Entropy of a gas of $N$ distinguishable Particles Suppose I have a container contain $N$ particles, all of which are distinguishable. How would I calculate the entropy of this gaseous system. The Hamiltonian of the system is simply the sum $\sum_{i=1}^N \frac{p_i^2}{2m_i}$ and I believe (not 100% sure) that the partition function is:
$Z = \frac{1}{h^{3N}}\int exp(-\beta H)d^3p_1...d^3p_N d^3q_1...d^3q_N = \frac{V^N}{h^{3N}}\int exp(-\beta H)d^3p_1...d^3p_N$
Is this correct so far?
Would the next logical step be to break the integral up into the product of $N$ integrals, each looking like $\int exp(-\beta \frac{p_i^2}{2m_i}) d^3p_i$, convert into spherical coordinates, and evaluate the gaussian integral?
Will this calculate the partition function?
 A: Assuming that one needs the entropy as a function of temperature ($T$), volume ($V$), and the number of particles ($N$), your starting point is almost correct (the reason for the almost will be explained at the end of this answer). For the moment, I will concentrate on formal manipulations. If one needs entropy as a function of other variables, other routes (and other ensembles) could allow getting more directly the answer.
Under such a hypothesis, the expression for $Z$ is correct. Being purely kinetic hamiltonian, the integration over the positions is trivial. Using cartesian coordinates, the integral over the $3N$ components of the momenta corresponds to a product of $3N$ gaussian integrals. There is no compelling reason to introduce spherical coordinates.
The reason I started writing that the expression of the partition function you are using is almost correct is related to the fact that, notwithstanding what is written in many textbooks, the partition function for a classical gas of distinguishable particles does not differ from the case of indistinguishable particles. If this were the case, the entropy of milk would not be extensive, and the entropy of a glass of milk would not be twice the entropy of half glass.
Unfortunately, clarification about such a point is quite recent, and I would be careful to use this result in front of an audience not aware of the current status of the discussion. The starting reference is a paper published in 2002 by Swendsen.
( Swendsen, R. H. (2002). Statistical mechanics of classical systems with distinguishable particles. Journal of statistical physics, 107(5), 1143-1166 ). A more recent and updated version is here.
A: Yes, exactly, this will be the partition function. However, there is no need to go to spherical coordinates. Just take the integral in Cartesian coorindates:
$$
\int e^{-\beta p_i^2/(2 m_i)} = \sqrt{\frac{2 \pi m_i}{\beta}}
$$
You will have the product $3 N$ $1D$ gaussian integrals:
$$
\prod_{i=1}^{N} \sqrt{\frac{2 \pi m_i}{\beta}}
$$
