---
title: "Isentropic relations for ideal gas and incompressible"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 5
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:isentropic-process-relations"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/isentropic-process-relations
site: BCT Engineering
---

# Isentropic relations for ideal gas and incompressible

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

- Asked in: 2072 Asoj, 6 marks; 2070 Magh, 6 marks (asked 2 times)
- Video: [Thermodynamics: Entropy for Ideal gas, incompressible and reversible processes](https://www.youtube.com/watch?v=tohdFx36d9w) — iChamander
- 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/isentropic-process-relations
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-5

## Question

Define entropy. Derive isentropic relations for an ideal gas and an incompressible substance.

## Short answer

Entropy: dS=(δQ/T)_rev. Isentropic ideal gas (constant k): T2/T1 = (v1/v2)^(k−1) = (P2/P1)^((k−1)/k), P v^k = constant. Incompressible substance (v const, specific heat c): ds = c dT/T, isentropic ⇒ T2=T1, no temperature change.

## Answer

**Entropy** is a property that measures the disorder of a system; for a reversible process dS = (δQ/T)_rev. An **isentropic process** is one with constant entropy (reversible and adiabatic).

The starting point is the two T ds relations (for a simple compressible substance, per unit mass):

T ds = du + P dv
T ds = dh − v dP

#### Ideal gas (constant specific heats)
Use du = cv dT and Pv = RT. With ds = 0:

0 = cv dT/T + R dv/v

Integrate between states 1 and 2, using R = cp − cv and k = cp/cv:

cv ln(T₂/T₁) + R ln(v₂/v₁) = 0
ln(T₂/T₁) = −(R/cv) ln(v₂/v₁) = (k−1) ln(v₁/v₂)

**T₂/T₁ = (v₁/v₂)^(k−1)**

Similarly, from T ds = dh − v dP: 0 = cp dT/T − R dP/P, so

**T₂/T₁ = (P₂/P₁)^((k−1)/k)**

Eliminating T using Pv = RT:

**P₂/P₁ = (v₁/v₂)^k**, i.e. Pv^k = constant

Example: air (k = 1.4) compressed isentropically from 100 kPa to 800 kPa at 300 K gives T₂ = 300 × 8^(0.2857) = 543.4 K.

#### Incompressible substance
For a solid or liquid, v = constant, so dv = 0 and du = c dT (c = cp ≈ cv). Then

ds = du/T = c dT/T

For constant specific heat:

s₂ − s₁ = c ln(T₂/T₁)

For an isentropic process, s₂ = s₁:

c ln(T₂/T₁) = 0, so **T₂ = T₁**

An isentropic process of an incompressible substance is also isothermal; there is no temperature change (the internal energy stays constant, u₂ = u₁).

## Check yourself

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

1. An isentropic process is one that is:
   A. isothermal and reversible
   B. irreversible and adiabatic
   C. reversible and adiabatic, with constant entropy
   D. constant pressure and reversible
   Answer: C. It has no entropy generation and no heat transfer, so s is constant. (Key point: Isentropic process)

2. For an ideal gas with constant specific heats, isentropic T-v relation is:
   A. T2/T1 = (v2/v1)^(k−1)
   B. T2/T1 = (v1/v2)^(k−1)
   C. T2/T1 = (v1/v2)^k
   D. T2/T1 = (P1/P2)^((k−1)/k)
   Answer: B. From cv ln(T2/T1) + R ln(v2/v1) = 0, with R/cv = k − 1. (Key point: Ideal gas T-v relation)

3. Eliminating T gives the isentropic relation:
   A. P v = constant
   B. P v^k = constant
   C. P/v^k = constant
   D. P^k v = constant
   Answer: B. Combining the T-v relation with Pv = RT gives P2/P1 = (v1/v2)^k. (Key point: Pv^k)

4. For an incompressible substance, an isentropic process means:
   A. temperature rises
   B. pressure stays constant
   C. temperature falls
   D. temperature stays constant (T2 = T1)
   Answer: D. ds = c dT/T = 0 requires dT = 0. (Key point: Incompressible isentropic)

## Related in Fundamental of Thermodynamics & Heat Transfer

- [Entropy relations: ideal gases and incompressible substance](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Aentropy-relations-ideal-gas-incompressible) — Fundamental of Thermodynamics & Heat Transfer, chapter 5
- [Entropy relations for an ideal gas](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Aentropy-ideal-gas-relations) — Fundamental of Thermodynamics & Heat Transfer, chapter 5