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?