An isochoric process is a thermodynamic process in which the volume of the system remains constant throughout the process. Since there is no change in volume, this process is also known as a constant volume process.
Isochoric processes are commonly observed when a gas is heated or cooled in a rigid, sealed container where expansion or compression is not possible.
Condition for Isochoric Process
In an isochoric process, the defining condition is that the volume remains constant.
\[dV = 0\]
As a result of this condition, pressure and temperature may change, but the volume of the system remains fixed.
Work Done in an Isochoric Process
Pressure–volume work is given by the general expression
\[\delta w = P_{ext}\,dV\]
For an isochoric process, since dV = 0, the work done becomes
\[\delta w = 0\]
Therefore, the total work done during an isochoric process is zero.
\[\boxed{w = 0}\]
Application of the First Law to Isochoric Process
The First Law of Thermodynamics is expressed as
\[dU = \delta q – \delta w\]
Substituting δw = 0 for an isochoric process, we obtain
\[\boxed{dU = \delta q_v}\]
This shows that in an isochoric process, the heat supplied to the system is entirely used to increase the internal energy of the system.
Heat Capacity at Constant Volume
The heat capacity at constant volume is defined as the amount of heat required to raise the temperature of the system by one degree at constant volume.
\[C_v = \left(\frac{\delta q}{dT}\right)_V\]
Since δq = dU for an isochoric process, we can write
\[C_v = \left(\frac{dU}{dT}\right)_V\]
This expression shows that the heat capacity at constant volume is directly related to the change in internal energy with temperature.
Isochoric Process for an Ideal Gas
For an ideal gas, internal energy depends only on temperature. Therefore, for an isochoric process involving an ideal gas, a rise in temperature results in an increase in internal energy.
The pressure of the gas increases with temperature according to Gay-Lussac’s law.
\[\frac{P}{T} = \text{constant}\]
PV Diagram for Isochoric Process
On a pressure–volume diagram, an isochoric process is represented by a vertical straight line, since the volume remains constant while pressure changes.
Solved Numerical Example (Exam Style)
Example
Question. One mole of an ideal gas is heated at constant volume so that its temperature rises from 300 K to 400 K. If the molar heat capacity at constant volume is 20.8 J mol⁻¹ K⁻¹, calculate the heat absorbed and the change in internal energy.
Given:
n = 1 mol
T₁ = 300 K
T₂ = 400 K
Cv = 20.8 J mol⁻¹ K⁻¹
Formula:
\[q_v = n C_v (T_2 – T_1)\]
Substitution:
\[q_v = 1 \times 20.8 \times (400 – 300)\]
Calculation:
\[q_v = 20.8 \times 100 = 2080\ \text{J}\]
Work Done:
\[w = 0\]
Change in Internal Energy:
\[\Delta U = q_v = 2080\ \text{J}\]
Answer:
\[\boxed{q = +2080\ \text{J}, \quad \Delta U = +2080\ \text{J}}\]
Practice Questions
1. Define an isochoric process.
2. Show that no work is done during an isochoric process.
3. Derive the relation between heat absorbed and change in internal energy for an isochoric process.
4. Explain the representation of an isochoric process on a PV diagram.