Describe the thermodynamic system of a wire.

[curiousvortex]

Homework Statement

http://i.imgur.com/2WKCR.jpg

Homework Equations

Potentially: L(final) = L[1 + tau/(YA) + alpha(T - To)]

The Attempt at a Solution

Frankly I have no idea where to begin.

 

[rude man]

What is F?

 

[francesco85]

Hi! In my opinion the answer is the following: the work done on the wire in order to stretch the wire of $\delta l$ is $\delta W=t \delta l$ where t is the tension of the wire (notice that \delta W is positive if $\delta l$ is positive, opposite behaviour with respect to the pressure and analogous behaviour with respect to the surface tension); indeed, one should reasonably make a positive work in order to stretch the wire; the first law of the thermodynamics in this situation is
$\delta U= T \delta S+ t \delta l$

Since F=U-TS (I mean with F the Helmoltz free energy) i obtain

$\delta F=- S \delta T+ t \delta l$


and so


$\frac{\partial F}{\partial l}|_T=t$

$\frac{\partial F}{\partial T}|_l=-S$

 

[rude man]

Well, you left out quite a few steps but it's right.

BTW my textbook uses "A" instead of "F" for the Helmholz function. And, if I may suggest, don't use t for temperature. Use T. t is almost always time.

 

[francesco85]

Well, you left out quite a few steps but it's right.

BTW my textbook uses "A" instead of "F" for the Helmholz function. And, if I may suggest, don't use t for temperature. Use T. t is almost always time.

t is the tension and T is the temperature :)

Anyway, can you tell me please what are the steps I left out? Thanks,
Francesco

 

[rude man]

Sorry, that got by me completely.

You gave an equation, whence did it come? How about starting from the 1st law?

 

[francesco85]

Sorry, that got by me completely.

You gave an equation, whence did it come? How about starting from the 1st law?

I'm sorry, but I think that I don't understand: let me go into the details of the derivation, also covering well-known thermodynamics result; is the equation you are talking about
$\delta U=T\delta S+t\delta l$ ?

If so, then I think that this is just the first law for the system we are talking about (with $\delta$ I mean an infinitesimal change; well, to be rigorous, an infinitesimal reversible change); for every thermodynamic system we are supposed to have a function of state U (the internal energy) and to be able to define an ordered work ($t\delta l$) and a disordered work ($T\delta S$); then, by definition, I define
$F=U-TS$ and then
$\delta F=\delta U-(\delta T)S-T(\delta S)=-S\delta T+t\delta l$.

Did I understand what you have asked or did you mean something else?

 

[rude man]

I'm sorry, but I think that I don't understand: let me go into the details of the derivation, also covering well-known thermodynamics result; is the equation you are talking about
$\delta U=T\delta S+t\delta l$ ?

If so, then I think that this is just the first law for the system we are talking about (with $\delta$ I mean an infinitesimal change; well, to be rigorous, an infinitesimal reversible change); for every thermodynamic system we are supposed to have a function of state U (the internal energy) and to be able to define an ordered work ($t\delta l$) and a disordered work ($T\delta S$); then, by definition, I define
$F=U-TS$ and then
$\delta F=\delta U-(\delta T)S-T(\delta S)=-S\delta T+t\delta l$.

Did I understand what you have asked or did you mean something else?

$\delta F=\delta U-(\delta T)S-T(\delta S)=-S\delta T+t\delta l$.

Maybe I'm being picky but
dU - SdT -TdS = -SdT + tdL was not obvious to me. Admittedly I did not identify 'orderd' and 'disorderd' work as tdL and TdS respectively. In fact I don't remember coming across those terms at all except that certainly they make sense, qualitatively at least.

You can do without that concept by
dU = dQ + dW 1st law, dw > 0 here.
dU = TdS + dW
dW = tdL
so dU = TdS + tdL
dF = dU - TdS - SdT
dF = TdS + tdL - TdS - SdT
dF = tdL - SdT
and the partials fall out as you indicated.

 

[francesco85]

[itex]

You can do without that concept by
dU = dQ + dW 1st law, dw > 0 here.
dU = TdS + dW
dW = tdL
so dU = TdS + tdL
dF = dU - TdS - SdT
dF = TdS + tdL - TdS - SdT
dF = tdL - SdT
and the partials fall out as you indicated.

Hello.

dQ=TdS for every thermodynamic system that undergoes an infintesimal reversible transformation, as it is well known form basic thermodynamics.

dW>0 doesn't make sense by itself; you should specify if dl is greater or smaller than 0.

dW=tdl is analogous to -pdV in a usual gas system; if you want to build a simple (the simplest?) model where you can see this, you can imagine the wire as composed by N+1 very small masses disposed on a row, each distant d from the next ones. The fact that the wire has a tension t means that every particle A exerts on the next one B a force t in the direction B "with the arrow pointing from B to A". we want to stretch this wire of a length dl; then, we can move the particle identified by the integer n of the following length : dl((n-1)/N). You can now easily conclude.

$\delta F=\delta U-(\delta T)S-T(\delta S)=-S\delta T+t\delta l$.

Maybe I'm being picky but
dU - SdT -TdS = -SdT + tdL was not obvious to me. Admittedly I did not identify 'orderd' and 'disorderd' work as tdL and TdS respectively. In fact I don't remember coming across those terms at all except that certainly they make sense, qualitatively at least.

take two generic functions f(x,y), g(x,y) (sufficiently regular): then, by some mathematical analysis course,
$d(f g)=g(\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy)+f(\frac{\partial g}{\partial x}dx+\frac{\partial g}{\partial y}dy)$

then we obtain the wantedd result if x=S, y=T, f(S,T)=S, g(S,T)=T.

dU - SdT -TdS =TdS+tdl - SdT -TdS=-SdT + tdL


Francesco