Topic 4

PV diagrams

A PV diagram turns a thermodynamics problem into a geometry problem. Work becomes an area, internal energy becomes a position, and a whole engine cycle becomes a loop you can measure with a ruler. Learn to read these and most of the unit stops requiring algebra.

Anatomy of the diagram

Pressure on the vertical axis, volume on the horizontal. Every point is a complete state of the gas: fix P and V and the ideal gas law fixes T too. Every curve is a process, and the arrow on it tells you which way time runs.

Point A is a state. Point B is a state. The curve between them is how the gas got there — and that route, not just the endpoints, determines Q and W.

Three things you can read at a glance

Temperature tracks the product PV. Points farther from the origin (up and right) are hotter. Along a single hyperbola, T is constant.
Internal energy for a monatomic gas is ³⁄₂PV — so ΔU depends only on the two endpoints, never on the path.
Work is the area under the curve, and it is the only quantity here that depends on the path.

Work is the area underneath

For a thin slice of the process where the volume changes by dV, the gas does P dV of work on its surroundings. Add up the slices:

W_{by gas} = ∫_{V_i}^V_f P dV = area under the curve W_{on gas} = −∫ P dV On a graph you rarely integrate — you decompose the area into rectangles and triangles and add.

Sign from direction

The area itself is always a positive number. The direction of travel supplies the sign:

Moving right (expanding): W_by = +area, W_on = −area
Moving left (compressing): W_by = −area, W_on = +area
Moving vertically: no area at all, so W = 0

Calculus corner

This is a genuine, non-decorative integral — one of the first places physics needs one. For an isobaric process P comes out of the integral and you get PΔV, the rectangle. For an isothermal process P = nRT/V, and ∫ nRT dV/V = nRT ln(V_f/V_i) — that is exactly where the logarithm in the isothermal work formula comes from. The lab has you evaluate this integral numerically on your own measured data.

The graph gallery

Four shapes. Learn to recognize them instantly.

Isobaric — constant pressure

Expansion at constant pressure. W_on = −PΔV, a plain rectangle. Moving right ⇒ the gas does work on the world. For a monatomic gas, Q = ⁵⁄₂nRΔT: some of the heat raises U, the rest is spent expanding.

Isochoric — constant volume

Heating at constant volume. A vertical segment encloses zero area, so W = 0 and the first law collapses to ΔU = Q. Every joule of heat goes into internal energy.

Isothermal — constant temperature

A hyperbola: PV = nRT = constant. Since ΔT = 0, ΔU = 0 and therefore Q = −W — all the heat absorbed leaves again as work. The gas is just a pass-through.

Adiabatic — no heat exchanged

Both start at the same state and expand. The adiabat falls faster, because the gas is also cooling as it expands — it is paying for the work out of its own internal energy, with no heat coming in to top it up. It therefore crosses to lower-temperature isotherms as it goes.

Shape on the PV plot	Process	Instant conclusion
Horizontal line	Isobaric	`W`_on = −`P`Δ`V`; area is a rectangle
Vertical line	Isochoric	`W` = 0, so Δ`U` = `Q`
Hyperbola	Isothermal	Δ`U` = 0, so `Q` = −`W`
Steeper-than-hyperbola curve	Adiabatic	`Q` = 0, so Δ`U` = `W`
Straight slanted line	(none of the four)	Work is a trapezoid: average pressure × Δ`V`
Closed loop	Cycle	Δ`U` = 0; net work = enclosed area

All four on one plot

Start every process from the same state and the differences become a matter of slope:

Moving up and/or right means PV grows and the gas gets hotter. The isotherm is the dividing line: anything below it is cooling, anything above it is heating. That places the adiabat unambiguously — expanding adiabatically must cool the gas, so it has to fall below the isotherm.

Path dependence

Same start, same finish, two routes. The endpoints fix ΔU, but the areas underneath are visibly different — so the work differs, and the heat must differ by exactly the same amount to compensate.

Path 1: expand first, at high pressure. Large area ⇒ a lot of work done by the gas.

Path 2: depressurize first, expand at low pressure. Much less work.

The one-sentence version

ΔU is a property of the endpoints; Q and W are properties of the route. If a question gives you two paths between the same states and asks which has the greater ΔU, the answer is “they are equal” — and it is asking whether you know that.

Cycles

Close the loop and the gas returns to its original state, so ΔU = 0 over the whole cycle. The first law then gives Q_net = W_net,by: over one cycle, whatever net heat went in came out as net work. And that net work is just the enclosed area.

Clockwise = heat engine. Net work done by the gas is positive: the expansion happens at high pressure, the compression at low. Net heat flows in.

Counterclockwise = refrigerator / heat pump. Work must be done on the gas, and the machine moves heat from cold to hot at that cost.

Fastest check on the exam

Glance at the loop direction. Clockwise ⇒ engine, net work out, net heat absorbed. Counterclockwise ⇒ refrigerator, net work in, net heat expelled. You can answer several multiple-choice questions from that alone.

How to read any PV problem

Label every state point with its P and V, and compute PV for each. That product is proportional to T, and for a monatomic gas U = ³⁄₂PV.
Identify each leg by its shape (horizontal, vertical, hyperbola, other).
Work, leg by leg, from the area. Vertical legs contribute nothing. Sign from the direction of travel.
ΔU, leg by leg, from ³⁄₂Δ(PV) — no path reasoning required.
Q = ΔU − W for each leg. Never compute Q directly if you can avoid it.
Check the totals. Around a full cycle, ΣΔU must be 0, and ΣW must equal the enclosed area with the right sign. If it does not, you have a sign error.

Worked examples

1. Work from a slanted line (trapezoid)

A gas expands along a straight line on the PV diagram from (1.0×10⁻³ m³, 4.0×10⁵ Pa) to (5.0×10⁻³ m³, 2.0×10⁵ Pa). Find the work done by the gas.

Do not use PΔV — the pressure is not constant. The region under a straight line is a trapezoid, so use the average height:

W_by = P_avgΔV = 4.0×10⁵ + 2.0×10⁵2 (4.0×10⁻³) = +1200 J And W_on = −1200 J. Expanding, so the gas did work on the world.

Units check: Pa · m³ = (N/m²)(m³) = N·m = J. Areas on a PV diagram are always energies — that is worth verifying once so you trust it forever.

2. Full cycle analysis (the big one)

A monatomic ideal gas is taken clockwise around the rectangular cycle A→B→C→D→A:

State	`P` (Pa)	`V` (m³)	`PV` (J)	`U` = ³⁄₂`PV` (J)
A	1.0×10⁵	2.0×10⁻³	200	300
B	3.0×10⁵	2.0×10⁻³	600	900
C	3.0×10⁵	6.0×10⁻³	1800	2700
D	1.0×10⁵	6.0×10⁻³	600	900

Computing U at every corner first is the whole trick — every ΔU below is then just a subtraction.

Leg	Type	`W`_on (J)	Δ`U` (J)	`Q` = Δ`U` − `W` (J)
A→B	Isochoric (up)	0	+600	+600
B→C	Isobaric (right)	−1200	+1800	+3000
C→D	Isochoric (down)	0	−1800	−1800
D→A	Isobaric (left)	+400	−600	−1000
Cycle		−800	0	+800

Checks, all three of which must pass:

ΣΔU = 600 + 1800 − 1800 − 600 = 0 ✓ (it is a closed cycle)
Enclosed area = ΔP × ΔV = (2.0×10⁵)(4.0×10⁻³) = 800 J, and W_by = +800 J ✓ (clockwise, so positive)
Q_net = +800 J = W_by ✓

Bonus — efficiency. Heat entered on A→B and B→C: Q_H = 600 + 3000 = 3600 J. So

e = |W_net||Q_H| = 8003600 = 0.22 = 22% Only heat absorbed goes in the denominator — the 1800 J and 1000 J expelled do not.

3. Identify the process from the graph

A gas moves along a curve on which PV is the same at both endpoints, and the curve bulges toward the origin. What kind of process is it, and what are Q, W, and ΔU if the gas expanded?

Equal PV at both ends means equal temperature at both ends, so ΔU = 0. The bulge-toward-the-origin shape is a hyperbola — an isothermal process. Expanding means the gas did work on the world, so W_on < 0, and the first law forces Q = −W_on > 0: heat flowed in.

Careful, though: equal PV at the endpoints alone only guarantees ΔU = 0. The path could still have wandered off the isotherm in between, in which case W would be a different number — but ΔU would still be zero.

4. Which path takes more heat?

Using the two paths from the path-dependence figure, both running A→B: which requires more heat input?

ΔU is the same for both (same endpoints). From Q = ΔU − W_on, and with W_on = −(area) for a rightward path,

Q = ΔU + area Bigger area ⇒ more heat required.

So Path 1 (expand at high pressure first) needs more heat. It also delivers more work. That is not a coincidence — the extra heat is precisely what pays for the extra work, since the internal energy change is fixed.

Checklist

I can name the process from the shape of the curve, instantly.
I get the sign of W from the direction of travel, not from memorized cases.
I compute U = ³⁄₂PV at every corner before doing anything else.
I know clockwise = engine and counterclockwise = refrigerator.
I check that ΣΔU = 0 around every closed cycle.