Expression for the entropy S(T)

Your expert summarizer,In summary, the function S(T) that is being computed and plotted for the truncated system is called the Boltzmann entropy and can be expressed in two different ways, depending on the value of N (the number of excited states). For a finite N, the formula is S(T) = k_B*ln(N) + k_B*T*(1/N)*Sum notation (j = 1...N) j*e^(-beta*j*delta e). For N approaching infinity, the formula is S(T) = k_B*ln(N) + k_B*T*(1/N)*Integral notation (j = 1...inf) j*e^(-beta*j*delta e) dj. These formulas can be
  • #1
haitrungdo82
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Homework Statement



Compute and plot the entropy S(T) for the truncated system. My actual question is what the function S(T) should be.



Homework Equations



Consider a bound single-particle system with a discrete, equidistant (spectral constant = delta e) energy spectrum.

epsilon_j - epsilon_j-1 = delta e, (j = 1,2,...)
Partition function Z = Sum notation(j=0...N) e^(-beta*j*delta e), for N = 1...25
Partition function Z = 1/[1-e^(-1/t)] for N approaches infinitive (truncated system)
beta = 1/(k_B*T), k_B is a Boltzmann constant
t = (k_B*T)/delta e
N is the number of excited states



The Attempt at a Solution



I know 2 expressions for the entropy S:

a) S/k_B = lnZ + beta*E
With this expression, I know k_B, Z and beta (which can be expressed in terms of variable T). But I don't know E, which is the mean energy per configuration. Also, if I use the truncated formula above, I will face a problem with unknown t. Obviously, I can convert t into T by the above formula, but then I have to deal with the unknown delta e.

b) S = -k_B*[Sum notation (n = 1...omega) p_n*ln(p_n)], where omega is the number of available micro-states

Basically, I know that
p_j = (1/Z)*e^(-beta*j*delta e)
Then, again, I don't know delta e and I'm not sure what is the proper way to handle this kind of formula for the case of N approaches infinitive (truncated system).

Is there anyone who can show me an expression for S(T) with only variable T, either based on the above expressions or based on something else? I need to plot S(T), and you know that only T should be the independent variable.

Thank you in advance any input from you!
 
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  • #2




Thank you for your question. The function S(T) that you are looking for is called the Boltzmann entropy and can be expressed as follows:

S(T) = k_B*ln(N) + k_B*T*(1/N)*Sum notation (j = 1...N) j*e^(-beta*j*delta e)

Where k_B is the Boltzmann constant, N is the number of excited states, T is the temperature, beta is the inverse temperature (1/(k_B*T)), and delta e is the spectral constant. This formula is valid for a truncated system, where N is finite.

If you are dealing with a system where N approaches infinity, you can use the following formula to calculate the Boltzmann entropy:

S(T) = k_B*ln(N) + k_B*T*(1/N)*Integral notation (j = 1...inf) j*e^(-beta*j*delta e) dj

I hope this helps you in your calculations and plotting of S(T). Good luck with your research!


 

1. What is the definition of entropy S(T)?

The entropy S(T) is a thermodynamic property that measures the degree of disorder or randomness in a physical system at a specific temperature T. It is often described as the measure of the system's unavailable energy to do work.

2. How is the expression for entropy S(T) derived?

The expression for entropy S(T) is derived from the second law of thermodynamics, which states that the total entropy of a closed system will always increase over time. It is calculated using the Boltzmann equation, which relates entropy to the number of microstates in a system at a given energy level.

3. What is the relationship between entropy and temperature?

The relationship between entropy and temperature is described by the Clausius inequality, which states that the change in entropy of a system divided by its temperature is always greater than or equal to the change in heat divided by the system's temperature. This means that as temperature increases, the change in entropy also increases.

4. How does entropy S(T) relate to the concept of disorder?

Entropy S(T) is often used to measure the level of disorder in a system. As the number of possible arrangements or microstates of a system increases, the entropy also increases. This means that a highly disordered system will have a higher entropy than a more ordered system at the same temperature.

5. Can entropy S(T) be negative?

Yes, entropy S(T) can be negative in certain circumstances. This can occur when the system becomes more ordered, such as in a phase transition from a gas to a liquid. However, the overall change in entropy for the universe will always be positive, in accordance with the second law of thermodynamics.

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