**Clausius’ theorem**

*The cyclic integral of ∂Q/T for a reversible cycle is always equal to zero.*

Proof of Clausius’ theorem:

Let’s consider a reversible heat engine.

From the absolute thermodynamic scale of temperature, we know that for a reversible process

Q_{1}/Q_{2} = T_{1}/T_{2}

Q_{1}/T_{1} – Q_{2}/T_{2} = 0

Q_{1}/T_{1} + (-Q_{2})/T_{2} = 0

Σ_{CYCLE} (Q/T) = 0

In terms of cyclic integral we can write above equation as

∫_{CYCLE} (∂Q_{R}/T) = 0 … (1)

Hence proved that the cyclic integral of ∂Q/T for a reversible cycle is always equal to zero.

**Clausius’ inequality**

*The cyclic integral of ∂Q/T for an irreversible process is always less than ∂Q/T for a reversible process (both operating between the same temperature limits).*

Proof of Clausius’ inequality:

Let’s consider two heat engines one is reversible and another is irreversible, operating between the same temperature limits.

Efficiency of reversible engine

η = 1 – (Q_{2}/Q_{1})

Efficiency of irreversible engine

η_{R} = 1 – (Q_{2R}/Q_{1R})

From the Carnot’s theorem we know that

*Efficiency of a reversible heat engine is always more than the efficiency of an irreversible heat engine (operating between the same temperature limits).*

η_{R} > η

1 – (Q_{2R}/Q_{1R}) > 1 – (Q_{2}/Q_{1})

Q_{2R}/Q_{1R} < Q_{2}/Q_{1} … (2)

From the absolute thermodynamic scale of temperature, we know that for a reversible process

Q_{2R}/Q_{1R} = T_{1}/T_{2}

Putting above values in equation (2), we get

T_{2}/T_{1} < Q_{2}/Q_{1}

(Q_{1}/T_{1}) – (Q_{2}/T_{2}) < 0

∫_{CYCLE} (dQ/T) < 0

Comparing above equation with equation (1) we get.

∫_{CYCLE} (dQ/T) < ∫_{CYCLE} (dQ_{R}/T) = 0

*Hence proved that the cyclic integral of ∂Q/T for an irreversible process is always less than ∂Q/T for a reversible process (both operating between the same temperature limits).*

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