Internal entropy change for nonquasistatic process

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SUMMARY

The internal entropy change (\Delta S_i) for a non-quasistatic process, where a reservoir influences a system with fixed N and V, is approximately equal to (\Delta T)^2 when the temperature difference between the reservoir (T_r) and the system (T_s) is minimal. The formula for the change in entropy for the system (\Delta S_s) is given by ∆S_s = 3/2(Nk(ln(Tr/Ts) - ((Tr – Ts )/ Tr))). By applying Taylor series expansion to the logarithmic and linear terms, the relationship simplifies to yield the desired approximation. This analysis is crucial for understanding entropy changes in thermodynamic processes.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically entropy.
  • Familiarity with Taylor series expansion in calculus.
  • Knowledge of fixed N and V conditions in thermodynamic systems.
  • Basic concepts of non-quasistatic processes in thermodynamics.
NEXT STEPS
  • Study the implications of non-quasistatic processes in thermodynamics.
  • Learn about the derivation and applications of the Taylor series expansion.
  • Explore the relationship between entropy and temperature in thermodynamic systems.
  • Investigate the role of fixed N and V in entropy calculations.
USEFUL FOR

This discussion is beneficial for thermodynamics students, researchers in physical chemistry, and professionals working with heat transfer and entropy calculations in engineering applications.

connor415
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Hi all,

I read that the internal entropy change of a system (\DeltaSi) 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 (\DeltaT)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 don't seem to be able to get this result.

Thanks everyone :)
 
Science news on Phys.org
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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