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Entropy change for reversible heat and work transfer
Short answer
Entropy dS=(δQ/T)_rev, a property from Clausius inequality. Reversible heat transfer: dS=δQ_rev/T, ΔS=∫δQ_rev/T; at constant T ΔS=Q/T. Reversible work transfer: δQ=0 and no dissipation → dS=0, ΔS=0; work carries no entropy, process isentropic.
Define entropy. Derive the expressions for change in entropy for reversible heat transfer and reversible work transfer process.
Entropy is a property of a system that measures the molecular disorder (randomness) of the system. It is defined for a reversible process by dS = (δQ/T)_rev, and its unit is kJ/K. For any process, the entropy of an isolated system never decreases.
Entropy as a property (Clausius)
For any cycle, Clausius inequality gives ∮ δQ/T ≤ 0, with equality for a reversible cycle. Take two reversible paths A and B between states 1 and 2. The cycle 1-A-2-B-1 is reversible, so
∮ δQ/T = ∫₁² (δQ/T)_A − ∫₁² (δQ/T)_B = 0
Hence ∫(δQ/T)_rev is the same for every reversible path. It depends only on the end states, so it is a property, called entropy:
dS = (δQ/T)_rev, S₂ − S₁ = ∫₁² (δQ/T)_rev
Reversible heat transfer
A reversible heat transfer takes place across an infinitesimal temperature difference, at the temperature T of the reservoir or system. Then
dS = δQ_rev/T
- For heat added (δQ > 0), entropy increases.
- For heat removed (δQ < 0), entropy decreases.
- At constant temperature T: ΔS = Q/T.
Example: a reservoir at 500 K supplying 100 kJ reversibly: ΔS = −100/500 = −0.2 kJ/K.
Reversible work transfer
In a reversible work transfer there is no heat transfer (δQ = 0) and no friction or dissipation (as in a frictionless piston, spring or flywheel). Then
dS = δQ_rev/T = 0, so ΔS = 0
Work carries no entropy. A reversible work transfer process is isentropic (S = constant). If friction or any irreversibility is present, entropy is generated, so ΔS > 0.
Summary: dS = δQ_rev/T for heat transfer, and dS = 0 for reversible work transfer.
✓ Answer checked by Claude Opus 5.5 (high effort)
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