Low Temp Physics: Why He (3) Needs Higher Pressure to Solidify

  • Thread starter Thread starter pleasehelpmeno
  • Start date Start date
  • Tags Tags
    Physics
pleasehelpmeno
Messages
154
Reaction score
0

Homework Statement


Why at absolute zero does He (three) have a larger molar volume and require a higher pressure to solidify than He(four)?



The Attempt at a Solution



I feel the answer lies in the Pauli exclusion principle, at low temps the ferions are nopt as easily confined as the bosons so they will have a larger molar volume, but i can't explain why this would then require a higher pressure to solidify:

Pv = nRT

where if v is increased surely P must decrease?
 
Physics news on Phys.org
I think I saw some pairing process for He3 - so you have to consider "molecules" (those are bosons again) instead of atoms.
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top