GR | # Login | Register
  GR8677 #88
GREPhysics.NET Official Solution    Alternate Solutions
Verbatim question for GR8677 #88
Statistical Mechanics}Pressure

Recall the following results for low temperatures,

where one realizes that the Fermi temperature, T_F tends to be on the order of thousands for most material, and that the low temperature regime temperatures are definitely far less than the Fermi temperature; one thus has P_F>P_B,P_C. Moreover, classical effects occur at around T=300K, while the problem specifies the temperature domain for bosons to be far lower than that, thus one deduces that P_B < P_C. Therefore, P_F>P_C>P_B.

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
There are no Alternate Solutions for this problem. Be the first to post one!
2023-07-11 19:01:55
This is very helpful thanks a lot for sharing it with the rest of us! wordle solver toolNEC
2014-10-17 02:22:50
Since we know that boson condensate at low temperature (like 4HE) PB must be the lowest pressure.
The only answer that takes consider this possibility is the right one!
2013-10-09 14:05:52
Electron degeneracy is one reason (along with neutron degeneracy pressure, which is even higher) why stars don't always collapse into a black hole. The more you compress a gas of electrons, they fill up all the energy states in pairs (by the Pauli exclusion principle) and this builds an immense pressure. On the other hand, bosons will all fall to the lowest energy level, and eventually form an Bose-Einstein condensate. Since they can all share the same energy state, the pressure they exert as they are compressed is extremely small, as they can keep going to lower energy states instead of pushing back against compression.NEC
2011-11-06 16:59:03
Don't you hate when you go too quickly and mess up the comparison signs? Alligator eats the bigger one.. :(NEC
pam d
2011-09-26 22:15:19
I just want to point out that all of these insights are really great, but there is always a way to get the question even if you don't have all of the information necessary. Specifically, if you know that bosons have to have contribute the weakest pressure (just think about photons as Almno10 pointed out) then you don't need to know whether or not the Fermi pressure is higher than the classical contribution. Choice (B) is the only one that puts bosons the lowest.

Also, lol @ (A).
2010-11-12 18:59:17
I recall that fermi pressure is so ridiculously high, that it is able to keep a star from collapsing under its own gravity. Bosons on the other hand...well, have you ever felt the pressure on your face from sunlight?
2012-04-12 20:12:47
it is called sunburn!
2010-04-09 15:08:48
Just would like to add a captivating factoid. The pressure due to the Pauli exclusion principle for electrons in a metal is so high that you can always treat metals as if they were at 0K!!! Or conversely, the electrons in a metal at 0K behave almost exactly as they do at 300K.NEC
2008-09-14 12:40:48
Bosons can occupy the same energy state as each other, where fermions have to obey the pauli-exclusion principle. Therefore, at cold enough temperatures, bosons would all be in the lowest energy state and it makes sense that they would have the lowest pressure, while fermions would be all over the place and have a higher pressure. If quantum effects were ignored, it makes sense that that pressure would be in between since fermions and bosons seem to be at two extremes.
2009-08-13 20:41:20
exactly how I did it too.
2010-11-15 19:38:27
Also, other things being equal, bosons tend to be closer together, and fermions farther apart, than distinguishable particles (which would obey Boltzmann statistics). Thus you would expect lower pressure for bosons and higher pressure for fermions.

Post A Comment!
You are replying to:
Since we know that boson condensate at low temperature (like 4HE) PB must be the lowest pressure. The only answer that takes consider this possibility is the right one!

Click here to register.
This comment is best classified as a: (mouseover)
Mouseover the respective type above for an explanation of each type.

Bare Basic LaTeX Rosetta Stone

LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces .
type this... to get...
$\langle my \rangle$
$\left( abacadabra \right)_{me}$
The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...