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I'm having some trouble understanding the inequality of Clausius. My treatment of the inequality applies to an engine operating in a cycle between two temperatures (i.e. like a Carnot engine). The inequality states that for an irreversible process:

$$ (1)\ \int _{closed\ path} \frac {dq_{irr}}{T}<0 $$

From what I've been reading, $ {dq_{irr}} $ is the heat added to the system (the engine) and $ T $ is the temperature of the reservoirs, $ T_{res} $. If this is the case, then $-{dq_{irr}}$ is the heat added to the reservoirs, $ dq_{irr,res} $, and

$$ (2)\ \int _{closed\ path} \frac {-dq_{irr}}{T_{res}}=\int _{closed\ path} \frac {dq_{irr,res}}{T_{res}}>0 $$

I read somewhere that if the reservoir is infinite then any heat added is actually occurring infinitely slowly. Is this true? If this is the case, then

$$ (3)\ dq_{irr,res} = dq_{rev, res} $$

$$ (4)\ \int _{closed\ path} \frac {dq_{rev, res}}{T_{res}}=∆S_{res}>0 $$

This seems reasonable. Since the engine itself is operating in a cycle, $ ∆S_{sys} = 0 $. This therefore satisfies the second law of thermodynamics: $$ (5)\ ∆S_{univ}=∆S_{sys}+∆S_{res}=0+∆S_{rev}>0 $$

Does this all look right? $ (5) $ seems to say that the entropy change for an irreversible process is 0 for the case of the system itself, but is > 0 for the case of the surroundings. Is this only the case for the situation I have described?

I am also unsure about $ (3) $. If the heat is being added to the reservoirs infinitely slowly, then wouldn't the same be true of the system? This would make $ ∆S_{sys} = -∆S_{res} ≠ 0 $ and is obviously not correct.

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  • $\begingroup$ Your question is confused in few aspects. First, for irreversible process, the Clausius inequality states only that the integral over cycle $\int \frac{dQ}{T_{res}} \leq 0$, not $<0$. Second, the cycle consists of non-equilibrium states, so there is no "path"; the process itself can be ascribed function $q(t)$, but there is no path in the state space. Consequently, the expression $dq_{irr,res} = dq_{rev,res}$ has no meaning - there is no way to directly compare heat infinitesimals when these correspond to different kinds of process. $\endgroup$ – Ján Lalinský Mar 3 '14 at 16:14
  • $\begingroup$ My book states that that the integral is < 0 and not ≤ 0. If it was equal to 0, wouldn't it be reversible? If I can't compare heat infinitesimals then what is the physical significance of the inequality? I wouldn't be able to relate it to entropy then, would I? $\endgroup$ – David Mar 3 '14 at 16:21
  • $\begingroup$ Which book uses <0? $\endgroup$ – Ján Lalinský Mar 3 '14 at 16:23
  • $\begingroup$ amazon.com/Physical-Chemistry-Keith-J-Laidler/dp/061815292X $\endgroup$ – David Mar 3 '14 at 16:24
  • $\begingroup$ I do not have the book. The Clausius inequality is, as far as I know, derived with $\leq$. There is no guarantee that for all irreversible processes one can use $<$. Most often, however, we are interested in such irreversible processes that indeed have $<$. $\endgroup$ – Ján Lalinský Mar 3 '14 at 16:44
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Exactly as you said: "(5) seems to say that the entropy change for an irreversible process is 0 for the case of the system itself, but is > 0 for the case of the surroundings." After the cycle is finished, by definition of the "cycle", the system returns to its original state irrespective of all the irreversibilities that may have taken place within it during some portions of the cycle. Entropy being a state variable it also assumes its initial value after the cycle. The only place therefore where the entropy will have increased after the cycle is in the reservoirs that do not perform any "cycle", they are sinks and sources of heat and their state can and will change per Clausius., i.e., $$\Delta S_{res} =\frac{q_{res}}{T_{res}}$$

As regards to equality vs. inequality I always learned and thought I understood that any time there is irreversible process it is inequality and never equality, and that goes for any part of the process. This can be understood intuitively by noting that irreversibility is caused by internal dissipation, i.e., heat generation, therefore during irreversible processes less heat is absorbed from the reservoir and more heat is rejected to the reservoir if compared to a reversible transition.

Incidentally, the absorbed or rejected $q$ of a reservoir is always reversible, that is why its entropy change is just the ratio of the heat and its temperature.

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  • $\begingroup$ Concerning your last paragraph, wouldn't that make my statement $ (3) $ correct? This again raises the question in the last paragraph of my original question. Isn't this also in opposition to Ján Lalinský's answer? $\endgroup$ – David Mar 4 '14 at 4:19
  • $\begingroup$ This answer is almost as linguistic as as substantive: you wrote $ (3)\ dq_{irr,res} = dq_{rev, res} $ but I wanted to avoid that for two reasons: (1) there is just no $dq$ because of $q$ is not function of state variables, (2) your notation implies that the reservoir may go through an irreversible process. As I said the reservoir is always reversible and formally is really a boundary condition. $\endgroup$ – hyportnex Mar 4 '14 at 15:35
  • $\begingroup$ If there is no such thing as an irreversible process for the reservoir, then why is the inequality of Clausius defined with $ dq_{irr} $ in $ (1) $? $\endgroup$ – David Mar 5 '14 at 15:00
  • $\begingroup$ I am not familiar with your book, but it may denote nothing more than that the process itself is irreversible. In other words, the system that is connected to the reservoir and exchanges the heat $dq_{irr}$ with it goes through the irreversible process not the reservoir. If you have a chance read this article by P.W. Bridgman: "The Thermodynamics of Plastic Deformation and Generalized Entropy". $\endgroup$ – hyportnex Mar 5 '14 at 15:34
  • $\begingroup$ I appreciate the help. I'll take a look into the article $\endgroup$ – David Mar 5 '14 at 15:38
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I read somewhere that if the reservoir is infinite then any heat added is actually occurring infinitely slowly. Is this true?

No, infinite reservoir does not imply infinitely slow transfer of heat. There is no direct relation.

Equation (4) is incorrect; if the integral is over closed path in the state space of the reservoir, its value is 0. If the integral corresponds to cycle made on the system where something is irreversible (like heat transfer across finite temperature difference), in the end, when the system returns to its original equilibrium state and the reservoir to some equilibrium state, the entropy of the reservoir will be higher than before (in the case of the heat transfer across finite temperature difference entropy is generated at the interface of the two systems with gradient of temperature).

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  • $\begingroup$ If I can't relate the inequality to ∆S, then were is this coming from: "... the entropy of the reservoir will be higher than before."? I showed this in $ (4) $ with the incorrect assumption of $ (3) $. $\endgroup$ – David Mar 3 '14 at 16:32
  • $\begingroup$ Because the entropy of the system is the same, and we assume that total entropy indeed increased in the process, we conclude that the entropy of the reservoir increased. $\endgroup$ – Ján Lalinský Mar 3 '14 at 16:47
  • $\begingroup$ I thought the inequality of Clausius could be used to prove the second law. This is essentially assuming the second law to be true. $\endgroup$ – David Mar 3 '14 at 17:34
  • $\begingroup$ The Clausius inequality is another statement of the second law. The second law states that the entropy of adiabatically isolated system cannot decrease. It can remain the same ( $\Delta S = 0$ ) or increase ($\Delta S > 0$). $\endgroup$ – Ján Lalinský Mar 3 '14 at 19:58

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