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In this video, the professor goes through a "derivation" for entropy using the Carnot cycle, in which he establishes that entropy is a state function for that process.

However, he then continues the lecture and then generalizes that entropy is a state function for any process. (e.x. around 49 minutes). The essence of his argument is that since entropy is a state function, we can calculate its change between any 2 points using a reversible process, even if it happened irreversibly in real life. However, my issue is, how do we know it is a state function? We were only able to say it is a state function when we went through a derivation using the Carnot cycle (a reversible process). It doesn't seem justified to say that is fact holds for any arbitrary process.

https://www.youtube.com/watch?v=ouSLRgkPzbI

To summarize, I am confused by the fact that our discovery that entropy is a state function was only arrived at when we looked at the Carnot cycle, a reversible process. However, how do we know that entropy is also a state function given by $\delta Q_{\mathrm{rev}}/T$ for any arbitrary (e.x. irreversible process)?

Qmechanic
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Milk Man
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1 Answers1

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However, my issue is, how do we know it is a state function? We were only able to say it is a state function when we went through a derivation using the Carnot cycle (a reversible process). It doesn't seem justified to say that is fact holds for any arbitrary process.

Entropy is usually thought of mathematically as a function of thermodynamic state variables.

"Entropy is state function" means that for given thermodynamic system (say gas in a cylinder with movable piston), all it suffices to determine value of entropy are values of all its state variables (say, internal energy and volume of the gas).

But this is indeed not clear from the original definition of entropy: entropy of state $X$ is often written or implied to be

$$ S(\mathbf X) = \int_{\mathbf X_0}^{\mathbf X} \frac{dQ}{T} $$ where $dQ$ is infinitesimal element of heat absorbed by the system and the integral is calculated along any continuous path connecting $X_0$ with $X$.

And you are right this does not make it clear why the expression does not depend on the kind of process that takes the state from $\mathbf X_0$ to $\mathbf X$.

The traditional way to show that the path does not matter and so the expression is a function of final $X$ only is based on the theoretical result that is true for the Carnot cycle with ideal gas. The result is that in such a cycle, sum of reduced heats is zero:

$$ \frac{\Delta Q_1}{T_1} + \frac{\Delta Q_2}{T_2} = 0. $$

This is then used to prove that for any reversible process that ends up at the same state it began, the integral along the closed path

$$ \oint \frac{dQ}{T} = 0. $$

This holds irrespective of the details of the path such as its shape or length. The usual way to do this proof is to replace the actual closed path (that is otherwise of arbitrary shape) by an alternating sequence of very tiny isotherms and adiabats that go very near the actual path. Those isotherms and adiabats can be thought of as part of fictive Carnot cycles and it can be shown that sum of reduced heats of those Carnot cycles approximates the above integral. But the sum of the reduced heats is 0 for those Carnot cycles. The smaller the Carnot cycles, the better the approximation to the original path, in the limit of zero size the isotherms and adiabats on the outer edge become the same as the original path.

After that is done, a mathematical theorem from multivariable calculus is used: if the loop integral is zero for any loop, the expression

$$ \int_{X_0}^{X} \frac{dQ}{T} $$

is a function of the points $X_0,X$ only, the actual path along which the integral is evaluated does not matter. One point ($X_0$) is usually set to be fixed and then the function is function of the final point ($X$) only.

Sean E. Lake
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Ján Lalinský
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    ay, there is the rub: "...replace the actual closed path (that is otherwise of arbitrary shape) by an alternating sequence of very tiny isotherms and adiabats that go very near the actual path". So what guarantees the existence of such replacement? Furthermore, that is a rather "jagged" way of approximating a Riemann integral, and thus its convergence is not obvious let alone convergence to the desired path. – hyportnex Mar 15 '18 at 12:46

  • "So what guarantees the existence of such replacement?" -- Good question. For simple systems whose thermodynamic space is two-dimensional and any state is determined by pressure $p$ and volume $V$ only, the replacement exists because any state point $[p,V]$ can be reached by a combination of isotherms and adiabats. For more dimensional spaces, it is not so clear why every possible state can be reached by isotherms and adiabats. The jaggednes is not so much of a problem, I think - the length of the replacement path does not matter, only that it is kept close to the original one.

     – Ján Lalinský Mar 15 '18 at 19:49
  • Exactly: for two variables the equation $\delta Q = A(x,y)dx + B(x,y)dy$ will always have an integrating factor so that observation does not have the same, if any, physical significance as the existence of an integrating for more than two variables has, which is what the 2nd law and Clausius claim. The existence of adiabatic surfaces is more or less equivalent (Caratheodory) to the 2nd law, so I have never understood what this "proof with jagged edges" contributes to the argument. Regarding convergence there is a famous example from mathematics showing the difficulty defining surface area. – hyportnex Mar 16 '18 at 12:28
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    The example is from Schwarz showing that one has to be very careful with inscribing simple flat pieces of triangles within a 3d circular cylinder otherwise the sum of the triangle areas do not converge to the surface area of the cylinder. – hyportnex Mar 16 '18 at 12:30
  • You are right that the construction of replacement path is not a general proof that the integral vanishes. But I do not think it is meant to be so: it is meant to motivate the introduction of entropy for simple systems. You are right this can be done purely mathematically as in 2D integrating factor always exists. But the advantage of this construction is for understanding where the entropy concept originally came from. Its generalization to higher dimensional systems is other matter. – Ján Lalinský Mar 16 '18 at 13:56
  • Hi, Thanks for the response. I'm having trouble understanding this part: "This holds irrespective of the details of the path such as its shape or length. The usual way to do this proof is to replace the actual closed path (that is otherwise of arbitrary shape) by an alternating sequence of very tiny isotherms and adiabats that go very near the actual path. ". The cause of my confusion is that I thought pressure and volume aren't well-defined for irreversible processes so how are we able to generalize this idea for irreversible processes? – Milk Man Mar 18 '18 at 00:15
  • The entropy is defined only for equilibrium states. If the irreversible process is so violent that the system goes out of equilibrium, entropy is not defined and the relation $\Delta S = \Delta Q /T$ does not hold. Only when the irreversible process comes back to a point where the system is in equilibrium state, the entropy is defined again. – Ján Lalinský Mar 18 '18 at 23:55
  • Hi, thanks for the response. For such a sitation, when the system comes back into equilibrium, then how is it that we can say that ΔS=ΔQrev/T since we weren't able to follow the process along any curve on a PV diagram? – Milk Man Mar 19 '18 at 03:46
  • The actual process may not have gone along any single curve in PV diagram (if the system was out of equilibrium, then there is no such curve). If that is the case, the relation $\Delta S = \Delta Q_{rev}/T$ cannot be applied to that actual process. Instead, the general Clausius inequality is valid: $\Delta S \geq \Delta Q / T_{reservoir}$, provided the reservoir can be assigned temperature (it must be big enough so that heat exchanges do not throw it out of equilibrium), When equality is used, a path (real or imaginary) is implied, where the system goes only through equilibrium states. – Ján Lalinský Mar 20 '18 at 22:38
  • @hyportnex From the argument you made, I guess you are not convinced yet. Did you find any other convincing proof?I'm looking for the same – Amsterdam6483 Apr 02 '20 at 12:08
  • @Vilvanesh here are three publications that give me confidence (not proof) that there is a proof somewhere, enjoy! [1] https://doi.org/10.1119/1.18158 ; [2] https://doi.org/10.1119/1.13423 ;[3] https://doi.org/10.1590/S1806-11172013000400006 – hyportnex Apr 02 '20 at 12:36