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## Main Question or Discussion Point

Hey can neone tell me why this is so? I have not got the ans in any book i read...

Thnx 4 replying

Thnx 4 replying

- Thread starter aniketp
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- #1

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Hey can neone tell me why this is so? I have not got the ans in any book i read...

Thnx 4 replying

Thnx 4 replying

- #2

Mapes

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One way of writing the First Law is

[tex]dU=q+w[/tex]

which means that the differential change in energy equals the heat transferred to a system plus the work done on a system. But another way is

[tex]dU=T\,dS-p\,dV[/tex]

where the work is expressed as a

If you haven't heard the terms generalized force and generalized displacement before, they're just ways to classify variables. Generalized forces are intensive, and they drive processes; a change in temperature drives heat flow, and a change in pressure drives mass flow. Generalized displacements, which are extensive, are the "stuff" that is transferred: entropy, volume.

[itex]S=Q_\mathrm{rev}/T[/itex] arises because at constant volume and temperature, and if the process is reversible (no excess entropy generated) we can integrate [itex]q=T\,dS[/itex] to give [itex]Q_\mathrm{rev}=T\Delta S[/itex].

Does this make sense?

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Hey mapes,

But isn't dU=TdS-PdV actually derived from dS=dQ/T?

But isn't dU=TdS-PdV actually derived from dS=dQ/T?

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nicksauce

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Entropy is defined as S = k*log(omega)

Temperature is defined as 1/T = dS/dU at constant V

Pressure is given by P = T * dS/dV at constant U

From the previous two relationships it follows that

dU = TdS - PdV, and S = Q/T is a special case.

For an intuitive discussion see, for example, An Introduction to Thermal Physics by Schroeder, sections 2&3.

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nicksauce

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But in k*log(omega) what do 'k' n 'omega represent?

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nicksauce

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Simple example:

Say I have a system of 3 "oscillators" (that are quantized), with a total energy of "q" units. Say q = 0, then there is only one possible arrangement (Omega=1). Say q = 1, then there are three possible arrangements (Omega = 2). Say q =2, then there are 6 possible arrangements, ie (2,0,0) three times, and (1,1,0) 3 times. So Omega = 6. Say q = 3, then there are 10 possible arrangements, (3,0,0) 3 times, (2,1,0) 6 times, and (1,1,1) once. So Omega = 10.

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Oh ,ok. got it now...thanks once more !

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http://en.wikipedia.org/wiki/Fundamental_thermodynamic_relation" [Broken]

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