JEE/NEET Physics · Thermodynamics series · Part 2 of 6 · All parts →
- Work done by gas = area under the P-V curve — always
- Isothermal (constant T): W = nRT ln(V₂/V₁) — slow, heat bath connected
- Isobaric (constant P): W = PΔV — simplest, and Q = nC_pΔT
- Isochoric (constant V): W = 0 — all heat goes to U
- Cyclic process: net work = area enclosed by the loop on P-V
A gas can change its state in four characteristic moods — hold temperature, pressure, or volume fixed, or hold heat out entirely. Each mood spends and stores energy differently, and the P-V graph tells you everything. Part 2 of the Thermodynamics series.
- Work = area under the curve
- Isothermal: the patient mood
- Isobaric and isochoric: the simple moods
- The four-mood table
- Cycles: loops pay net
- Solved examples
- Common mistakes
- This physics in your daily life
- Practice set
- Recap
Work = Area Under the Curve
W = ∫P dV: on a P-V diagram, the work done by the gas is literally the area under the path. Higher path, more work; steeper path, different trade. This geometric picture solves half of all thermodynamics questions.
Isothermal: The Patient Mood
Temperature pinned means ΔU = 0, so every joule of heat in converts to work out — the gas is a perfect conduit. But it must happen slowly enough for heat to keep flowing in from the bath.
Isobaric and Isochoric: The Simple Moods
Constant pressure: W = P(V₂ − V₁), a rectangle’s area — and the gas needs extra heat for both work and warming: Q = nC_pΔT with C_p = C_v + R. Constant volume: W = 0 (no piston motion), so Q = ΔU = nC_vΔT — the purest heating.
The Four-Mood Table
| Process | Held fixed | Work | Heat |
|---|---|---|---|
| Isothermal | T | nRT ln(V₂/V₁) | = W exactly |
| Isobaric | P | PΔV | nC_pΔT |
| Isochoric | V | 0 | nC_vΔT (all to U) |
| Adiabatic | Q (heat!) | (next part) | 0 |
Cycles: Loops Pay Net
Return the gas to its start (a closed loop on P-V) and ΔU = 0 — the net work over the cycle is the area enclosed by the loop. Clockwise loop: gas does net work (engines). Anticlockwise: work done ON gas (refrigerators).
Solved Examples
W = PΔV = 10⁵ × 2×10⁻³ = 200 J.
✔
Answer: 200 J
W = nRT ln(V₂/V₁) = 2 × 8.3 × 300 × 1.1 ≈ 5.5 kJ.
And Q = 5.5 kJ too — the isothermal identity. ✔
Answer: ≈5.5 kJ (= Q in)
W_net = +40 J (clockwise = engine direction); ΔU = 0 over the cycle → Q_net = +40 J.
The gas converted 40 J of heat to work over one lap — a miniature engine. ✔
Answer: W = Q = +40 J
- Using PΔV for isothermal work. Pressure changes throughout — only the logarithm formula survives.
- Reading area on the wrong side. Work is area UNDER the curve (down to the V-axis), not between arbitrary lines.
- C_p vs C_v swap. Constant pressure uses C_p = C_v + R; constant volume uses C_v — the ‘+R’ is the piston’s share.
- Assuming fast = isothermal. Fast processes are adiabatic (no time for heat); isothermal needs glacial slowness.
This Physics in Your Daily Life
- Petrol engine strokes — intake/expansion/exhaust at roughly constant pressure phases: isobaric-ish bookkeeping under your bonnet.
- A syringe capped and squeezed — volume barely changes: near-isochoric, pressure climbs steeply (feel it).
- Cooking gas expanding through a regulator — roughly isothermal thanks to heat from surroundings: the patient mood on demand.
- Weather balloons rising — air packets expand isothermally-ish in slow rise (or adiabatically in fast convection: the next part).
- Every engine rating in kW — the loop area per cycle × cycles per second: power is literally enclosed P-V area, spun fast.
A gas has three dials — pressure, volume, temperature — linked by PV = nRT. Hold each in turn (or hold heat itself) and you get the characteristic processes: not four arbitrary recipes, but the four natural ways to let a three-dial machine evolve. Every real process is a blend of these four pure moods.
Isothermal doubling at 300 K, 1 mol: W = 2490 ln2 ≈ 1727 J, fully paid by heat. Same doubling adiabatically (next part): gas does MORE work per initial pressure but pays from savings — temperature drops ~120 K. Two moods, two invoices.
On the P-V canvas: isotherms are hyperbolas (PV = const); steeper-than-isotherm curves are adiabatics. Isochoric is a vertical line (zero area — no work); isobaric a horizontal one (rectangular area). The four moods are four SHAPES, and the shapes do the arithmetic.
Practice set (answers hidden — try first)
(NEET-level) Isochoric process: work =
(JEE Main-level) 3 mol at 400 K double volume isothermally: W ≈
(NEET-level) C_p − C_v =
(Concept) Clockwise P-V loop: the system
(JEE Main-level) Isobaric P = 2×10⁵ Pa, ΔV = 10⁻³ m³: W =
- W = area under P-V curve
- isothermal: W = nRT ln(V₂/V₁), Q = W
- isobaric: W = PΔV, Q = nC_pΔT
- isochoric: W = 0
- cycle: net W = enclosed loop area
- 🔁 four processes table
- 🔁 work formulas for each
- 🔁 area reading on P-V
- 🧠 Chant: ‘T-slow, P-rectangle, V-nothing, Q-nothing (next!)’.
- 🧠 C_p = C_v + R — ‘the +R is the piston’s tip’.
- 🏠 Daily: engine kW = loop area × laps per second.
- 🏠 Daily: capped syringe = isochoric squeeze.
Quick revision
- Work done by gas = area under the P-V curve — always
- Isothermal (constant T): W = nRT ln(V₂/V₁) — slow, heat bath connected
- Isobaric (constant P): W = PΔV — simplest, and Q = nC_pΔT
- Isochoric (constant V): W = 0 — all heat goes to U
- Cyclic process: net work = area enclosed by the loop on P-V
- Work = area under the curve
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