IB Physics HLThermodynamicsPaper 1 & 2First Law~10 min read
First Law of Thermodynamics
Last time we found the two ways to change a gas’s energy: heat it, or do work on it. The first law simply keeps the books. It says energy is never lost — so if you add up the heat and the work, you know exactly how the gas’s internal energy changed. It’s the accountant’s rule of thermodynamics.
📚 What you need to know
The first law is just conservation of energy applied to a gas
Q = ΔU + W — heat in equals the rise in internal energy plus the work the gas does
Q = heat added to the gas, ΔU = change in internal energy, W = work done by the gas
Usually handier rearranged: ΔU = Q − W
Signs: heat in → +Q, heat out → −Q; gas expands → +W, gas compressed → −W
Constant volume: no work, so W = 0 and Q = ΔU
Constant pressure:W = pΔV
The law is just energy bookkeeping
Pour some heat into a gas. That energy can’t vanish, so it has to go somewhere. It splits between two jobs: some of it stays inside the gas as extra internal energy (the molecules speed up), and some of it leaves as work when the gas expands and pushes on its surroundings. Add those two together and you get back exactly the heat you put in:
The first law of thermodynamicsQ = ΔU + W
where Q is the heat added to the gas (J), ΔU is the change in its internal energy (J), and W is the work done by the gas (J).
The heat you add doesn’t disappear. Part of it raises the internal energy (ΔU), and part of it leaves as work done by the gas (W). Together they equal Q.
Think of it like a bank account. Heat is money coming in; internal energy is your balance; work is money spent. Whatever comes in is either saved or spent — nothing goes missing. That’s all the first law says.
Most exam questions ask for the change in internal energy, so it’s worth rearranging the law into the form you’ll reach for again and again:
The form you’ll use most
ΔU = Q − W
Getting the signs right
Here’s the only genuinely tricky part. Each of the three quantities can be positive or negative, and one wrong sign wrecks the answer. Keep two simple rules in your head: Q is positive when heat goes IN, and W is positive when the gas pushes OUT (expands).
Heat into the gas is +Q; heat out is −Q. The gas expanding is +W (work by the gas); the gas being squashed is −W (work done on it).
Quantity
Positive (+)
Negative (−)
Q (heat)
heat added to gas
heat removed
W (work)
gas expands (work by gas)
gas compressed (work on gas)
ΔU (internal energy)
gets hotter
gets cooler
The classic slip: forgetting that W is the work done by the gas. If the gas is compressed, work is done on it, so W is negative. Put that into ΔU = Q − W and the double negative pushes the internal energy up — which makes sense, because squashing a gas warms it.
WE 1
A gas absorbs 800 J of heat and does 300 J of work as it expands. Find the change in its internal energy.
Step 1 — sort the signs: heat in → Q = +800 J; gas expands → W = +300 JStep 2 — use ΔU = Q − WΔU = 800 − 300ΔU = +500 JInternal energy rises by 500 J — the other 300 J left as work.
WE 2
A gas is compressed: 250 J of work is done on it, and it loses 100 J of heat to the surroundings. Find the change in its internal energy.
Step 1 — the signs. Heat lost → Q = −100 J. Work done ON gas → W = −250 JStep 2 — use ΔU = Q − WΔU = (−100) − (−250)ΔU = −100 + 250ΔU = +150 JEven though heat left, squashing the gas did enough work on it to warm it up.
Two easy cases: constant volume and constant pressure
Two situations come up so often they’re worth memorising, and both come straight from “work is the area under a p–V line”.
At constant volume the line is vertical, so there’s no area and no work: every joule of heat becomes internal energy. At constant pressure the line is horizontal, and the rectangle underneath is the work, W = pΔV.
Constant volume. If the gas can’t change size, it can’t push anything, so it does no work: W = 0. The first law collapses to a lovely simple result — all the heat goes straight into internal energy:
Constant volumeW = 0 ⇒ Q = ΔU
Constant pressure. Here the gas can expand, doing work W = pΔV (from the last page). So the heat has to cover both the rise in internal energy and that work:
Constant pressureQ = ΔU + pΔV
WE 3
A gas is heated at a constant pressure of 1.0 × 105 Pa. It absorbs 500 J of heat and expands by 2.0 × 10−3 m3. Find the change in its internal energy.
Step 1 — find the work done by the gas, W = pΔVW = (1.0×10⁵) × (2.0×10⁻³) = 200 JStep 2 — use ΔU = Q − W (Q = +500 J, W = +200 J)ΔU = 500 − 200ΔU = 300 JOf the 500 J of heat, 200 J left as work and 300 J stayed as internal energy.
Heat in Q
− work done by gas W
Rise in internal energy ΔU
💡 Top tips
Learn it as ΔU = Q − W — that’s the form most questions want.
Two sign rules: +Q = heat IN, +W = gas expands (work OUT). Everything else follows.
Constant volume ⇒ W = 0 ⇒ Q = ΔU — all the heat becomes internal energy.
Constant pressure ⇒ W = pΔV — then feed it into the first law.
Compression? Work is done on the gas, so W is negative.
⚠ Common mistakes
Forgetting that W is work done by the gas — so a compression gives a negativeW
Writing ΔU = Q + W — in this convention it’s Q−W
Assuming heat always warms a gas — if it does enough work while expanding, its temperature can even fall
Forgetting that W = 0 at constant volume (no volume change, no work)
Mixing up which reservoir is which — keep signs tied to the gas, not the surroundings
Quick recap: The first law is conservation of energy for a gas: Q = ΔU + W, or ΔU = Q − W. Heat in is +Q; work done by an expanding gas is +W; a squashed gas gives −W. At constant volume W = 0 so Q = ΔU; at constant pressure W = pΔV.
Brilliant — you can now follow energy around any change a gas goes through. But there’s a catch the first law can’t explain: it happily allows a cold drink to warm your hand or your hand to warm the drink, yet only one of those ever happens on its own. The direction that changes naturally run is set by a brand-new idea — entropy — which we meet next.
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