# Definition of first law of thermodynamics

1. Nov 14, 2012

### Outrageous

The total work is the same in all adiabatic processes between any two equilibrium states having the same kinetic and potential energy.

That is another way to describe first law of thermodynamics , and define internal energy.
My question is what does "the same kinetic and potential energy " mean ?
Is that mean the kinetic energy of the gas system before adiabatic compression is same as kinetic energy of the gas system after adiabatic compression?
But I though when adiabatic compression, the temperature of the system will increase , and the kinetic energy will increase as well. Correct?

Thank you

2. Nov 15, 2012

### Andrew Mason

It means that if you take two adiabatic processes, one between state I1 and F1 and the other between I2 and F2, the work done on/by the system will be the same IF the internal energy of the system at I1 and at I2 are the same AND the internal energy of the system at F1 and F2 are the same.

This is not particularly useful as a general statement of the first law. There are states of a system between which no adiabatic path exists.

AM

3. Nov 16, 2012

### Outrageous

So it is the internal energy for the adiabatic process remains same, then it has nothing to do with the kinetic energy and potential energy?
Is that mean the kinetic energy will not necessary remain constant?
But that book say,
The total work is the same in all adiabatic processes between any two equilibrium states having the same kinetic and potential energy.

4. Nov 16, 2012

### DrDu

For example?

5. Nov 16, 2012

### Studiot

Well how about a heat exhanger?
Or a process where two liquids at different temperatures are mixed?
Or perhaps the passage of electric current through a resistor?

6. Nov 16, 2012

### DrDu

These seem to be processes rather than states of a system.
E.g. the resistor without current forms a system in equilibrium. The equilibrium states of the resistor can be labeled by temperature (among other, e.g. volume). Clearly the temperature (and internal energy) of the resistor can be changed by doing work even when it is thermally insulated.
Hence the different states are adiabatically accessible (although eventually only unidirectional).

7. Nov 16, 2012

### Studiot

I can't speak for what Andrew had in mind, but I have often pointed out that to define a thermodynamic system you have to specify three things.

The system
The system boundaries
The system process.

With the resistor I have done that, so long as we accept it is not surrounded by an adiabatic enclosure.

Yes the system is the resistor,
Yes, the process is the passage of current,
and yes you cannot prevent the generation of heat and its transfer to the surroundings under these conditions.

8. Nov 16, 2012

### Staff: Mentor

A process is the method by which you move between states.

9. Nov 16, 2012

### DrDu

Ok, but the point is that if the resistor is in contact with a heat reservoir, I assume it to be one of constant temperature, then the the resistor will be in the same equilibrium state after the current has flown as it has been before the current has flown. So evidently the two equilibrium states whose internal energy you want to compare are adiabatically accessible from each other trivially.

10. Nov 16, 2012

### DrDu

In case of mixing two fluids of different temperature, the argument is even more obvious, as the exchange of heat between the two sub systems is still in accordance with the complete system not exchanging heat with the surrounding.

11. Nov 16, 2012

### Studiot

I'm sorry I don't follow what you are getting at.

The term adiabatic refers exclusively to transfers across the system boundary and the variables involved.

Thus we can talk of an adiabatic container or an adiabatic process, but never an adiabatic variable for a system (state) variable.

Thermodynamic states refer exclusively to the values of system variables. These may be instantaneous values or they may persist in time - which we refer to as steady state even if they are the result of continuous transfers.

Either way it makes no sense to talk of an 'adiabatic state'.
I think this was just a loose bit of terminology from Andrew; I have the greatest respect for his thermodynamic knowledge.

12. Nov 16, 2012

### DrDu

Who does so?

13. Nov 16, 2012

### Studiot

Actually looking back I note that Andrew actually spoke of an adiabatic path between states.

As russ_watters said, the path is the track of a system process between states.

But none of this is relevent to your post #10

I still don't see what you are getting at there, what "argument is obvious"?

14. Nov 16, 2012

### DrDu

In the case of the resistor I understand that the equilibrium states before and after flow of the current are identical. Obviously there is also an adiabatic path linking two identical states, i.e. doing nothing.
Mixing of two liquids of different temperatures is already an adiabatic process, or do you consider some heat being transfered to the surrounding?

15. Nov 16, 2012

### Studiot

I don't recall mentioning (thermodynamic) equilibrium - quite the reverse I mentioned temperature difference which is the thermodynamic definition of disequilibrium.

This thread is about the First Law and you cannot apply the First Law within a system, only across the system boundary.

It is important to correctly define the system.

In the case of the resistor, no law of thermodynamics tells us that any particular proportion of the electrical energy in the resistor is converted to heat. We assume near 100% conversion from other theory.
Once we have this heat, it raises the resistor temperature above that of the surroundings so we cannot prevent heat transfer across the boundary from the system to the surroundings.
So there is no adiabatic process available to the system.

In the case of the two liquids mixing strictly, you should define one liquid as the system, the other as the surroundings and then you can apply the First Law to obtain the heat flow between the two liquids.
Further, again as with the resistor this heat flow cannot be prevented.
So there is no adiabatic path available to the process.

Last edited: Nov 16, 2012
16. Nov 16, 2012

### DrDu

My patience explaining basic thermodynamics is exhausted.
I would be more interested in an example for two states of a system not possibly linked by an adiabatic path by Andrew.

17. Nov 16, 2012

### Studiot

Science Advisors should not be able to get away with nonsense like the above by making highly insulting statements like this

Last edited: Nov 16, 2012
18. Nov 16, 2012

### Andrew Mason

Internal energy a system is the sum of the kinetic and potential energy of the system. For a system undergoing an adiabatic expansion in which work is done by the system, there must be a reduction in kinetic + potential energy of the system. For an ideal gas, potential energy does not exist - internal energy is all kinetic. So the kinetic energy and temperature would decrease.

The statement is rather poorly worded, in my view. It does sound like they are saying that the kinetic and potential energy of the beginning and end states are the same. But if that were the case, it seems to me that there would be no change in internal energy. So the statement would be meaningless. So I conclude that the statement is intended to mean what I said in my earlier post.

AM

19. Nov 16, 2012

### Andrew Mason

I was thinking about two states in which V is the same for each state but P and T differ: eg for an ideal gas where: State 1 = (P0,T0,V0); State 2 = (3P0, 3T0, V0)

But, I suppose you could adiabatically compress and then allow adiabatic free expansion back to the original volume to reach the final state. There is certainly no reversible adiabatic path between the two states.

AM

Last edited: Nov 16, 2012
20. Nov 16, 2012

### Outrageous

The total work is the same in all adiabatic processes between any two equilibrium states having the same kinetic and potential energy.