---
title: "Entropy relations for an ideal gas"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 5
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:entropy-ideal-gas-relations"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/entropy-ideal-gas-relations
site: BCT Engineering
---

# Entropy relations for an ideal gas

Fundamental of Thermodynamics & Heat Transfer · Chapter 5: Second Law of Thermodynamics

- Asked in: 2071 Magh, 6 marks
- Video: [Entropy Change of Ideal Gases | Thermodynamics | (Solved Examples)](https://www.youtube.com/watch?v=IHXCdTVAHho) — Question Solutions
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/entropy-ideal-gas-relations
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-5

## Question

Define entropy. Derive in detail mathematical expression for entropy relations of 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).

## Answer

**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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The ideal gas entropy relations start from:
   A. T ds = du + P dv and T ds = dh − v dP
   B. T ds = du − P dv
   C. T ds = dh + v dP
   D. ds = dQ
   Answer: A. These come from the first law with δq_rev = T ds. (Key point: Starting Tds equations)

2. The T-v entropy relation for constant cv is:
   A. s2 − s1 = cv ln(T2/T1) − R ln(v2/v1)
   B. s2 − s1 = cp ln(T2/T1) + R ln(v2/v1)
   C. s2 − s1 = cv ln(T2/T1) + R ln(v2/v1)
   D. s2 − s1 = cv ln(v2/v1)
   Answer: C. From ds = cv dT/T + R dv/v. (Key point: T-v relation)

3. The T-P entropy relation is:
   A. s2 − s1 = cp ln(T2/T1) + R ln(P2/P1)
   B. s2 − s1 = cv ln(T2/T1) − R ln(P2/P1)
   C. s2 − s1 = cp ln(P2/P1)
   D. s2 − s1 = cp ln(T2/T1) − R ln(P2/P1)
   Answer: D. From ds = cp dT/T − R dP/P. (Key point: T-P relation)

4. When cp varies with T, entropy change uses:
   A. the mean cp only
   B. s2 − s1 = s2° − s1° − R ln(P2/P1) with tabulated s°
   C. R ln(T2/T1)
   D. constant cv
   Answer: B. s° is the tabulated absolute entropy at 1 atm. (Key point: Variable specific heats)

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