Thermodynamics: Entropy of 2 large Einstein Solids

[Bosley]

Homework Statement

Consider a system of 2 large, identical Einstein solids. Each solid has N=10^23 oscillators, and the total energy units in the combined system is 2N.

a) Assuming that all of the microstates are allowed, compute the entropy of this system. This is the entropy over long time scales.
b) Compute entropy again assuming the system is in its most likely macrostate. This is the entropy over short time scales.
c) Is the issue of time scales really relevant here?
d) Suppose that at a moment when the system is near its most likely macrostate, you insert a partition between the solids so they can't exchange energy anymore. Now, even over long time scales, the entropy is the answer from part a. So the second law of thermodynamics has, in a sense, been violated. Is this significant? Should we lose sleep over it?

Homework Equations

We are given that the multiplicity of the combined system (that is, all possible microstates) is $$\frac{2^{4N}}{\sqrt{8 \pi N}}$$.

We're given that the multiplicity of the most likely macrostate (where Energy A = Energy B = N units) is $$\frac{2^{4N}}{4 \pi N}$$.

The Attempt at a Solution

a) $$S = k*log(W)$$
$$S = k*log(2^{4*10^{23}}/\sqrt{8*\pi*10^{23}})$$
Problem: This is incalculably large. Is there some different way that I should be approximating a value for this? Mathematica overflows. Wolfram gives back a nonsensical "power of 10 representation" that is incorrect. What to do?

b) Likewise, $$S = k*log(2^{4*10^{23}}/(4*\pi*10^{23})$$ is incalculably large. Does the problem just want me to say "Both are basically infinite"? This is confusing, though, because I thought we were supposed to usually get tractable values for entropy.

c) Since both numbers for a and b are apparently intractably large (unless I'm doing something terribly wrong), it seems that the time scale doesn't matter?

d) I don't know.

Advice please?

 

[superspartan9]

Where did you get your multiplicity of the combined system equation?

 

[superspartan9]

Nevermind. I feel dumb now.

 

[Kribzon]

Try using log rules. So, for example,

$$S = k*ln(2^{4*10^{23}}/\sqrt{8*\pi*10^{23}})$$ becomes
$$S=k*[(4*10^{23})*ln(2)-(1/2)ln(8*\pi*10^{23})]$$

And
$$S = k*log(2^{4*10^{23}}/(4*\pi*10^{23})$$
Becomes
$$k*[(4*10^{23})*ln(2)-ln(4*\pi*10^{23})]$$

Can you evaluate this now?

 

[paranormal]

Hello,

I would like to provide some guidance on how to approach this problem and how to think about thermodynamics and entropy.

First, it is important to understand that the equations given in the homework statement are not meant to be directly calculated. They are there to provide a theoretical framework for understanding the behavior of the system. In this case, the equations describe the multiplicity of the system, which is related to the number of microstates that the system can occupy.

Second, it is important to understand the concept of entropy. Entropy is a measure of the disorder or randomness of a system. In thermodynamics, it is related to the number of possible microstates that a system can occupy at a given energy level. In other words, a system with a higher entropy has more possible ways to arrange its energy among its particles.

Now, let's address your specific questions:

a) The equation for calculating entropy is S = k*log(W), where W is the multiplicity of the system. In this case, the multiplicity is given by \frac{2^{4N}}{\sqrt{8 \pi N}}. This is a very large number, but it does not mean that the entropy is infinite. Remember that the equation for entropy includes a logarithm, which means that the value of entropy will depend on the logarithm of the multiplicity. In this case, the entropy will be a very large number, but it is still a finite value.

b) The most likely macrostate for this system is when the energy is evenly distributed between the two solids, with each solid having N units of energy. The multiplicity in this case is given by \frac{2^{4N}}{4 \pi N}. Again, this is a very large number, but it is still a finite value. It is important to note that this is the most likely macrostate, not the only possible macrostate. There are still other macrostates that the system can occupy, but they are less likely.

c) The issue of time scales is relevant here because the equations for entropy are based on statistical mechanics, which deals with the behavior of systems over long time scales. The equations do not take into account the short-term fluctuations of the system, which can occur over short time scales.

d) The second law of thermodynamics states that the entropy of a closed system will either remain constant or increase over time. In this case, when the partition is inserted between