# 1st law of thermodynamics

1. Apr 1, 2007

### Kolahal Bhattacharya

1. The problem statement, all variables and given/known data

Sears and Salinger develops the concept of 1st law of thermodynamics by stating that workdone along all adiabatic paths between two eqlbm. states(in same KE & PE) are the same iff dissipative work does not change the configuration of the system.From this,they claims the existence of a state property called internal energy of the system which is essentially an exact differential.It is given by
This much is OK.Now,they define heat flow Q into the system as Q=W-W_ad where W is the total work.
What do they mean by the 'total work'?I suppose,thie total work includes:
1.configuration work;be it reversible,irreversible,free expansion etc.
2.Dissipative work;at constant configuration,or changing the configuration.
Specifically,I am interested in the bold typed case mentioned last of all.Please let me know if I am correct.

2. Apr 1, 2007

### Mentz114

Unfortunately there's no bold emphasis in your post so we can't see which bit is troubling you.

3. Apr 2, 2007

### Kolahal Bhattacharya

OK,I want to know if the dissipative work that can change the config. is also included in the total work W.This means the dissipative work will change the volume and thereby will do work.The book does not make it clear.

4. Apr 2, 2007

### Mentz114

$$Q=W-W_{ad}$$

Look at the other way around - anything that is not reversible and changes the system must be heat. So the heat is the irreversible part of the total work.

I'm using 'reversible' to mean 'adiabatic' here which may be a liberty.

Last edited: Apr 2, 2007
5. Apr 2, 2007

### Kolahal Bhattacharya

I agree.Yet, for the sake of understanding,I am insisting to clarify that the
'anything that is not reversible and changes the system' includes non-quasistatic irreversible configuretion work as well as dissipative configuration work.Is it?

6. Apr 2, 2007

### Andrew Mason

In a reversible adiabatic process, the system is quasi-static and always at equilibrium. A non-reversible adiabatic process is dynamic and can produce kinetic energy as the result of work being done on or by the gas.

Consider a free expansion of a gas into a vacuum. The gas does no work on the surroundings in expanding, since the pressure the gas is acting against is 0. But it does work on itself and as a result, the gas moves outward with kinetic energy. Since this is non-random motion, this is not an increase in temperature. So the adiabatic work dW_adb = PdV = 0 but total dW is not 0. Temperature is not well defined, however, since the expanding gas is not in thermodynamic equilibrium.

AM

7. Apr 2, 2007

### Kolahal Bhattacharya

I am afraid we are gradually shifting out from my question.

Let me clarify it once more.Sears Salinger consider a system undergoing an adiabatic process between two eqlbm. points 'a' & 'b' having the same kinetic & potential energy.First a free expansion(configuration work zero) from 'a' to 'c'.In this path,they assume no dissipative work.Next,from 'c' to 'b',a reversible adiabatic process.Since it is an experimental fact that total workdone in all adiabatic paths are equal, they say that "total work W_ad is the same in all adiabatic process between any two equilibrium states having the same kinetic & potential energy".

As d'(W_ad) is constant,they define a state property U such that

[Then, total work W_ad includes even irreversible adiabatic configuration work,even if it cannot be shown in PV diagram]

Now,as I got introduced to the concept of heat flow Q=W-W_ad,in the context of non-adiabatic processes,I found the following unclarified in the book:

Does W include dissipative configuration work?Note that we assumed none in the path 'a'-'b'.This is the central question for which I am getting late to bed.So,please understand this question and clarify.

AM's post gives some interesting points.For an ideal gas, temperature change in a free expansion is zero.So,Joule co-efficient (du/dv)_T=0.i.e du=0.
=>d'(W_ad)=0 in this part.Actually,this follows from the general form of 1st law of thermodynamics,where the change in the (KE+PE) between state i and f are accounted.

8. Apr 3, 2007

### Andrew Mason

In the compression, dW_ad = -dU since dQ = 0. So you can rewrite dQ = dW - dW_ad as dQ = dW + dU, which is just the first law.

The problem is that the distinction between work and heat is somewhat artificial. If I compress a gas adiabatically, the gas will get hotter due to the addition of energy (work). If I compress a gas slowly (quasi-statically) so that the gas is always in equilibrium, without allowing heat to flow into or out of the gas, I have a reversible adiabatic compression. If, however, I compress it rapidly to the same final volume by applying more pressure than the gas, I do more work.

In the rapid compression, I add additional kinetic energy to the gas. So the gas, when it settles down, must end up at a higher temperature. You can think of this as a non-reversible adiabatic compression or as a non-adiabatic compression where the excess work W - W_ad is added heat (work in excess of the reversible adiabatic work). It makes more sense to me to restrict the term 'adiabatic' to reversible 'adiabatic' processes, but it seems that I am in a minority.

Another way of looking at this is:

1. there is only one adiabatic path between two identical states. It may or may not be a reversible adiabatic path. If it is a reversible adiabatic path, the adiabatic condition: $PV^\gamma = K$ will apply.

2. there are any number of adiabatic paths between two identical volumes of a gas. The difference is the amount of work done on or by the gas and the final temperature.

AM

9. Apr 3, 2007

### Kolahal Bhattacharya

I hope I got your point.Because when we apply more external pressure,workdone is greater,as mentioned by you,but the process is irreversible,irreversible W_ad> reversible W_ad.So,there may be arbitrary no. of ways of irreversible W_ad where irreversible work done will be different.
So,W_ad between two paths will not be constant.

What about my other question?Does W include dissipative configuration work?

10. Apr 4, 2007

### Andrew Mason

I am not sure what "dissipative configuration work" means.

Also, please note that in my previous post, my reference to "paths between two identical volumes of a gas" may be confusing. I meant that there are many adiabatic paths (from V1 to V2 of the gas) where the V1 for all the paths are identical and the V2 for all the paths are identical but where V1 and V2 are different.

AM

11. Apr 4, 2007

### Kolahal Bhattacharya

Let me approach systematically:
1.dissipative configuration work means a dissipative work that changes the confoguration,say,volume.Suppose,in the path 'a' to 'c' as referred to in my fourth thread we assumed free expansion without a dissipative work that changes configuration.However,if we assumed the presence of such dissipative work,that would be an example of dissipative configuration work.
2.OK,I see...[context:identical volumes...]

3.I am sorry,I need to talk about the reversible/irreversible adiabatic process once again.
From the first law,dU=d'Q-d'W.
Here,d'Q=0
According to you,(also I supposed soin my last thread),d'(W_ad) refers to only reversible processes.How?After all,first law is valid for all processes-be it reversible or irreversible.So,there is a contrast.Even,Sears-Salinger says that "total workdone",not the 'W_ad in reversible paths'.

4.Next,you said in a previous thread,free expansion is a non-random motion.Is it?I suppose not.

12. Apr 4, 2007

### Andrew Mason

I am still not clear on this. Are you speaking about work that is done on the gas but does not go into the internal energy of the gas?

dW_ad has to be the reversible adiabatic work done/by the gas in going from V1 to V2 because that is the only meaningful quantity. As I explained, you can have an infinite number of adiabatic paths, each with different work done/by the gas, in going from one volume to another. There is only one reversible adiabatic path (where $PV^\gamma = K$).

The first law is valid for all paths, reversible or not. dQ = dU + dW. If dQ = 0 then dW = dW_ad = -dU. If the definition of "dissipative configuration work" refers to work that is not equal to the internal energy change, then there must be another quantify of Work/energy (dW) that escapes the system. The first law becomes dQ + dW_dc = dW + dU. Since dQ = 0, dW - dW_dc + dU = 0, so dW = dW_dc + dW_ad or dW_dc = dW - dW_ad

There is random motion of the gas molecules initially only because molecules are colliding with each other. At equilibrium, the average kinetic energy of these molecules determines the temperature.

But this motion becomes a less random motion of molecules as the gas expands freely into the vacuum. Think of a ball of gas that is beginning to expand into the vacuum. The molecules on the outer edge of that ball that happen to be going outward continue to go outward and keep going, leaving the other gas molecules behind. That process repeats as the next outer layer of molecules moves into the vacuum. This makes the motion less random as the gas moves in a particular direction.

AM

13. Apr 4, 2007

### Kolahal Bhattacharya

1.As W_ad refers only to reversible (or, irreversible,for the sake of arguement,say) adiabatic process,dissipative configuration work will not be included in d'(W_ad).If included,d'(W_ad) will be non-unique by virtue of non-uniqueness of dissipation.So,it is not.Then,it is not also included in dU as

On an accepatble ground,you got d'(W_dc) = d'W - d'(W_ad)
It shows that d'(W_dc) is a quantity equivalent to Q.So,it may be included in W or Q...OK.

2.Even there are infinite no. of adiabatic paths between V1 and 2,did you consider that after all we are to achive a state f starting from a state i?Even if I start in irreversible manner,it may be that to achive the state (Pf,Vf),I am forced to compromize to run another irreversible process. So,that d'(W_ad)= d'(W_ad)_1+d'(W_ad)_2 is the same.Likewise,it may be that for all adiabatic processes(no matter whether reversible or not) W_ad=constant.I do not know if it can be proved.But,it looks more consistent with 1st law where we have no bar on the naure of paths.

3.What will you say if you allow the gas contained in a closed cylinder to expand freely in vacuum?You just open the lead, and the gas simply spreads itself over a room,not in any particular fashion/isotropically.Does it moves non-randomly?

14. Apr 7, 2007

### Kolahal Bhattacharya

Maximum,I may thank you for the 1st part.
The 2nd part cannot be possible as as self work is always zero,and a gas cannot work on itself.The d(W_ad) is same for all paths.Be it reversible or be it irreversible.
The 3rd part was even miserable.You said that free expansion is not a random process.

Last edited by a moderator: Apr 7, 2007
15. Apr 7, 2007

### Andrew Mason

I think we may have a language problem here.

A gas can do work on itself just as water under pressure can do work on itself. A sudden rapid free expansion of a gas results in the gas moving outward with kinetic energy. That energy is evident when it slams into the far wall. Where does the kinetic energy come from if not from the gas?

If free expansion was a random process, the molecules entering the vacuum would be travelling in a random direction. Since there is a rapid net movement of molecules from the high pressure to the low pressure, the molecules are not moving completely randomly.

AM

16. Apr 7, 2007

### Kolahal Bhattacharya

AM,I may say sorry as I insulted you in my last thread.

However,I am NOT going to agree with what you have said.

What you are talking about the kinetic energy is not clear.Ofcourse,the gas molecules moves in a bigger space,but that does not help at all.Kinetic energy=(1/2)mv^2.How can you show me that their velocities are also increased?That simply cannot be.When you are considering a free expansion,the gas molecules do not get sudden kinetic energy.They were already in random motion.Rather,in free expansion,their inner pressure get reduced.This shows that on a kinetic theory basis,that their velocity reduces.The gas CANNOT work on itself as always in clasical physics.

Regarding the randomness,you have changed your opinion from 'non random' to 'not completely random'.Free expansion is an irreversible non-quasistatic process and any book should be able to make you understand that in this process,the entropy increases.So,the system get more randomized.It can be nothing but a random process.

The moderator of this forum may leave me outside,but,I understand one thing very clearly.If the physics and concept is the thing I want,I will not compromize with anything.And a true lover of the subject will not try to justify himself at any cost,he will try to understand the basic laws from even the ignorant people.When I insulted you,I had fallen from that religion of physics,so I am shameful about my act.You should also aprreciate the correct concept even if you are not shameful about your attempt to make physics go for you.

Good bye to this forum.

17. Apr 8, 2007

### Andrew Mason

When a gas freely expands into a vacuum it is not in thermodynamic equilibrium! It is dynamic. Not all of the gas has the same kinetic energy. The part that is flowing out into the vacuum has more energy because it is being pushed by the gas on one side with a net force (into the vacuum).

Let's start with a chamber of gas under pressure P and temperature T. There is a small hole in the chamber wall controlled by a valve which opens to a large empty chamber (vacuum). Consider an element of gas dm at that hole. When the valve opens, $dm$ is pushed out. This is because the molecules inside the chamber are pushing it out and there is nothing on the other side pushing back. So this element of gas of volume dV in the chamber blasts out into the vacuum. Then the valve is closed.

What is the temperature of the gas, dm? Is it greater than T, equal to T, or less than T?

The temperature of the gas element dm was T when it was in the chamber under pressure. But then the gas in the chamber kicked dm out into the vacuum it added additional kinetic energy. When things settle down and reach equilibrium again, dm will have greater energy than before - so its temperature will be greater than T. Where does this energy come from? From the gas in the chamber. The gas in the chamber performed work on dm and increased its energy. So the gas in the chamber must have lost energy. It has cooled to something less than T.

This is how a gas does work on itself.

Eventually, when all the gas is let out, the temperature of the gas will even out and it will be T.

If the gas is always in thermodynamic equilibrium, it cannot do work on itself. But if it propels itself into a dynamic state, it does do work on itself. The total energy content does not change, but the distribution of energy within the gas changes.

If it is in a dynamic state (not in thermodynamic equilibrium) the motion of the molecules is not completely random - ie. the distribution of velocities is not uniform throughout the gas.

It goes from a random state (thermodynamic equilibrium) to a dynamic state (not completely random) as it is expanding into the vacuum to a random state when things settle down and again reach thermodynamic equilibrium.

The free expansion of a gas is not a simple thing to analyse. See "http://www.dufourlaw.com/physics/entropypuzzle.pdf" [Broken], for example, regarding the entropy of a freely expanded gas.

AM

Last edited by a moderator: Apr 22, 2017 at 4:51 PM
18. Apr 8, 2007

### Kolahal Bhattacharya

Agree.But I believe the "dm" molecules have the same range of energy as their inmates behind.I will justify it later in this post.

OK.Lets proceed assuming the total system is isolated.

Do not agree.
Let us first consider the "chember" molecules.In equilbm,they were colliding with each other.They had certain mean free path.From Maxwell's velocity and energy distribution we know that maximum of the "chember" molecules have the same velocity/energy range.So,just at the instant the valve is opened,in that time range t_0 to t_0+dt,the distribution has not been changed remarkably,the outgoing 'faster' (called by you) "dm" moleules must have almost the same kinetic energy as that of their inmates.You cannot insist that your "dm" molecules are the ones having high velocity.You must justify.
(The distribution changes with temperature only.Since temp cannot change in that small duration,the distribution will look like the same).

You claimed the "dm" molecules move faster as they were pushed.If this push is arising from collisions,it simply assymetric.Why should the inner molecules collide "dm"s to impart greater KE on them?How do the inmates know when and whom to collide?Remember atleast in the beginning,the KE range is the same,as I showed.If the push is from interactions then the blow to you will be-for ideal gas,no intermoleculer interaction exist.Hence the molecules in the chember cannot push "dm"s.
For the sake of arguement,let me say what I said earlier is wrong.That is the "dm" molecules truly have greater KE and those left at the "chember" have lower KE.Former has higher temp,the latter has lower.So what?What does it prove?You have made two systems different on closing the valve and you cannot say the "chember" gas has done work on itself,because that has been the part of a different system.
Even,if when all the gas is let out, P,v will change, and T will change... I do not see why t will be the same.

If it is in a dynamic state,and if a gas can work on itself in dynamic state,what is the use of talking about this intermediate energy distribtion?Remember, what I asked was a free expanson within a single enclosure,where we were concerned about the two eqlbm. points,not any intermediate states.But there you said the gas will work on itself and that you took into consideration in 1st law.

Maxwell's distribution shows the distribution of velocities is not uniform throughout the gas even in equilibrium.So,be precise...what you are saying as dynamic state is nothing but a simple non-equilibrium state.

OK,I think I am going to like this part...

I thank you very much as you replied...I might have left this forum...I must praise your commitment.

19. Apr 9, 2007

### Kolahal Bhattacharya

AM,I am believing that you have left the discussion.If you are less interested,
I have nothing to say.But,you should confirm whether you approved or were not satisfied by my logical development in my last thread.

20. Apr 10, 2007

### denverdoc

he will return. Innocent bystander I am, isee both of your points of view. I'm a layman. And as such often have to fall back on cartoon images in the what if category involving extremes. My first thought is what if the aperture was so small as to prevent escape of any molecule? Could one reach equilibrium? Obviously yes.

And if there is significant pressure gradient and by allowing an aperture of just one molecules diameter, the guys on the outside have best odds of achieving egress. Now as they disappear, randomly moving molecules nearby have better chance of occupying same space as their predecessors. I think one can continue argument with result that it is a wave of negative entropy proceding from outside in. I don't have the vocabulary to express it mathematically, but hope we can at least continue the debate without the OP leaving PF.