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The formula for entropy change states that change in entropy is not possible if heat change is nil.

∆S=∆Q/T

But what if we adiabatically expand a gas? Won't the molecules get more space to vibrate in, hence increasing the randomness of the system, therefore increasing the entropy?

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    $\begingroup$ The thinking and searching comment from your other question applies here as well. A good start is e.g. reading your textbooks and relevant articles on Wikipedia or educationnal sites and following their links. $\endgroup$
    – Poutnik
    Nov 16, 2020 at 8:13
  • $\begingroup$ The formula you provide is applicable during isothermal (constant T) processes. In addition, it can be applied only when the process is carried out reversibly, that is, q is the heat exchanged during a reversible process. $\endgroup$
    – Buck Thorn
    Nov 16, 2020 at 16:33
  • $\begingroup$ Related : What is the difference between reversible and irreversible adiabatic expansion? $\endgroup$
    – J...
    Nov 16, 2020 at 17:13

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The answer to your question is yes and no. You are correct in supposing that as an ideal gas expands the entropy will increase as it has more space to occupy and so the number of ways the molecules can be placed in the total space has increased. (Try to not think of entropy in terms of randomness but in the number of ways molecules can be positioned in space, or occupy different energy levels). However, accompanying expansion is cooling and if the adiabatic process is carried out reversibly the decrease in entropy due to cooling exactly matches that increase due to expansion, and so the entropy change is zero.

If the expansion is irreversible and adiabatic, the gas does less work, looses less internal energy and so there is entropy increase as the cooling is now not as large as when expansion is adiabatic and reversible, and so the two process, entropy increase due to expansion and decrease due cooling no longer match.

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