Internal entropy change for nonquasistatic process

  • Thread starter connor415
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Hi all,

I read that the internal entropy change of a system ([tex]\Delta[/tex]Si) for a non-quasistatic process where a reservoir cools or heats the system, where N and V are fixed and no work is done is approximately equal to ([tex]\Delta[/tex]T)2, when the difference between TReservoir and Tsystem is small. The book said that expanding the logarithm in:

∆Ss= 3/2(Nk(ln(Tr/Ts) - ((Tr – Ts )/ Tr )))

can achieve this.

Note:

∆Ss= change in entropy for system
Tr=Temperature of reservoir
Ts=Temperature of system

I have played around with it for a while but dont seem to be able to get this result.

Thanks everyone :)
 

Answers and Replies

  • #2
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Let $$T_r=T_s+\Delta T$$ Then $$\ln{(T_r/T_s)}-((T_r-T_s)/T_r)=\ln{(1+\frac{\Delta T}{T_s}})-
\frac{\Delta T}{T_s+\Delta T}$$Expanding both terms in Taylor series yields$$\ln{(T_r/T_s)}-((T_r-T_s)/T_r)=x-\frac{x^2}{2}...-x+x^2...=\frac{x^2}{2}...$$where ##x=\Delta T/T_s##
 

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