---
title: "Heat vs work transfer; isothermal work"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 2
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:heat-vs-work-isothermal-work"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/heat-vs-work-isothermal-work
site: BCT Engineering
---

# Heat vs work transfer; isothermal work

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

- Asked in: 2072 Magh, 4 marks
- Video: [Work & Heat Transfer](https://www.youtube.com/watch?v=1qI7Ys1Of6o) — TutorialsPoint
- 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/heat-vs-work-isothermal-work
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-2

## Question

Differentiate between heat transfer and work transfer. Derive the mathematical expression for work transfer for an isothermal process.

## Short answer

Heat transfer is boundary energy transfer due only to temperature difference, disorganized; work transfer is from non-thermal potentials, organized. Both are path functions. For an ideal gas isothermal process, work W = P₁V₁ ln(V₂/V₁) = mRT ln(V₂/V₁), derived from W = ∫ P dV with PV = constant.

## Answer

#### Heat transfer vs work transfer

| Basis | Heat transfer | Work transfer |
|---|---|---|
| Cause | Temperature difference | Force acting through a distance (non-thermal potential) |
| Energy form | Disorganised, microscopic | Organised, macroscopic |
| Conversion | Not fully convertible to work in a cycle | Fully convertible to heat |
| Sign convention | Into system +ve | By system +ve |
| Mode | Conduction, convection, radiation | Boundary, shaft, electrical |
| Property? | No, path function | No, path function |

#### Work for an isothermal process (ideal gas)
Boundary work W₁₂ = ∫₁² P dV.  
For T = constant, PV = mRT = C, so P = mRT/V.

W₁₂ = mRT ∫₁² dV/V = mRT ln(V₂/V₁)

Since P₁V₁ = P₂V₂ = mRT:

**W₁₂ = P₁V₁ ln(V₂/V₁) = P₁V₁ ln(P₁/P₂)**

Work is positive for expansion (V₂ > V₁) and negative for compression. Because ΔU = 0 for an ideal gas at constant T, the first law gives Q₁₂ = W₁₂.

## Check yourself

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

1. Work transfer differs from heat transfer in that its driving potential is
   A. a temperature difference
   B. a non-thermal potential such as force or voltage
   C. always a pressure difference only
   D. none
   Answer: B. Work arises from force through a distance or other non-thermal potentials. (Key point: Heat vs work cause)

2. For an isothermal ideal-gas process, P in terms of V is
   A. P = mRT V
   B. P = mRT/V^2
   C. P = mRT/V
   D. P = constant
   Answer: C. PV = mRT = C, so P = mRT/V. (Key point: Isothermal pressure)

3. The isothermal work expression is
   A. mRT (V2 - V1)
   B. mRT/(V2 - V1)
   C. P1V1 ln(V1/V2)
   D. P1V1 ln(V2/V1)
   Answer: D. W = mRT ∫dV/V = mRT ln(V2/V1) = P1V1 ln(V2/V1). (Key point: Isothermal work)

4. Why is Q12 = W12 for an isothermal ideal gas?
   A. Because ΔU = 0 at constant T for an ideal gas
   B. Because work is a property
   C. Because pressure is constant
   D. Because volume is constant
   Answer: A. Internal energy of an ideal gas depends only on T, so ΔU = 0 and Q = W. (Key point: First law link)

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