Thermodynamics: Einstein solid (simple step in derivation)

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iScience
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S=kln([itex]\frac{eq}{N}[/itex])N --->= S=Nkln([itex]\frac{q}{N}+1[/itex])

i understand that the e goes away and the N exponent comes down but where does the +1 come from?
 
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iScience said:
S=kln([itex]\frac{eq}{N}[/itex])N --->= S=Nkln([itex]\frac{q}{N}+1[/itex])

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

No, I think it's just wrong.