Canonical partition function of an ideal gas (unit analysis)

[Somali_Physicist]

Homework Statement

Basically the units of the Canonical Partition Function within the logarithms should be zero

Homework Equations

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.

(F_u((mk_bTV^2/3)/(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.

k_bT => J (Thermal Energy)
m => Kg
ħ² => J²s²
V^2/3 => J^2/3

Now ignoring the power to -3/2 =>
Kg*J*J^2/3*s^-2*J^-2
=> kg*s^-2*J^-1/3
=> kg^2/3s^-8/3m^-2/3
Now Even if we introduce -3/2 we don't acquire a unitless solution

 

[vela]

Your units for the volume V are wrong.

A useful combination to know is $\hbar c = 197~{\rm eV\,nm}$. If you multiply the top and bottom by $c^2$ where $c$ is the speed of light, you get
$$\frac{(mc^2)(k_\text{B} T) V^{2/3}}{2\pi(\hbar c)^2}.$$ In that form, it's pretty easy to see that both the numerator and denominator have units of (energy x length)^2.

 

[Somali_Physicist]

Your units for the volume V are wrong.

A useful combination to know is $\hbar c = 197~{\rm eV\,nm}$. If you multiply the top and bottom by $c^2$ where $c$ is the speed of light, you get
$$\frac{(mc^2)(k_\text{B} T) V^{2/3}}{2\pi(\hbar c)^2}.$$ In that form, it's pretty easy to see that both the numerator and denominator have units of (energy x length)^2.

Volume ! i thought it was potential energy!
Gracias!

well for completeness
Kg * J * m^3*2*3-1 *J^-2*S^-2
=> Kg*J^-1*m²*s^-2
=> J*J^-1 => 1
Note - Kg*m²*s^-2 = J