IB Physics HLThermodynamicsPaper 1 & 2Internal Energy & Work~11 min read
Thermodynamic Systems
A “thermodynamic system” is just a chunk of gas we’ve decided to keep an eye on. There are only two ways to change its energy: heat it, or push on it (do work). This page sets up both ideas — the energy stored inside a gas, and the work a gas does when it expands. Get these two straight and the whole topic clicks into place.
📚 What you need to know
The internal energyU of a gas is the total energy of all its molecules added together
An ideal gas has no forces between molecules, so no potential energy — its internal energy is all kinetic
Change in internal energy depends only on temperature: ΔU ∝ ΔT
ΔU = ³⁄₂NkBΔT = ³⁄₂nRΔT (temperature in kelvin)
When a gas changes volume at constant pressure it does work: W = pΔV
On a pressure–volume graph, the work done is the area under the line
Expansion → work done by the gas (+); compression → work done on the gas (−)
Internal energy: the energy stored inside a gas
Every molecule in a gas is on the move, so it has kinetic energy. Add up the energy of all the molecules and you get the gas’s internal energy, U. In a solid or liquid the molecules are also held together by bonds, so they store energy in two ways: kinetic (they still jiggle) and potential (in the bonds).
An ideal gas is simpler. We assume its molecules feel no forces between one another. No forces means no bonds — and no bonds means no potential energy. So for an ideal gas, internal energy is purely kinetic.
A solid or liquid stores both kinetic and potential energy. An ideal gas has no bonds, so all of its internal energy is kinetic — pure molecular motion.
This is why, for an ideal gas, “internal energy” and “total kinetic energy” mean exactly the same thing. If an exam asks for the total internal energy of an ideal gas, you can safely say it’s just the total kinetic energy of the molecules.
Here’s the payoff. Kinetic energy depends on how fast the molecules move, and how fast they move depends on temperature. So the internal energy of an ideal gas depends on nothing but its temperature:
Internal energy tracks temperature
ΔU ∝ ΔT
Put a number on it with either of these two forms — they’re the same equation counted two different ways:
Change in internal energy of an ideal gasΔU = ³⁄₂NkBΔT = ³⁄₂nRΔT
Use the first form when you know the number of moleculesN (with the Boltzmann constant kB), and the second when you know the number of molesn (with the gas constant R). They match because NkB = nR.
The number-one trap in this whole topic: temperature must be in kelvin. Always add 273 to a Celsius value before you put it into any of these equations.
WE 1
An ideal gas is heated so its temperature rises from 27°C to 327°C. Show that its internal energy exactly doubles, and explain the trap a careless student falls into.
Step 1 — turn both temperatures into kelvin27°C → 27 + 273 = 300 K327°C → 327 + 273 = 600 KStep 2 — compare, using U ∝ T (kelvin)600 ÷ 300 = 2.0Internal energy doublesThe trap: in °C it looks like 327 ÷ 27 ≈ 12×. That’s wrong — only the kelvin ratio counts.
WE 2
3.0 mol of an ideal gas is warmed from 300 K to 400 K. Find the increase in its internal energy. (R = 8.31 J mol−1 K−1.)
Step 1 — use ΔU = ³⁄₂ nRΔT, with ΔT = 400 − 300 = 100 KΔU = 1.5 × 3.0 × 8.31 × 100ΔU ≈ 3.7 × 10³ JAlready in kelvin, so no conversion needed here — but always check.
Work done by a gas
Now the second way to change a gas’s energy. When a gas expands, it pushes the walls of its container outward. Pushing something and making it move is the very definition of doing work. So an expanding gas does work on its surroundings.
Let’s build the formula from a gas trapped behind a piston. The gas is at pressure p, and the piston has area A. Pressure is force per area, so the gas pushes the piston with a force F = pA. If the piston slides out a distance s, the work done is force × distance:
The gas pushes the piston (area A) with force F = pA. The piston moves a distance s, sweeping out a volume A×s = ΔV. So work = pA×s = pΔV.
The clever last step: the distance s multiplied by the area A is just the extra volume the gas gained, ΔV. Swap As for ΔV and the piston disappears from the formula entirely:
Work done at constant pressureW = pΔV
where W is the work done (J), p is the pressure the gas pushes against (Pa), and ΔV is the change in volume (m3). This assumes the pressure stays the same while the gas expands.
WE 3
A gas expands at a constant pressure of 2.0 × 105 Pa. Its volume grows from 1.0 × 10−3 m3 to 4.0 × 10−3 m3. How much work does the gas do?
Step 1 — find the change in volumeΔV = (4.0 − 1.0) × 10⁻³ = 3.0 × 10⁻³ m³Step 2 — use W = pΔVW = (2.0×10⁵) × (3.0×10⁻³)W = 600 JThe volume grew, so this work is done by the gas on its surroundings.
Reading work off a p–V graph
There’s a lovely shortcut hiding in that formula. On a graph of pressure against volume, the work done is simply the area under the line. If the pressure is constant, that area is a rectangle (p × ΔV). If the pressure changes as the gas expands, the area is a trapezium — but it’s still just “the area underneath”.
The shaded area under the line from A to B equals the work done by the gas. For a sloped line, that area is a trapezium: average pressure × change in volume.
The direction tells you the sign. Whether work comes out positive or negative depends only on whether the gas got bigger or smaller:
What the gas does
Volume
Work
Expands
increases, +ΔV
+W (by gas)
Is compressed
decreases, −ΔV
−W (on gas)
WE 4
A gas is compressed. On its p–V graph the pressure rises from 100 kPa to 300 kPa as the volume falls from 0.050 m3 to 0.010 m3. Find the work done.
Step 1 — the area is a trapezium: average pressure × volume change|ΔV| = 0.050 − 0.010 = 0.040 m³area = ½ × (100 + 300)×10³ × 0.040area = ½ × 400 000 × 0.040 = 8000 JStep 2 — put in the signW = −8.0 × 10³ JThe gas was squashed, so work is done on it — that’s why W is negative.
F = pA
× distance s
W = pAs
A×s = ΔV
W = pΔV
💡 Top tips
Kelvin, always. Convert every temperature to kelvin before using any ΔU equation.
Two forms, same equation. Use ³⁄₂NkBΔT if you’re given molecules, ³⁄₂nRΔT if you’re given moles.
Ideal gas = all kinetic. No forces means no PE, so internal energy is total kinetic energy.
Work = area under the p–V line. Rectangle if pressure is constant, trapezium if it changes.
Mind the sign: expansion is +W (by the gas), compression is −W (on the gas).
⚠ Common mistakes
Leaving temperature in °C instead of kelvin
Thinking a “×3 rise in °C” triples the internal energy — only the kelvin ratio matters
Mixing up N (molecules) and n (moles) in the two ΔU forms
Forgetting that work is only done when the volume changes — no volume change means W = 0
Giving a compression a positive work value — it should be negative (work done on the gas)
Quick recap: Internal energy U is the total energy of a gas’s molecules. An ideal gas has no bonds, so it’s all kinetic and depends only on temperature: ΔU = ³⁄₂NkBΔT = ³⁄₂nRΔT (in kelvin). A gas that changes volume at constant pressure does work W = pΔV, which is the area under a p–V line — positive when it expands, negative when it’s compressed.
Nicely done — you now have the two levers that change a gas’s energy: heating it and doing work on it. Next we tie them together with the First Law of Thermodynamics, Q = ΔU + W — the simple bookkeeping rule that makes sure energy is never lost, only moved around.
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