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# Re: [Phys-l] Thermodynamics question

If your system includes a battery which can interact with the outside, your dw must be more than just pdv - dw must include a term to account for the energy transferred by that battery.

Bob Sciamanda
Physics, Edinboro Univ of PA (Em)
treborsci@verizon.net
http://mysite.verizon.net/res12merh/

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From: "LaMontagne, Bob" <RLAMONT@providence.edu>
Sent: Sunday, January 10, 2010 4:42 PM
To: "Forum for Physics Educators" <phys-l@carnot.physics.buffalo.edu>
Subject: [Phys-l] Thermodynamics question

For better or worse, my engineering thermodynamics text is wedded to the

du = dq - dw

approach to the first law. Here, dq is the energy transferred to the system due to a demperature difference and dw is energy transferred out of the system by all other means - both are inexact differentials.

Also, it takes the traditional approach of relating ds and dq in any process (reversible or irreversible) as

T ds >= dq

The Gibbs relation amongst the state variables is (P is pressure and dv is change in specific volume)

T ds = du + P dv

one can put this together algebraically to obtain the following

P dv = T ds - du > dq - du = dw

Therefore

dw <= P dv

I am puzzled by this because I can think of a system where dv = 0 yet the system sends energy out to its environment by a battery inside the system attached by wires to a resistor outside in the environment. The energy, dw, transferred out is definitely > 0.

As I stated in the beginning , disavowing du = dq - dw is not going to help me here.

Bob at PC
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