- #1
SW VandeCarr
- 2,199
- 81
As a practical matter, the internal energy of a system is treated as a state function of a the system and is concerned only with the kinetic energy of particles, not the potential energy. So for thermodynamic entropy, S=E/T, (S=entropy, T= absolute temperature) I'm considering E to be kinetic energy.
Now thermodynamic temperature has been defined as the average kinetic energy per unit particle. If so, we can say T=E/N where E is the internal (kinetic) energy of a system. Then, S=E/T=E/E/N=N. That is, S becomes a dimensionless number equal to (or direct function of) the number of particles in the system. The Boltzmann equation: S=k log W considers only the number of particles in a system since [tex]W=N!/\Pi N_{i}![/tex].
If S=f(N), then the entropy of a system of N particles would seem to be constant. That would mean it doesn't matter if you add or remove energy or dilute or concentrate the particles. S is always constant unless you add or remove particles. Is this true?
Now thermodynamic temperature has been defined as the average kinetic energy per unit particle. If so, we can say T=E/N where E is the internal (kinetic) energy of a system. Then, S=E/T=E/E/N=N. That is, S becomes a dimensionless number equal to (or direct function of) the number of particles in the system. The Boltzmann equation: S=k log W considers only the number of particles in a system since [tex]W=N!/\Pi N_{i}![/tex].
If S=f(N), then the entropy of a system of N particles would seem to be constant. That would mean it doesn't matter if you add or remove energy or dilute or concentrate the particles. S is always constant unless you add or remove particles. Is this true?
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