Question

I need a bit of help getting this started and I can hopefully figure out the rest!
I'm need to find an expression so that dW = P*dL

Answer 1

i'll maybe have a look-see. Q for you, though, what's the context of this question, and, by the looks of it it's undergraduate standard, possibly thermodynamics ?
http://perendis.webs.com

Answer 2

Yes @osprey, it is Thermo of Materials.
The question wants me to find the combined first and second law for dU(Internal Energy) for a reversible change.

Answer 3

Where only mechanical work is done denoted by dW = P*dL

Answer 4

I attempted to solve for P(force) and this is what I have come up with. @osprey

Answer 5

As you stretch a solid rod, it gets warmer ? (I remember that tungsten did last time I yanked a small rod apart ... And very hot, actually.)
Just scratched out my birdbrain - feathers everywhere ! encl is pptx which MAY help. "Top of my head" stuff, or is that beak ?
Keep me posted, thermo is one of my ... hmm (I can't write it here)

Answer 6

I think that sooner or later there's got to be some differentials taken, and the variables look like just about everything in sight .. temp, force, area possibly (for stress) material constants, possibly a bit of heat capacity ... I'm guessing ... maybe, on the blunderbuss theory, I'll get somewhere near the mark.
Just reread and seen "only mechanical work is done". That seems to fly in the face of one of my previous posts, where I reckoned that the tungsten got hot ('cos I didn't allow it to cool down ADIABATIC ?)

Answer 7

Thanks, Ill keep trying at it and yes I was definitely expecting differentials at some point! @osprey

Answer 8

couple of points. I've got a bet on with myself that this q may stem from Illinois ? Second would be to obtain some further information. For what it's worth, I usually "expect" to see dW = pdV in thermo. I accept that that could just be notation, but, as with a lot of this stuff, it doesn't help the situation. Putting it another way, I find it "unsighting", as it were. And there's an awful lot of symbols flying around here ! All that said, SOMETHING may just plop out (actually it usually works that way - things do plop out; I often wonder if the source questioners design the questions to look awful ((and succeed)) but there's a trick to them. Applied maths is particularly good at it).

Answer 9

I'm sorry to say you'd be on the losing end of that bet! Yes your're right to think that PdV is expected in thermo, however that is mostly the case when dealing with gases(P-V Work) but here we have a solid which we are told is stretched Uni-axially by a force P. @osprey

Answer 10

oehernand
I need a bit of help getting this started and I can hopefully figure out the rest! I'm need to find an expression so that dW = PdL

I got my ferret out, and I've been ferreting. I found this problem ...
The tension F in a metal wire 100cm and area 0.001cm sq is increased quasi-statically and isothermally from zero to 100kgf. If the area and Young's modulus are constant, what's the work done in the process ?

Looking at the above precis of the problem, it looks less promising to help than it did. But, I'll post it anyway ...

Answer 11

My ferret is clearly feeling frisky, 'cos I've been having a noss at a text book by ZEMANSKY on stretched wires under "special topics". We could be up to our necks in partial differentiations here. I get the feeling that since temp is involved in your post, that your system is "non isothermal" (tautological, I know). I hope that it's adiabatic, then 'cos oooo deeeeeah meeeeee ! There could be squiggles everywhere ...
:)

Answer 12

As T is not stated to be a constant, I'm afraid we can assume a non-isotherm case unfortunately. I was thinking of just having everything in terms of PdL. I will attach the full problem so that you may see the full picture. @osprey

Answer 13

I say that because If you take a look at part two, you can see the only variables they care for are U,P,L,T and S.

Answer 14

This is what I have so far and I've got a terrible feeling that all of part 2 is wrong.