---
title: "Gas work: constant P, constant T, PV^1.35"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 2
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:gas-work-three-processes-150kpa"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/gas-work-three-processes-150kpa
site: BCT Engineering
---

# Gas work: constant P, constant T, PV^1.35

Fundamental of Thermodynamics & Heat Transfer · Chapter 2: Energy and Energy Transfer

- Asked in: 2070 Bhadra, 6 marks
- 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/gas-work-three-processes-150kpa
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-2

## Question

A gas is contained in a piston cylinder device initially at a pressure of 150 kPa and a volume of 0.04 m³. Calculate the work done by the gas when it undergoes the following processes to a final volume of 0.1 m³, (i) Constant-Pressure (ii) Constant temperature (iii) PV^1.35 = constant.

## Short answer

Constant pressure: W = P(V₂-V₁) = 9.0 kJ. Isothermal: W = P₁V₁ ln(V₂/V₁) = 5.50 kJ. Polytropic (n=1.35): P₂ = P₁(V₁/V₂)^1.35, then W = (P₂V₂ - P₁V₁)/(1-n) = 4.70 kJ.

## Answer

**Given:** P₁ = 150 kPa, V₁ = 0.04 m³, V₂ = 0.1 m³. Using W = ∫P dV (kPa × m³ = kJ).

#### (i) Constant pressure
W = P(V₂ − V₁) = 150 × (0.1 − 0.04) = **9.0 kJ**

#### (ii) Constant temperature (PV = C)
W = P₁V₁ ln(V₂/V₁) = 150 × 0.04 × ln(0.1/0.04)  
 = 6.0 × ln(2.5) = 6.0 × 0.9163 = **5.50 kJ**

#### (iii) PV^1.35 = constant
P₂ = P₁ (V₁/V₂)^1.35 = 150 × (0.4)^1.35 = 150 × 0.2903 = 43.54 kPa

W = (P₂V₂ − P₁V₁)/(1 − n) = (43.54 × 0.1 − 150 × 0.04)/(1 − 1.35)  
 = (4.354 − 6.0)/(−0.35) = **4.70 kJ**

**Answer: W = 9.0 kJ (isobaric), 5.50 kJ (isothermal), 4.70 kJ (polytropic, n = 1.35).** The areas under the curves on the P-V diagram decrease in this order, since the pressure falls faster for larger n.

## Check yourself

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

1. At 150 kPa from 0.04 to 0.1 m^3, constant-pressure work is
   A. 15 kJ
   B. 9.0 kJ
   C. 6.0 kJ
   D. 60 kJ
   Answer: B. W = 150 x (0.1 - 0.04) = 9.0 kJ. (Key point: Constant pressure work)

2. Isothermal work is P1V1 ln(V2/V1) = 6.0 x ln 2.5, which is about
   A. 5.50 kJ
   B. 9.0 kJ
   C. 15 kJ
   D. 2.4 kJ
   Answer: A. ln 2.5 = 0.9163, so W = 5.50 kJ. (Key point: Isothermal work)

3. For PV^1.35 = C, P2 = 43.54 kPa. Work is (P2V2 - P1V1)/(1 - n), which gives
   A. 4.70 kJ with both numerator and denominator negative
   B. -4.70 kJ
   C. 9.0 kJ
   D. 47.0 kJ
   Answer: A. (4.354 - 6.0)/(-0.35) = 4.70 kJ, positive for expansion. (Key point: Polytropic work)

4. Why is the work smallest for the polytropic n = 1.35 process?
   A. Pressure falls faster with larger n, so the area under the curve is smaller
   B. Its volume change is smaller
   C. It has constant pressure
   D. It has a higher initial pressure
   Answer: A. Larger n makes pressure drop faster, reducing the area under the P-V curve: 9.0 > 5.50 > 4.70 kJ. (Key point: Ordering)

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