Classical/quantum transition for an ideal gas in 1m^3 box

[Dixanadu]

Homework Statement

Hey guys,

So I have to estimate the temperature at which the classical / quantum transition happens for a 1m^3 box of air. This is done by comparing the de Broglie wavelength with the average distance between the particles - so basically the transition happens when they are comparable.

Homework Equations

Average distance between the particles: $(\frac{V}{N})^{\frac{1}{3}}=n^{-\frac{1}{3}}$ where n is the number density

de Broglie wavelength: $\lambda = \frac{h}{\sqrt{2\pi m k T}}$ where k is the Boltzmann constant.

The Attempt at a Solution

Right, so here's what I did. Since the volume V = 1, we can say that
$N=n$, using the distance between the particles equation.

We also know that $pV=NkT=nkT$ since n = N, so assuming standard pressure (since the question asks me to estimate), we can say that $n=\frac{10^{5}}{kT}$.

The transition happens when $\lambda ≈ n^{-\frac{1}{3}}$. So replacing λ with our expression for n, we get this
$\frac{h}{\sqrt{2\pi m k T}}=(\frac{10^{5}}{kT})^{-\frac{1}{3}}$

Using the approximation that the mass m of air is around 30 kg / mol, the mass of one air molecule is around 5 x 10^-23kg. Plugging that in and solving for T gives me

$T≈5.6\times 10^{-4}K$, which seems wrong for some reason..

Can you guys help me out? thanks!

 

[TSny]

Using the approximation that the mass m of air is around 30 kg / mol, the mass of one air molecule is around 5 x 10^-23kg.

kg??

 

[Dixanadu]

is that wrong? i just divided 30 by avogadro's number o.o

 

[Dixanadu]

oh wait i think its 5 x 10^-26 kg...its 30 g/mol not 30kg/mol XD

 

[Dixanadu]

So after that correction i get 0.035 K...is that better?