- 2071 Magh · 6 marks
Entropy relations for an ideal gas
Short answer
Entropy is a property measuring disorder; for a reversible process dS = (δQ/T)_rev. For an ideal gas: s₂−s₁ = cv ln(T₂/T₁) + R ln(v₂/v₁) = cp ln(T₂/T₁) − R ln(P₂/P₁) = cv ln(P₂/P₁) + cp ln(v₂/v₁). With variable specific heats, s₂−s₁ = s₂°−s₁° − R ln(P₂/P₁).
Define entropy. Derive in detail mathematical expression for entropy relations of an ideal gas.
Entropy is a property that measures the molecular disorder of a system; for a reversible process dS = (δQ/T)_rev (unit kJ/K).
Starting equations
For a pure compressible substance (per unit mass), from the first law δq = du + P dv and δq_rev = T ds:
(1) T ds = du + P dv (2) T ds = dh − v dP (since h = u + Pv, dh = du + P dv + v dP)
For an ideal gas: Pv = RT, du = cv dT, dh = cp dT, and R = cp − cv.
Relation 1: T and v
From (1): ds = du/T + (P/T) dv = cv dT/T + R dv/v
Integrating with constant cv:
s₂ − s₁ = cv ln(T₂/T₁) + R ln(v₂/v₁)
Relation 2: T and P
From (2): ds = dh/T − (v/T) dP = cp dT/T − R dP/P
s₂ − s₁ = cp ln(T₂/T₁) − R ln(P₂/P₁)
Relation 3: P and v
From T₂/T₁ = P₂v₂/(P₁v₁) in relation 1 (and cv + R = cp):
s₂ − s₁ = cv ln(P₂/P₁) + cp ln(v₂/v₁)
Variable specific heats
When cp changes with T, define the absolute entropy at 1 atm: s°(T) = ∫₀ᵀ cp dT/T. Then
s₂ − s₁ = s₂° − s₁° − R ln(P₂/P₁)
with s° read from gas tables (e.g. air tables).
Special cases
- Isothermal: s₂ − s₁ = R ln(v₂/v₁) = −R ln(P₂/P₁).
- Isentropic (s₂ = s₁): T₂/T₁ = (v₁/v₂)^(k−1) = (P₂/P₁)^((k−1)/k).
Example: air heated from 300 K to 600 K at constant pressure: Δs = 1.005 ln 2 = 0.697 kJ/kg·K.
✓ Answer checked by Claude Opus 5.5 (high effort)
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