Clausius’ Inequality and Clausius’ Theorem

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

Q₁/Q₂ = T₁/T₂

Q₁/T₁ – Q₂/T₂ = 0

Q₁/T₁ + (-Q₂)/T₂ = 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 – (Q₂/Q₁)

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 – (Q₂/Q₁)

Q_2R/Q_1R < Q₂/Q₁   … (2)

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

Q_2R/Q_1R = T₁/T₂

Putting above values in equation (2), we get

T₂/T₁ < Q₂/Q₁

(Q₁/T₁) – (Q₂/T₂) < 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).

Also read:

Concept of entropy

Carnot’s Heat Engine

What is a irreversible process?

What is Thermodynamic equilibrium?