Carnot cycle — four stages with PV diagram interpretation

Question

What are the four stages of the Carnot cycle? How do we interpret each stage on a PV diagram, and what is the efficiency formula?

Solution — Step by Step

The gas expands slowly while in contact with the hot reservoir at temperature $T_1$ . Heat $Q_1$ is absorbed from the reservoir.

Since temperature is constant ( $\Delta T = 0$ ), internal energy does not change ( $\Delta U = 0$ ). All absorbed heat converts to work:

Q_1 = W_1 = nRT_1 \ln\left(\frac{V_B}{V_A}\right)

On the PV diagram, this is a hyperbolic curve (isothermal) from A to B.

The gas is insulated (no heat exchange, $Q = 0$ ) and continues to expand. It does work at the expense of its internal energy, so temperature drops from $T_1$ to $T_2$ .

W_2 = nC_v(T_1 - T_2)

On the PV diagram, this is a steeper curve (adiabatic) from B to C. The gas reaches the cold temperature $T_2$ .

The gas is compressed while in contact with the cold reservoir at $T_2$ . Heat $Q_2$ is released to the cold reservoir.

Q_2 = W_3 = nRT_2 \ln\left(\frac{V_C}{V_D}\right)

On the PV diagram, this is a hyperbolic curve from C to D (going left — compression).

The gas is insulated again and compressed further. Temperature rises from $T_2$ back to $T_1$ , returning the gas to its initial state.

W_4 = nC_v(T_2 - T_1)

The cycle is complete. The net work done is the area enclosed by the four curves on the PV diagram.

graph TD
    A["A: Start - high T, low V"] -->|Isothermal expansion at T1| B["B: high T, high V"]
    B -->|Adiabatic expansion| C["C: low T, highest V"]
    C -->|Isothermal compression at T2| D["D: low T, lower V"]
    D -->|Adiabatic compression| A

Why This Works

The Carnot efficiency is:

\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{Q_2}{Q_1}

where $T_1$ and $T_2$ are in Kelvin (absolute temperature).

This is the maximum possible efficiency for any heat engine operating between temperatures $T_1$ and $T_2$ . No real engine can exceed Carnot efficiency — this is a direct consequence of the second law of thermodynamics.

Stage	Process	Heat	Work	Temperature
A $\to$ B	Isothermal expansion	$+Q_1$ (absorbed)	Positive (by gas)	Constant $T_1$
B $\to$ C	Adiabatic expansion	0	Positive (by gas)	Drops $T_1 \to T_2$
C $\to$ D	Isothermal compression	$-Q_2$ (released)	Negative (on gas)	Constant $T_2$
D $\to$ A	Adiabatic compression	0	Negative (on gas)	Rises $T_2 \to T_1$

Alternative Method

For JEE numericals, if only temperatures are given:

\eta = 1 - \frac{T_2}{T_1} \times 100\%

Example: Hot source = 500 K, cold sink = 300 K.

\eta = 1 - \frac{300}{500} = 0.4 = 40\%

If the engine absorbs 1000 J from the hot source, it does 400 J of work and rejects 600 J to the cold sink.

A common JEE question: “How to increase Carnot efficiency?” Either increase $T_1$ or decrease $T_2$ . Lowering $T_2$ is more effective — reducing it to 0 K (impossible but theoretical) would give 100% efficiency.

Common Mistake

Students use Celsius instead of Kelvin in the Carnot efficiency formula. If the source is at 200 degrees C and the sink is at 50 degrees C, the efficiency is NOT $1 - 50/200 = 75\%$ . Convert to Kelvin first: $T_1 = 473$ K, $T_2 = 323$ K, so $\eta = 1 - 323/473 = 31.7\%$ . This error appears in almost every thermodynamics exam and is one of the easiest marks to lose.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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