Suppose a system has two parts, $A$ & $B$. They were initially at different temperatures and hadn't achieved mechanical equilibrium. After attainting thermodynamic equilibrium, do they continue to exchange heat energy? ie. after attaining thermal equilibrium, ie. Is the heat that $A$ gains from $B$ the same as the heat gained by $B$ from $A$ so that the temperature remains constant ? Meaning to say, is thermodynamic equilibrium dynamic like chemical equilibrium? Or is it static? What is the reason for any of the two?
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1$\begingroup$ I'm not sure I get your question. It's hard to imagine a completely static thermodynamic equilibrium, since small parts can still move from one part to the other, but in such a way that the average net flow is practically zero. $\endgroup$– Phoenix87Commented Dec 27, 2014 at 16:22
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$\begingroup$ In thermal equilibrium every part of the system is at the same temperature. In thermal equilibrium nothing changes. $\endgroup$– CuriousOneCommented Dec 27, 2014 at 16:22
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$\begingroup$ Thermodynamic equilibrium can be dynamic in cases of chemical equilibrium.A dynamic equilibrium exists once a reversible reaction ceases to change its ratio of reactants/products, but substances move between the chemicals at an equal rate, meaning there is no net change(closed system). $\endgroup$– GowthamCommented Dec 27, 2014 at 16:51
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$\begingroup$ What's the difference? D'Alembert's principle, for example, established dynamics upon statics. $\endgroup$– GeremiaCommented Dec 28, 2014 at 4:48
2 Answers
It is a dynamic equilibrium. To see this notice that the Boltzmann distribution law is statistical in nature, for instance, the molecules that have energies between $E_1$ and $E_1+\Delta E_1$ at time $t_1$ are not necessarily the same than the ones in between those same energies at time $t_2$. Particles collide (and thus exchange kinetic energy) all the time, and while the distribution stays the same, the particles that contribute to different parts of the distribution keep changing.
Update: It is dynamic, like chemical equilibrium, because the molecules do not all have the same kinetic energy. If, at equilibrium, you randomly pick a molecule, there some probability (from the Boltzmann distribution) that it will have its energy between $E_1+\Delta E_1$. If after some time $\Delta t$ you pick those same molecules again, there is again the same probability to find those molecules with an energy between $E_1$ and $E_1+\Delta E_1$ is the same, but those same molecules will likely have a different energy than the one you measured earlier. this is because they molecules collide all the time and exchange energy with each other. From a macroscopic point of view the two states, at times $t_1$ and $t_1+\Delta t$, are the same; from a microscopic point of view, the two states are different. Thus it is a macroscopic equilibrium not a microscopic one. Same when have ice and water at equilibrium. The molecules that form part of the ice or part of the water keep changing all the time. At equilibrium, you keep the same amounts of water and ice, but the molecules that make up the ice and the water keep being exchanged between the ice and the water.
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$\begingroup$ Sir, I saw my book's definition of dynamic equilibrium. In it, the rate of opposite processes must be equal. In ice-water example, the rate at which water freezes = rate at which ice melts. $\endgroup$– user36790Commented Dec 31, 2014 at 7:57
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$\begingroup$ So,sir can I use this equal-rate definition for "thermal" & "mechanical" equilibrium? If so,how?? $\endgroup$– user36790Commented Dec 31, 2014 at 8:02
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$\begingroup$ @user36790 when you have a lot of particles instead of two, it becomes correct that the rate of opposite processes must be equal. Thus, for every particle that leaves some energy interval after a collision there is another (on average) that enters that interval from another collision $\endgroup$– user65081Commented Dec 31, 2014 at 15:59
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$\begingroup$ @user36790 I am not sure what you are asking in your second comment. $\endgroup$– user65081Commented Dec 31, 2014 at 15:59
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$\begingroup$ So, the rate at which particles exit from the distribution= the rate at which particles enter the distribution,right? Sir,how does one microstate make a transit to another microstate?? And sir,if everything is happening so smoothly in equilibrium,why do thermal fluctuations occur that defy equilibrium? $\endgroup$– user36790Commented Jan 1, 2015 at 5:03
Equilibrium is dynamic microscopically but macroscopically it is static because net flow is zero.