# Describe the thermodynamic system of a wire.

• curiousvortex
In summary, the conversation discusses the derivation of an equation for the work done on a wire under tension in order to calculate the change in internal energy and temperature. The conversation also covers the concepts of ordered and disordered work, and the derivation of the equation using mathematical analysis. The equation is then used to calculate the partial derivatives of the Helmholtz function.
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.

What is F?

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

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

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

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.

rude man said:
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

Sorry, that got by me completely.

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

rude man said:
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?

francesco85 said:
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.

rude man said:
$You can do without that concept by dU = dQ + dW 1st law, dw > 0 here. dU = TdS + dW dW = tdL so dU = TdS + tdL dF = dU - TdS - SdT dF = TdS + tdL - TdS - SdT dF = tdL - SdT and the partials fall out as you indicated. Hello. dQ=TdS for every thermodynamic system that undergoes an infintesimal reversible transformation, as it is well known form basic thermodynamics. dW>0 doesn't make sense by itself; you should specify if dl is greater or smaller than 0. dW=tdl is analogous to -pdV in a usual gas system; if you want to build a simple (the simplest?) model where you can see this, you can imagine the wire as composed by N+1 very small masses disposed on a row, each distant d from the next ones. The fact that the wire has a tension t means that every particle A exerts on the next one B a force t in the direction B "with the arrow pointing from B to A". we want to stretch this wire of a length dl; then, we can move the particle identified by the integer n of the following length : dl((n-1)/N). You can now easily conclude. rude man said: [itex]\delta F=\delta U-(\delta T)S-T(\delta S)=-S\delta T+t\delta l$.

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

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## 1. What is a thermodynamic system?

A thermodynamic system is a physical system that is studied under the principles of thermodynamics, which is the science of energy and its transformations. It can be a closed system, where no energy or matter is exchanged with its surroundings, or an open system, where energy and matter can be exchanged.

## 2. How does a wire fit into the thermodynamic system?

A wire can be considered as a part of a thermodynamic system because it is a physical object that has properties such as temperature and energy that can be studied and analyzed using the principles of thermodynamics.

## 3. What is the significance of a wire in thermodynamics?

A wire is significant in thermodynamics because it plays a crucial role in energy transfer and conversion. Wires are commonly used to carry electricity, which is a form of energy, and they also have properties that are affected by thermodynamic processes, such as electrical resistance and heat conduction.

## 4. How is the thermodynamic system of a wire described?

The thermodynamic system of a wire can be described by analyzing its properties, such as temperature, energy, and entropy, and how they change under different thermodynamic processes. This can help in understanding the behavior of the wire and its role in energy transfer and conversion.

## 5. Can the thermodynamic system of a wire be affected by external factors?

Yes, the thermodynamic system of a wire can be affected by external factors such as temperature, pressure, and the presence of other substances. These factors can change the properties of the wire and impact its behavior in thermodynamic processes.

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