First law - ΔU = q + w (chemist sign convention): Visual walkthrough

Step 1 — Draw the box and name what is inside it

WHAT. Picture a sealed box of gas. We put an imaginary bag — a boundary — around exactly the stuff we care about. Everything inside the bag we call the system. Everything outside (the room, the flame, the water bath) we call the surroundings.

WHY. Before we can talk about energy changing, we have to say whose energy. Drawing the boundary is not a formality — it decides the sign of every number later. "In" and "out" only mean something once you know where the line is.

PICTURE. In the figure, the yellow dashed line is the boundary. Inside, the little chalk-blue dots are gas molecules jiggling. That jiggling — the total energy stored in all the molecular motion and molecular forces — is the one quantity we will track. We give it the name $U$ for internal energy.

Figure — First law - ΔU = q + w (chemist sign convention)

The change is written $Δ U$ . The symbol $Δ$ (Greek "delta") simply means "final minus initial":

$Δ U = U_{final} - U_{initial}$

$U_{final}$ ::: the internal energy after the process — the box at the end
$U_{initial}$ ::: the internal energy before the process — the box at the start
$Δ$ ::: the instruction "subtract start from end"

Step 2 — The one law we refuse to break

WHAT. Zoom out so the box and the whole room are inside a giant bag. Nothing enters or leaves this giant bag — it is an isolated universe. For it, we assert one thing: its total energy is a fixed number that never changes.

WHY. This is the whole engine of the derivation. Everything else is just careful bookkeeping applied to this one refusal: energy of the universe = constant. No experiment has ever violated it.

PICTURE. Two chalk boxes sit inside one big dashed bag: "system" and "surroundings". A pink arrow shows energy sliding from one to the other. The total (blue number in the corner) stays frozen no matter which way the arrow points.

$E_{universe} = E_{system} + E_{surroundings} = constant$

$E_{universe}$ ::: the total energy of the giant isolated bag — frozen
$E_{system}$ ::: energy of our box (this is $U$ )
$E_{surroundings}$ ::: energy of everything else

If the whole thing is frozen but the two parts can trade, then whatever one part gains, the other must lose by exactly the same amount. In symbols, if the total does not change, the two changes must cancel:

$Δ E_{system} + Δ E_{surroundings} = 0 ⟹ Δ E_{system} = - Δ E_{surroundings}$

That minus sign is the seed of every sign convention to come: your gain is their loss.

Step 3 — Only two doors in the boundary

WHAT. Energy can cross the boundary in only two distinct ways. Not three, not ten. Two.

Heat — energy crossing because the two sides are at different temperatures.
Work — energy crossing because an organised force pushes through a distance (like a piston moving).

WHY. At the molecular level these are the only mechanisms that exist. Heat is disorganised: hot molecules on one side randomly slam into cold ones on the other, speeding them up — no marching in step. Work is organised: a whole wall (a piston) moves together, every molecule pushed the same direction at once. Because these are physically different, we give them different letters and track them separately.

PICTURE. The boundary now has two labelled doors. Through the red door labelled $q$ , chalk-pink squiggly arrows (random collisions) carry heat. Through the blue door labelled $w$ , a solid piston pushes inward with one straight blue arrow (organised push).

Step 4 — Choose "into the system = positive" (the chemist's ledger)

WHAT. We must decide which direction counts as a "+" number. The chemist's rule: anything flowing INTO the system is positive.

Heat absorbed by the system: $q > 0$ . Heat released: $q < 0$ .
Work done on the system (compression, piston pushes in): $w > 0$ . Work done by the system (expansion, piston pushes out): $w < 0$ .

WHY. A chemist watches the system like a bank account: deposits (energy in) are positive, withdrawals (energy out) are negative. This is a choice, but a consistent one. (Physicists sometimes flip the work sign and write $Δ U = q - w$ ; we do not, because we want every "in" to read as "+".)

PICTURE. A chalk bank ledger. Two green "+" arrows point inward: "heat IN" and "work ON". Two pink "−" arrows point outward: "heat OUT" and "work BY". The rule "into is plus" is boxed in yellow.

Step 5 — Add up the deposits: the equation appears

WHAT. The change in the system's internal energy is just the net of the two doors: heat that came in, plus work that came in.

WHY. There are no other doors (Step 3) and no other place for energy to hide inside a rigid boundary. So the system's energy change must equal exactly what crossed the boundary — nothing more, nothing less. That is the accounting.

PICTURE. A single chalk equation with the two doors from Step 3 flowing into a " $Δ U$ " jar. Every arrow that entered the jar is summed.

That is the whole law. It is conservation of energy (Step 2) filtered through "only two doors" (Step 3) and "into is plus" (Step 4).

Step 6 — Why $Δ U$ is honest but $q$ and $w$ are not (state vs path)

WHAT. $Δ U$ depends only on where you start and where you end. But $q$ and $w$ each depend on the route you took between those two states.

WHY. Think of hiking between two points on a hill. Your change in altitude depends only on the two endpoints — that is $Δ U$ , a state function. But the distance you walked depends on whether you zig-zagged — that is like $q$ and $w$ , path functions. Two different routes from state A to state B can have wildly different $q$ and different $w$ , yet $q + w$ always lands on the same $Δ U$ . See State functions vs path functions.

PICTURE. Two chalk paths (a straight yellow one, a wiggly pink one) climb between the same two dots A and B. The vertical rise $Δ U$ is identical for both; the path lengths ( $q$ , $w$ ) differ.

This is why we may add reaction energies like Lego bricks (Hess's Law) and why $Δ U$ deserves a clean $Δ$ while $q$ and $w$ do not.

Step 7 — Every case the box can be in

WHAT. Now we run the box through all four combinations of " $q$ in/out" and "work on/by", plus the degenerate cases where a term is zero. No situation is left unshown.

PICTURE. A 2×2 chalk grid of boxes, each with its arrows and its resulting $Δ U$ sign, with the three zero-cases drawn along the edge.

Case	$q$	$w$	Result	Physical picture
Heat in, compressed	$+$	$+$	$Δ U > 0$ (definitely rises)	Both doors deposit
Heat out, expanded	$-$	$-$	$Δ U < 0$ (definitely falls)	Both doors withdraw
Heat in, expands	$+$	$-$	$Δ U = q + w$ (could be anything)	Deposit vs withdrawal — compare sizes
Heat out, compressed	$-$	$+$	$Δ U = q + w$ (could be anything)	Same tug-of-war

Degenerate (zero) cases — these must never surprise you:

Rigid box, no volume change (Bomb calorimetry): no PV work, so $w = 0$ and $Δ U = q$ . All the heat becomes stored energy.
Thermally insulated (Adiabatic processes): no heat door, so $q = 0$ and $Δ U = w$ . All the work becomes stored energy.
Isothermal ideal gas (Isothermal processes): for an ideal gas $U$ depends on temperature only, so constant $T$ means $Δ U = 0$ , forcing $q = - w$ (heat in exactly pays for work out).

Step 8 — Four numbers to make it real

The one-picture summary

One boundary, two doors, one rule: whatever crosses in, added up, is the change in stored energy. Heat door plus work door feed one jar labelled $Δ U$ , and "into is plus" fixes every sign.

Recall Feynman retelling — say it back in plain words

Draw a bag around your gas: inside is the system, everything else is the surroundings. The universe's total energy is frozen, so any energy the system gains, the surroundings lost — that minus sign is where all the "plus and minus" comes from. Energy can sneak across the bag in only two ways: heat, which is random hot-molecule collisions, and work, which is an organised push like a piston. We agree to call anything heading into the bag positive, like a bank deposit. Add the two deposits and you get the change in the box's stored energy: $Δ U = q + w$ . The stored energy $Δ U$ only cares where you started and stopped (a state function, like altitude), while $q$ and $w$ each depend on the route (path functions, like distance walked) — yet they always add back to the same $Δ U$ . If the box is rigid, the work door is shut ( $w = 0$ ) and $Δ U = q$ . If the box is insulated, the heat door is shut ( $q = 0$ ) and $Δ U = w$ . If it is an ideal gas held at constant temperature, nothing is stored ( $Δ U = 0$ ) so heat in must equal work out. That is the entire first law, drawn.

Recall Quick self-test

A gas expands and does 300 J of work while absorbing 500 J of heat. What is $Δ U$ ? ::: $q = + 500$ , $w = - 300$ , so $Δ U = 500 - 300 = + 200 J$ . Why is $w$ negative when the gas expands? ::: The system pushes outward — energy leaves the boundary, a withdrawal, so it is negative in the chemist convention. In a bomb calorimeter, why does the measurement give $Δ U$ and not $Δ H$ ? ::: Constant volume shuts the work door ( $w = 0$ ), so $Δ U = q$ directly.