Internal entropy change for nonquasistatic process

Since ##x## is small, we can ignore higher order terms and thus the entropy change for the system is approximately equal to $$\frac{3}{2}Nk\frac{\Delta T^2}{2T_s^2}=\frac{3}{4}Nk\frac{\Delta T^2}{T_s^2}$$.In summary, the book states that for a non-quasistatic process with fixed N and V, and no work done, the internal entropy change of the system is approximately equal to the square of the temperature difference between the reservoir and the system, when the difference is small. This can be achieved by expanding the logarithm in a Taylor series, where higher order terms can be
  • #1
connor415
24
0
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 don't seem to be able to get this result.

Thanks everyone :)
 
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  • #2
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##
 

1. What is "internal entropy change"?

"Internal entropy change" refers to the change in the amount of disorder or randomness within a system. It is a measure of how much the system's energy is distributed among its particles or components.

2. What does "nonquasistatic process" mean?

A nonquasistatic process is a type of thermodynamic process where the system is not in equilibrium during the entire process. This means that the system does not go through a series of small, incremental changes and instead experiences significant changes in a short period of time.

3. How is internal entropy change calculated for a nonquasistatic process?

The internal entropy change for a nonquasistatic process is calculated using the second law of thermodynamics, which states that the change in entropy of a closed system is equal to the sum of the heat transferred to or from the system divided by its temperature. This can be represented mathematically as ΔS = Q/T.

4. Why is the internal entropy change important?

The internal entropy change is important because it is a fundamental concept in thermodynamics and helps us understand how energy is transferred and transformed within a system. It also allows us to predict the direction and efficiency of processes, and is crucial in the design and optimization of various technologies and systems.

5. Can the internal entropy change for a nonquasistatic process be negative?

Yes, the internal entropy change for a nonquasistatic process can be negative. This would mean that the disorder or randomness within the system has decreased, which is possible if energy is transferred out of the system or the system experiences a decrease in temperature. However, this would violate the second law of thermodynamics and is not a common occurrence in nature.

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