21

You're right as far as it goes -- if you can come up with a Newtonian system that reaches a stationary state from a non-stationary one, then the system must be non-deterministic. The point (to the extent there is a point here) is that this is not as easy as you seem to assume it is. The vast majority of nice smooth Newtonian systems cannot reach any ...


19

Revamped Answer. 2017-07-01 There is no contradiction because your analysis only includes what happens to the gaseous working substance in the Stirling engine, and it neglects a crucial component of the engine called the regenerator. If the regenerator is not included as a component of the engine when we perform the efficiency analysis, then we don't have ...


19

Most quasi-static processes are irreversible. The issue comes down to the following: the term quasi-static applies to the description of a single system undergoing a process, whereas the term irreversible applies to the description of the process as a whole, which often involves multiple interacting systems. In order to use the term quasi-static, one has to ...


18

There's a distinction between microscopic reversibility and macroscopic reversibility. Or if you will, a difference between something being irreversible in theory versus irreversible in practice. (Or absolutely irreversible versus probabilistically irreversible.) A hopefully relatable analogy: Imagine that you have a large number of coins in front of you. ...


12

This is a very interesting, important and at the same time subtle matter which is useful not only in thermodynamics, but other areas as well, and not everyone quite understands it. First off, two statements that you'd better just memorize for now: In a reversible process, equilibrium is never left. Yes, that is a paradox. That's why all real processes are ...


11

The bottom line is that hot water loses heat at high temperature, giving a small negative entropy change while the cold water gain heat at low temperature resulting in a high entropy change. The net entropy change is positive. We can explicitly see this: At any instant, the infinitesimal change in the entropy of the system is $$dS=\frac{dQ_H}{T_H}+\frac{...


11

Your question goes right in the kernel of the meaning of the term state function. A state function is a function defined over all possible states of the system such that its value for every state does not depend on how the system reached the state. Each state has a definite and unique value for the given state function. The state $A$ has a definite value ...


10

Although this isn't obvious, the system doesn't return to its initial state. If you were to very slowly remove the weight from the piston, then the gas would do work on the piston as you removed it, which means that its internal energy would be reduced. If you remove the weight very quickly then the gas still does work on it, but it will do less work than it ...


10

To get the entropy change for a system experiencing an irreversible process, the first step is to forget entirely about the actual irreversible process and, instead, devise a reversible process that takes the system between the same initial and final equilibrium states. That is what is meant by $dq_{rev}/T$. The reversible process that you devise does not ...


10

It is a standard practice in textbook thermodynamics analyses to tacitly equip the surroundings with equipment that does not generate entropy during a process. One such piece of equipment is the so-called ideal constant temperature reservoir. The fluid in such a reservoir is assumed to have an infinite mass times heat capacity so that the temperature of ...


10

Its almost true. Of course, Jamie would beg to differ. However, the general logic is simple. If you have something that's irreversible, energy is lost. That energy was work that could have been used to produce valuable work. If you consider this gas expansion as part of a cycle, then it is more clear that that irreversability calls for more work to be ...


10

Irreversibility comes from the thermodynamics: the probability that we return to the same state in any reasonable amount of time is extremely small. In more technical terms: the entropy is increasing. The proof that the irreversible macroscopic behavior can emerge from the reversible microscopic behavior is known as the Boltzmann H-theorem. (At the time of ...


9

In your question you mentioned two examples -- (1) slowly moving something that has friction, and (2) gradually mixing two chemicals that react spontaneously ($\Delta G\gg0$). Then you said neither of these count as quasi-static because of (1) stiction, and (2) minimum droplet size due to surface tension. I see your objections as pointless nitpicking. ...


8

Many processes cannot be drawn on a p-V diagram because the pressure is not always defined. Those processes that can be drawn are called "quasi-static". However, you cannot look at a certain path and say whether it represents a reversible or irreversible process for sure. For example, imagine a vertical line on the p-V plot, corresponding to adding heat to ...


8

Pretty much the entire field of Classical Mechanics comes down to one thing: prediction. Given the initial conditions of a system, and a set of mathematical laws that model reality, we want to be able to tell what state the system will be in after a given time. The principal of reversibility that Susskind mentions is essential to Classical Mechanics ...


8

Squeezing the wavefunction means confining it to a smaller space. It takes more energy to confine something within a small space than within a big one. 2, 3: These are consequences of the quantum adiabatic theorem: if you take a system in state $n$ of some system, and act on the system sufficiently slowly, it ends up still in state $n$ of the new system. ...


8

To do it reversibly, you can heat the body from $T_1$ to $T_2$ (i.e., over a finite temperature change) using an infinite sequence of constant temperature reservoirs, in which each reservoir in turn is only dT higher in temperature than the body at any time (and also only dT higher in temperature than the reservoir before it in the sequence). Each increment ...


8

The temperature appearing the the Clausius inequality is definitely the temperature of the "boundary interface (with the surroundings)", or simply the temperature of the sources. One of the best places I have seen this discussion is in Fermi's book, chapter 5, section 11. He is explicit about it. To see this you have to recapitulate the steps in obtaining ...


8

In $dS = \frac{dQ}{T}$, the $dQ$ is the heat exchange on a reversible path from the initial state to the final state, irrespective of how the process is actually carried out.


8

The practical definition of a reversible process is one for which the system (no mass entering or leaving) passes through a continuous sequence of thermodynamic equilibrium states. No, the entropy change of a system does not have to be zero for a reversible process. But, the entropy change for the combination of a system plus its surroundings does have to ...


7

The difference is that one expansion is quasi-static (the reversible one) while the other is spontaneous because of a dramatic change of the external constraints (the irreversible one). In the quasi-static case, you start off indeed in the state where gas pressure equates external pressure. An external operator then slightly decreases the outside pressure ...


7

Reversible processes are important because they are related to the efficiency of a process. Take for examples a pair hotplates, one at 100C and one at 0C. In a theoretically ideal setting you could extract some work $W_0$ from this system until the two hotplates reached equilibrium. Then we would say the process is reversible, because in the same ideal world ...


7

One place to look for relatively direct evidence is in the cross sections of time-reversed nuclear and particle physics reactions. For instance comparing $$A + n \to B + \alpha $$ with $$B + \alpha \to A + n$$ consistently shows the same (energy dependent) cross-sections for both directions where these reaction can be done between ground states of the ...


7

The long story short is "we don't know." We actually don't know any laws of physics, if you get down to it. None of them. The universe is what it is, and it behaves the way it behaves. That's really all we can say for sure. Now we can say that we have found laws which we have no evidence to suggest they're ever broken. We have laws that we have put to ...


7

THE RECIPE Apply the First Law of Thermodynamics to the irreversible process to determine the final thermodynamic equilibrium state of the system Totally forget about the actual irreversible process (entirely), and focus instead exclusively on the initial and final thermodynamic equilibrium states. This is the most important step. Devise a reversible ...


7

You went wrong in interpreting the Kelvin-Plank statement of the second law. There is no violation. The statement is (the details of the statement may differ depending on the references)(italics done by me for emphasis): "It is impossible to devise a cyclically operating heat engine, the effect of which is to absorb energy in the form of heat from a single ...


7

I think the confusion can be removed if we change the position of the subscripts: Let us say we give a certain amount of heat $\delta Q$ to the system. If we do this via a reversible process, the resultant change in entropy of the system is given as $\delta S_{rev} = \frac{\delta Q}{T}$ Now, consider if we provide the same amount of heat to the system, but ...


7

The reason why more work is done in a reversible process than an irreversible process is in an irreversible process entropy is generated within the gas whereas in a reversible process entropy is not generated. That additional entropy has to be transferred to the surroundings in the form of heat, leaving less heat available to be converted to work. To ...


6

Yes. For a reversible process, we have the relation \begin{align} dS = \frac{\delta Q}{T} \end{align} and for an adiabatic process, we have (by definition) \begin{align} \delta Q = 0, \end{align} which implies that \begin{align} dS=0. \end{align}


6

Let's look at your first statement: A thermodynamic transformation that has a path (in its state space) that lies on the surface of its equation of state (e.g., $PV=NkT$) is always reversible I don't think this is right, but there may be some implicit qualifiers in your statement that I'm missing. Here's an example to illustrate why. Consider a system ...


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