- #1
Bobhawke
- 144
- 0
I can't figure out how to show that the Clausius entropy
[tex] \Delta S= \int_{T_1}^{T_2} \frac{dQ}{T}[/tex]
is equivalent to the Boltzmann entropy
[tex] S=kln \Omega [/tex]
at thermal equilibrium.
Does anyone know how to do this?
[tex] \Delta S= \int_{T_1}^{T_2} \frac{dQ}{T}[/tex]
is equivalent to the Boltzmann entropy
[tex] S=kln \Omega [/tex]
at thermal equilibrium.
Does anyone know how to do this?