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Re: dW turns into dQ



Yes, once you have chosen a reversible path for the calculation of
dS=dQ/T, you must correctly (and uniquely) specify dQ vs. dW (to
correctly evaluate dS this way). In such idealized quasi-static
processes, however, there is (I think) never much to quibble about as to
what is dW and what is dQ.

But I really think that logically it goes the other way. Once the final
state is determined (in part by applying the first law, with wide
discretion in dW vs dQ), then the change in entropy is determined and
known from its state function - this then determines dQ=TdS. In a
reversible process it is the entropy change which defines dQ.

IE.: for reversible processes the first law reads: dE = dW +TdS. Here
only W is not a state variable and is defined by this relation. There is
no question in defining everything else (and adding dQ=TdS).

But be careful! This dQ is for this reversible path only - the specified
end states can be connected by other processes involving different values
of dQ.

Bob

Bob Sciamanda (W3NLV)
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
http://www.velocity.net/~trebor

----- Original Message -----
From: Jim Green <JMGreen@SISNA.COM>
To: <PHYS-L@lists.nau.edu>
Sent: Monday, November 01, 1999 4:40 PM
Subject: Re: dW turns into dQ


At 01:17 PM 01-11-99 -0500, you wrote:
This might be compared to asking, in F=mA, "if F includes the sum of
electrical, gravitational, etc forces and only their (vector) sum
matters
for calculating A, why the partitioning?" Of course it is because this
partitioning is our way of enabling the calculation of that sum, but
the
resulting A wouldn't care if you used a different partitioning
(taxonomy
of F) so long as you applied the same vector sum.

Yes, Bob, but not quite. As far as dE goes mox nix. BUT dQ also
changes
the entropy! So it is requisite to at least consider some sort of
partition between dQ and dW -- surely so in the cases of reversible
actions.

Jim Green
mailto:JMGreen@sisna.com
http://users.sisna.com/jmgreen