First law - ΔU = q + w (chemist sign convention): Worked examples

Before we solve anything, let us lay out every kind of case the topic can throw at you — so that when you finish, you have seen them all.

The scenario matrix

Recall the meaning of each letter (built in the parent note):

$Δ U$ = change in internal energy — energy kept by the system.
$q$ = heat: $q > 0$ if heat flows into the system, $q < 0$ if it flows out.
$w$ = work: $w > 0$ if work is done on the system (compression), $w < 0$ if done by the system (expansion).

Here is the full grid of case-classes. Each worked example below is tagged with the cell it fills.

Cell	Case class	$q$	$w$	Typical trigger words
A	Rigid container, heat in	$+$	$0$	"constant volume", "rigid", "bomb"
B	Isothermal expansion, ideal gas	$+$	$-$	"isothermal", "expands", " $Δ U = 0$ "
C	Isothermal compression, ideal gas	$-$	$+$	"isothermal", "compressed"
D	Exothermic reaction, constant $V$	$-$	$0$	"releases heat", "bomb calorimeter"
E	Adiabatic compression	$0$	$+$	"insulated", "no heat", "compressed"
F	Adiabatic expansion	$0$	$-$	"insulated", "expands"
G	Both $q$ and $w$ non-zero, same sign	$+$	$+$	heat in and compressed
H	Degenerate: nothing happens	$0$	$0$	"closed", "no change" / cyclic
I	Real-world word problem	mixed	mixed	engines, breathing, kettles
J	Exam twist: non-PV work at constant $V$	$+$	$+$	"stirrer", "electrical heater", rigid

Now the examples. Each fills at least one cell; together they cover A–J.

Worked example Example 1 — Cell A: rigid container, heat in

Statement: $0.50$ mol of gas sits in a sealed rigid steel flask. You supply $750$ J of heat. Find $Δ U$ and $w$ .

Forecast: Rigid means the walls cannot move. Guess: does any work get done? Will $Δ U$ be less than, equal to, or greater than $750$ J?

Sign of $q$ . Heat is supplied → flows in → $q = + 750 J$ . Why this step? "Into is plus" — we fix $q$ from the direction of flow before touching the equation.
Sign and size of $w$ . PV work is $w = - P_{ext} Δ V$ . Rigid flask → $Δ V = 0$ → $w = 0$ . Why this step? No boundary can move, so the coordinated force-times-displacement is zero.
Apply the law. $Δ U = q + w = 750 + 0 = + 750 J$ . Why this step? With two of three known, the equation delivers the third.

Verify: All the heat is kept, none leaks out as expansion. $Δ U = q$ exactly, as it must at constant volume. Units: J $+$ J $=$ J. ✓

Worked example Example 2 — Cell B: isothermal expansion of an ideal gas

Statement: $2$ mol of ideal gas expands isothermally at $310$ K and does $1500$ J of work pushing a piston out. Find $Δ U$ and $q$ .

Forecast: Temperature is held fixed. For an ideal gas, what does that tell you about $Δ U$ before you compute anything?

$Δ U$ from the isothermal fact. For an ideal gas $U = \frac{3}{2} n R T$ depends only on $T$ . Isothermal → $Δ T = 0$ → $Δ U = 0$ . Why this step? Internal energy of an ideal gas is pure molecular kinetic energy; no temperature change means no change in $U$ .
Sign of $w$ . Gas pushes the piston out → work done by the system → $w = - 1500 J$ . Why this step? The system spends energy; in chemist convention that is a negative deposit.
Solve for $q$ . $0 = q + (- 1500) \Rightarrow q = + 1500 J$ . Why this step? Heat must flow in to replace exactly the energy lost as work, keeping $T$ constant.

Verify: $q + w = 1500 + (- 1500) = 0 = Δ U$ . ✓ Physically: expansion tends to cool the gas; heat in cancels that so $T$ stays $310$ K. (See Isothermal processes.)

Worked example Example 8 — Cell H: degenerate / cyclic (nothing net)

Statement: A gas is taken around a full cycle and returns to its exact starting state. Over the cycle it absorbs $2000$ J of heat. Find $Δ U_{cycle}$ and the total work $w$ .

Forecast: It ends where it began. What does that force $Δ U$ to be — regardless of the messy path?

$Δ U$ from the state-function idea. $U$ is a state function; same start and end → $Δ U_{cycle} = 0$ . Why this step? $Δ U = U_{final} - U_{initial}$ and here $U_{final} = U_{initial}$ .
Solve for $w$ . $0 = q + w = 2000 + w \Rightarrow w = - 2000 J$ . Why this step? The equation still holds over the whole cycle.

Verify: $w = - 2000$ J means the system does $2000$ J of net work on the surroundings — exactly the heat it took in. This is the engine idea behind the Carnot cycle. ✓ ( $q + w = 2000 - 2000 = 0$ .)

Worked example Example 9 — Cell I: real-world word problem (kettle → steam)

Statement: $1.0$ mol of water boils at $373$ K and $1$ atm, turning to steam. It absorbs $40, 700$ J of heat, and the expanding steam does $3100$ J of work pushing back the atmosphere. Find $Δ U$ of vaporisation.

Forecast: Heat goes in to boil it, and it pushes out against the air. Two opposite-sign contributions — will $Δ U$ be less than the $40, 700$ J of heat?

Sign of $q$ . Absorbed to boil → $q = + 40, 700 J$ . Why this step? Boiling is endothermic.
Sign of $w$ . Steam expands against the atmosphere → work done by system → $w = - 3100 J$ . Why this step? The vapour pushes the surroundings out (PV work against $P_{ext}$ ).
Apply the law. $Δ U = 40, 700 + (- 3100) = + 37, 600 J$ . Why this step? Only what is left after paying the expansion work is stored inside.

Verify: $Δ U = 37, 600$ J $<$ $Δ H = q_{p} = 40, 700$ J, and the gap is exactly $∣ w ∣ = 3100$ J. This matches $Δ H = Δ U + P Δ V$ — the reason enthalpy exists. ✓

Worked example Example 10 — Cell J: exam twist, non-PV work at constant volume

Statement: A rigid insulated flask contains a solution. An electric heater does $1200$ J of electrical work on it, and a stirrer adds another $80$ J of shaft work. No heat crosses the wall. Find $Δ U$ .

Forecast: The trap: "rigid, so $w = 0$ ." Is that true here?

$q$ . Insulated → no heat crosses → $q = 0$ . Why this step? Definition of adiabatic wall.
Is PV work zero? Rigid → $Δ V = 0$ → PV work $= 0$ . But electrical and stirrer work are non-PV work and still count. Why this step? The equation $w = - P Δ V$ covers only expansion/compression; other organised energy transfers are separate.
Total work. $w = w_{elec} + w_{shaft} = 1200 + 80 = + 1200 + 80 = + 1280 J$ (both done on the system → positive).
Apply the law. $Δ U = 0 + 1280 = + 1280 J$ . Why this step? All the injected work is retained; nothing escapes.

Verify: $Δ U = 1280$ J. If you had wrongly set $w = 0$ , you would get $0$ J — a common exam mistake. Constant volume kills only the PV part of $w$ . ✓

Recall Quick self-test

Isothermal ideal-gas expansion: what is $Δ U$ ? ::: $0$ — temperature fixed, so internal energy unchanged. Adiabatic compression, $500$ J of work on the gas: $Δ U$ ? ::: $+ 500$ J — $q = 0$ , all work stored. A gas does $300$ J of work on the surroundings. Sign of $w$ ? ::: $w = - 300$ J (work by system). Full cycle absorbing $2000$ J of heat: net work $w$ ? ::: $- 2000$ J, because $Δ U_{cycle} = 0$ . Rigid insulated flask, heater does $1000$ J: is $w = 0$ ? ::: No — that's non-PV electrical work; $w = + 1000$ J.