.5204, so 20^{0.4} \approx 10^{0.5204} \approx 3.31. Therefore 20^{1.4} \approx 20 \times 3.31 = 66.3.

P_2 \approx 1.0\times 10^5 \times 66.3 = 6.63\times 10^6\,\text{Pa} \approx 66\,\text{atm}.

Why: the compression ratio 20 raised to \gamma = 1.4 gives about 66. So the pressure jumps by a factor of 66 — from 1 atm to 66 atm. If the process had been isothermal, the pressure would have risen only by the compression ratio itself, so only to 20 atm. The adiabat is 3.3× higher.

Step 3. Compute the final temperature from TV^{\gamma - 1} = const.

T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma - 1} = 310 \times 20^{0.4} \approx 310 \times 3.31 \approx 1027\,\text{K}.

Why: TV^{\gamma - 1} = const gives T_2/T_1 = (V_1/V_2)^{\gamma - 1} = 20^{0.4} \approx 3.31. So the temperature rises from 310 K to about 1027 K — a jump of over 700 K. Diesel auto-ignites around 480 K, so the air is hot enough to ignite the fuel spray as soon as it is injected. This is how a diesel engine works: no spark plug, just high-enough compression to trigger ignition.

Step 4. Compute the work done on the gas.

Number of moles: n = P_1 V_1 / (R T_1) = (10^5 \times 6\times 10^{-4})/(8.314 \times 310) = 60/2577 \approx 0.0233 mol.

W_\text{adi} = \frac{n R (T_1 - T_2)}{\gamma - 1} = \frac{0.0233 \times 8.314 \times (310 - 1027)}{0.4} = \frac{0.0233 \times 8.314 \times (-717)}{0.4}.
W_\text{adi} \approx \frac{-139}{0.4} \approx -347\,\text{J}.

Why: the gas does -347 J of work on the piston, which is the same as saying the piston does +347 J of work on the gas. That 347 J went entirely into raising the internal energy of the air, since Q = 0. Check: \Delta U = n C_v \Delta T = 0.0233 \times (5/2)\times 8.314 \times 717 \approx 347 J — matches.

Result: Final pressure 66 atm, final temperature 1027 K, work done on the gas 347 J (per cycle, per cylinder, roughly). The air is hot enough to auto-ignite diesel fuel without a spark.

What this shows: adiabatic compression is how diesel engines work. The Rudolf Diesel insight of 1893 was that a high-enough compression ratio makes the compressed air hot enough to ignite any injected fuel — no electric spark, no high-voltage ignition coil, just a very robust compression stroke. A petrol engine has a lower compression ratio (about 10:1) because petrol has a lower auto-ignition temperature and would ignite prematurely at diesel ratios; it therefore uses a spark plug and gets less thermodynamic efficiency.

Example 3: The speed of sound in air

Sound is a pressure wave travelling through air, and its speed depends on how stiffly the air responds to compression. Newton originally computed the speed of sound assuming the compressions and rarefactions in a sound wave were isothermal. He got 280 m/s — about 15 per cent too low. Laplace corrected the mistake by realising the compressions are too fast to be isothermal; they are adiabatic. Recompute the speed of sound in air at 300 K using the adiabatic bulk modulus, and compare with the measured value of 347 m/s.

Sound wave compression regionsA horizontal row of dots showing a sound wave passing through air. Alternating high-density and low-density bands are drawn. Each compression region is labelled as too fast for heat exchange, so adiabatic. Two curves below indicate the pressure and density oscillations. A speed arrow labelled v equals square root of gamma P over rho points to the right.compressionrarefactioncompressionP(x,t)v = √(γP/ρ)~347 m/s
A sound wave compresses and rarefies air too fast (microsecond timescales) for heat to flow between regions — so the compressions are adiabatic, not isothermal. The bulk modulus that sets the sound speed is therefore the adiabatic modulus $\gamma P$, not the isothermal modulus $P$.

Step 1. The general formula for the speed of a longitudinal (compression) wave in a fluid is

v = \sqrt{\frac{B}{\rho}},

where B is the bulk modulus and \rho is the density. The bulk modulus depends on whether the compression in the wave is isothermal or adiabatic.

Step 2. For an ideal gas, the isothermal bulk modulus is B_T = P and the adiabatic bulk modulus is B_\text{adi} = \gamma P.

Why: by definition, B = -V (dP/dV). On an isotherm, (dP/dV)_T = -P/V, so B_T = P. On an adiabat, (dP/dV)_Q = -\gamma P/V, so B_\text{adi} = \gamma P. The adiabatic bulk modulus is \gamma times the isothermal — gas is stiffer when you compress it fast, because the warming boosts its pressure response.

Step 3. Sound-wave compressions last about a microsecond at audio frequencies — too fast for heat to flow between the compressed and rarefied regions. So the compressions are adiabatic, and you use B = \gamma P:

v = \sqrt{\frac{\gamma P}{\rho}}.

Step 4. Plug numbers for air at T = 300 K, P = 1.013\times 10^5 Pa, \rho = 1.18 kg/m³, \gamma = 1.4:

v = \sqrt{\frac{1.4 \times 1.013\times 10^5}{1.18}} = \sqrt{\frac{1.418\times 10^5}{1.18}} = \sqrt{1.202\times 10^5}\,\text{m/s}
v \approx 347\,\text{m/s}.

Why: the arithmetic gives \sqrt{1.2\times 10^5} \approx 346.8, which rounds to 347 m/s. This matches the measured speed of sound in dry air at 300 K. Newton's isothermal prediction would give v = \sqrt{P/\rho} = 293 m/s — 15 per cent too low. The correction is the factor \sqrt{\gamma} = \sqrt{1.4} \approx 1.183, exactly the observed discrepancy.

Result: The speed of sound in air at 300 K is v = \sqrt{\gamma P/\rho} \approx 347 m/s — in excellent agreement with measurement. The adiabatic, not the isothermal, bulk modulus sets the number.

What this shows: sound speed is direct experimental evidence that fast compressions are adiabatic. The \gamma in the sound-speed formula is not a cosmetic factor; it was the 15-per-cent puzzle that stood for a century after Newton until Laplace identified adiabaticity as the missing ingredient. The same \gamma you use in the adiabatic equation shows up in the speed of a Diwali firecracker's pressure pulse, the shock wave off an ISRO rocket, and every car horn on the G.T. Road.

Common confusions

Going deeper

If you came here to understand the two idealised processes and apply them on exams, you have what you need. What follows fleshes out the polytropic generalisation, the reversibility argument, and the mechanical work done by a real gas (i.e. with the van der Waals correction).

Polytropic processes — one family, one index

A polytropic process obeys PV^n = const for some real index n. Special cases:

  • n = 0: P = const — isobaric.
  • n = 1: PV = const — isothermal.
  • n = \gamma: PV^\gamma = const — adiabatic.
  • n \to \infty: V = const — isochoric (squish the equation: V^{1/n}P^{1/n \cdot n} = const simplifies to V = const in the limit).

The work done in a general polytropic process (for n \ne 1) is

W_\text{poly} = \frac{P_1 V_1 - P_2 V_2}{n - 1},

derived by direct integration of P = C/V^n in exactly the same way as the adiabatic integral — just with n replacing \gamma. For n = 1, the formula diverges and the correct answer is the isothermal logarithm nRT \ln(V_2/V_1).

Real engine cylinders are modelled with a polytropic index n measured from the \log P vs \log V slope of real indicator cards. Typical values: n \approx 1.3 for the compression stroke of a petrol engine (so between isothermal and pure-air adiabatic, reflecting some heat loss). n \approx 1.35–1.4 for diesel compression (closer to adiabatic because less cooling time).

Reversibility — what "slow enough" means quantitatively

The isothermal derivation assumed the gas was in equilibrium with the reservoir at every instant. If you compress faster than the gas can conduct heat to its wall, local hot spots form — the gas is not at a single temperature during the process. It is doing something irreversible.

The scale that determines whether a process counts as isothermal is the ratio of the process time \tau_\text{proc} to the thermal equilibration time \tau_\text{therm}. \tau_\text{therm} is roughly the time for heat to diffuse across the gas: \tau_\text{therm} \sim L^2 / \alpha, where L is the gas's characteristic length and \alpha is the thermal diffusivity. For air at room temperature, \alpha \approx 2\times 10^{-5}\,\text{m}^2/\text{s}. For a 10 cm cylinder, \tau_\text{therm} \approx (0.1)^2 / 2\times 10^{-5} = 500 s — eight minutes for complete equilibration. So truly isothermal compression of a 10 cm cylinder requires strokes lasting several minutes. A 4 ms diesel stroke is a factor of \sim 10^5 faster than that — firmly in the adiabatic regime.

The symmetric criterion for adiabaticity is that \tau_\text{proc} \ll \tau_\text{therm}. The \sim sign and the factor of 5 or so in L and \alpha shift the criterion by an order of magnitude, but the logic is firm: real processes are isothermal when slow and adiabatic when fast, with a fuzzy polytropic middle ground when the two times are comparable.

Reversibility and the second law

Even a slow isothermal compression is reversible only if the temperature difference between the gas and the reservoir is infinitesimal throughout. A finite temperature difference drives heat flow irreversibly, generating entropy. The idealised isothermal discussed above is actually a quasi-static reversible isothermal; that distinction matters for the second law and for computing efficiencies.

An adiabatic process is reversible only if carried out quasi-statically (infinitely slowly, so that the gas is uniform throughout). A truly reversible adiabatic is sometimes called an isentropic process because the entropy of the system does not change. A fast adiabatic — a sudden expansion into vacuum, say — is adiabatic (no heat flow) but not reversible (entropy goes up). The PV^\gamma = const equation applies to reversible adiabats; irreversible adiabats (free expansion) do not obey it.

A real gas on an adiabat — van der Waals correction

For a real gas obeying (P + a/V^2)(V - b) = RT, the adiabatic equation is modified. The derivation follows the same first-law route, but dU now has a term a/V^2 \, dV coming from the volume dependence of internal energy (a real gas has U = U(T, V), not just U(T)). The resulting adiabatic equation (for one mole) is

\left(T - \frac{a}{V^2 C_v}\right) (V - b)^{\gamma - 1} = \text{const},

an obvious generalisation of T V^{\gamma - 1} = const that reduces to the ideal form when a \to 0 and b \to 0. For most gases at moderate conditions, the correction is small, and the ideal-gas adiabat is used even though the gas is technically van der Waals. Near the critical point the correction becomes important.

Isothermal and adiabatic efficiency bounds

A heat engine that operates in a cycle including two isothermal strokes and two adiabatic strokes is a Carnot engine, and its efficiency is the maximum allowed by the second law for a given pair of reservoir temperatures. The isothermal is where the engine exchanges heat with the reservoirs; the adiabatic is where the gas's temperature is changed without heat exchange. The specific combination is no accident: any other process would either waste heat to a wrong-temperature reservoir (irreversibility from finite \Delta T heat flow) or do less work (area-under-curve sub-optimal). The heat-engine article will build this out fully — but it all rests on the two processes derived above.

Where this leads next