1
$\begingroup$

I am currently watching a thermodynamics course offered by MIT. Here is the link:

https://www.youtube.com/watch?v=kLqduWF6GXE&list=PLA62087102CC93765&index=2

From minute 18:00 until 19:10, the lecturer talks about how thermodynamics is only concerned with systems in equilibrium, then he defines a system in equilibrium to be one whose properties do not change in space or time....(1) Later on in these lectures, the lecturer says that these assumptions are good if we assume compression/expansion of a gas happen "slowly".

I have two questions:

Question 1: Later on in this MIT course, one considers paths plotted in a P-V diagram. I assume that these paths are parametrized by time, so that pressure and density become a function of time. This seems to contradict the lecturer's remark above (see (1) above). Can someone clarify what's going on? Do we want our model to assume that the intensive properties of the system are time-dependent or not time-dependent?

Question 2: I recall from my mechanical engineering classes, internal combustion engines were studied using thermodynamics. These engines obviously operate at high speeds, and so I wouldn't expect their intensive properties to be space-time independent. Why is it valid to study them using thermodynamics? Can one for example take the average of space-time varying intensive properties and work with them?

Question 3: In an attempt to wrap my mind around the importance of time independence for equilibrium to occur, I have a third question: Can someone give me an example of a system (possibly open system) that is not in equilibrium, but whose intensive properties are space-independent (but possibly time-dependent)? If the answer is that there can not be such a system, then we can just redefine the state of equilibrium as a system whose intensive properties are space independent (i.e. drop the time independence condition).

$\endgroup$
3
  • $\begingroup$ Mr. downvoter, do you care to clarify? :) $\endgroup$
    – Amr
    Nov 23, 2021 at 16:02
  • $\begingroup$ " but whose intensive properties are state independent (but possibly time dependent) ? " Did you mean space (or spatially) independent? $\endgroup$
    – Bob D
    Nov 23, 2021 at 19:31
  • $\begingroup$ @Bob D Yes, it's a typo. thanks $\endgroup$
    – Amr
    Nov 24, 2021 at 15:20

4 Answers 4

2
$\begingroup$

Question 1:

A P-V diagram, without an explanation of the particulars of the process, gives no information on the time of a process. The pressure in the diagram is always the external pressure on the system. It is only equal to the system pressure if the process is carried out very slowly (quasi-statically) such that the system is always in equilibrium with the surroundings.

As an example, in an isochoric (constant volume) heat addition or extraction processes, where the P-V diagram is a vertical line, the addition or extraction of heat can be be very gradual (reversible), in which case the change in pressure occurs very slowly, or very rapid (irreversible), in which case the change in pressure occurs quickly. You can't tell which is the case just by looking at the diagram.

Question 2:

The reason the IC engine cycle is modeled as an ideal reversible process, which of course it isn't, is to show what the maximum possible theoretical efficiency that such a cycle can have. Just like the Carnot heat engine cycle gives the maximum possible efficiency of any cycle operating between two thermal reservoirs.

Question 3:

As far as I know, any equilibrium state has to be time independent. The question of spatial independence depends on how the system is defined.

For example, consider a system that is the contents of a rigid, insulated container, making it not only a closed system, but an isolated system. The system consists of two subsystems of ideal gases separated by a rigid partition that are each internally in equilibrium, but have different pressures. We can say that the state of such a system is time independent but not spatially independent, given the pressure differences and the disequilibrium that would result if the partition was removed..

Hope this helps.

$\endgroup$
2
  • $\begingroup$ Thanks for your answer, its still hard for me to wrap my mind around whether our model requires time independence or not, but feel free to ignore me at this point. One more question, Given a closed system whose properties are space independent (but possibly time dependent), in what scenario can such a system be out of equilibrium ? I will edit the body of my question to include this new question $\endgroup$
    – Amr
    Nov 23, 2021 at 18:59
  • $\begingroup$ See update to. my answer $\endgroup$
    – Bob D
    Nov 23, 2021 at 20:48
1
$\begingroup$

First, thermodynamic equilibrium is not exactly analogous to mechanical equilibrium, in which a system is truly in a mechanical constancy. At any given point within in the system at a given time, there will be small fluctuations from true equilibrium. However, such fluctuations are vanishingly small for large macroscopic systems in what we describe as equilibrium.

Second, a state of thermodynamic equilibrium is a highly relative property. It is a valid approximation under the proper temporal and spatial ranges. For example, a new copper pipe may be well approximated as in an equilibrium state with it's environment on the order of years, but will corrode on the over of decades. A puddle on a cloudy day may be approximated as in thermodynamic equilibrium, but over several (even cloudy) days, will evaporate. In terms of spatial equilibrium, at a given time, a certain gas may approximate equilibrium on small, localized volumes, but may exhibit global nonequilibrium. For a rough example of this, take the atmosphere directly around your home on a mild day. At a given time of day, the pressure may remain at a stable value, however, if we were to record the pressure for your country at the same time, the assumption of equilibrium would be invalid.

All of this is to say that, when we define a state of thermodynamic equilibrium as a state of a system where certain state variables remain static in space and time, we mean, over appropriate spatial and temporal scales, we can approximate these macroscopic variables as stationary.

As far as your question about slow expansion/compression of a gas, this is an entirely different assumption. In thermodynamics, a system's state doesn't change unless we alter its state externally (i.e., add heat or perform work). Now, thermodynamics is strictly concerned with a system in equilibrium. If the time it takes the system to relax to an equilibrium state once an external interaction is applied (call this $\tau_{r}$) is much larger than the time is takes to apply the interaction ($\tau_{i}$), than the system will be out of equilibrium significantly longer than the change that caused it does. However, if $\tau_{i}>> \tau_r$, than the system can be approximated to instantly return to a state of equilibrium as the change is applied. This approximation is known as a quasi-static process. For an example of a quasi-static process, imagine a cylindrical container of gas with a pump handle at the top. The gas starts with an equilibrium pressure $P$. The gas also has a specific wave speed $v$, at which waves will propagate through it. If the speed at which you compress the gas at is much slower than the wave speed, the pressure in the gas will roughly stabilize at each increment of the compression. This is a good approximation of a quasi-static process.

As for Question 1, the concept of a quasi-static process is integral. The assumption that a P-V curve is parameterized by time is incorrect, as the system's state variables must be static over time and space, unless acted on externally. What the P-V curve on a P-V diagram shows is the state evolution of the system as a quasi-static process is enacted on the system. That process could be implemented over any arbitrary time interval, so long as the assumption of quasi-staticity.

As for Question 2, I'm a physicist, so internal combustion engines are not my expertise, but I believe it all comes down to idealizations again. It's not the motion of the pistons or other engine parts that we consider in equilibrium, but rather its thermodynamic state variables. As you mention, this is of course not reality. What we typically use thermodynamics for in relation to engines are idealized engines, in order to quantify upper bounds on operating limits. We often assume quasi-static cycles on P-V diagrams to model the engine cycle and quantify properties such as heat exchange and work performed. The idealized engine is the Carnot engine, from which a multitude of information is gleaned, such as maximum engine efficiency between two temperatures, the thermodynamic temperature scale, etc.

$\endgroup$
1
$\begingroup$

The main concept that you need to acquire is the concept of a quasistatic process. A quasitatic process is one which is followed sufficiently slowly that the sequence of states can be regarded, to adequate approximation, as a sequence of equilibrium states. You can imagine this by supposing that the system is moved a little in its state space (by the adjustment of an external constraint such as pressure) and then there is a pause while it equilibrates, and then it is moved again. But a continuous process will give the same result if it is slow enough.

Concerning the claim that 'thermodynamics is only concerned with equilibrium states': this is a misuse of terminology I'm afraid. The correct usage is 'equilibrium thermodynamics is only concerned with equilibrium states'. But thermodynamics more generally is very often concerned with systems that are out of equilibrium. Indeed, many of the central results are important because their range of applicability is very wide: they can treat the outcomes of far-from-equilibrium processes as well as quasistatic ones. An example is the explosive expansion of a gas into vacuum (called free expansion or Joule expansion). We let the process happen and then make statements about overall conservation and things like that. The first and second laws still apply.

One can also model many (not all) non-equilibrium states by dividing the system up into small regions such that each small region is in a good approximation to an equilibrium state, but the different regions are not in equilibrium with one another.

The relaxation time for a litre of gas at room temperature and standard pressure is about ten seconds (I am quoting here from my own book). The relaxation time of each small region of diameter ten microns in such a gas is about 100 nanoseconds. Therefore, to model the internal combustion engine, we can, as a first approximation, adopt a quasistatic treatment of each small cell of gas. This will not be a good model of the spark itself and the explosion, but it can reach reasonable precision for the rest of the cycle. (And for a very rough model you can just treat the whole cylinder contents at once and ignore all the complicated fluid dynamics).

$\endgroup$
2
  • $\begingroup$ Thanks for your answer, its still hard for me to wrap my mind around whether our model requires time independence or not, but feel free to ignore me at this point. One more question, Given a closed system whose properties are space independent (but possibly time dependent), in what scenario can such a system be out of equilibrium ? I will edit the body of my question to include this new question $\endgroup$
    – Amr
    Nov 23, 2021 at 18:59
  • $\begingroup$ the concept of relaxation time in your answer seems to answer my third question $\endgroup$
    – Amr
    Nov 23, 2021 at 19:09
0
$\begingroup$

Question 1: these P-V plots can be observations. if they are then that dousn't contradict the lecturer. it just shows the results of some experiment. the calculations learned in this course are only concerned with systems in equilibrium (without change over time). so if you leave out the time component and only do the calculations on a specific setup (sometimes in relation to a different setup) then everything works out. it dousn't really concern how one setup moves to the other one.

Do we want our model to assume that the intensive properties of the system are time dependent or not time dependent ? yes

Question 2: Why is it valid to study them using thermodynamics ? like i said above each state/setup can be analysed using thermodynamics. just not how they change from one to the next.

Can one for example take the average of space-time varying intensive properties and work with them ? Somewhat you don't get accurate results though (maybe ok for a rough estimate). Continuum mechanics is the one i know of that is concerned with change over time. but those calculations get way, way more difficult and some other property's have to be removed from consideration.

Question 3: these calculations don't work if you want to know how stuff changes over time. all you can get is what is it now and how will it eventually be when a thing has been changed a long time ago.

in my comment i mentioned a vacuum chamber connected to a pressure vessel by a valve. lets consider the ideal gas law $$PV=nRT$$ when you open the valve and instantly measure the pressure at the vacuum side you will find the sensor reading 0 pressure. but the one on the other side in the pressure vessel reads a lot of pressure. neither of these measurements are matching our calculations for what it will eventually be. this is why you need the system to be in equilibrium.

$\endgroup$
2
  • $\begingroup$ Thanks for your answer, its still hard for me to wrap my mind around whether our model requires time independence or not, but feel free to ignore me at this point. One more question, Given a closed system whose properties are space independent (but possibly time dependent), in what scenario can such a system be out of equilibrium ? I will edit the body of my question to include this new question $\endgroup$
    – Amr
    Nov 23, 2021 at 18:59
  • $\begingroup$ if you have a vacuum chamber and a pressure vessel connected by a valve. the moment the valve is opened the system is out of equilibrium. after some time it would find a new equilibrium and stay in that until a thing is done about it. $\endgroup$ Nov 23, 2021 at 20:39

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.