# Can a gas decrease its Temp. without doing Work nor exchanging Heat?

1. May 22, 2014

### rollingstein

Two ways of cooling a (real) gas are exchanging heat or doing external work. Let's assume we have a fixed volume of gas that does no work nor exchanges any heat (with surroundings i.e. adiabatic).

Is it possible for such a gas parcel to cool at all? Or heat up?

My intuition says no. An isolated, adiabatic, isochoric parcel of gas cannot change T.

But what prevents it fundamentally? First law constraints would say U = const. (since W & Q both zero). But for a real gas U=fn(T,P) right? So theoritically T could change & P changes to compensate & keep U a constant?

So can it? Or can't it?

What gives? Is it a second law constraint?

2. May 22, 2014

### Staff: Mentor

Not in equilibrium, as your gas fills the whole available volume then.

You can have chemical reactions or other non-equilibrium conditions to get changes.
You can also have stable temperature differences within the gas if you account for gravity (or any other source of a non-uniform potential).

3. May 22, 2014

### dauto

It is possible if the gas is not initially at equilibrium. For instance, you might have the chamber subdivided into two separate chambers connected to each other by a narrow opening. If the pressure in the chambers are initially different from each other then gas will flow from one to the other and the gas temperature might change by Joule–Thomson effect.

4. May 22, 2014

### 256bits

Is the narrow opening necessary for the chambers being at different pressures?

It isn't for a gas expanding into an evacuated chamber to observe the Joule-Thompson effect on real gases.

5. May 23, 2014

### rollingstein

To clarify, let's assume:
(1) No chemical reactions
(2) No internal gradients. (i.e. internal properties initially are homogeneous)
(3) No subdivisions etc.
(4) No gravitational or other potentials

Now in this situation, can the gas change it's T & P such that U remains unchanged?

Obviously the first law isn't being violated here if it did change T,P in a consistent manner. i.e. The gas could spontaneously cool & adjust P to keep U unchanged?

My intuition says it cannot. But what fundamental law prevents it.

6. May 23, 2014

### dauto

No, the narrow opening isn't necessary. It just makes easier (at least for me) to visualize the processes.

7. May 23, 2014

### dauto

Even with homogeneous initial conditions it is possible that the gas was in a metastable configuration. For instance, you might have supersaturated water vapor which eventually might spontaneously start condensing itself, and temperature will change when that happens. That's an example of spontaneous symmetry breaking.

8. May 23, 2014

### rollingstein

Ok, let's exclude phase transitions. Can it still happen? If not, what law prohibits it?

9. May 23, 2014

### rollingstein

Out of curiosity, if the system is closed, adiabatic & cannot do external work where does the energy difference between the high energy vapor & low energy condensed phase go to? Will the vapor condense & heat up the liq?

Can the metastable vapor even condense in a strict adiabatic, isochoric enclosure?

10. May 23, 2014

### AlephZero

Conservation of energy says the internal heat energy has to go somewhere (unless the specific heat of the gas is zero for some strainge reason).

If you keep changing the question by excluding every suggestion of where the energy might go, you will clearly end up with the answer that it's impossible.

11. May 23, 2014

### rollingstein

With all due respect, I don't think I changed it, in any significant way. That's an unfair characterization.

I started with an adiabatic, isolated gas, and that's where it still stands. The clarifications only excluded corner cases like metastable states, external potentials, reactions, internal partitions etc.

Yes, I didn't frame a lawyer-like question with ten clauses to make it water tight against any eventuality.

It may indeed be impossible, that's hardly surprising. The more interesting question is what law prevents it. There's no violation of conservation of energy as far as I can see. Q,W both being zero requires dU = 0.

U=fn(T,P) for a real gas. The question remains why cannot T & P both change & yet keep U a constant.

12. May 23, 2014

First law of thermodynamics says no

This is explained by the second law of thermodynamics which introduces two systems that are at equilibrium with themselves but not each other that are within close proximity to each other are then allowed to interact with each other will reach a mutual thermal equilibrium. The transfer of thermal energy from one system to the other to reach the state of equilibrium and any transfer of energy between systems is considered work.

13. May 23, 2014

### AlephZero

Another way to look at it is the comment I made earlier that it implies $C_V == 0$. For an ideal gas $C_V$ is related to the number of degrees of internal freedom of the particles and is positive (unless you want to postulate that the gas constant $R$ could hypothetically be zero without violating any physical "laws").

I think that is a pretty good justification of why the answer is "no" for a real gas, if you exclude anything that can change the number of internal degrees of freedom per unit mass (e.g. chemical reactions etc).

14. May 23, 2014

### dauto

Yes, the vapor condenses and heats up itself and the liquid hence the temperature change.

Yes, a metastable can decay without any outside stimulus. It's called spontaneous symmetry breaking.

15. May 23, 2014

### namanjain

16. May 24, 2014

### rollingstein

Ah yes. I think that's the explaination that makes sense.

Just to clarify: Q = m Cv ΔT & Q=0 since adiabatic.

So if m & Cv are both non zero ΔT must be zero. Do I read you correctly?