- #1

- 83

- 0

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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter aniketp
- Start date

- #1

- 83

- 0

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

Science Advisor

Homework Helper

Gold Member

- 2,593

- 20

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?

- #3

- 83

- 0

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?

- #4

nicksauce

Science Advisor

Homework Helper

- 1,272

- 5

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.

- #5

nicksauce

Science Advisor

Homework Helper

- 1,272

- 5

- #6

- 83

- 0

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

- #7

nicksauce

Science Advisor

Homework Helper

- 1,272

- 5

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.

- #8

- 83

- 0

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

- #9

- 1,851

- 7

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

Last edited by a moderator:

Share: