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Ok, so the question is about the concept of increasing entropy. We obtain the result (by utilizing the Clausius inequality theorem) $dS = \frac{dQ_{rev}}{T} \geq \frac{dQ_{irr}}{T}$.

Then it's stated that for a closed system, $dQ_{irr}$ is zero and therefore $dS \geq 0$. Fair enough, the total energy in a closed system is constant and therefore no heat (thermal energy in transit) can flow in or out. The thing that bothers me though, is that I cannot imagine any process where an amount $dQ_{rev}$ can transferred to or from the system if the system is closed. And that leads me to $dQ_{rev}=0$ as well which results in $dS=0$ .

Now I know that there is a problem, as entropy is in fact generated when heat flows between subsystems in the isolated system, which one could calculate. The problem originates from the statement that $dQ_{irr}$ is zero. In the isolated system an irreversible amount of heat can (and will be) transferred between subsystems of different temperatures. Even though the net heat transfer is zero, $\frac{dQ_{irr}}{T_1} + \frac{-dQ_{irr}}{T_2}$ should also be zero in an isolated system with two subsystems of temperatures $T_1$ and $T_2(>T_1)$ for this to work out, which is incorrect. The explanation is much appreciated.

Kind regards!

Sam Jefferson
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2 Answers2

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Your statement

Even though the net heat transfer is zero, $dQ_{irr}/T_1 - dQ_{irr}/T_2$ should also be zero in an isolated system with two subsystems of temperatures T1 and T2

is incorrect. The total entropy of an isolated system may increase. The Universe is an isolated system, and its entropy is increasing all the time.

Indeed, when heat $Q$ is transferred from a warmer to a colder part of the system, the total entropy of the system increases by $Q/T_1 - Q/T_2$.

A correct statement is: the total entropy of an isolated system in thermodynamic equilibrium remains the same. A system with internal temperature variations is not in equilibrium.

Jim Danner
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  • Yes that's exactly what I figured out myself, but then why exactly is stated $\frac{dQirr}{T} = 0$ (in a closed system)? These things appear contradictory. – Sam Jefferson Dec 29 '18 at 10:20
  • Without knowing the full context, my guess is that $Q_{irr} = 0$ refers to heat exchange between the isolated system and its surroundings. That is 0 by definition. Or perhaps they are talking about a system that is already in thermodynamic equilibrium ($T_1 = T_2$ for all pairs of subsystems) – Jim Danner Dec 29 '18 at 10:24
  • It's not exactly stated but I assume it's between the isolated system and its surroundings outside the system: ''Consider a thermally isolated system. In such a system $dQ = 0$ for any process, so that the above inequality becomes $dS \geq 0''$ (page 141 Blundell & Blundell, Concepts in Thermal Physics). It's not necessarily stated that the subsystems are in thermal equilibrium. That's the confusing part, as then $dQirr = 0$ but $\frac{dQirr}{T} \neq 0$ for the whole isolated system. – Sam Jefferson Dec 29 '18 at 10:32
  • Yes, looks like they’ve written it down confusingly. – Jim Danner Dec 29 '18 at 10:40
  • What is the right way to interpret it though? Why aren't we saying $dQrev=0$ aswell if we were to be talking about the isolated system and its surroundings? – Sam Jefferson Dec 29 '18 at 10:43
  • I think when they write “In such a system $dQ=0$ for any process”, they refer to the total heat for the entire system; and because the system can’t lose heat (with a negative $dQ$), it can only have internal processes that increase the net entropy. This contrasts with an open system, whose internal entropy may go down as it releases heat to the surroundings (e.g. a fridge). – Jim Danner Dec 29 '18 at 11:57
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Even if you have no heat transferred for an irreversible process in an isolated system, for the reversible path between the same two end states, the system does not have to be treated as isolated, and heat can be transferred into or out of the system reversibly. All the term "reversible path" means is that you have to identify an alternative path between the same two end states of the system that is reversible. The reversible path does not need to bear any resemblance whatsoever to the actual irreversible path that the system suffered.

For a cookbook recipe (with examples) on how to determine the entropy change for an irreversible path suffered by a system of constant mass, see the following link: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/ One of the examples in this link is your sub-system example.

Chet Miller
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