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A common thermodynamics question shows a cycle on a PT or VT graph and asks for the work done, or asks which PV graph matches it. Students who can read PV graphs well often stall here, because the shapes change. This post gives you the rules for each pair of axes, converts one cycle across all three graphs with numbers, and shows the trap in the clockwise rule.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus list the first law of thermodynamics and isothermal and adiabatic processes. Section 11.8 of the NCERT Class 11 Physics textbook (2026-27 reprint) defines isothermal, adiabatic, isochoric and isobaric processes and the work in each. Graph conversion is not a separate syllabus line; it is the skill exams use to test these processes. For the work formulas themselves, see our guide to isothermal and adiabatic processes.
What does each process look like on each graph?
Everything follows from pV = nRT. Hold one variable fixed and see what is left.
Isochoric (V constant).
PV: vertical line.
PT: P = (nR/V)T, a straight line through the origin. A steeper line means a smaller volume.
VT: horizontal line.
Isobaric (P constant).
PV: horizontal line.
PT: horizontal line.
VT: V = (nR/P)T, a straight line through the origin. A steeper line means a lower pressure.
Isothermal (T constant).
PV: a rectangular hyperbola, PV = constant.
PT: vertical line.
VT: vertical line.
Worked example: one cycle on three graphs
One mole of an ideal monatomic gas goes round a cycle. Take R = 8.314 J mol⁻¹ K⁻¹, as given by NIST's CODATA value to four significant figures.
A: P = 100 kPa, T = 300 K, so V = nRT/P = 8.314 × 300 / 10⁵ = 0.0249 m³ (24.9 L).
A → B: heated at constant volume to 600 K. Pressure doubles to 200 kPa.
B → C: expands isothermally at 600 K until P = 100 kPa, so V doubles to 49.9 L.
C → A: cooled at constant pressure back to A.
On PV axes: a vertical line up from A to B, a hyperbola down and to the right from B to C, and a horizontal line left from C to A. The path runs clockwise.
On PT axes (T across, P up): A → B is a straight line through the origin (isochoric), from (300 K, 100 kPa) to (600 K, 200 kPa). B → C is a vertical line at 600 K, from 200 down to 100 kPa. C → A is horizontal at 100 kPa. This path also runs clockwise.
On VT axes (T across, V up): A → B is horizontal at 24.9 L. B → C is vertical at 600 K, from 24.9 up to 49.9 L. C → A is a straight line through the origin (isobaric). This path runs anticlockwise.
Work, from the PV graph only:
A → B: no volume change, W = 0.
B → C: W = nRT ln(V_C/V_B) = 8.314 × 600 × ln 2 = 3458 J.
C → A: W = PΔV = 10⁵ × (0.0249 − 0.0499) = −2494 J.
Net work by the gas: 3458 − 2494 = 963 J, positive, as the clockwise PV loop says.
Check with heat. For a monatomic gas, Cv = 3R/2 and Cp = 5R/2. Q(A → B) = 1 × 12.47 × 300 = 3741 J. Q(B → C) = W = 3458 J, since ΔU = 0 on an isotherm. Q(C → A) = 1 × 20.79 × (−300) = −6236 J. Net heat = 3741 + 3458 − 6236 = 963 J. Over a full cycle ΔU = 0, so net heat must equal net work, and it does.
Why does the clockwise rule fail on other graphs?
The area enclosed on a PV graph is the work, because work is the integral of P dV. On PT or VT axes, the enclosed area has no such meaning, and the sense of rotation can flip, as it did on the VT graph above. If you see a cycle on PT or VT axes, redraw it on PV first, or work out each leg's work from the change in volume.
This is not a small slip. In a study of 653 students in three university physics courses, Meltzer (2004) found "a widespread and persistent tendency" to treat work and heat as if they were state functions. A large majority of the students he interviewed thought that the net work or net heat in a cyclic process must be zero. The worked example shows why that is wrong: the gas returns to A, but it has done 963 J of net work.
Which mistakes cost marks?
Calling a sloping line isobaric on a PT graph. On PT axes, a line through the origin is isochoric.
Reading work as an area on a VT or PT graph. Only the PV area is work.
Assuming net work in a cycle is zero. Only ΔU is zero over a cycle.
Mixing up steepness. On a PT graph, steeper isochores mean smaller volume. On a VT graph, steeper isobars mean lower pressure.
Using litres with kPa and R = 8.314. Convert to m³ and Pa, or remember that 1 kPa × 1 L = 1 J.
A practice routine
Draw the three-by-three grid of process shapes from memory and check it against the lists above.
Take a cycle from your textbook drawn on PV axes and redraw it on PT and VT axes. Mark which way it runs on each.
Find the net work for each cycle from the PV graph, then check with Q = W over the full cycle.
Finish with one question from our thermodynamics sign conventions guide so the signs stay straight.
If you want to watch a state point move along an isotherm while the other graphs update, the Fermi Simulation Lab has interactive simulations for ideas easier to see than read.
Today, redraw the worked cycle on all three axes without looking, then recompute its net work.
