Work done in each process — derivation: Worked examples

This is a drill page for the derivation note. There we built the one master idea $W = \int_{V_{1}}^{V_{2}} P d V$ Here we throw every kind of case at it: each sign, each degenerate input, each limit, a word problem, and an exam trap. Guess before you compute — that "Forecast" habit is how you catch nonsense answers.

Recall What each symbol means (built once, used everywhere)

$P$ = pressure = force per area the gas pushes with (unit: pascal, $Pa = N/m^{2}$ ).
$V$ = volume the gas fills (unit: $m^{3}$ ; note $1 L = 1 0^{- 3} m^{3}$ ).
$W = \int P d V$ = the running sum of tiny pushes $P d V$ as volume changes. On a $P$ – $V$ picture it is the area under the path.
$n$ = moles of gas, $R = 8.314 J mol^{- 1} K^{- 1}$ (the gas constant), $T$ = absolute temperature in kelvin.
$γ$ (gamma) = ratio of heat capacities, $= 5/3$ for a monatomic gas, $= 7/5$ for a diatomic gas. It appears only in adiabatic work.
$ln$ = natural logarithm — the function that undoes $e^{(\cdot)}$ ; it shows up because $\int \frac{d V}{V} = ln V$ .

The scenario matrix

Every question this topic can throw is one cell of this table. The worked examples below are tagged [Cell n] so you can see the whole space gets covered.

Cell	Case class	What makes it tricky
1	Isobaric expansion ( $V_{2} > V_{1}$ )	positive work, pull $P$ out
2	Isobaric compression ( $V_{2} < V_{1}$ )	work comes out negative
3	Isochoric ( $d V = 0$ ) — degenerate	area is zero, $W = 0$ no matter the heat
4	Isothermal expansion	$P$ varies → log appears, $W > 0$
5	Isothermal compression	same log, $V_{2} < V_{1}$ → $ln < 0$ → $W < 0$
6	Adiabatic expansion (gas cools)	power-law integral, $W > 0$ , sign trap
7	Limiting behaviour	$V_{2} \to V_{1}$ (all $W \to 0$ ); $γ \to 1$ (adiabatic $\to$ isothermal)
8	Cyclic loop — sign by direction	net $W$ = enclosed area; clockwise $> 0$
9	Real-world word problem	translate words → which formula
10	Exam twist	mixed path / first-law combination

Figure — Work done in each process — derivation

The picture above is the whole matrix in one glance: same two endpoints, different coloured paths, different shaded areas. Work is the area, and the area depends on the path.

Example A — Isobaric expansion [Cell 1]

Example B — Isobaric compression [Cell 2]

Example C — Isochoric, the degenerate case [Cell 3]

Example D — Isothermal expansion [Cell 4]

Example E — Isothermal compression [Cell 5]

Example F — Adiabatic expansion [Cell 6] + sign trap

Example G — Limiting behaviour [Cell 7]

Worked example

Two limits at once. (i) An isothermal process with $V_{2} \to V_{1}$ (barely any expansion). (ii) An adiabatic curve as $γ \to 1$ : show it approaches the isothermal answer.

Forecast: (i) No volume change → $W \to 0$ . (ii) When $γ = 1$ , $P V^{γ} = K$ becomes $P V = K =$ constant, which is the isothermal condition — so the two formulas should merge.

Step 1 — limit (i) numerically. Why this step? Take $1$ mol at $300 K$ , $V_{2} / V_{1} = 1.001$ . $W = 1 \times 8.314 \times 300 \times ln (1.001) = 2494.2 \times 0.0009995 = 2.493 J$ As $V_{2} / V_{1} \to 1$ , $ln (1) = 0$ , so $W \to 0$ ✓ — smoothly, not abruptly.

Step 2 — limit (ii) conceptually. Why this step? The adiabatic work $\frac{P _{1} V _{1} - P _{2} V _{2}}{γ - 1}$ has $γ - 1 \to 0$ in the denominator, but the numerator also $\to 0$ because when $γ = 1$ the curve is $P V =$ const so $P_{1} V_{1} = P_{2} V_{2}$ . The $0/0$ resolves (L'Hôpital) exactly to $n R T ln (V_{2} / V_{1})$ — the isothermal result.

Verify: Both limits behave. The Adiabatic Process and PV^gamma curve is steeper than the Isothermal Process and Boyle's Law curve for $γ > 1$ ; at $γ = 1$ they coincide (figure above).

Example H — Cyclic loop, sign by direction [Cell 8]

Worked example

A gas runs the rectangular loop $A \to B \to C \to D \to A$ shown above: $A = (V = 1 m^{3}, P = 1 \times 1 0^{5} Pa)$ , $B = (3, 1 \times 1 0^{5})$ , $C = (3, 3 \times 1 0^{5})$ , $D = (1, 3 \times 1 0^{5})$ , traversed in that order. Find $W_{net}$ .

Forecast: Trace the order $A \to B \to C \to D \to A$ : it runs anticlockwise here (expand at low pressure, compress at high pressure). Anticlockwise ⇒ $W_{net} < 0$ . Magnitude = box area.

Step 1 — leg by leg. Why this step? Only the horizontal legs have $d V \neq = 0$ ; vertical legs ( $V$ fixed) give $0$ .

$A \to B$ (expand at $P = 1 0^{5}$ ): $W = 1 0^{5} \times (3 - 1) = + 2 \times 1 0^{5} J$ .
$B \to C$ (isochoric): $0$ .
$C \to D$ (compress at $P = 3 \times 1 0^{5}$ ): $W = 3 \times 1 0^{5} \times (1 - 3) = - 6 \times 1 0^{5} J$ .
$D \to A$ (isochoric): $0$ .

Step 2 — sum. Why this step? Net work is the algebraic sum around the loop. $W_{net} = 2 \times 1 0^{5} + 0 - 6 \times 1 0^{5} + 0 = - 4 \times 1 0^{5} J$

Verify: Enclosed area $= Δ P \cdot Δ V = (2 \times 1 0^{5}) (2) = 4 \times 1 0^{5} J$ ; sign negative because anticlockwise ✓. Matches Carnot Cycle logic: loop direction sets the sign.

Example I — Real-world word problem [Cell 9]

Example J — Exam twist: mixed with first law [Cell 10]

Worked example

In an isothermal expansion, a gas absorbs $Q = 5000 J$ of heat. A student claims "the work done is zero because temperature is constant, so $Δ U = 0$ ." Correct them and find the actual work.

Forecast: The claim confuses " $Δ U = 0$ " with " $W = 0$ ". At constant $T$ the piston still moves, so $W \neq = 0$ . In fact first law forces $W = Q$ .

Step 1 — apply the first law. Why this step? $Δ U = Q - W$ . Isothermal ⇒ $T$ constant ⇒ $Δ U = 0$ (internal energy depends only on $T$ ). Therefore $0 = Q - W \Rightarrow W = Q = 5000 J .$

Step 2 — where the student went wrong. Why this step? $Δ U = 0$ says heat and work are equal, not that either is zero. The gas quietly converts all $5000 J$ of absorbed heat into work.

Verify: $W = 5000 J$ , positive (expansion) ✓. Consistent with the log formula: whatever $n R T ln (V_{2} / V_{1})$ equals, it must here equal the $5000 J$ absorbed.

Recall Rapid self-test

Isobaric compression $0.05 \to 0.02 m^{3}$ at $2 \times 1 0^{5}$ Pa, work by gas ::: $- 6000 J$ Isothermal, $2$ mol, $400$ K, volume $10 \to 30$ L ::: $+ 7307 J$ Same but compressed $30 \to 10$ L ::: $- 7307 J$ Adiabatic diatomic, $500 \to 380$ K, $1$ mol ::: $+ 2494 J$ Anticlockwise rectangular loop, $Δ P = 2 \times 1 0^{5}$ , $Δ V = 2$ ::: $- 4 \times 1 0^{5} J$ Isothermal expansion absorbing $5000$ J of heat, work ::: $5000 J$ (since $Δ U = 0 \Rightarrow W = Q$ )

Connections

Parent: the derivation
First Law of Thermodynamics — used in Examples C and J
P-V Diagrams — every figure lives here
Adiabatic Process and PV^gamma — Examples F, G, I
Isothermal Process and Boyle's Law — Examples D, E, J
Internal Energy and Cv — why adiabatic gas cools
Carnot Cycle — cyclic sign rule (Example H)