Partition function of a gas of $N$ identical classical particles Partition function of a gas of $N$ identical classical particles is given by
$$ Z~=~\frac {1}{N! h^{3N}} \int \exp[-\beta H(p_1.......p_n, x_1....x_n)]d^3p_1...d^3p_n,d^3x_1...d^3x_n $$
in this above equation we use $N!$ as the total number of sub-systems of a system of identical particles. and $ h^{3N} $ to make the partition function dimensionless. I am not clear how $ h^{3N} $ is used to make it dimensionless.
 A: Notice that the $e^{-\beta H}$ is dimensionless, while each factor of $dp$ contributes one factor with the dimensions of momentum while each $dx$ contributes one factor with the dimensions of length.  Therefore each factor $dp dx$ contributes a factor with dimensions of angular momentum.  Since there are $3N$ of these factors (N particles and 3 dimensions) in the integration measure, the integral has a total dimension of angular momentum to the power of $3N$.  On the other hand, $h$ has dimensions of angular momentum, so dividing by $h^{3N}$ makes the full expression dimensionless.
A: The easiest way to think about it is that $\exp(\dots)$ is just a number and doesn't affect the dimension.  However, you still have $3N$ factors of the momenta and the position lying around that will give you dimensions of [Length x Momentum]${}^{3N}$.  Planck's constant has the units of Length x Momentum, so the $3N$ factors of $h$ cancel the $3N$ factors coming from the integral.
A: The phase space of coordinates and momenta components of the N particles has certain size in that space, there is a number of microstates inside this size determined by the cell size in the quantized phase space (due to uncertainity h for each degree of freedom dx dp and for 3N dergrees of freedom of the N particles) it will be  h^3N. The number of possible arrangements of these N distinguishable,  particles is N! repeated in the numerator), but they are indistigwishable we must correct it by dividing the quantity b N!.
