4
$\begingroup$

A text says:

The laws of thermodynamics deal with energy changes of macroscopic systems involving a large number of molecules rather than microscopic systems containing a few molecules. Thermodynamics is not concerned about how and at what rates these energy transformations are carried out but is based on initial and final states of system undergoing the change. Laws of thermodynamics apply only when a system is in equilibrium or it moves from one equilibrium state to another equilibrium state. Macroscopic properties like pressure and temperature fo not change with time for a system in equilibrium state.

Why do the laws of thermodynamics not apply when the system is microscopic or not at equilibrium but in intermediate stage? How can we say that thermodynamics cannot tell anything about rate and method of energy transformations?

$\endgroup$
  • $\begingroup$ en.wikipedia.org/wiki/Non-equilibrium_thermodynamics $\endgroup$ – getafix Jan 3 '17 at 15:10
  • $\begingroup$ You have a reaction $\ce{A <=> B}$. You have one molecule of $\ce{A}$ in a closed container. Does the reaction quotient have any meaning here? $\endgroup$ – Zhe Jan 3 '17 at 15:23
  • $\begingroup$ Where is that text from? Especially the sentence starting "Laws of thermodynamics apply ..." is utter nonsense. $\endgroup$ – Karl Jul 30 at 7:03
3
$\begingroup$

I'd agree with Li Zhi that this is a very broad question. Let me make a few points and see if they clear up the confusion.

Macroscopic systems - The gist here is that thermodynamics is based on statistics so you need a large enough system (i.e. enough molecules or atoms) for statistics to apply. Thus you can't get the Maxwell-Boltzmann distribution from 3 molecules of gas sealed in some container. You need millions of molecules - a macroscopic system.

initial and final states of system - The foundations of thermodynamics works with the energy difference between some starting point and some end point.

Think of it sort of like rolling a rock downhill. At the top of the hill the rock has some potential energy. Part way down the hill the rock stops and has a lower potential energy. So the difference in potential energy can be calculated by the height difference.

Naively using the rock and hill analogy, thermodynamics says that rocks roll downhill not uphill. Also in order to move the rock uphill you have to input energy into the rock-hill system.

Dynamics - Since thermodynamics is about the energy difference between the initial state and the final state, (basic) thermodynamics doesn't deal with chemical kinetics. But (basic) thermodynamics and chemical kinetics can be combined to become a specialized field of study such as non-equilibrium thermodynamics.

$\endgroup$
  • $\begingroup$ So without chemical kinetics just thermodynamics, is it always valid at equilibrium only ? $\endgroup$ – Matt Jan 3 '17 at 18:00
  • 2
    $\begingroup$ Going back to my rock-hill analogy, the difference in potential energy depends only only the height of the rock. The rock is at "equilibrium" only when it is still. $\endgroup$ – MaxW Jan 3 '17 at 18:13
  • $\begingroup$ Why is it valud at just equilibrium what is problem with non equilibrium state ? Has it got to do something with entropy ,S $\endgroup$ – Matt Jan 3 '17 at 18:18
  • $\begingroup$ I'm oversimplifying this, but consider the rock. If I want to measure the "total energy" of the rock while moving I need to know how fast the rock is moving the slope of the hill, the direction of the rock, the friction between the rock and the hill and so on. So it is a much more complicated problem. But is easy to measure the difference in potential energy of the still rock at two different heights on the hill. $\endgroup$ – MaxW Jan 3 '17 at 19:11
  • $\begingroup$ So thermodynamics works by starting with some "standard state" and then calculating the energy differences from that standard state to some other state. Chemists could do that 100 years ago. It isn't practical to measure thermodynamic properties in an absolute sense since it isn't practical to start with all substances at absolute zero. $\endgroup$ – MaxW Jan 3 '17 at 19:20
3
$\begingroup$

The mathematics of equilibrium systems are far simpler than others

It isn't strictly true that thermodynamics doesn't apply to systems that are not at equilibrium only that the simple theorems and formulae of equlibrium thermodynamics only apply to systems at equilibrium.

The reason why most discussion and most teaching is about those equilibrium systems is because the mathematical results are far simpler and far easier to teach. They are also widely useful even with systems that are not exactly at equilibrium. Chemists can understand the key features of many reactions, for example, by understanding the equilibrium thermodynamics of the reaction (like the heat of formation of the products). Exceptions can be explained by understanding kinetic barriers in the reaction pathway.

But when details of non-equilibrium behaviours are required, the behaviour of many systems is so complicated there are no easy rules to teach or to talk about.

Non-equilibrium thermodynamics is actually a very large field and key results have won Nobel prizes (Prigogine won the 1977 Nobel in Chemistry for work on key aspects of it). Arguably a great deal of the complexity of the universe and life itself is the result of the complexity and structures that can arise when systems are not at equilibrium. So you could even argue that our ability to understand any thermodynamics is a result of the complexity the can arise from non-equilibrium systems. If the subject is that complex, there are not going to be many simple formulae or generalisations to teach or talk about.

And that is probably the real reason why some people think thermodynamics only applies to systems at equilibrium: anything else is too complicated to talk about so educators and textbook writers tend to avoid the subject.

$\endgroup$
2
$\begingroup$

Your questions are really too broad to answer in this forum. One of the strengths of Classical Thermodynamics is that it is "path independent". As your textbook stated, this means it is primarily focused on the starting and end points and not any intermediate states. This could be looked at as a weakness, since it doesn't help much distinguish between alternative paths, or it could be (and is) looked at as a strength since your answer/prediction isn't dependent on (doesn't rely on) the intermediate states (path). In other words, it gains predictive power by limiting itself to path independent states.

There's a lot you can know about a process, given only its initial and final conditions. One of the things this allowed was the development of Classical Thermodynamics BEFORE the discovery of the atomic nature of matter! Just think about that! Before we even knew about atoms, we could predict what the energy balance (not to mention entropy) was of a particular reaction!

Another part of your question states Thermodynamics "doesn't apply" to non-equilibrium systems. First, there's a whole sub-category of Thermodynamics which does deal with systems far from equilibrium, if you're interested. Second, think about what you're asking. You should know (if you think about it) that nothing, nothing is "at" equilibrium...yet. Give it another 10E15 years. So if you actually think that Thermodynamics can't be applied until the Universe comes to thermal (and energetic) equilibrium, you're being foolish. It does apply to systems "near" equilibrium – which is enormously, stupendously useful (if it didn't it would be virtually useless, imho).

The way I look at it you give up the details (the intermediate steps) because otherwise the results would depend on them and if they did then we'd have to know what those steps (states or mechanisms) are in order to apply Thermodynamics – and every time we were wrong about the intermediate steps (which is quite often, they're much harder to pin down because they are occurring on much shorter time scales and are (by definition) much less stable) we'd be wrong about the Thermodynamics – we exchange the intermediate details with more powerful predictions about results.

It's like you asking why the Ideal Gas Law only applies to point particles without volume or with no interactions between them. Short answer: it applies, but not perfectly, to real world gasses. Likewise, thermodynamics applies, but not perfectly, to real world systems near equilibrium.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.