Re: calculus
- From: "erehwon" <duff@xxxxxxxxxxx>
- Date: Tue, 13 Dec 2005 21:33:59 +0000 (UTC)
ok thx for the help. I shall go read up some more...
take care,
JJ
"Robert Low" <mtx014@xxxxxxxxxxxxxx> wrote in message
news:405f47F1846abU1@xxxxxxxxxxxxxxxxx
> erehwon wrote:
> > Can someone help with the following logic? I think I am wrong...
> > If free energy of a system is a function of pressure and temperature, we
can
> > write
> > R = G(p,t)
> > Using the chain rule, we
> > can write:
> > (1) dR/dx = dR.dt/dt.dx + dR.dp/dp.dx
>
> This is pretty messed up: I think that
> you're saying
>
> dR = R_p dp + R_t dt
>
> where _p and _t mean partial derivatives wrt p and t
> respectively.
>
> You can interpret this as saying that if p and and t are
> functions of some parameter q, then
>
> dR/dx = R_p dp/dx + R_t dt/dx
>
> where the first is thinking about R as a function of q
> via the dependence of p and t on q.
>
>
> > (2) comparing like terms gives us:
> > (3)dR/dt = -S (is this correct?)
> > (4) dR/dp = V (is this correct?)
>
> This is (or at least can be made) correct, though
> you really ought to think about what the various
> terms mean.
>
> > ON THE OTHER HAND Working another way by setting (a) = (b) we get:
> > dR.dt/dt + dR.dp/dp = Vdp -Sdt
> >
> > => dR/dt + dR/dt = Vdp/dt - S
>
> How did dR.dp/dp (whatever that's supposed to mean)
> divided by dt become dR/dt?
>
> You really need to distinguish partial and ordinary derivatives,
> or you're just going to get more and more confused. I'd recommend
> a good revision of several variable calculus before you go
> much further with thermodynamics.
.
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