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.
Key equation
Δ V=0
Key equation
W=PΔ V=P(0)=0
Key equation
Q=Δ U
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
Key equation
PT=constant
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.
1
Working
Constant volume: ΔV = 0
2
Step 2
W = PΔV = 0
3
Step 3
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.
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Exam preparation
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.
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.
Quick revision
Revision Zone
Review the key facts, formulas and checks before you move on.
Isochoric Process
Sticky Notes · Quick Revision
1.
Key Point 1
Volume remains constant: ΔV = 0.
2.
Key Point 2
Pressure-volume work is zero: W = PΔV = 0.
3.
Key Point 3
The first law reduces to Q = ΔU.
4.
Key Point 4
Heating raises internal energy, temperature and pressure.
