Isochoric Process

The Isochoric Process

When a thermodynamic system undergoes a change while its volume remains strictly constant, the process is called isochoric (from Greek isos = equal, chora = space). The defining constraint is $V = \text{constant}$, which means $\Delta V = 0$ and therefore no work is done by or on the system.

Since $W = 0$, the first law of thermodynamics $\Delta U = Q - W$ simplifies dramatically. With $W = 0$, we get:

This is the central result: all heat added to the system goes entirely into increasing its internal energy, and all heat removed comes entirely from a decrease in internal energy. There is no "leakage" into mechanical work.

Applying the Ideal Gas Law

For an ideal gas undergoing an isochoric process, the equation of state $PV = nRT$ becomes particularly simple. Since $V$ is constant, we can write:

This gives the direct proportionality between pressure and temperature:

If you heat a gas at constant volume, its pressure rises proportionally with temperature. If you cool it, pressure falls proportionally. This is why a sealed container left in sunlight builds up pressure — the volume is fixed, so temperature increase forces pressure increase.

Heat and Molar Specific Heat at Constant Volume

The heat exchanged in an isochoric process is related to the temperature change through the molar specific heat at constant volume, $C_V$. For $n$ moles of an ideal gas:

Since $\Delta U = Q$ for an isochoric process, we also have:

This is a powerful result: the change in internal energy of an ideal gas depends only on the temperature change, regardless of the path taken. The constant-volume specific heat $C_V$ is the bridge between temperature change and internal energy change.

For a monatomic ideal gas, $C_V = \frac{3}{2}R$. For diatomic gases at moderate temperatures, $C_V = \frac{5}{2}R$. These values come from the equipartition of energy theorem.

Properties of an Isochoric Process

The textbook lists three key properties. Each follows directly from the definition $V = \text{constant}$.

Property (I): The work done is zero.

This is immediate from the definition of thermodynamic work for a gas: $W = \int_{V_i}^{V_f} P \, dV$. Since $V_i = V_f$, the integral is over a zero range, giving $W = 0$.

›Proof

Proof of Property (I)

The work done by a gas during any process is:

$W = \int_{V_i}^{V_f} P \, dV$

For an isochoric process, $V$ is constant throughout, so $V_i = V_f$. The limits of integration are identical:

$W = \int_{V_i}^{V_i} P \, dV = 0$

The integral over a zero-width interval is zero regardless of the integrand. Hence $W = 0$.

Property (II): The first law reduces to $\Delta U = Q$.

With $W = 0$, the first law $\Delta U = Q - W$ becomes $\Delta U = Q - 0$, so:

All heat transfer changes only the internal energy. No energy is diverted to mechanical work.

Property (III): The pressure is directly proportional to the absolute temperature.

From the ideal gas law $PV = nRT$, with $V$ constant:

The quantity in parentheses is constant for a fixed amount of gas in a fixed volume. Therefore $P \propto T$, or equivalently:

This is sometimes called the pressure law or Gay-Lussac's law.

Graphical Representation

On a $P$-$V$ diagram, an isochoric process appears as a vertical line (constant volume). The area under the curve — which represents work — is zero, since the line has no horizontal extent. …