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When doing problems in thermodynamics, there is often a situation where a gas is thermally insulated inside a container.

To me, this would mean that there is no heat transfer to the gas ($Q=0$), so the gas can do work at the expense of it's internal energy.

I was doing Aptitude Problems in Physics by S.S. Krotov. In a certain problem, we have a gas under a thermally isolating piston, yet somehow the author says that, due to this, the internal energy of the gas doesn't change ($\Delta U=0$).

I am trying to find an error in the reasoning, and I am not quite sure what does thermally insulated mean.

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I am terribly sorry for not posting earlier, but this is the first chance I got. I figured it would be best to post the problem & solution directly. Maybe I am misinterpreting the solution after all. Problem (part 1) Problem (part 2)

Here is the solution. The sentence in red seems to be putting me off.

Solution (part 1) Solution (part 2) Solution (part 3)

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  • $\begingroup$ You should give the complete wording of the problem. It is possible that this "thermally insulating piston" divides two gases in a cylinder, so the total work is zero, heat supply is zero, so the total internal energy is not changed. $\endgroup$
    – akhmeteli
    Commented Feb 18, 2017 at 20:14
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    $\begingroup$ -1. Unclear what you are asking. You obviously know what the term means. If you are querying the reasoning in a textbook, we need to see exactly what it says, in context. $\endgroup$
    – sammy gerbil
    Commented Feb 19, 2017 at 0:39
  • $\begingroup$ @akhmeteli Once again sorry for the late edit. $\endgroup$
    – Knight1805
    Commented Feb 21, 2017 at 16:15

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Dont be too hard on a transaltion of a russian book (I assume?) :)

You are right, the marked sentence is nonesense; both parts are right, but there is no connection. The temperature in the left part does not change - but not because of the insulation, but because there is a thermostat; and that no heat is conducted is only important for the temperature in the right part.

Was this the only part that confused you? Is everything clear if you forget this sentence?

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  • $\begingroup$ Yep. That was the additional piece of information I needed. Thank you! $\endgroup$
    – Knight1805
    Commented Feb 22, 2017 at 15:09
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So I don't see where they say that the internal energy of the gas does not change. And the temperature in the left-hand part does not change because this part is in constant thermal contact with the thermostat.

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  • $\begingroup$ But since the internal energy of that gas is U=nC_v*T this means that since T is constant, the change in U is zero. This is what confuses me. Could you please give an exact explanation of the term "thermostat"? $\endgroup$
    – Knight1805
    Commented Feb 22, 2017 at 0:02
  • $\begingroup$ @Knight1805: So, OK, the temperature does not change and the internal energy does not change. A thermostat is typically a large system whose temperature is kept constant, so systems that are in contact with the thermostat stay at the temperature of the thermostat. $\endgroup$
    – akhmeteli
    Commented Feb 22, 2017 at 0:20
  • $\begingroup$ So I guess that was the misleading term... one more question: what would happen if the piston did conduct heat? How would we write the 1st law of TD (where would that energy go)? $\endgroup$
    – Knight1805
    Commented Feb 22, 2017 at 0:26
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The red part looks like a typo. The point is that the left part is at constant temperature by a constraint and the role of the piston being insulating is to let the temperature of the right side change. I think if the piston were not insulating then both sides would have the same temperature that is constrained to remain the same and therefore you would have the spring not move at all and $Q=Q'$.

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