Understanding State Functions in Thermodynamics: Consequences and Clarifications

[member 392791]

I am having difficulty understanding what a state function is. Yes, I know its a function of the state of the parameter, but what are the consequences of state functions? Just beginning thermodynamics and have difficulties keeping everything straight here.

 

[Simon Bridge]

The state function describes the relationship between the different state variables.

 

[Studiot]

The state function describes the relationship between the different state variables.

Are you sure?

I have always understood that a state function is another name for a state variable.

http://chemwiki.ucdavis.edu/Physical_Chemistry/Thermodynamics/State_Functions

State functions (or variables) are connected by an 'equation of state' such as PV=nRT.

@Woopydalan

If the reference does not clear up your question come back here for more and we can look at the relationship between the maths and the physics.

 

[Matterwave]

The state function describes the relationship between the different state variables.

I think you mean equation of state when you write state function there.

A state function is a function that is only dependent on the current state of the system and not on how that system got to be in that state. Things like T, P, V, E, and S are all state functions. Things that aren't state functions, are things that depend on the path a system takes to get to a state. These include things like Heat (the amount of heat into/out of a system depends on the process) and Work.

 

[member 392791]

Well, it says that a state function is only dependent on the current state of the system, so why are they evaluated using integrals, which would imply where the state started and began by evaluating at the upper and lower limit.

Is an easy way to remember the state functions is by the things you calculate with Δ? Meaning, we usually look for ΔE_internal, ΔH, ΔS, ΔT, etc. These are all state functions. The only exception I can think of is pressure, which I haven't seen as anywhere solving for ΔP.

Also, when calculating the work done on a system W = -PΔV, suppose the pressure and volume are changing. Which P is to be used?

 

[Studiot]

so why are they evaluated using integrals

They are only evaluated using integrals when we have to.

The important physics to understand is that the thermodynamic state of a system can be completely specified by knowing the values of certain state functions. You do not have to know them all since missing ones can be calculated by using equations connecting them.

So if we have a list of values at state A and at state B we do not need to integrate we simply take the difference. This difference is the gross difference and we use the symbol Δ. Calculus is not needed.

For example volume is a state function so the if the volume in state A is V_a and in state B is V_b

Then ΔV = V_b - V_a

Also, when calculating the work done on a system W = -PΔV, suppose the pressure and volume are changing. Which P is to be used?

If both P and V are changing you cannot use PΔV.
This is where we must introduce calculus and evaluate an integral as the sum of as series of infinitesmal changes PδV from V_a to V_b. To evaluate this integral we must know P as a function of V.

We can come back to this once you have understood the first part.

 

[andrien]

You need to know only two state function to identify the state of system(in most cases).Also state function is independent of path like for an ideal gas internal energy is state(point) function and so it only depends on temperature(for ideal gas,no path dependence).

 

[member 392791]

Ok, I thought the ΔV was a consequence of calculus..i.e using integral of P dV from Vi to Vf.

 

[Studiot]

Take a simpler example from mechanics.

Work = Force x Displacement
If I carry a mass m up the hill from my house to the monument at the top of the hill the work done

W = mgΔH

Where ΔH = H_monument - H_house

Where H is the hieght (above sea level).

I am assuming the the force is constant and equal to mg throughout.

The symbol Δ is also known as the difference operator and is best reserved for gross differences.

If I did work extending a spring the force is not constant.

In fact for each infinitesimal extension δe I can consider the force constant and deduce that the infinitesimal work is

δW = Fδe

To obtain the total work from extension A to extension B I sum (integrate)

∫δW = ∫Fδe

In order to do this I need either e as a function of F or F as a function of e.

Conventionally we have F = ke where k is the spring constant so

∫δW = ∫keδe = k∫eδe = k[e²/2]

It is the same with the product of pressure and volume change - the product has the dimensions of work or energy.

Does this help?

 

[member 392791]

Yes, so is gravitational potential energy a state function? If so, why is the word "state function'' not introduced until thermodynamics.

 

[Studiot]

Notice that backalong in post#6 I said

"The thermodynamic state"

Yes you could consider potential energy a mechanical state function. Indeed there is a correspondence between Hamiltonian/Lagrangian mechanics and Thermodynamics.
This is explored by Story in his book

Dynamics on Differential OneForms

This is a very modern view.

As to why the nomenclature grew up as it did, you will need to consult the history books.