Thermodynamics: Einstein solid (simple step in derivation)

[iScience]

S=kln($\frac{eq}{N}$)^N --->= S=Nkln($\frac{q}{N}+1$)

i understand that the e goes away and the N exponent comes down but where does the +1 come from?

 

[Dick]

S=kln($\frac{eq}{N}$)^N --->= S=Nkln($\frac{q}{N}+1$)

i understand that the e goes away and the N exponent comes down but where does the +1 come from?

That's not even right. kln((eq/N)^N)=Nk(1+ln(q/N)). Use more parentheses to show what you really mean. Just use rules of logarithms, and show how you are using them.

 

[iScience]

there are no more parentheses that's what's in my book. is this even an approximation??

 

[Dick]

there are no more parentheses that's what's in my book. is this even an approximation??

No, I think it's just wrong.

 

[paranormal]

The +1 comes from the fact that in the Einstein solid model, each particle has two possible energy states: either it is in a low energy state or a high energy state. This means that the total number of possible energy states for N particles is N+1. Therefore, when we take the natural logarithm of the ratio of the total number of possible energy states to the number of particles, we get ln(q/N+1). This is then multiplied by the Boltzmann constant, k, to give the entropy, S.