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Re: [Phys-L] work versus mechanical transfer of energy



In the context of
dE = T ds - P dV [1a]

on 01/06/2016 12:09 PM, Carl Mungan wrote:
Heat and work have no intrinsic value or importance; some times they
help us calculate what we really want to know (such as the final
temperature of an object) but when they don’t there’s no need to try
to calculate them.

Exactly. We're allowed to talk about them, but we're not
required to.

Heat and work are always useful in reversible processes.

If they are always useful for you, that's commendable. It means
you are smart and careful. However, pedagogically speaking, you
probably don't want to tell students they are always useful, if
only because there are so many mutually-inconsistent definitions
of heat and work. Some day some chemist will talk to your student
about "heat of reaction" and things will get ugly fast.

On another note, I think formula [1a] (plus any other terms to
account for mass transfer, electrical work, etc) is called the
thermodynamic identity and it only reduces to the first law of
thermodynamics under quasistatic conditions, which is only an
approximation. The identity is indeed just a Taylor expansion of the
energy in terms of its relevant variables.

It seems to me that calling it "the" thermodynamic identity
is a bit of a misnomer. The only identity involved comes from
calculus, not from thermo. We are just expanding the exterior
derivative of E in terms of thermodynamically-interesting
variables.

If you write it in terms of finite deltas, it's a Taylor
series ... but if you write it in terms of the differential
operator "d" it's quite a bit more robust than that. With
a Taylor series, you have to worry about how fast it converges
and about the radius of convergence ... but a derivative such
as dE or dV exists in a zero-sized neighborhood, no radius
required.

Similarly, if you think of it as a Taylor series, you have
to deal with the fact that ΔE is not a function of state.
Rather, it is a function of two states (initial and final).
This stands in contrast to dE, which really is a function
of state. It is a vector-valued function of state, but we
can cope with that relatively easily.

You can use the exterior derivative to construct a Taylor
series, but you're not obliged to, and they're not the same
thing.

Last but not least, it seems to me that [1a] has got little or
nothing to do with the first law of thermodynamics, i.e. with
conservation of energy. It has more to do with the fact that
E is a differentiable function of state than the fact that it is
conserved. None of the steps used to derive [1a] used the fact
that E is conserved. You could perhaps "connect" this equation
to conservation of energy, but you would have to do a bunch of
extra work and bring in a bunch of additional information,
including things like the third law of motion, et cetera. To
appreciate what I’m saying, it may help to apply the same
calculus identity to some non-conserved function of state,
perhaps the Helmholtz free energy F. You can go through the
same steps and get an equation that is very similar to [1a].
If you did not already know what’s conserved and what not,
you could not figure it out just by glancing at the structure
of these equations.

Bottom line: To me, the first law of thermodynamics states that
energy is conserved. Nothing more, nothing less.

Equations such as [1a] are just examples of using calculus (i.e.
the chain rule) to expand the derivative of this-or-that state
function in terms of the chosen variables.

On 01/06/2016 02:27 PM, Diego Saravia wrote:

I use in that eq dU as internal energy not dE or energy

gravitational energy is outside dU

To me that seems like one step forward and two steps back. For
one thing, U is not conserved, so to do anything useful you need
to constantly switch back and forth between U and E. Secondly,
the definition of E is reasonably well known, whereas it’s hard
to learn and hard to remember what’s supposed to be included in
U. In situations where the height h is relevant, you will have
to keep track of it one way or another, so the only question is
whether you will need one new variable or two (h alone versus
h and U both).