12 min readPathway sheetSection B long answer (5 marks), often with a choice: Carnot cycle with indicator diagram and efficiency, and why it cannot be built; Section A MCQ (1 mark) numericals on η = 1 − T₂/T₁; also a NEET favourite

Carnot cycle and its efficiency

An ideal Carnot engine runs isothermal and adiabatic expansion, then compression. Its efficiency η = 1 − T₂/T₁ is the highest any engine between two temperatures can reach.

Four stepsIsothermal expansion (A→B), adiabatic expansion (B→C), isothermal compression (C→D), adiabatic compression (D→A)

Efficiencyη = 1 − H₂/H₁ = 1 − T₂/T₁, with T in kelvin; for a Carnot engine H₂/H₁ = T₂/T₁

Net workW = H₁ − H₂ = area ABCD enclosed by the cycle on the P-V diagram

Key resultη does not depend on the working substance; η would be 100% only if T₂ = 0 K, which is impossible

Carnot's ideal engine

Sadi Carnot imagined a perfect, reversible engine to find the best that any engine could do. The working substance, one mole of an ideal gas, sits in a cylinder with insulating walls, a perfectly conducting base and a frictionless piston. Three things are needed: a source at high temperature T₁, a sink at low temperature T₂, and an insulating stand. Placing the cylinder on each of these in turn takes the gas through four steps, called the Carnot cycle.

The Carnot cycle in four steps

back to A (P₁, V₁) at T₁

Net work in one cycle = area ABCD on the P-V diagram.

The four operations

1
A to B, isothermal expansion: the cylinder stands on the source. The gas expands slowly at T₁ and takes in heat H₁. Work done by the gas W₁ = area under AB.
2
B to C, adiabatic expansion: the cylinder is moved to the insulating stand. The gas expands further, using its own internal energy, so it cools to T₂. Work W₂ = area under BC.
3
C to D, isothermal compression: the cylinder stands on the sink. The gas is compressed slowly at T₂ and gives out heat H₂. Work done on the gas W₃ = area under CD.
4
D to A, adiabatic compression: back on the insulating stand, the gas is compressed until it warms to T₁ and returns to A. Work done on the gas W₄ = area under DA.
5★
Net work W = (W₁ + W₂) − (W₃ + W₄) = area ABCD. The gas is back at its start, so ΔU = 0 and W = H₁ − H₂.

★ key step

How to draw the Carnot cycle on a P-V diagram

1
Draw axes: pressure P up the side and volume V along the bottom.
2
Mark A high on the left. Draw a gently falling curve to B and label it isothermal, T₁, with an arrow H₁ in.
3
From B, draw a steeper falling curve to C and label it adiabatic.
4
From C, draw a gently rising curve back to the left, to D. Label it isothermal, T₂, with an arrow H₂ out.
5
From D, draw a steep rising curve back to A and label it adiabatic.
6
Put arrows round the loop clockwise (A, B, C, D) and shade the inside: net work = area ABCD.

Efficiency of the Carnot engine, and why a real one is not possible

1
Efficiency η = work done ÷ heat taken from the source = (H₁ − H₂)/H₁ = 1 − H₂/H₁.
2
For a Carnot engine, H₂/H₁ = T₂/T₁, with the temperatures in kelvin. So η = 1 − T₂/T₁.
3
η does not depend on the working substance, only on the two temperatures.
4
η would be 1 (100%) only if T₂ = 0 K. That would turn all the heat into work, which breaks the second law. So η is always less than 1.
5
No engine working between the same two temperatures can be more efficient than a Carnot engine.
6★
Why a real Carnot engine is not possible: the isothermal steps need the piston to move extremely slowly, so heat has time to flow. The adiabatic steps need it to move extremely fast, so no heat flows. Perfect insulation, a perfectly conducting base and zero friction are also impossible. So every real engine is less efficient.

★ key step

Worked example: a Carnot engine

Problem

A Carnot engine works between a source at 227 °C and a sink at 27 °C. It takes 5000 J of heat from the source in each cycle. Find (a) its efficiency, (b) the work done per cycle and (c) the heat given to the sink. (d) What source temperature would give 50% efficiency with the same sink?

Working

Change to kelvin: T₁ = 227 + 273 = 500 K, T₂ = 27 + 273 = 300 K.
(a) η = 1 − T₂/T₁ = 1 − 300/500 = 0.40, that is 40%.
(b) W = ηH₁ = 0.40 × 5000 J = 2000 J.
(c) H₂ = H₁ − W = 5000 J − 2000 J = 3000 J.
(d) 0.50 = 1 − 300/T₁, so T₁ = 300 ÷ 0.50 = 600 K, that is 327 °C.

Answer

(a) 40%; (b) 2000 J; (c) 3000 J; (d) 600 K (327 °C)

In real life

Coal-fired thermal power stations turn heat into electricity through steam turbines. The Carnot result explains why engineers use the hottest safe steam and cool river or cooling-tower water as the sink.

Exam angle

5 marks: “Describe the four operations of a Carnot cycle with the help of an indicator diagram. Obtain its efficiency, and explain why a Carnot engine cannot be made in practice.” Isothermal expansion (H₁ in at T₁), adiabatic expansion (cools to T₂), isothermal compression (H₂ out at T₂), adiabatic compression (back to T₁). Net work = area ABCD = H₁ − H₂. η = 1 − H₂/H₁ = 1 − T₂/T₁. The isothermal steps need a very slow piston and the adiabatic steps a very fast one, with no friction or heat leak: impossible.

2 marks: “A Carnot engine works between 400 K and 300 K. Find its efficiency.” η = 1 − T₂/T₁ = 1 − 300/400 = 0.25, that is 25%.

2 marks: “A Carnot engine takes 300 cal of heat from a source at 500 K and rejects 180 cal to its sink. Find the temperature of the sink and the efficiency.” H₂/H₁ = T₂/T₁, so T₂ = 500 K × 180/300 = 300 K. η = 1 − 180/300 = 0.40, that is 40%.

Remember it

Carnot goes I-A-I-A: Isothermal and Adiabatic expansion, then Isothermal and Adiabatic compression. η = 1 − cold/hot, in kelvin.