Topic 8.4.37 min read
Isochoric Process
An isochoric process occurs at constant volume. Because the boundary does not move, no pressure-volume work is done and all net heat changes internal energy.
Choose Your Level
Beginner Level
Heat a gas in a rigid sealed container and its volume cannot change. With no expansion, the gas does no pressure-volume work, so heating raises its internal energy and pressure.
Recommended for Class 11
Isochoric Process
Overview
An isochoric process occurs at constant volume. Because the boundary does not move, no pressure-volume work is done and all net heat changes internal energy.
Learning Outcomes
- Volume remains constant: ΔV = 0.
- Pressure-volume work is zero: W = PΔV = 0.
- The first law reduces to Q = ΔU.
- Heating raises internal energy, temperature and pressure.
- The P–V graph is a vertical line called an isochor.
- A rigid closed container approximates an isochoric system.
Intermediate Level
Condition and First-Law Reduction
An isochoric or constant-volume process occurs when the system boundary cannot move. Since there is no volume change, pressure-volume work vanishes.
Heating and Cooling
| Energy transfer | Q | ΔU | T and P |
|---|---|---|---|
| heating | positive | increases | increase |
| cooling | negative | decreases | decrease |
P–V Graph
Because volume has one fixed value while pressure changes, the P–V graph is a vertical line. Its area width is zero, consistent with zero work.
Connection with Gay-Lussac’s Law
For a fixed amount of ideal gas at constant volume, pressure rises in direct proportion to kelvin temperature.
Heating a Rigid Container
A rigid container receives 320 J of heat. Neglect other forms of work. Find W and ΔU.
Constant volume: ΔV = 0
W = PΔV = 0
ΔU = Q − W = 320 J − 0
Final answer: W = 0 and ΔU = +320 J
Examples
- Heating gas in a fixed thermally conducting container
- Idealized rapid combustion in a fixed engine volume
- A sealed rigid vessel during heating or cooling
Common Mistakes
- Constant volume does not mean constant pressure.
- W = 0 because ΔV = 0, even if pressure changes.
- Heat supplied changes internal energy rather than expansion work.
- An isochor is vertical on a P–V graph.
Short-Question Preparation
Q. Define an isochoric process.
A. A thermodynamic process in which volume remains constant.
Q. Why is work zero?
A. W = PΔV and ΔV = 0.
Q. How does the first law reduce?
A. With W = 0, Q = ΔU.
Q. What is an isochor?
A. The vertical P–V line representing a constant-volume process.
Sticky Notes
- Volume remains constant: ΔV = 0.
- Pressure-volume work is zero: W = PΔV = 0.
- The first law reduces to Q = ΔU.
- Heating raises internal energy, temperature and pressure.
- The P–V graph is a vertical line called an isochor.
- A rigid closed container approximates an isochoric system.
60-Second Revision
Isochoric means ΔV = 0. Hence W = PΔV = 0 and Q = ΔU. Heating raises internal energy, temperature and pressure. The P–V graph is a vertical isochor.
Summary
An isochoric process occurs at constant volume. Because the boundary does not move, no pressure-volume work is done and all net heat changes internal energy.
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
- Volume remains constant: ΔV = 0.
- Pressure-volume work is zero: W = PΔV = 0.
- The first law reduces to Q = ΔU.
- Heating raises internal energy, temperature and pressure.
- The P–V graph is a vertical line called an isochor.
- A rigid closed container approximates an isochoric system.
Important formulas
- ΔV = 0
- W = 0
- Q = ΔU
- PT = constant
Common exam traps
- Constant volume does not mean constant pressure.
- W = 0 because ΔV = 0, even if pressure changes.
- Heat supplied changes internal energy rather than expansion work.
- An isochor is vertical on a P–V graph.
Long question structure
Definition → condition → first-law reduction → graph → physical explanation → example.
Common Mistakes
Common Mistakes
- Mistake
Constant volume does not mean constant pressure.
- Mistake
W = 0 because ΔV = 0, even if pressure changes.
- Mistake
Heat supplied changes internal energy rather than expansion work.
- Mistake
An isochor is vertical on a P–V graph.
Revision Zone
60-Second Revision
Quick cues from this topic — use them for last-minute revision.
Volume remains constant: ΔV = 0.
Pressure-volume work is zero: W = PΔV = 0.
The first law reduces to Q = ΔU.
Heating raises internal energy, temperature and pressure.
An isochoric process occurs at constant volume. Because the boundary does not move, no pressure-volume work is done and all net heat changes internal energy.
- Definition
- An isochoric process occurs at constant volume. Because the boundary does not move, no pressure-volume work is done and all net heat changes internal energy.
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Exam Practice
Important Question
Q1. Define an isochoric process.
Important Question
Q2. Why is work zero?
Important Question
Q3. How does the first law reduce?
Important Question
Q4. What is an isochor?