IB Physics HLThermodynamicsPaper 1 & 2The Four Processes~12 min read
Thermodynamic Processes
A gas can change in endless ways, but four clean changes cover almost everything you’ll meet in the exam. Each one holds a single quantity fixed — and their names all start with “iso“, Greek for “same”. The trick is lovely and simple: once you know what stays constant, the first law (Q = ΔU + W) hands you everything else. (Throughout, the gas is assumed ideal.)
📚 What you need to know
The four processes: isovolumetric (ΔV = 0), isobaric (Δp = 0), isothermal (ΔT = 0), adiabatic (ΔQ = 0)
Isovolumetric: no work, W = 0, so Q = ΔU
Isobaric:W = pΔV, so Q = ΔU + pΔV
Isothermal: ΔU = 0, so Q = W
Adiabatic:Q = 0, so W = −ΔU
On a p–V graph: isobaric is horizontal, isovolumetric is vertical, isothermal is a curve, adiabatic is a steeper curve
For a monatomic ideal gas, an adiabatic change obeys pV5/3 = constant
Four processes, one rule each
Every one of these keeps a single quantity constant. Spot which one, set the matching term in the first law to zero, and read off the rest:
Process
Held constant
First law gives
Isovolumetric
volume (ΔV = 0)
Q = ΔU
Isobaric
pressure (Δp = 0)
Q = ΔU + pΔV
Isothermal
temperature (ΔT = 0)
Q = W
Adiabatic
no heat (ΔQ = 0)
W = −ΔU
Here’s the story behind each one:
Isovolumetric (constant volume): the gas is trapped in a rigid container, so it can’t expand and does no work. Every joule of heat goes straight into internal energy — the gas just gets hotter.
Isobaric (constant pressure): the gas is free to expand while the pressure holds steady, so it does work W = pΔV. The heat you add splits between raising internal energy and doing that work.
Isothermal (constant temperature): kept at a steady temperature (in slow contact with a big reservoir), the internal energy can’t change, so ΔU = 0. All the heat added comes straight back out as work.
Adiabatic (no heat flow): no heat gets in or out (the change is fast, or the gas is insulated). Any work the gas does must come entirely from its own internal energy — so its temperature changes.
WE 1
Heat is supplied to three ideal gases. Gas P is kept at constant volume, gas Q at constant temperature, and gas R at constant pressure. For each, state whether the work done W, the change in internal energy ΔU, and the temperature change ΔT are positive, zero, or negative.
P (constant volume): no work → W = 0. Then Q = ΔU so ΔU > 0, and ΔT > 0Q (constant temperature): ΔT = 0, so ΔU = 0. Then Q = W so W > 0R (constant pressure): gas expands so W > 0; heat also raises ΔU > 0, so ΔT > 0Same starting move every time: pick what’s constant, zero that term, then use Q = ΔU + W.
Seeing them on a p–V graph
Each process has its own signature shape on a pressure–volume graph — and since the work done is the area under the line, the shape tells you the work at a glance.
Isobaric runs horizontal (constant pressure); isovolumetric runs vertical (constant volume, no area, so no work); isothermal and adiabatic are curves that fall as the gas expands.
Isothermal vs adiabatic: the confusing pair
These two look alike — both are downward curves — and students mix them up constantly. The key difference is steepness: the adiabatic curve always drops faster.
Starting from the same point A, the adiabatic curve falls below the isothermal one. In an isothermal expansion the temperature is held up by incoming heat; in an adiabatic expansion no heat comes in, so the gas also cools, and its pressure drops faster.
Think of it this way: in an isothermal expansion, heat keeps flowing in to keep the temperature up, propping the pressure. In an adiabatic expansion there’s no such help — the gas pays for its own work out of its internal energy, cools down, and so its pressure sinks more steeply. Steeper curve = the adiabatic one.
The adiabatic equation (monatomic gas)
Adiabatic changes are special enough to get their own equation. For a monatomic ideal gas (helium, neon, argon), pressure and volume are tied together by:
Combine this with the ideal gas law pV = nRT and the pressures drop out, leaving a handy version in terms of temperature and volume:
Temperature formT1V12/3 = T2V22/3
WE 2
A monatomic ideal gas is compressed adiabatically from a pressure of 1.0 × 105 Pa and volume 6.0 × 10−3 m3 to a volume of 2.0 × 10−3 m3. Find the new pressure.
Step 1 — rearrange p₁V₁5/3 = p₂V₂5/3 for p₂p₂ = p₁ × (V₁/V₂)5/3p₂ = (1.0×10⁵) × (6.0/2.0)5/3 = (1.0×10⁵) × 35/3p₂ ≈ 6.2 × 10⁵ PaCompression squeezes the gas, so the pressure jumps up — sensible.
WE 3
A monatomic ideal gas at 600 K expands adiabatically from 2.0 × 10−3 m3 to 5.0 × 10−3 m3. Find the new temperature.
Step 1 — use T₁V₁2/3 = T₂V₂2/3, so T₂ = T₁(V₁/V₂)2/3T₂ = 600 × (2.0/5.0)2/3 = 600 × (0.4)2/3T₂ ≈ 326 KExpanding with no heat in, the gas cools from 600 K to about 326 K — exactly why adiabatic curves are steeper.
One caution: the exponent 5⁄3 is for monatomic gases, whose atoms only move in straight lines. Gases made of bigger molecules can also spin, so their exponent is different — but for IB, the monatomic case is the one you need.
Entropy in each process
Because ΔS = ΔQ/T, entropy simply follows the heat. Any process that takes heat in raises the gas’s entropy; any that lets heat out lowers it. The special case is adiabatic: no heat flows, so ΔQ = 0 and the entropy doesn’t change at all.
Process
Heat ΔQ
Entropy ΔS
Isothermal / isobaric / isovolumetric — heat in
> 0
increases
… the same, heat out
< 0
decreases
Adiabatic (either direction)
= 0
no change
Tuck this away for later: a reversible adiabatic change keeps entropy exactly constant. That “no entropy change” step is one of the ingredients of the perfect engine cycle — the Carnot cycle — you’ll meet at the end of the topic.
Always start from Q = ΔU + W and zero one term: ΔV = 0 ⇒ W = 0; ΔT = 0 ⇒ ΔU = 0; ΔQ = 0 ⇒ Q = 0.
On a p–V graph: horizontal = isobaric, vertical = isovolumetric, and the adiabatic curve is steeper than the isothermal.
Adiabatic maths (monatomic):pV5/3 = constant; for temperatures use TV2/3 = constant.
Adiabatic ⇒ ΔS = 0 — no heat, no entropy change. Remember this for Carnot.
⚠ Common mistakes
Muddling isovolumetric (W = 0) with isothermal (ΔU = 0) — a different term vanishes in each
Thinking adiabatic means constant temperature — it’s the opposite: with no heat to hold it steady, the temperature changes
Drawing the adiabatic curve the same as the isothermal — the adiabatic is steeper
Using pV5/3 for a change that isn’t adiabatic, or for a non-monatomic gas
Forgetting kelvin in TV2/3 = constant
Quick recap: Four “iso” processes, each fixing one quantity: isovolumetric (W = 0, Q = ΔU), isobaric (W = pΔV), isothermal (ΔU = 0, Q = W), adiabatic (Q = 0, W = −ΔU). On a p–V graph they’re horizontal, vertical, and two curves — the adiabatic steeper. A monatomic adiabatic obeys pV5/3 = constant, and adiabatic changes keep entropy constant.
That’s the whole toolkit of gas changes. Real machines don’t use them one at a time — they string several into a repeating loop to turn heat into work over and over. Next, in Heat Engines, we join these processes into a cycle and work out exactly how much useful work you can get.
Want this to actually click before the exam?
Book a free meeting and let’s work through the tricky bits together.
