Foundations — Work done in each process — derivation

This page assumes you know nothing. Every letter in the parent note — $P$, $V$, $T$, $dV$, $\int$, $\ln$, $\gamma$, $n$, $R$ — is built here from a picture before it is ever used in a formula. Read top to bottom; each block earns the next.

0. The physical scene (the picture everything hangs on)

Before any symbol, look at the machine we are describing.

A cylinder holds gas. A piston (the sliding lid) can move. Gas molecules constantly bang the piston from inside and push it outward. If the piston moves out a little, the gas has done work — it moved something a distance against a force. Everything below is just careful bookkeeping of that push.

1. Force $F$ and distance $x$ — what "work" even means

• Plain words: pushing hard over a long way = lots of work; a tiny nudge = tiny work.
• The picture: an arrow (force) pointing the way the piston slides, times how far it slides.
• Why the topic needs it: work is the output we are trying to compute. The gas's whole job is to push the piston, and "how much push × how far" is work.

$x$ is just position — how far along the cylinder the piston sits. $dx$ (read "dee-ex") means a tiny change in $x$ — the piston creeps out by a hair.

2. Pressure $P$ and area $A$ — turning the push into pressure

• Plain words: the gas doesn't push at one point — it pushes on the whole face of the piston. Pressure tells you how hard per unit of surface; multiply by the face area $A$ to get the total force.
• The picture: many little molecule-arrows hitting the piston face; add them all and you get one big force arrow $F=PA$.
• Why the topic needs it: we never measure the piston's force directly. We measure pressure (a gauge reading) and the piston area. So $F=PA$ is how we convert "the thing we know" into "the thing work needs."

$A$ = area of the piston face, in square metres. It stays constant (the cylinder is a fixed width).

3. Volume $V$ and the swept-volume trick $Adx=dV$

• Plain words: when the piston slides out by a hair $dx$, the gas gains a thin disc of new space. That disc's volume = its face area $A$ times its thickness $dx$.
• The picture (red disc): a coin-thin slab swept out by the moving piston. Its volume is $Adx$.
• Why the topic needs it — the key swap: this lets us trade distance for volume:

$dW=Fdx=(PA)dx=P(Adx)=PdV$

That is the whole master formula born: the tiny work is pressure times the tiny volume swept. We swap $dx$ for $dV$ because pressure and volume are what a gas actually gives us — you never read a piston's position off a gauge, but you always know $P$ and $V$.

$dV$ = a tiny sliver of volume. $V_1$ = starting volume, $V_2$ = final volume.

4. The integral sign $\int$ — "add up all the slivers"

• Plain words: the tall skinny "S" is just a fancy "Sum." The little numbers $V_1$ (bottom) and $V_2$ (top) say where the summing starts and stops.
• The picture: the area under the pressure curve, chopped into thin vertical rectangles. Each rectangle is $P$ tall and $dV$ wide, so its area is $PdV$ — exactly one tiny bit of work. Add every rectangle → total work → the whole shaded area.
• Why this tool and not plain multiplication? Because $P$ changes as the gas expands (it drops as the gas spreads out). If pressure were frozen we could just do $P\times (V_2-V_1)$. Since it isn't, we must use rectangles thin enough that $P$ is steady inside each one — that is precisely what an integral is for.

5. Temperature $T$, moles $n$, gas constant $R$ — the ideal gas law

• $T$ = temperature in kelvin (K): a measure of how fast the molecules jiggle. Hotter = harder banging = more push.
• $n$ = number of moles: a count of gas particles (one mole ≈ $6\times 10^{23}$ of them). More gas = more pushers.
• $R$ = universal gas constant $=8.314{\text{J K}}^{-1}{\text{mol}}^{-1}$: a fixed conversion number nature hands us so the units of pressure×volume come out as energy.

Why the topic needs it: the master formula wants $P$ as a function of $V$. The gas law is the machine that gives it. Rearranged, $P=\frac{nRT}{V}$ — and that is what we substitute into the integral in the isothermal case.

6. The logarithm $\ln$ — the answer to "add up $1/V$"

• Plain words: when the pressure sliver height is $\frac{nRT}{V}$, the thing left to sum is $\frac{1}{V}$ over each sliver. The running total of $\frac{1}{V}$ is not a normal power — it is the logarithm.
• The picture: the area under the curve $y=1/V$ between $V_1$ and $V_2$; that area equals $\ln V_2-\ln V_1=\ln \frac{V_2}{V_1}$.
• Why the topic needs it: it is the only new maths tool the isothermal case demands. Without it you cannot express the area under a $1/V$ curve.

$\ln \frac{V_2}{V_1}$ is positive when $V_2>V_1$ (expansion → positive work), zero when nothing changes, negative when compressed. That sign behaviour is exactly what the physics needs.

7. Gamma $\gamma$ and $P{V}^{\gamma }=K$ — the adiabatic curve

• Plain words: when no heat enters or leaves (a gas wrapped in a perfect blanket), pressure and volume no longer obey the gentle $PV=\text{const}$ curve — they follow the steeper rule $P{V}^{\gamma }=K$.
• The picture: on a $P$–$V$ diagram the adiabatic curve dives more steeply than the isothermal one, because as the gas expands it also cools, so pressure drops faster.
• Why the topic needs it: to integrate $\int PdV$ adiabatically we again need $P$ as a function of $V$. Here $P=K{V}^{-\gamma }$ — a power of $V$ — so the integral is a power-rule integral, harder than a log but still doable.

Prerequisite map

Equipment checklist

Test yourself — reveal only after you have answered aloud.

Connections

• Parent: Work done in each process — this page builds every symbol it uses
• Hinglish version →
• P-V Diagrams — where "area under the curve" lives
• Isothermal Process and Boyle's Law — where the $\ln$ tool is used
• Adiabatic Process and PV^gamma — source of $\gamma$ and $P{V}^{\gamma }=K$
• Internal Energy and Cv — defines $C_v$ behind $\gamma$
• First Law of Thermodynamics — where these $W$ values plug in
• Carnot Cycle — combines all four processes