System vs surroundings; open, closed, isolated: Visual walkthrough

We assume nothing. We start from a jar on a table.

Step 1 — Draw the boundary: everything splits into two

WHAT. Pick the thing you care about — a gas in a jar. Draw a line around it. Everything inside that line is the system. Everything outside is the surroundings. Together they make the whole universe.

WHY. Energy accounting needs a fence. Without a fence, "how much energy came in" has no meaning — in from where, to what? The fence — the boundary — is the accountant's dividing line. It is the single most important object in all of thermodynamics, because its properties (not the gas) decide what is allowed to happen.

PICTURE. Look at the figure. The blue blob is the system, the pale region around it is the surroundings, the black ring is the boundary. Two arrows sit ready at the boundary: an orange one for matter (little particles) and a red one for energy (a wavy heat squiggle). The whole game is: which of these two arrows is allowed through the black ring?

Figure — System vs surroundings; open, closed, isolated

Step 2 — Write the ONE law we never break: conservation

WHAT. State the one axiom everything rests on: the total energy of (system + surroundings) never changes. Any energy the system gains, the surroundings lost, and vice-versa.

$change in the system’s internal energy Δ U_{sys} + change in the surroundings’ internal energy Δ U_{surr} = 0$

Here $Δ$ (Greek "delta") just means "final minus initial" — how much a quantity changed. And $U$ , the internal energy, is the total energy locked inside the system: the kinetic energy of every jiggling molecule plus the energy stored in their bonds. $Δ U_{sys}$ is therefore "how much the jar's inner energy went up or down."

WHY. This is the whole engine. We are not assuming the First Law yet — we are assuming something even more basic (energy is conserved for the universe) and will derive the First Law from it. Why start here? Because the universe has no outside — nothing can leak away — so its energy books must balance to exactly zero. That single fact seeds everything.

PICTURE. The figure shows a see-saw. Energy leaving the surroundings tips down on the right; the exact same amount tips the system up on the left. The bar is perfectly balanced — the sum is pinned at zero.

Step 3 — Name the TWO doors energy can use: heat and work

WHAT. Energy doesn't cross the boundary as a formless blob. It crosses through exactly two doors: as heat or as work. Give them symbols and a sign rule.

$Δ U_{sys} = heat absorbed by the system q + work done on the system w$

$q$ = heat: energy crossing because of a temperature difference — random, disordered molecular collisions passing energy across the wall.
$w$ = work: energy crossing because of an organised push — a force moving through a distance, like a piston sliding.
Sign rule: $q > 0$ when heat flows in, $w > 0$ when work is done on the system. Energy arriving is positive; energy leaving is negative.

WHY two doors and not one? Because they are physically different and we can control them separately with the boundary. A wall can let heat through but forbid work (fixed wall, conducts heat), or allow work but forbid heat (moving but insulated piston). Splitting energy into $q$ and $w$ is precisely what lets the boundary act like two independent switches. This equation is the First Law of Thermodynamics.

PICTURE. The system box now has two labelled gateways in its wall: a red "HEAT $q$ " door (wavy arrow) and an orange "WORK $w$ " door (piston arrow). Arrows point inward with $+$ signs, outward with $-$ signs.

Step 4 — Isolated system: bolt EVERY door shut

WHAT. Build the strictest boundary: it blocks matter, blocks heat ( $q = 0$ ), blocks work ( $w = 0$ ). Feed those into the master equation.

$Δ U = = 0 (adiabatic) q + = 0 (rigid) w = 0 \Rightarrow Δ U_{isolated} = 0$

Two boundary properties do this:

Adiabatic wall (perfect insulation) → no heat crosses → $q = 0$ .
Rigid wall (volume cannot change) → the wall never moves → no $P Δ V$ work → $w = 0$ .
Impermeable wall → no matter crosses → $Δ m = 0$ .

WHY start with isolated? Because it is the cleanest case — every door shut, so the answer is the plainest possible: the internal energy simply cannot change. This is also the setting for Entropy and Second Law: only for an isolated system is the entropy guaranteed to never decrease, precisely because $Δ U = 0$ removes all outside influence.

PICTURE. The thermos flask. Every gateway from Step 3 is drawn with a red ✗ struck through it. Inside, a label reads $Δ U = 0$ , $Δ m = 0$ . Nothing gets in or out.

Step 5 — Closed system: open ONLY the energy doors

WHAT. Loosen the boundary by one notch: keep matter locked out, but let energy through. Now $q$ and $w$ are live again; only mass is frozen.

$Δ U = q + w, matter door still shut Δ m = 0$

The wall is impermeable (no matter) but diathermal (heat crosses) and possibly movable (work crosses). Nothing in the master equation simplifies away — but $Δ m = 0$ means the amount of stuff, $n$ moles, is a fixed number we can rely on.

WHY this is chemistry's favourite boundary. With $n$ fixed we can write internal energy changes as $Δ U = n C_{V} Δ T$ (see Heat Capacity at Constant Volume) and, at constant pressure, use Enthalpy and Constant Pressure. A sealed flask that still trades heat is exactly what a lab bench gives you — see State Functions vs Path Functions for why $Δ U$ doesn't care how the heat and work arrived, only the endpoints.

PICTURE. The sealed steel jar. The matter door has the red ✗; the heat door (red squiggle) and work door (orange piston) are wide open with green ✓. The molecule count $n$ is stamped as constant.

Step 6 — Open system: throw EVERY door open

WHAT. Remove the last lock. Matter, heat and work can all cross. The master equation still holds, but now the moving matter carries its own energy in and out, so we can no longer treat $n$ as constant.

$Δ U = q + w + new term: mass flow drags energy along \sum (energy carried by matter in/out)$

The last term is why open-system energy books need an extra column — each parcel of matter that enters brings its internal energy (and, in flow, its enthalpy) with it.

WHY it matters. Almost all real chemistry and biology is open: a test tube venting steam, a candle inhaling $O_{2}$ and exhaling $C O_{2}$ , your body eating and excreting. The boundary here is permeable to everything.

PICTURE. The open beaker. All three doors are green ✓; an extra orange arrow of little particles streams both in and out, tagged "carries energy." $Δ U$ can be anything.

Step 7 — The switchboard: one equation, three settings

WHAT. Line up all three boundaries as three settings of the same master equation. Reading a boundary's properties tells you which terms survive.

System	Matter door	Heat $q$	Work $w$	What survives
Isolated	✗ shut	$= 0$	$= 0$	$Δ U = 0$
Closed	✗ shut	live	live	$Δ U = q + w, n$ fixed
Open	✓ open	live	live	$Δ U = q + w + flow$

WHY. This is the payoff: you never memorise three separate laws. You memorise one ( $Δ U = q + w$ ) and read the walls to see which switches are off. The special zero-cases live here too:

Adiabatic wall alone ( $q = 0$ , work still allowed) gives $Δ U = w$ — the world of Adiabatic Processes.
Rigid wall alone ( $w = 0$ , heat still allowed) gives $Δ U = q$ .
Both off → isolated → $Δ U = 0$ .

PICTURE. A literal switchboard: three rows, each with a MATTER / HEAT / WORK switch flipped on or off, and the resulting simplified equation glowing beside it.

The one-picture summary

Below is the whole derivation on one canvas: the conservation see-saw at the top feeds the master equation $Δ U = q + w$ ; three boundary "filters" (open / closed / isolated) sit below it, each greying-out the doors it blocks, each producing its own simplified law. Trace top to bottom and you have re-derived every case from the single seed "energy is conserved."

Recall Feynman retelling — say it back in plain words

Imagine drawing a fence around a jar of gas: inside is the system, outside is the surroundings, and the fence is the boundary. Nature keeps one unbreakable rule — energy is never lost, so whatever the system gains the surroundings lost; their total change is exactly zero. Energy can only sneak across the fence through two doors: heat (random jostling from a temperature difference) and work (an organised push, like a piston). That gives the master law $Δ U = q + w$ — the change in the jar's internal energy equals heat coming in plus work done on it. Now the fence itself is a set of switches. Weld it rigid and it forbids work; insulate it and it forbids heat; seal it and it forbids matter. Isolated flips all switches off, so $Δ U = 0$ — nothing changes. Closed locks only the matter door, so energy still flows but the amount of stuff stays fixed — this is the lab flask, the calorimeter, the case chemistry loves. Open unlocks everything and matter itself carries energy along, so anything can happen — this is the candle, the pot, your body. One law, three switch settings. Read the walls, and the physics writes itself.

Recall Quick self-test

Which door does a "closed" boundary shut? ::: Only the matter door; heat and work can still cross. Why is $Δ U = 0$ for an isolated system? ::: Because $q = 0$ (adiabatic) and $w = 0$ (rigid), so $Δ U = q + w = 0$ . Rigid vs impermeable — what does each block? ::: Rigid blocks work ( $w = 0$ , volume fixed); impermeable blocks matter. Different properties. For the $N_{2}$ example, why did all heat become $Δ U$ ? ::: The rigid wall made $w = 0$ , so $Δ U = q$ entirely. Is the universe closed or isolated? ::: Isolated — it has no surroundings, so no energy can be exchanged either.

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