Visual walkthrough — First law of thermodynamics — dU = dQ − dW, sign conventions

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Step 1 — Draw the system and give every arrow a name

• The yellow arrow pointing in is heat. We give it the name $Q$ (from the Latin quantitas caloris). It is energy that crosses the wall because outside is hotter than inside.
• The red arrow on the piston is work. We call it $W$. It is energy the gas spends by pushing the piston out.
• The blue jiggling dots are the molecules. The total energy of all that jiggling is the internal energy, named $U$.

$$\underset{\text{heat in, yellow}}{\underbrace{Q}}\underset{\text{work out, red}}{\underbrace{W}}\underset{\text{energy stored inside, blue}}{\underbrace{U}}$$

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Step 2 — Energy is never lost: write the ledger

Picture a tank of water (the figure). Water pours in from a tap ($Q$). Water is stored, raising the level ($\Delta U$). Water flows out through a spout ($W$). The iron rule of a tank:

$$\underset{\text{poured in}}{\underbrace{Q}}=\underset{\text{level rise}}{\underbrace{\Delta U}}+\underset{\text{poured out}}{\underbrace{W}}$$

• $Q$ — everything that entered (the tap).
• $\Delta U$ — the change in what is stored. The little triangle $\Delta$ means "change in" — final minus initial.
• $W$ — everything that left (the spout).

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Step 3 — Rearrange into the famous form

Starting from the ledger $Q=\Delta U+W$, subtract $W$ from both sides:

$$\boxed{\Delta U=\underset{\text{heat added}}{\underbrace{Q}}-\underset{\text{work done BY gas}}{\underbrace{W}}}$$

For a tiny sliver of the process we write the same law with tiny changes:

$$dU=dQ-dW$$

• $d$ in front means "an infinitesimally small amount of." Instead of the whole journey, this is one microscopic step of it.

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Step 4 — Where does the work actually come from? $dW=PdV$

Look at the piston close-up. Let the piston have area $A$. The gas presses on it with pressure $P$ (force per unit area). So the total push is:

$$\underset{\text{force}}{\underbrace{F}}=\underset{\text{pressure}}{\underbrace{P}}\times \underset{\text{piston area}}{\underbrace{A}}$$

Now let the piston slide out a tiny distance $dx$ (the red arrow). Work is force times distance:

$$dW=Fdx=PAdx$$

But watch the shaded green slab the piston sweeps out — its volume is area times thickness:

$$\underset{\text{swept slab}}{\underbrace{Adx}}=dV\text{(tiny volume change)}$$

Substitute and the geometry disappears, leaving pure thermodynamics:

$$\boxed{dW=PdV}⟹W={\int }_{V_1}^{V_2}PdV$$

• $P$ — pressure at that instant.
• $dV$ — the sliver of volume added in that instant.
• The $\int$ (integral) is a running total — it adds up every sliver $PdV$ from the start volume $V_1$ to the end volume $V_2$. We need it because $P$ can change as the gas expands, so we cannot just multiply once.

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Step 5 — The sign of $W$ from the sign of $dV$

Two panels in the figure:

• Expansion (left): the piston moves out, volume grows, so $dV>0$. Then $dW=PdV>0$. The gas does work on the surroundings — energy leaves. In $\Delta U=Q-W$ this subtracts from stored energy.
• Compression (right): the piston is pushed in, volume shrinks, so $dV<0$. Then $dW=PdV<0$. The surroundings do work on the gas — energy enters. In $\Delta U=Q-W$, subtracting a negative $W$ adds energy.

$$dV>0\Rightarrow W>0\text{ (gas pushes out)}dV<0\Rightarrow W<0\text{ (gas gets squeezed)}$$

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Step 6 — The four special processes: kill one term at a time

Each panel is a $P$–$V$ path (volume across, pressure up):

Process	What's held fixed	Term that dies	Law becomes
Isochoric	volume $V$	$dV=0\Rightarrow W=0$	$\Delta U=Q$
Isobaric	pressure $P$	none	$\Delta U=Q-P\Delta V$
Isothermal	temperature $T$	$\Delta U=0$ (ideal gas)	$Q=W$
Adiabatic	no heat crosses	$Q=0$	$\Delta U=-W$

• Isochoric (vertical line): the piston is bolted. It cannot move, so no volume sweeps out, $W=0$. All heat becomes stored energy — the gas just gets hotter.
• Isobaric (horizontal line): pressure constant, so $W=P\Delta V$ (the integral of a constant is just $P$ times the width).
• Isothermal (gentle curve): for an ideal gas, $U$ depends only on temperature $T$ (see Internal energy and degrees of freedom). Hold $T$ fixed and $U$ cannot change: $\Delta U=0$, so every joule of heat leaves as work.
• Adiabatic (steep curve): walls insulated, $Q=0$. Now work and internal energy trade directly — compress it and it heats up, with nowhere for the heat to escape.

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Step 7 — The degenerate case: free expansion into vacuum

Picture a box split by a wall: gas on the left, vacuum on the right (the figure). Puncture the wall. The gas rushes to fill the whole box — volume doubles.

• No piston, no push: the gas expands into empty space, so there is no external pressure to push against. Work done $=0$, even though volume changed. (In $dW=PdV$, the relevant opposing pressure is zero.)
• Insulated walls: no heat crosses, $Q=0$.

Feed both into the law:

$$\Delta U=Q-W=0-0=0$$

So $U$ is unchanged — and for an ideal gas that means temperature does not change at all, even though the gas expanded! This is the counter-intuitive gem that proves $\Delta U$ tracks state, not the drama of the volume changing.

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The one-picture summary

Everything on one canvas: heat $Q$ pours in (yellow), work $W$ leaves out the piston (red), and whatever is left raises the internal energy $U$ (blue). The ledger balances exactly:

$$\underset{\text{stayed}}{\underbrace{\Delta U}}=\underset{\text{came in}}{\underbrace{Q}}-\underset{\text{went out}}{\underbrace{W}},W=\int PdV$$

Recall Feynman retelling — the whole walkthrough in plain words

A box of gas has three doors: heat in, work out, and energy stored inside. Nothing can appear from nowhere (Step 1–2), so what comes in = what stays + what goes out — that's the whole law, just a tank of water with a tap and a spout (Step 3). The "work out" is the gas shoving the piston: force is pressure times area, distance is how far it slides, and multiplying gives $PdV$ — the volume it sweeps (Step 4). Whether that work is plus or minus is just whether the volume grew or shrank (Step 5). To feel the law, kill one term: bolt the piston (no work), or insulate the walls (no heat), or hold the temperature so the stored energy can't change (Step 6). And the trick that proves it all: let the gas burst into vacuum — volume doubles but it pushes nothing and loses no heat, so it doesn't even change temperature (Step 7). What you eat equals what you store plus what you spend.

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Connections

• Parent topic
• Conservation of energy (mechanics)
• Work done in thermodynamic processes (PV diagrams)
• Isothermal, adiabatic, isobaric, isochoric processes
• Internal energy and degrees of freedom
• Heat capacities Cp and Cv
• Second law of thermodynamics