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The definition of entropy contains the term $Q_\text{rev}$ which means the heat supplied or taken out reversibly. I thought yes it can be after all only the initial & final states are important as entropy is a state function irrespective of the process heat is transferred. However I was baffled when I first read Clausius' theorem where it is written that $dS \geq \dfrac{Q}{T}$. If $Q$ is transferred irreversibly, then $dS$ is greater than $\dfrac{Q_\text{irrev}}{T}$; if the heat transfer is reversible, then only $dS$ equals $\dfrac{Q_\text{rev}}{T}$. So, does that mean entropy depends on the process heat energy is transferred?? Then, how can it be a state function? Where am I mistaking? Please explain.

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  • $\begingroup$ since $dS = \frac {\delta Q_{rev}}{T}$ you also have $TdS=\delta Q_{rev}$. As $\Delta S = S-S_0$ for fixed $S_0$ is a state function, $\Delta Q_{rev}$ also behaves like a state function (note the difference between $\delta$ and $\Delta$) $\endgroup$ – hyportnex Jul 17 '15 at 11:27
  • $\begingroup$ @user31748: You are proving it circularly! $\endgroup$ – user36790 Jul 17 '15 at 11:28
  • $\begingroup$ I am not proving anything, I am trying to explain by illustrating it from a different angle. $\endgroup$ – hyportnex Jul 17 '15 at 11:29
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    $\begingroup$ it does not matter for the entropy how you got there, be it reversible or irreversible, but if you got there reversibly then usually it is easier to calculate the entropy change from the reversibly supplied heat transfer $\delta Q_{rev}$ via the equation $S_1-S_0 =\int_0^1 \frac {\delta Q_{rev}}{T}$ $\endgroup$ – hyportnex Jul 17 '15 at 11:38
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    $\begingroup$ Would this question and answer be useful? $\endgroup$ – march Jul 19 '15 at 17:19
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For an infinitesimal heat transfer $\delta Q$ the inequality of Clausius states that $$\Delta S = S_1-S_0 = \int_0^1 {\dfrac {\delta Q_\text{rev}}{T}} > \int_0^1 {\dfrac {\delta Q_\text{irrev}}{T}}$$

Here $\delta Q_\text{rev}$ and $\delta Q_\text{irrev}$ denote reversible and irreversible heat transfers, respectively. Thus if the process is reversible and we know what $\delta Q = \delta Q_\text{rev}$ is at each step then we can calculate the entropy change from the integral $\Delta S = \int_0^1 {\dfrac {\delta Q}{T}} $. But if the process is irreversible the integral $\int_0^1 {\dfrac {\delta Q}{T}}$ only gives a lower bound for the entropy change not the actual change.

The difference between the entropy change and the integral is the internally generated entropy by the process $$\Delta S - \int_0^1 {\dfrac {\delta Q}{T}} = \sigma_\text{irrev} $$ and is characteristic to it. For example, a resistor $R$ with a dc current $I$ through it and kept at constant temperature in steady state generates $\dot \sigma = \dfrac {I^2R}{T}$ entropy per unit time and sheds the same to its environment along with ${\dot q = I^2R}$ heat flux.

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  • $\begingroup$ +1. Thanks for answering, sir. However, sir, Though, I have understood the mathematical relation that $\Delta S . \int_0^1 \dfrac{dQ_\text{irrev}}{T}$, can you tell me what actually goes on during the irreversible process that makes this inequation true? $\endgroup$ – user36790 Jul 18 '15 at 2:44
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    $\begingroup$ "dissipation" in all its forms and manifestations (friction, stiction, resistance, etc.,). For example, the magnetic domain walls between the neighboring domains are sticky and when are moving relative to each other, then magnetic energy dissipates and heat evolves and if you were to keep it at a constant temperature then you will dump net heat and entropy in the environment over a "BH" cycle. $\endgroup$ – hyportnex Jul 18 '15 at 18:48
  • $\begingroup$ So, all due to these non-conservative forces, $$dS = \dfrac{dQ_\text{actual}}{T} + \dfrac{dW_\text{lost}}{T}$$, which ultimately implies that $$dS > \dfrac{dQ_\text{irrev}}{T} $$ as $$\dfrac{dW_\text{lost due to irreversibility}}{T}$$ is always positive & greater than zero for irreversible processes. Now, sir, can't we measure this additional entropy & add to the $\dfrac{dQ_\text{irrev}}{T}$ in order to get the total entropy change? Then we can also measure the entropy change due to irreversible process without going to any reversible path. .... $\endgroup$ – user36790 Jul 19 '15 at 7:12
  • $\begingroup$ ....So, why is it always advised to find the entropy change by "dreaming" a reversible path between the initial & final states? After all, we can measure the entropy change even for irreversible path by adding the additional entropy due to the work lost, right? $\endgroup$ – user36790 Jul 19 '15 at 7:17
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    $\begingroup$ Tolman & Fine core.ac.uk/download/pdf/4872759.pdf $\endgroup$ – hyportnex Jul 19 '15 at 16:56
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Suppose you start with a system in some state $P_1, V_1, T_1$ and you add some quantity of heat $\Delta Q$ to it so the system changes to a different state $P_2, V_2, T_2$. The final state will depend on how you added the heat $\Delta Q$. Adding the heat $\Delta Q$ in a reversible process will result in different values for $P_2, V_2, T_2$ compared with adding the same amount of heat $\Delta Q$ in an irreversible process.

Entropy is indeed a state function, so if you know $P_2, V_2, T_2$ you can calculate the entropy change. Since reversible and irreversible processes will result in different values for $P_2, V_2, T_2$ they will also result in different values for the entropy change.

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  • $\begingroup$ @user36790: a good example is the Joule expansion. In this process $\Delta Q = 0$ but $\Delta S > 0$ i.e. $\Delta S > \Delta Q/T$. $\endgroup$ – John Rennie Jul 17 '15 at 7:28
  • $\begingroup$ See, sir, the Joule Expansion can be carried on isothermally(reversible) or by free expansion(irreversible) leading to the same final state. So, is it not contradictory to your answer that for different processes, final states are different & so the entropy is different? $\endgroup$ – user36790 Jul 17 '15 at 8:11
  • $\begingroup$ @user36790: A Joule epansion is by definition a free expansion. In a Joule expansion the temperature doesn't change. If you did the same expansion adiabatically the expanding gas would have to do work on something, and as a result the internal energy decreases and the final temperature is lower. The two final states are different. $\endgroup$ – John Rennie Jul 17 '15 at 8:32
  • $\begingroup$ Sir, I really feel sorry for my behaviour & I apolozise for that. Yours was helpful but the other's was the one that cleared my confusion. Thanking you for posting the answer. Please forgive me if it was a bad behaviour:) $\endgroup$ – user36790 Jul 19 '15 at 17:38

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