# Internal entropy change for nonquasistatic process

1. Feb 13, 2010

### connor415

Hi all,

I read that the internal entropy change of a system ($$\Delta$$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 ($$\Delta$$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 :)