Carnot cycle — four stages with PV diagram interpretation

Question

Describe the four stages of the Carnot cycle. What happens during each stage, and what is the shape of the PV diagram? Derive the expression for Carnot efficiency.

(JEE Main 2023 asked about the efficiency formula; NEET tests the process identification)

Solution — Step by Step

The gas is in contact with the hot reservoir at temperature $T_H$ . It expands slowly, absorbing heat $Q_H$ from the reservoir. Temperature stays constant because heat absorbed = work done.

W_1 = nRT_H \ln\frac{V_B}{V_A} = Q_H

The gas is now insulated (no heat exchange). It continues to expand, doing work at the expense of internal energy. Temperature drops from $T_H$ to $T_C$ (cold reservoir temperature).

W_2 = nC_v(T_H - T_C)

The gas contacts the cold reservoir at $T_C$ . It is compressed, releasing heat $Q_C$ to the cold reservoir. Temperature stays constant.

W_3 = nRT_C \ln\frac{V_D}{V_C} = -Q_C

The gas is insulated again and compressed further. Temperature rises from $T_C$ back to $T_H$ , returning to the starting state.

Carnot efficiency:

\eta = 1 - \frac{T_C}{T_H}

This is the maximum possible efficiency for any heat engine operating between $T_H$ and $T_C$ . No real engine can exceed this.

flowchart TD
    A["A: Start<br/>High T, Low V"] -->|"Stage 1: Isothermal expansion<br/>Absorbs Q_H at T_H"| B["B: High T, Higher V"]
    B -->|"Stage 2: Adiabatic expansion<br/>T drops to T_C"| C["C: Low T, Highest V"]
    C -->|"Stage 3: Isothermal compression<br/>Releases Q_C at T_C"| D["D: Low T, Lower V"]
    D -->|"Stage 4: Adiabatic compression<br/>T rises to T_H"| A
    style A fill:#ff6b6b,stroke:#333
    style C fill:#87CEEB,stroke:#333

Why This Works

The Carnot cycle is an idealised cycle — all processes are reversible (quasi-static). It consists of two isothermal processes (where heat exchange occurs) and two adiabatic processes (where temperature changes without heat exchange).

The efficiency depends ONLY on the temperatures — not on the working substance, the gas, or anything else. This universality is what makes the Carnot theorem so powerful. It sets an absolute upper limit that no real engine can surpass.

For a cycle operating between 500 K and 300 K: $\eta = 1 - 300/500 = 0.4 = 40\%$ . Even the most perfectly designed engine between these temperatures cannot exceed 40% efficiency.

Alternative Method

The Carnot efficiency formula can be rearranged to find the minimum work needed for a refrigerator. The coefficient of performance (COP) of a Carnot refrigerator is: $\text{COP} = \frac{T_C}{T_H - T_C}$ . JEE problems sometimes ask for the refrigerator version — just remember that a refrigerator is a reverse Carnot cycle.

Common Mistake

Students use Celsius temperatures in the Carnot efficiency formula. The formula $\eta = 1 - T_C/T_H$ requires absolute temperatures (Kelvin). Using Celsius gives a completely wrong answer. For example, between 227°C and 27°C: correct answer is $1 - 300/500 = 40\%$ . Using Celsius: $1 - 27/227 = 88\%$ — completely wrong. Always convert to Kelvin first.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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