- #1
Andrew_
- 13
- 0
Hello,
I'm still trying getting familiar with the concepts of statistical mechanics and thermodynamics , and there's this equation for S of an ideal system, which somehow perplexes me. Suppose the ideal gas is composed of N indistinguishable atoms , then :
[tex]S = \frac{U - U_{0} }{T} + k Ln \frac{q^N}{N!}[/tex]
At T = 0 this expression becomes :
[tex]S = \frac{3}{2} Nk + k Ln \frac{{g_0}^N}{N!}[/tex]
where g0 is level of degeneracy of the ground state.
Is there anything wrong with this ? If not , why doesn't the entropy equal to zero at T = 0 kelvin ? Doesn't that contradict the 3rd law of thermodynamics ?
S = kLnW = 0 because at T = 0 , W = 1 .
What am I doing wrong ?
I'm still trying getting familiar with the concepts of statistical mechanics and thermodynamics , and there's this equation for S of an ideal system, which somehow perplexes me. Suppose the ideal gas is composed of N indistinguishable atoms , then :
[tex]S = \frac{U - U_{0} }{T} + k Ln \frac{q^N}{N!}[/tex]
At T = 0 this expression becomes :
[tex]S = \frac{3}{2} Nk + k Ln \frac{{g_0}^N}{N!}[/tex]
where g0 is level of degeneracy of the ground state.
Is there anything wrong with this ? If not , why doesn't the entropy equal to zero at T = 0 kelvin ? Doesn't that contradict the 3rd law of thermodynamics ?
S = kLnW = 0 because at T = 0 , W = 1 .
What am I doing wrong ?