Topic 8.4.49 min read
Isobaric Process
An isobaric process occurs at constant pressure. Heat transfer can change internal energy and produce expansion work, so Q = ΔU + PΔV.
Choose Your Level
Beginner Level
When a gas is heated under a freely moving piston with a constant load, it expands while pressure stays fixed. The heat both raises internal energy and pushes the piston.
Recommended for Class 11
Isobaric Process
Overview
An isobaric process occurs at constant pressure. Heat transfer can change internal energy and produce expansion work, so Q = ΔU + PΔV.
Learning Outcomes
- Pressure remains constant: P₂ = P₁.
- Work done by the gas is W = PΔV.
- The first law becomes Q = ΔU + PΔV.
- During heating, both volume and internal energy may increase.
- The P–V graph is a horizontal line called an isobar.
- The rectangular area under an isobar equals work.
Intermediate Level
Condition and Energy Balance
An isobaric process keeps pressure constant while volume and temperature may change. A slowly moving piston under a constant external load is a useful model.
Piston Derivation of Work
For piston area A, the gas force is F = PA. If the piston moves by Δy, then ΔV = AΔy. Therefore W = FΔy = PAΔy = PΔV.
P–V Graph
An isobar is horizontal because pressure is unchanged. The area of the rectangle under the path has height P and width ΔV, so it equals PΔV.
| Change | ΔV | W by gas |
|---|---|---|
| expansion | positive | positive |
| compression | negative | negative |
| no volume change | zero | zero |
Constant-Pressure Expansion
A gas expands from 0.020 m³ to 0.035 m³ at 200 kPa while its internal energy increases by 1.5 kJ. Find W and Q.
ΔV = 0.035 − 0.020 = 0.015 m³
W = PΔV = (200,000 Pa)(0.015 m³) = 3000 J
Q = ΔU + W = 1500 J + 3000 J
Final answer: W = 3.0 kJ and Q = 4.5 kJ
Textbook Sign Example
A gas has ΔU = −200 J and does W = +50 J. Find the heat transfer.
Q = ΔU + W
Q = −200 J + 50 J
Final answer: Q = −150 J
Examples
- Boiling water to steam in an open container
- Slow expansion beneath a piston with constant load
- Phase change under constant atmospheric pressure
Common Mistakes
- Constant pressure does not make work zero if volume changes.
- Convert pressure to pascals when volume is in m³ to obtain joules.
- Use ΔV = V₂ − V₁ with its sign.
- An isobar is horizontal on a P–V graph.
Short-Question Preparation
Q. Define an isobaric process.
A. A thermodynamic process in which pressure remains constant.
Q. Write work done at constant pressure.
A. W = PΔV = P(V₂ − V₁).
Q. How does the first law appear?
A. Q = ΔU + PΔV.
Q. What does the area under an isobar represent?
A. The work done by the gas.
Sticky Notes
- Pressure remains constant: P₂ = P₁.
- Work done by the gas is W = PΔV.
- The first law becomes Q = ΔU + PΔV.
- During heating, both volume and internal energy may increase.
- The P–V graph is a horizontal line called an isobar.
- The rectangular area under an isobar equals work.
60-Second Revision
Isobaric means P is constant. Work is W = PΔV, so Q = ΔU + PΔV. Expansion gives positive work. The P–V path is a horizontal isobar whose rectangular area equals work.
Summary
An isobaric process occurs at constant pressure. Heat transfer can change internal energy and produce expansion work, so Q = ΔU + PΔV.
Ask-the-Tutor Modes
- Explain the energy transfers simply.
- Apply the textbook sign convention.
- Derive the reduced equation.
- Solve a numerical step by step.
- Explain the P–V graph.
- Test me with an MCQ.
Exam Focus
Exam Focus
Important points
- Pressure remains constant: P₂ = P₁.
- Work done by the gas is W = PΔV.
- The first law becomes Q = ΔU + PΔV.
- During heating, both volume and internal energy may increase.
- The P–V graph is a horizontal line called an isobar.
- The rectangular area under an isobar equals work.
Important formulas
- P₂ = P₁
- W = PΔV
- Q = ΔU + PΔV
Common exam traps
- Constant pressure does not make work zero if volume changes.
- Convert pressure to pascals when volume is in m³ to obtain joules.
- Use ΔV = V₂ − V₁ with its sign.
- An isobar is horizontal on a P–V graph.
Long question structure
Definition → condition → first-law reduction → graph → physical explanation → example.
Common Mistakes
Common Mistakes
- Mistake
Constant pressure does not make work zero if volume changes.
- Mistake
Convert pressure to pascals when volume is in m³ to obtain joules.
- Mistake
Use ΔV = V₂ − V₁ with its sign.
- Mistake
An isobar is horizontal on a P–V graph.
Revision Zone
60-Second Revision
Quick cues from this topic — use them for last-minute revision.
Pressure remains constant: P₂ = P₁.
Work done by the gas is W = PΔV.
The first law becomes Q = ΔU + PΔV.
During heating, both volume and internal energy may increase.
An isobaric process occurs at constant pressure. Heat transfer can change internal energy and produce expansion work, so Q = ΔU + PΔV.
- Definition
- An isobaric process occurs at constant pressure. Heat transfer can change internal energy and produce expansion work, so Q = ΔU + PΔV.
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Exam Practice
Important Question
Q1. Define an isobaric process.
Important Question
Q2. Write work done at constant pressure.
Important Question
Q3. How does the first law appear?
Important Question
Q4. What does the area under an isobar represent?