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Does entropy remain constant in when no heat is transferred between the system and surroundings?

If I do only mechanical or electrical work to change the properties like the volume of the system, would entropy be affected?

Because I understand entropy as the number of possible arrangements of atoms in a system. The following is sourced from the book - Physical chemistry by Atkins and Paula

Heat transfer

The above picture displays transfer of heat in an exothermic reaction which causes thermal motion in surroundings.

Work done on surroundings

The above picture shows mechanical work done by the system on the surroundings.

Now that got me thinking, if the atoms move in an "organized way", won't the final and initial state of entropy of the system remain constant?

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    $\begingroup$ Review this search site:stackexchange.com Entropy change isolated system $\endgroup$
    – Poutnik
    Commented 15 hours ago
  • $\begingroup$ Might want to state which book is "this book" (provide a citation). $\endgroup$
    – Buck Thorn ♦
    Commented 12 hours ago
  • $\begingroup$ Entropy is constant if the process is reversible and no heat is exchanged $\endgroup$
    – Piyush
    Commented 6 hours ago
  • $\begingroup$ Entropy is a state function so it does not matter how one goes from start to end only the starting and ending conditions matter. (Just like potential energy, its always the same at the top of a hill no matter how you get there). To calculate $\Delta S$ always use $\Delta S=q_{rev}/T$ so imagine for example isothermally expanding a gas into a vacuum from $V_1\to V_2$ then a reversible path gives $q_{rev}=nR\ln(V_2/V_1)$ for $n$ moles and so you find the change in entropy. $\endgroup$
    – porphyrin
    Commented 5 hours ago

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Does Entropy remain constant in when no heat is transferred between the system and surroundings?

The absence of a heat transfer does not guarantee that the entropy change of the system is zero. The following expression of the second law of thermodynamics refers to an exchange under reversible conditions:

$\text{d} S = \frac{\text{d}q_\text{rev}}{T}$

The general case

$\text{d} S \ge \frac{\text{d}q}{T}$

leaves it unclear whether the entropy change is zero unless it is known that the process is in fact reversible.

In other words, an adiabatic process can result in a change in the entropy of the system if it is performed irreversibly.

The differential form for a pure substance (absent work other than mechanical)

$\text{d} S = \frac{C_V}{T} \text{d} T + \left( \frac{\partial p}{\partial T} \right)_V\text{d}V$

indicates that an increase in T tends to increase S, as does an increase in V provided $\left( \frac{\partial p}{\partial T} \right)_V \ge 0$. For an ideal gas $\left( \frac{\partial p}{\partial T} \right)_V = \frac{nR}{V}\gt 0$. In a reversible adiabatic process these contributions cancel, whereas in the irreversible case they don't. Compression work during a reversible adiabatic process results in a higher temperature ("more disorder") but also in a smaller volume ("less disorder").

Work and heat are ways of transferring energy to or from a system. However it is how processes ultimately change how energy is distributed through a sytem that matters. Compressing results in less ways of distributing energy, whereas increasing temperature creates more ways.

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