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My textbook answers this question as $\Delta S> 0$ but I really don't know why. If system is isolated, then $dQ=0$, i.e., $S=0$ ($S=dq/T$).

I don't really get why question has provided an additional information about $\Delta U = 0$. What is its use? And why is my answer incorrect?

physicopath
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mehulmpt
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3 Answers3

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There may be a chemical reaction or a change of state going on inside the system.
The $\Delta U=0$ might be there to help you ie to remind you that for an isolated system $Q$ and $W$ are both zero and so must $\Delta U$ be zero or to hinder you (a distractor) to make you worried as to why that statement was made you knowing full well what $\Delta U$ is for an isolated system?.

Farcher
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Let's dive into some simple mathematics a first: $$\Delta U=\int { dU } $$ and $$\Delta S=\int { \frac{dU}{T} } $$ now you are claiming that $\Delta S= 0$. if so , we will try to draw a graph of $U-T$. As U is a function of T only when n is fixed, both $\Delta U = 0$ and$\Delta T = 0$. We see that any graph we draw must retrace the path from which it came to satisfy your condition $\Delta S= 0$.

Now we see that if one path is spontaneous , the the retracing path is non-spontaneous as they are exactly reverse and vice-versa.

We know that non-spontaneos processes won't occur in isolated systems. so your claim is wrong.

NIkhil Reddy Ramolla
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if du=0 For isolated system it means , q=o And acc to. 1st law of thermodynamics , du=q+w ,q=0 for isolated system du=0+w , (given du =0 ) 0=0+w, w=o=pdv dh= du + pdv , dh=0+0 , dh=0 . dg= dh-tds. (gibbs equation) dg= 0-tds, means rxn spontaneous

ds = -dg/t .......(1) Acc to eq 1 if t dec then ds= -dg/-t. therefor ds will be positive . If temp inc. Then ds will be negative

So, there will be two answer

Prakash Shukla
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