Basically the units of the Canonical Partition Function within the logarithms should be zero
The Attempt at a Solution
N here is a number so we ignore the left logarithms, applying a "Unit function " for the terms within the logarithm.
(Fu((mkbTV2/3)/(2πħ2)) ) -3/2 :
The above function can work on each individual portion and spit out the unit values , assuming all the operations act in the same way.
kbT => J (Thermal Energy)
m => Kg
ħ2 => J2s2
V2/3 => J2/3
Now ignoring the power to -3/2 =>
Now Even if we introduce -3/2 we don't acquire a unitless solution