KAHIBARO
Table of Contents
Constant Volume Processes
An isochoric process is a thermodynamic process that occurs at constant volume. The word comes from roots meaning "same volume". If the volume of a gas does not change, then the container is rigid, or the system is constrained so that expansion and compression cannot occur.
Because the volume stays fixed, isochoric processes are especially important for understanding the relation between heat, temperature change, and internal energy.
What Stays Constant
In an isochoric process,
$$\Delta V = 0$$
This means the initial and final volumes are the same:
$$V_i = V_f$$
A gas in a sealed rigid container is the standard example. The gas may be heated or cooled, so its pressure and temperature can change, but the volume cannot.
Work Done in an Isochoric Process
For a gas, the work done during a volume change is
$$W = \int P \, dV$$
But in an isochoric process, $dV = 0$, so the work is zero:
$$W = 0$$
This is the defining mathematical consequence of constant volume.
For any isochoric process,
$$\Delta V = 0 \qquad \Rightarrow \qquad W = \int P\,dV = 0$$
No boundary work is done because the system does not expand or contract.
This does not mean that nothing happens. Heat can still enter or leave the system, and the pressure and temperature can still change.
First Law in an Isochoric Process
The first law of thermodynamics is
$$\Delta U = Q - W$$
Since $W = 0$ in an isochoric process, it becomes
$$\Delta U = Q$$
So any heat added to the system increases its internal energy, and any heat removed decreases its internal energy.
In an isochoric process,
$$W = 0 \qquad \Rightarrow \qquad \Delta U = Q$$
All the heat transfer goes into changing the internal energy.
For an ideal gas, internal energy depends only on temperature. So if heat is added at constant volume, the temperature rises. If heat is removed, the temperature falls.
Pressure and Temperature at Constant Volume
For an ideal gas,
$$PV = nRT$$
If $V$ and $n$ are constant, then
$$\frac{P}{T} = \frac{nR}{V} = \text{constant}$$
So pressure is directly proportional to absolute temperature:
$$\frac{P_1}{T_1} = \frac{P_2}{T_2}$$
This means that if the temperature increases in a rigid container, the pressure also increases.
For an ideal gas in an isochoric process,
$$\frac{P}{T} = \text{constant}$$
or equivalently,
$$\frac{P_1}{T_1} = \frac{P_2}{T_2}$$
Temperatures must be in kelvin.
Heat Capacity at Constant Volume
In an isochoric process, the relevant heat capacity is the heat capacity at constant volume. For $n$ moles of an ideal gas,
$$Q = n C_V \Delta T$$
Since $\Delta U = Q$ for an isochoric process, we also have
$$\Delta U = n C_V \Delta T$$
This formula is very useful because it connects temperature change directly to internal energy change.
For a monatomic ideal gas,
$$C_V = \frac{3}{2}R$$
and so
$$\Delta U = \frac{3}{2} nR \Delta T$$
More generally, the value of $C_V$ depends on the type of gas.
Graph on a Pressure-Volume Diagram
On a $P$-$V$ diagram, an isochoric process is shown as a vertical line because the volume remains fixed while the pressure changes.
The area under a curve on a $P$-$V$ diagram represents work done. Since the line is vertical, there is no horizontal width, so the area under the process is zero. This matches the result $W = 0$.
Physical Interpretation
An isochoric process is easiest to picture in a rigid metal tank. If the gas inside is heated, the molecules move faster. They strike the walls more often and with greater force. The volume stays unchanged because the walls do not move. As a result, the pressure rises.
If the gas is cooled, the molecules move more slowly, and the pressure drops.
The key idea is that energy changes the microscopic motion of the particles, not the size of the container.
Common Examples
Some common examples of approximately isochoric processes are shown below.
| Situation | Volume change | What changes |
|---|---|---|
| Gas in a rigid sealed container being heated | Zero | Temperature and pressure increase |
| Gas in a rigid container being cooled | Zero | Temperature and pressure decrease |
| Early stage of combustion in a closed rigid chamber | Approximately zero | Pressure rises rapidly |
Real systems are often only approximately isochoric, because containers may expand slightly. But the ideal model is still very useful.
Summary Relations
The essential formulas for an isochoric process are collected here.
For an isochoric process:
$$\Delta V = 0$$
$$W = \int P\,dV = 0$$
$$\Delta U = Q$$
For an ideal gas:
$$\frac{P_1}{T_1} = \frac{P_2}{T_2}$$
$$Q = nC_V \Delta T$$
$$\Delta U = nC_V \Delta T$$
These relations make isochoric processes one of the simplest thermodynamic processes to analyze.