## Thermodynamics

water2
Posts: 1
Joined: Sun Nov 22, 2009 5:26 pm

### Thermodynamics

How to prove the zeroth law of thermodynamics?
Last edited by water2 on Mon Nov 23, 2009 5:34 pm, edited 2 times in total.

grae313
Posts: 2297
Joined: Tue May 29, 2007 8:46 pm

### Re: Thermodynamics

t3chn0n3rd
Posts: 11
Joined: Sat Jun 21, 2008 4:43 pm

### Re: Thermodynamics

I guess you could approach this either experimentally or theoretically.

Experimentally you could go to the lab and perform experiments to determine if 2 systems are in equillibrium with another system they are in equillibrium with each other.

But I enjoy the math, and think it would be more fun to theoretically prove this law.

grae313
Posts: 2297
Joined: Tue May 29, 2007 8:46 pm

### Re: Thermodynamics

lol, he edited his post to make it not so blatant.

Isn't the zeroth law if A=B and B=C then A=C? Isn't that an axiom?

t3chn0n3rd
Posts: 11
Joined: Sat Jun 21, 2008 4:43 pm

### Re: Thermodynamics

Yeah if a=b and b=c then a=c. But I was saying you could go into a lab and perform an experiment. Maybe boil 3 or so pans of water, and get out a thermometer.

But I was saying it would be more fun to prove it theoretically using the 2nd law of thermo and some differential equations or something.

Anytus
Posts: 2
Joined: Sun Sep 12, 2010 8:05 pm

### Re: Thermodynamics

Bit of an old question but I wanted to provide what I think is a solution (I'm sure the OPs homework deadline has LONG since passed).

We should be able to prove the zeroth law by assuming the counter-factual.

Assume we have 3 systems A, B, and C. A is in thermodynamic equilibrium with C and B is in thermodynamic equilibrium with C but A and B and NOT in thermodynamic equilibrium. We know that when two systems are in equilibrium they share certain characteristics, one of which is that they have the same temperature. System A has temperature T1 and B has temperature T2. Clearly then C must have BOTH temperature T1 and T2. This is a contradiction (one system cannot have 2 temperatures) and so it must be the case that T1 = T2 and A and B are in thermodynamic equilibrium (or at least they ahve the same temperature and you chould check all other properties similarly).

kroner
Posts: 218
Joined: Fri Sep 25, 2009 1:58 am

### Re: Thermodynamics

Anytus wrote:Bit of an old question but I wanted to provide what I think is a solution (I'm sure the OPs homework deadline has LONG since passed).

We should be able to prove the zeroth law by assuming the counter-factual.

Assume we have 3 systems A, B, and C. A is in thermodynamic equilibrium with C and B is in thermodynamic equilibrium with C but A and B and NOT in thermodynamic equilibrium. We know that when two systems are in equilibrium they share certain characteristics, one of which is that they have the same temperature. System A has temperature T1 and B has temperature T2. Clearly then C must have BOTH temperature T1 and T2. This is a contradiction (one system cannot have 2 temperatures) and so it must be the case that T1 = T2 and A and B are in thermodynamic equilibrium (or at least they ahve the same temperature and you chould check all other properties similarly).

The possibility of an objective temperature scale depends on the zeroth law. Your argument is completely circular.

kroner
Posts: 218
Joined: Fri Sep 25, 2009 1:58 am

### Re: Thermodynamics

grae313 wrote:lol, he edited his post to make it not so blatant.

Isn't the zeroth law if A=B and B=C then A=C? Isn't that an axiom?

To be fair, I have to take issue with this too. The zeroth law is not just a restatement of math definitions. It has actual physics content, which what you said misses: that we can assign a number to every object that's in thermal equilibrium and then that these numbers will predict how heat flows between various objects.

As always, whether you take the zeroth law to be an axiom depends on which facts you want to start with as "most fundamental" and which you want to prove from those. Pretty sure t3chn0n3rd is right that you can prove it from the second law, but I don't actually know anything about thermodynamics.

quizivex
Posts: 1035
Joined: Tue Jan 09, 2007 6:13 am

### Re: Thermodynamics

yeah, what kroner said.

The concept of temperature is well defined only if the zeroth law holds. It's an assumption inherent in the rest of the theory, which I guess is why they added it as another "law". Perhaps you could call it an axiom since the rest of the theory relies on it, yet even if it has been justified experimentally, it can't be proven unless you've tested it with every possible substance in the universe.

Thermo isn't fresh in my mind now, but I don't think the zeroth law can be proven from the second because the statement of the second uses temperature.

In stat mech (also not fresh in my mind), the zeroth law is a result, not an axiom, since temperature is defined as a number in terms of microstates, so A=B & B=C -> A=C is a trivial result.

cesascencio
Posts: 7
Joined: Tue Jun 28, 2011 8:25 pm

### Re: Thermodynamics

I am so surprised that no one has mentioned entropy. The derivation of all equilibrium conditions require the second law of thermodynamics. Ok, so here is a "proof" of the zeroth law:

Second Law: The universe acts spontaneously so as to maximize its entropy.
First Law: The amount of energy in the universe is conserved (that is, constant).

Ok, here we go. Suppose you have a composite system made of two subsystems A & B in thermal contact. Also suppose that this composite system is isolated from the rest of the universe (subsystems can only interact with each other) and that the subsystems can only exchange units of energy q (no particle exchange, volume exchange...). In such a composite system changes in total entropy can only come about via exchanges in energy units. Agree?

Since the composite system is isolated from the rest of the universe, energy is conserved. Thus, the total number of energy units is conserved.

N = NA + NB ----> total energy in the composite system is then U = N*q = (NA + NB) *q

where

NA = number of energy units belonging to A ---> not constant due to energy exchange
NB =number of energy units belonging to B ---> not constant due to energy exchange
N = total number of energy units of composite system ---> constant

The energy of each subsystem is then ---

UA = NA*q UB = NB*q so that NA = UA/q
NB = UB/q

where U = conserved = UA + UB, but UA and UB individually can change due to energy exchanges between subsystems as long as U is conserved, right?

So! We are allowing these subsystems to spontaneously exchange energy. Will the spontaneous transfer of energy between these subsystems every cease? Yes! The second law tells us that once the total entropy is maximized, spontaneous changes in the total entropy of the universe (in our case, the isolated composite system) will cease and we know that for this composite system changes in entropy are governed by energy exchanges. Thus, energy exchanges will cease once the total entropy is maximized. What we need to look at are changes in the total entropy of the composite system due to the changes in energy of either one of the systems. The reason is as follows: The maximum total entropy of the composite system describes a unique microstate of the composite system (this microstate corresponds to one of the configurations of the macrostate with the largest multiplicity) in which each subsystem contains a particular number of energy units such that spontaneous energy exchanges no longer occur. In other words, there is a unique NA and NB for which the total entropy will be maximum. We need to look for the maximum of the entropy vs. NA curve (or NB curve). We are looking at Stot = f(NA) and we want to find the condition for which Stot = Stotmax:

From elementary calculus we know that extremal points will occur when dStot/dNA = 0, but we also know that an extremal point can be a stationary point, minimum or maximum. But the second law tells us that the total entropy will spontaneously increase so that we need not to worry about minima or stationary points on the Stot vs NA curve (there is a particular NA for which Stot is a maximum, right?) . Lets look for the maximum.

the total entropy of the composite system is Stot = SA + SB, we need to find the condition for which dStot/dNA = 0:

dStot/dNA = d(SA + SB)/dNA = 0 ----> dSA/dNA + dSB/dNA = 0 -----> dSA/dNA = - dSB/dNA, but any loss of energy units by A is
the gain in energy units in B
dNA = -dNB

----> dSA/dNA = dSB/dNB , but NA = UA/q and NB = UB/q
----> dSA/d(UA/q) = dSB/d(UB/q) , but q is a quantum of energy and is therefore constant so that it cancels from both sides and we are left with the thermal equilibrium (no more spontaneous energy exchanges between A and B) condition:

** dSA/dUA = dSB/dUB**

So what if we have a third subsystem, C, and it is in thermal equilibrium with A. That is, it satisfies the condition dSC/dUC = dSA/dUA . Ok then, if we were to assume that A was also in thermal equilibrium with B, we have dSA/dUA = dSB/dUB.

So we have dSA/dUA = dSC/dUC and dSA/dUA = dSB/dUB, so this implies that dSC/dUC = dSB/dUB MUST be true. In words, if system A is in thermal equilibrium with system C and system B is also in thermal equilibrium with system A, then, by way of the thermal equilibrium conditions, system C must also be in thermal equilibrium with system B. This is the zeroth law of thermodynamics. Therefore the zeroth law is a result of the second law. One more thing:

dS/dU is the definition of the inverse temperature in thermodynamics so that we have the equilibrium condition for two systems:

dSA/dUA = dSB/dUB ----> 1/TA = 1/TB ----> TA = TB in thermal equilibrium

If there is a third system such that TC =TA and TB = TA, then it follows that TB = TC, again the zeroth law!

If I screwed up anywhere, you think my logic is flawed, or you have a question, please feel free to send me a message.

Hope it helps! ....and it's not too late