Is Entropy Constant in an Adiabatic Process?

Entropy in Reversible and Irreversible Adiabatic Processes

An adiabatic process is a transformation in which no heat is exchanged with the surroundings:

\[q=0.\]

From the First Law of Thermodynamics,

\[\Delta U=q+w,\]

and therefore, for an adiabatic process,

\[\Delta U=w.\]

The absence of heat exchange does not by itself determine the entropy change:

• Reversible adiabatic process: $\Delta S_{\text{sys}}=0$.
• Irreversible adiabatic process in a closed system: $\Delta S_{\text{sys}}>0$.

• Entropy change of the universe:

  \[\Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}}.\]

For a reversible process,

\[\Delta S_{\text{univ}}=0,\]

whereas an irreversible process produces entropy:

\[\Delta S_{\text{univ}}>0.\]

Important distinction.
The entropy change of the system is calculated by comparing its initial and final equilibrium states, even when the actual transformation is irreversible. The entropy change of the universe indicates whether the transformation is reversible or irreversible.

Worked Example

One mole of ideal gas expands adiabatically from

\[V_1=10.0\,\text{L}\]

to

\[V_2=20.0\,\text{L}.\]

We will determine:

1. the entropy change in a reversible adiabatic expansion;
2. the entropy change in an irreversible adiabatic free expansion;
3. the entropy change of the universe in both cases.

Reversible Adiabatic Expansion

For a reversible transformation, the entropy change is defined by

\[\Delta S_{\text{sys}} = \int\frac{\delta q_{\text{rev}}}{T}.\]

Since the process is both reversible and adiabatic,

\[\delta q_{\text{rev}}=0,\]

and therefore

\[\Delta S_{\text{sys}}=0.\]

No heat is transferred to the surroundings, so

\[\Delta S_{\text{surr}}=0.\]

Consequently,

\[\Delta S_{\text{univ}}=0.\]

A reversible adiabatic process is therefore also called an isentropic process.

Irreversible Adiabatic Free Expansion

Suppose that the gas expands freely into a vacuum.

The process is adiabatic:

\[q=0.\]

Since the gas expands against zero external pressure, no work is performed:

\[w=0.\]

It follows that

\[\Delta U=0.\]

For an ideal gas, internal energy depends only on temperature. Therefore,

\[\Delta T=0.\]

The entropy change of one mole of ideal gas between the initial and final equilibrium states is

\[\Delta S_{\text{sys}} = R\ln\left(\frac{V_2}{V_1}\right).\]

Substituting the given volumes,

\[\Delta S_{\text{sys}} = 8.314\ln\left(\frac{20.0}{10.0}\right) = 8.314\ln 2.\]

Thus,

\[\Delta S_{\text{sys}} = 5.76\,\text{J K}^{-1}.\]

No heat is exchanged with the surroundings, so

\[\Delta S_{\text{surr}}=0.\]

Therefore,

\[\Delta S_{\text{univ}} = 5.76\,\text{J K}^{-1}.\]

The transformation is adiabatic, but its entropy increases because the process is irreversible.

Why Can Entropy Increase When $q=0$?

The relation

\[dS=\frac{\delta q_{\text{rev}}}{T}\]

refers to a reversible path connecting two equilibrium states.

In a reversible adiabatic process, the actual path is reversible and $\delta q_{\text{rev}}=0$, so the entropy remains constant.

During an irreversible free expansion, the actual process cannot be used directly in the integral that defines entropy change. A hypothetical reversible path must be considered between the same initial and final equilibrium states. Along that path, the entropy change is

\[\Delta S_{\text{sys}} = R\ln\left(\frac{V_2}{V_1}\right)>0.\]

Thus, $q=0$ does not imply $\Delta S=0$. It implies constant entropy only when the adiabatic transformation is also reversible.

Results

For the reversible adiabatic expansion,

\[\Delta S_{\text{sys}} = \Delta S_{\text{surr}} = \Delta S_{\text{univ}} = 0.\]

For the irreversible adiabatic free expansion,

\[\Delta S_{\text{sys}} = \Delta S_{\text{univ}} = 5.76\,\text{J K}^{-1},\]

while

\[\Delta S_{\text{surr}}=0.\]

This example illustrates the Second Law of Thermodynamics:

\[\Delta S_{\text{univ}}\geq 0.\]

Equality holds for a reversible process; a strict inequality characterizes an irreversible process.