Adiabatic Process on a PV Diagram

An adiabatic process is one where no heat is exchanged with the surroundings. $Q = 0$. The gas expands or compresses, but it's perfectly insulated — any change in internal energy has to come from work alone.

Adiabatic processes are the mirror image of isothermal processes in a certain sense. Isothermal lets heat flow freely to keep $T$ constant; adiabatic blocks heat completely, so $T$ changes as the gas does work (expansion cools it, compression heats it).

You feel adiabatic processes all the time without realizing it:

• A bicycle pump heats up when you rapidly inflate a tire. That's adiabatic compression — you're doing work on the air, it has nowhere to dump the extra energy, so the internal energy rises, and internal energy of an ideal gas means temperature.
• A spray can feels cold when you spray it, because the propellant inside is adiabatically expanding as it exits. Expansion cools it.
• A rising warm air mass in the atmosphere cools as it expands into lower pressure aloft — and if it cools enough for water vapor to condense, you get clouds. This is the whole basis of cloud formation.

The key equation

For an ideal gas undergoing an adiabatic process:

$P V^{\gamma} = \text{constant}$

Here $\gamma$ (gamma, the "adiabatic index") is the ratio of specific heats:

$\gamma = \dfrac{C_{P}}{C_{V}}$

Its value depends on the number of degrees of freedom of the gas molecules:

• Monoatomic gases (He, Ne, Ar) — particles can only move in 3 directions, no rotation, no vibration. $\gamma = 5/3 \approx 1.667$.
• Diatomic gases (H₂, N₂, O₂, air) — particles can move AND rotate about 2 axes (ignoring vibrations at normal temperatures). $\gamma = 7/5 = 1.4$.
• Polyatomic gases (CO₂, H₂O vapor, CH₄) — even more degrees of freedom. $\gamma$ closer to $9/7 \approx 1.286$ or lower.

Air is mostly N₂ and O₂, so air is well-approximated as diatomic with $\gamma = 1.4$.

Why is the adiabatic curve steeper than the isothermal?

On a PV diagram, an adiabatic curve through a given point is always steeper than the isothermal curve through the same point. Here's why.

For an isothermal process: $PV = \text{const}$, so $P \propto 1/V$. Slope in log-space: $-1$.

For an adiabatic process: $PV^{\gamma} = \text{const}$, so $P \propto 1/V^{\gamma}$. Slope in log-space: $-\gamma$.

Since $\gamma > 1$, the adiabatic slope is steeper (more negative) than the isothermal slope. Visually, the adiabatic curve falls faster as $V$ increases.

Geometric intuition: when you expand a gas isothermally, heat flows in to keep $T$ (and therefore the "height" of the isotherm) from dropping too fast. With no heat flowing in, the gas has to pay for the expansion work out of its own internal energy, so it cools faster. Lower $T$ means lower $PV$ product at any given $V$, so the adiabatic curve dives below the isothermal curve as volume grows.

Relating temperature to volume

For an adiabatic process, you can derive a clean relationship between $T$ and $V$:

$T V^{\gamma - 1} = \text{constant}$

So if you expand a gas adiabatically from $V_{1}$ to $V_{2}$:

$\dfrac{T_{2}}{T_{1}} = \left(\dfrac{V_{1}}{V_{2}}\right)^{\gamma - 1}$

For air ($\gamma = 1.4$, so $\gamma - 1 = 0.4$), doubling the volume drops the temperature by a factor of $(1/2)^{0.4} \approx 0.758$. Starting from 300 K, you'd end at 227 K — a drop of 73 K just from doubling the volume.

Work done during an adiabatic process

Since no heat flows ($Q = 0$), the first law $\Delta U = Q - W$ reduces to:

$\Delta U = -W$

Work done BY the gas equals the drop in internal energy. Integrating:

$W_{\text{adi}} = \dfrac{P_{1} V_{1} - P_{2} V_{2}}{\gamma - 1}$

For the default problem ($P_{1} = 100\;\text{kPa}$, $V_{1} = 30\;\text{L}$, expanding to $V_{2} = 60\;\text{L}$, $\gamma = 1.4$):

Step 1: Find $P_{2}$ from $PV^{\gamma}$ = const.

$P_{2} = P_{1}\left(\dfrac{V_{1}}{V_{2}}\right)^{\gamma} = 100{,}000 \times (30/60)^{1.4}$

$= 100{,}000 \times 0.3789$

$\approx 37{,}892\;\text{Pa} \approx 37.9\;\text{kPa}$

Compare to isothermal ($P_{1} V_{1} = P_{2} V_{2}$): $P_{2,\text{iso}} = 50\;\text{kPa}$. The adiabatic pressure at the same volume is lower — the gas has cooled.

Step 2: Find $T_{2}$.

$T_{2} = T_{1} \left(\dfrac{V_{1}}{V_{2}}\right)^{\gamma - 1} = 300 \times (1/2)^{0.4} \approx 300 \times 0.7579 \approx 227.4\;\text{K}$

The gas cooled by about 73 K purely from doing work against the piston. No heat was extracted — the energy came out of the gas's own internal thermal store.

Step 3: Compute the work.

$W = \dfrac{P_{1}V_{1} - P_{2}V_{2}}{\gamma - 1}$

$P_{1}V_{1} = 100{,}000 \times 0.030 = 3000\;\text{J}$

$P_{2}V_{2} = 37{,}892 \times 0.060 = 2273.5\;\text{J}$

$W = \dfrac{3000 - 2273.5}{0.4} = \dfrac{726.5}{0.4} \approx 1816.3\;\text{J}$

Compare to isothermal work for the same expansion: $W_{\text{iso}} = n R T \ln 2$. For 1 mol at 300 K: $W_{\text{iso}} \approx 1729\;\text{J}$.

Wait — the adiabatic work is larger than the isothermal work? That's because starting from the same initial state, the adiabatic expansion ends at a different (lower) pressure at $V_{2}$. The comparison is tricky because "same expansion" doesn't uniquely specify a process.

A cleaner comparison: if both start at the same initial state AND end at the same final volume, the isothermal process does more work (the isothermal curve sits above the adiabatic for any volume larger than $V_{1}$). The adiabatic does less work because it has less "pressure" to push with after it's cooled down. The area under the adiabatic curve is smaller.

Look at the visualization — toggle Show isothermal comparison to see both curves overlaid starting from the same point, and you'll see the adiabatic curve diving below the isothermal.

Why adiabatic expansion cools and compression heats

The first law for an adiabatic process says $\Delta U = -W$. In an expansion, $W > 0$ (gas does positive work on surroundings), so $\Delta U < 0$ — internal energy decreases, meaning temperature drops.

In a compression, the environment does work on the gas — so $W < 0$ in our convention — and $\Delta U > 0$. Internal energy increases, meaning the gas gets hotter.

This is why a bicycle pump gets warm during use: you're compressing the air inside, the process is fast enough that heat can't escape through the walls in time (nearly adiabatic), and the work you do goes straight into heating the gas. Touch the pump cylinder right after pumping — it's noticeably warm.

And this is why spray cans feel cold: the liquid propellant adiabatically expands as it exits the nozzle. With no time for heat to flow in from the surroundings, the fluid cools dramatically. If you spray long enough, frost can form on the can's outside.

Adiabatic in the real world

• Internal combustion engines. The compression stroke in a gasoline engine is nearly adiabatic. Air fuel-mixture is squeezed to about 1/10th its original volume in a fraction of a second — too fast for heat to escape. The temperature jumps to ~700 K, and the spark plug ignites the hot mixture. Diesel engines push compression even further (~1/20) and the air gets hot enough to ignite the fuel without a spark at all.
• Atmospheric physics. Rising air cools adiabatically at about 10°C per kilometer of altitude (the "dry adiabatic lapse rate"). Once the rising air cools enough for water vapor to condense, clouds form. The condensation releases latent heat, and the moist air cools more slowly — about 6°C/km — which is why thunderstorm updrafts can rise very high.
• Refrigeration. Refrigerators and air conditioners use the adiabatic expansion of a compressed refrigerant to produce cold temperatures. The refrigerant is compressed (gets hot), then loses heat to the outside through condenser coils, then expanded adiabatically through a valve — the sudden expansion drops the temperature dramatically, and the cold refrigerant cools the inside of the fridge.
• Gas turbines / jet engines. Hot, high-pressure combustion gases expand adiabatically through the turbine blades, doing work to spin them. The expansion cools the gas, and the energy extracted powers the compressor (and the fan, if there is one).

Try it

• Drag $V_{2}$ to watch both the adiabatic and isothermal curves update. Notice how the adiabatic curve drops faster.
• Toggle between $\gamma = 5/3$ (monoatomic, noble gases like helium) and $\gamma = 7/5$ (diatomic, air). The 5/3 curve is steeper — monoatomic gases cool faster on expansion.
• Watch the temperature readout. For an isothermal expansion it stays locked at 300 K. For an adiabatic expansion it drops visibly.
• Enable "Show isothermal overlay" to see both curves on the same axes starting from the same point. They cross at $V = V_{1}$ (both pass through the starting state) and diverge as you move away.
• Try adiabatic compression (set $V_{2} < V_{1}$). The adiabatic curve rises above the isothermal — the gas is heating up faster than it would in an isothermal compression.