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Thermodynamics · Part 3 of 10

Indicator Diagrams, State Variables and Cycles

A P–V graph of a process, the indicator diagram, shows a gas's states, the path between them and the work done, all at once. This part reads it carefully: why work and heat depend on the path but internal energy doesn't, and how a closed loop gives the net work and heat of a cycle, with a sign set by its direction.

Builds on: Part 2 · First Law of Thermodynamics.

Video coming soon

Indicator diagram

Point, curve, area

Each point is a state ( $P$ , $V$ , and $T = P V / n R$ ). A curve is the process $P = f (V)$ . The work done by the gas is the area under it: $W = \int_{V_{1}}^{V_{2}} f (V) d V$ , positive for expansion and negative for compression.

A curve between two states on a P–V diagram, area shaded. — The shaded area is the work done by the gas.

Two paths between the same states with different areas. — Paths I and III between the same two states.

State and path variables

Same ends, different W

Path I (high pressure, then down) encloses more area than path III (down, then low pressure), so W differs; Q differs too. But $Δ U = Q - W$ is the same for every path: U is a state variable, Q and W are path variables.

Cyclic process

$Q_{n e t} = W_{n e t}$ = loop area

A cycle returns the gas to its start, so $Δ U = 0$ and $Q_{n e t} = W_{n e t}$ , the area enclosed. Clockwise: work done by the gas, heat absorbed (engine). Anticlockwise: work done on the gas, heat rejected (refrigerator).

A closed clockwise loop on a P–V diagram. — Clockwise loop: net work done by the gas.

Triangle cycle ABC traced anticlockwise. — Leg by leg: +20 J, 0, −40 J.

Sign of a cycle

Anticlockwise: $W_{n e t} < 0$

A (1 m³, 10 Pa) → B (3 m³, 10 Pa) → C (3 m³, 30 Pa) → A is anticlockwise. Area $= \frac{1}{2} \times 2 \times 20 = 20$ J, so $W_{n e t} = - 20$ J: done on the gas.

Summary

Key formulas

W = \int_{V_{1}}^{V_{2}} f (V) d V

Properties of an Indicator Diagram

Δ U_{12} = Q - W (any path)

State Variable and Path Variables

Δ U_{cycle} = 0

Cyclic Processes

Q_{net} = W_{net} = loop area

Cyclic Processes

Worked examples

One for every idea

Properties of an Indicator Diagram

1. A gas is taken from state 1 to state 2 along two different paths on a P–V diagram. Is the work done by the gas necessarily the same along both paths?

Work is the area under the path on the P–V diagram.
Two different paths between the same states generally enclose different areas.

No: work depends on the path, not just the end states.

State Variable and Path Variables

2. A system goes from state I to state F along path IAF, absorbing 55 J of heat and doing 25 J of work. Find the change in internal energy between I and F. If path IBF does 10 J of work, how much heat does it absorb?

$Δ U = Q - W = 55 - 25 = 30$ J.
ΔU is the same along IBF: $Q = Δ U + W = 30 + 10$ .

$Δ U = 30$ J for any path; along IBF, $Q = 40$ J. (Illustrative Example 3.6)

Cyclic Processes

3. One mole of an ideal gas goes round the cycle ABCA on a V–T diagram: AB at constant volume from 300 K to 500 K; BC isothermal at 500 K until the volume is 5/3 of its starting value; CA a straight line through the origin back to A. Find the net heat supplied in one cycle.

CA is a line through the origin on a V–T graph, so V ∝ T: constant pressure. (5V₀/3 at 500 K matches V₀ at 300 K.)
AB: $W = 0$ . BC: $W = R T ln \frac{V _{C}}{V _{B}} = 500 R ln \frac{5}{3}$ . CA: $W = R (T_{A} - T_{C}) = - 200 R$ .
$W_{n e t} = R (500 ln \frac{5}{3} - 200) = 8.314 \times 55.4$ ; ΔU = 0 over the cycle, so $Q_{n e t} = W_{n e t}$ .

$Q_{n e t} \approx 461$ J. (The chapter JSON's version of Illustrative Example 3.9 describes legs that can't form a cycle; its 2146 J uses a volume ratio of 5/2 from the book's figure. This consistent cycle teaches the same method.)

Positive and Negative Cycles

4. In a cycle ABCA, A = (1 m³, 10 N/m²), B = (3 m³, 10 N/m²), C = (3 m³, 30 N/m²), traced A → B → C → A. Find the net work done and say whether it is done by or on the gas.

Area of the triangle $= \frac{1}{2} \times 2 \times 20 = 20$ J.
A → B → C → A is anticlockwise on the P–V plane.
Check: AB +20 J, BC 0, CA −40 J.

$W_{n e t} = - 20$ J: 20 J of net work is done on the gas. (Illustrative Example 3.7)

JEE-style question

Your turn

A gas goes round a cycle that is anticlockwise on its P–V diagram and encloses an area of 50 J. Over one complete cycle:

(a)the gas does 50 J of work

(b)the gas absorbs 50 J of net heat

(c)the gas gives out 50 J of net heat

(d)ΔU = −50 J

Show the answer and the traps

Anticlockwise: 50 J of work is done on the gas. ΔU = 0, so Q = W = −50 J. The gas gives out 50 J of heat: option c.

Option a is the clockwise cycle. Option b flips the sign of the heat.

And ΔU is zero over any complete cycle, never −50 J.

Watch out

Common mistakes

Assuming equal ΔU means equal Q and WOnly

Q - W

is path-independent; Q and W separately change from path to path.

Assuming every cycle does positive workClockwise: work by the gas. Anticlockwise: work on the gas, heat rejected.

Forgetting ΔU = 0 over a full cycleBack at the start means the same T, so

Δ U = 0

and

Q_{n e t} = W_{n e t}

Practice

Try these

1. A gas expands from 2 m³ to 5 m³ at a constant pressure of 100 Pa. Find the work done by the gas.

$W = P Δ V = 300$ J.

2. From state A to B along path 1, a gas absorbs 80 J and does 30 J of work. Along path 2 it does 10 J of work. Find the heat absorbed along path 2.

$Δ U = 50$ J, so $Q_{2} = 50 + 10 = 60$ J.

3. A gas goes round a clockwise rectangular cycle between V = 1 and 3 m³ and P = 100 and 300 Pa. Find the net heat absorbed per cycle.

Area $= 2 \times 200 = 400$ J $= Q_{n e t}$ .

4. Over one cycle a gas absorbs 900 J of heat and rejects 600 J. Find the net work done by the gas.

$Δ U = 0$ , so $W_{n e t} = 900 - 600 = 300$ J.

Indicator Diagrams, State Variables and Cycles

Point, curve, area

Same ends, different W

Qnet​=Wnet​ = loop area

Anticlockwise: Wnet​<0

Key formulas

One for every idea

Your turn

Common mistakes

Try these

$Q_{n e t} = W_{n e t}$ = loop area

Anticlockwise: $W_{n e t} < 0$