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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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    The expression $dS = dQ/T$ only holds for reversible processes, so since the textbook says that $\Delta S \neq 0$ while $dQ = 0$, that must mean that the process is not reversible, and something is going on inside the system. That means that you haven't given us enough information about the problem, and $\Delta U = 0$ is not enough extra information. Can you be more specific about the details of the problem? – march Mar 07 '16 at 07:45
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    This is the exact question language: "For an isolated system, $\Delta$U =0, what will be $\Delta$S ?" – mehulmpt Mar 07 '16 at 08:02
  • Actually the questions are related to definition of 'terms'- for an isolated system one can mean that no thermal energy can enter or leave the system , but no qualifying statement is there about the "equilibrium' nature of the system- if it is in equilibrium then entropy will remain constant but if its in non-equilibrium 'work -energy'process may increase its entropy-common example is our universe -whose entropy is increasing though its isolated. – drvrm Mar 07 '16 at 08:23
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    For an isolated system, the equation should be stated as $$\Delta S\geq0$$. Certainly, if nothing happens in the isolated system, both $\Delta U$ and $\Delta S$ are zero. – Chet Miller Mar 07 '16 at 16:31
  • Objects fall downwards. Analogously, the isolated system will minimize its free energy (or the appropriate potential). Thus $\Delta F \leq 0$. Thus, $\Delta U=0$ is a valuable information for $\Delta S$ – Bort Oct 14 '16 at 10:37
  • Maybe helpful https://physics.stackexchange.com/questions/119387/why-can-the-entropy-of-an-isolated-system-increase – SRS Feb 09 '18 at 16:51

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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  • This feels like the right answer, I think. Isolated means $\Delta U = 0$, $W=0$, and $Q=0$, and so if any process actually goes on inside the system, it is necessarily irreversible, and so $\Delta S > 0$. Certainly it is true that $\Delta S \geq 0$. – march Mar 07 '16 at 22:23
  • What if their is no reaction going on inside the system or change in state ? –  Jul 23 '20 at 09:57
  • Kindly reply. I am confused a lot in the same problem. –  Jul 23 '20 at 09:58
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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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