Thermodynamics basics.

Urmi Roy

No,I mean suppose we heat something up,suppose water....I could easily visualise the heat making the water boil...this is a result of increasing the internal kinetic energy...but I couldn't imagine the water starting to flow,like in a pipeway because of the heat!

P.S I found something called the two property rule which says that the state of a thermodynamic system in equilibrium can be completely defined by 2 variables.

George Jones

Urmi Roy, there is an ambiguity in question 2. from your first post. Did you mean:

2. How many state variables (Voume,pressure,temperature etc.) are needed to specify an arbitrary (possibly non-equilibrium) state (at a fixed time) of a thermodynamic system?

If you meant this, then I think the answer is 5, as Andy has said.

Or did you mean

2. How many state variables (Voume,pressure,temperature etc.) are needed to specify an equilibrium state of a thermodynamic system?

If you meant this, then I think the answer is 2, as DrDu and the_house have said. So, I think Andy and DrDu have been answering different questions.

Things are still fuzzy for me, but here is what I think is going on. It takes 5 variables, pressure, volume, temperature, entropy, and U, (in the simple, but general systems we are considering) to pin down an arbitrary state. Since Q = TdS for quasi-static processes, the (hyper)surface on which dU = TdS - PdV holds is the set of all equilibrium states. I think it can be shown that this surface (in our case) is 2-dimensional. To find a specific 2-dimensional manifold of equilibrium states, an equation of state is used. The equation of state together with dU = TdS - PdV is used to generate two Maxwell(-like) relations. The equation of state together with the two Maxwell relations act as three equations of constraint that restrict the 5-dimensional manifold of arbitrary states to the 2-dimensional manifold of equilibrium states.

Urmi Roy

Oops! Sorry for the ambiguity....I didn't realise the gravity of my mistake.

Yes,I did mean the number of state variables needed to define an equilibrium state,as that's all we've been taught to think about yet!
So I guess that answer has been given by DrDu.However,I think I gained a lot in looking through their arguments...It's interesting to know that even a system in a non-equilibrium state can be defined by a fixed number of variables i.e 5.

the_house

Don't get too attached to that statement. I admit I don't know too much about non-equilibrium thermodynamics, but I don't think it's accurate. George himself admitted things are still "fuzzy" for him, so unless he wants to give a reference that describes the out-of-equilibrium part of this 5 dimensional state space, I would take it with a grain of salt. I'm not convinced it even has any meaning, let alone that it uniquely specifies a "thermodynamic system" (whatever that means when the system is far from equilibrium).

Anyway, it does seem we have agreement on the equilibrium case. There are only 2 independent variables that uniquely specify an equilibrium system (plus another for each conserved quantity, but in the simplest case we can ignore that.)

Urmi Roy

Yeah...things do still seem a bit fuzzy,but I got a concrete answer to what I was actually looking for!

Sorry if my questions seem never-ending,but could you people please look into what I put forward in the first post of this page too(and the rest of my brief converstaion with Studiot about it),please? It's something that has got me thinking!
(It's about the SFEE).

George Jones

Yes, I see what you mean. I am struggling to find, in this context, a question that has answer "5." Clearly, such a question should exist. I'll continue on with my method of successive approximations , and modify what I wrote in my previous post.

How many state variables are needed to accommodate (in one space) all possible equilibrium states of all thermodynamic systems?

It takes 5 variables, pressure, volume, temperature, entropy, and U, to accommodate (in one space) all possible equilibrium states of all thermodynamic systems. The relations 0 =dU - TdS + PdV shows that set of equilibrium states for any one thermodynamic system system forms a 2-dimension surface in the 5-dimensional space. To find a specific 2-dimensional manifold of equilibrium states, an equation of state for a specific system is used. The equation of state together with dU = TdS - PdV is used to generate two Maxwell(-like) relations. The system's equation of state together with the two Maxwell relations act as three equations of constraint that restrict the 5-dimensional manifold of equilibrium states for all arbitrary systems to the 2-dimensional manifold of equilibrium state for the specific system.

the_house

Sounds good to me. I think we're making progress.

Sorry, Urmi, I don't think I have anything intelligent to add about your other question, but don't let that stop you from asking more of them.

DrDu

I want to add two points. 1. even for an equilibrium system, there may be more variables to specify the state than two. Two is probably the minimum for most realistic systems.
2. I don't think that any non-equilibrium state can be specified completely by the five variables listed by Andy. Think of a heat driven turbulent flow. You need a whole fields of variables to define the state.

the_house

Exactly. At the risk of sounding repetitive, the number is 2+n, where n is the number of relevant conservation laws for the system in question. 2 is the minimum and the simplest case, but there are many systems that require the treatment of various chemical potentials.

Studiot

All that elasticity theory we did last term?

Let the expansion work against some sort of spring. Energy will be stored in the spring.

Let one side of the expanding object push against a fixed object (the ground?).
The expanding object will expand and push itself bodily upwards, gaining potential energy against gravity.

Andy Resnick

I agree it's not '5'. However, the total number required is an *odd* number.

Andy Resnick

Here's a few:

http://iopscience.iop.org/0305-4470/36/17/301

http://ieeexplore.ieee.org/Xplore/lo...hDecision=-203

http://jmp.aip.org/jmapaq/v39/i1/p32...sAuthorized=no

Edit: Upon reading my posts, I think they may come across as a little 'gruff'. Honestly, I'm just excited to be learning something new.

Urmi Roy

I'm sorry if I'm missing something really obvious!

Okay,so I see what you mean....we have a container full of an ideal gas and one side of it is movable,and attached to a spring....now we heat the gas,and it makes the container's side move and compress the spring....in that way,we sort of see the gas flowing...but we calculate this work as the PdV work(flow work),don't we? Not as the (m(v squared))/2!.

...again,suppose we have a container in which,for the sake of simplicity,the top is open...so now when we heat the gas,it flows upward...we could say the gas molecules acquire potential energy from the heat we supplied it(actually it's because we normally assume the ideal gas molecules to be point masses,really light, that all my confusions arise).

Studiot

I was just trying to answer this question.

If you heat the water in the boiler of a steam locomotive some of that heat is used to impart velocity to the water as it travels along with the train.

Urmi Roy

You mean as in overcoming the inertia? That's an interesting point! But then,wouldn't it be analogous to a train carrying a big box(or something of that kind) just as heavy as the water....
I think I still need an explanation to help me understand what's going on in this example

EDIT: I think I've found another example in which heat energy could be converted to the m(vsquared)/2 i.e kinetic energy of the liquid as a whole...in convection!

Urmi Roy

Hi,
I've run into something else....
We have a heat engine that is operated between two energy reservoirs which have finite heat capacities,my book says the work would continue to be produced till the temperatures of the two reservoirs became the same....but the two reservoirs are not directly in contact with eachother...then why do they wait for the entire heat engine to come to a particualr temperature?

Urmi Roy

This time I'm really really really confused...
Here's what it is..please help...

The clausius inequality says that for an irreversible process,(sum of Q/T)<del S....where S is the entropy change of a reversible process acting between the same two points on the T -S diagram....however,the (sum of Q/T)is basically the entropy change of the irreversible process(acting between two specific points on T-S diagram)...so shouldn't that be greater than the S (entropy change in the reversible process acting between the same two pints on the T-S diagram)?

On the other hand,entropy is a point function.....so the (sum of Q/T) for irreversible process which is basically the entropy change for the irreversible process should be equal to the entropy change delS of the reversible process acting between the same two points on the T-S diagram...