T08 1(c), 2(a)
T08 by Carlos Daniel Cervantes García
For the first problem we can search search "for
For the first problem we can search search "for
the corresponding expression in your notes", we can find the expression for the entropy for the case of a one dimensional harmonic oscilator, using this expression we can calculate the chemical potential using a partial derivative of the energy, so, the result is as follows:
For the case of the problem T08.2 (a), we can calculate the entropy by simply integrating, then, using a method for diferential equations we can find the rest of the equation g(v).
In principle, the calculation seems correct. But it is carried out with some lack of clarity. To start, you could have simply started from the differential, T dS = dU + p dV :
ResponderBorrardS = T^{-1} (partial U/partial T) dT + T^{-1} [ (partial U/partial V) + p ] dV , integrating the right hand side
Then, you suddenly include the thermal wavelength lambda = h/sqrt(2pi m k_B T) , without explanation. One would really expect an arbitrary constant to be added, the entropy of a given reference state.
The second comment, is about the problem with mu dN. Since the problem as a whole also deals with the chemical potential, one should worry about the term "mu dN" , i.e.
T dS = dU + p dV - mu dN. The major reason for the confusion about this term, was of course my fault. But still, I want to state, what would have been the correct argument. That would be based on replacing the extensive variables, S,U,V by intensive variables s,u,v for which the same fundamental equation holds, but naturally without the term mu dN. Then one can calculate the intensive "s" in the standard way -- and then multiply with "N" to obtain the extensive "S".