I was under the impression that if a gas expands isothermally then energy must be transferred from the surroundings to the gas as it is performing work. But in that case surely using $$\Delta S= \frac{\Delta Q}{T}$$ if the change in energy for the surroundings is negative then the change in entropy must also be negative?
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1$\begingroup$ Entropy change can be negative, no problem in that. The total entropy change just will always be positive. $\endgroup$– SteevenCommented May 10, 2017 at 9:50
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$\begingroup$ @Steeven The question I'm looking at asks "Which process will increase the entropy of the local surroundings?" and the answer is "The isothermal expansion of a gas". Surely in this case only the entropy of the gas will increase, and the entropy of surroundings will decrease (with overall entropy change being positive)? $\endgroup$– Alex JonesCommented May 10, 2017 at 9:54
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$\begingroup$ I agree with you. The book answer must be wrong. $\endgroup$– Chet MillerCommented May 10, 2017 at 12:00
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$\begingroup$ @ChesterMiller yeah I'd think that but it's from a past paper for IB exams so I'm not sure they can have got it wrong. In the examiner comments they find the answer by eliminating the other 3 options (multiple choice) but don't actually explain why it's that particular answer :/ $\endgroup$– Alex JonesCommented May 10, 2017 at 13:18
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$\begingroup$ Well, we know that in the isothermal reversible expansion of an ideal gas, the entropy of the surroundings decreases. So maybe you can provide us with the exact wording of the question, and the exact wording of their comments. $\endgroup$– Chet MillerCommented May 10, 2017 at 14:09
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1 Answer
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The $Q$ term that you used in your formula, represent the heat absorbed (or evolved) for reversible processes only. For irreversible processes the term for change in entropy is different.
In an isothermal process,
$\Delta T = 0 \Rightarrow \Delta U =0$,
Therefore, $P\Delta V = q$
When the gas expands against external pressure it uses some of its internal energy and to compensate for the loss in the internal energy it absorbs heat from the surrounding.
But the thing about reversible processes is that, $\Delta {S_{universe}}=0$
$\Delta {S_{system}}=-\Delta {S_{surrounding}}$.
For all irreversible processes, the entropy of the universe increases. It doesn't matter if the surrounding's entropy decreases and if it does, the entropy change for the universe will either be 0 (reversible processes) or positive (irreversible processes) .
For irreversible processes, the entropy change associated with the state change is
$dS=\frac{Q_{actual}}{T}+\frac{(dW_{reversible}-dW_{actual})}{T}$
The subscript 'actual' refers to an actual process i.e, irreversible process.
Since, $dW_{reversible} > dW_{actual}$
$dS > \frac{dQ_{actual}}{T}$.
For more check this out : http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node48.html
