The atoms in gas ( gas constant R=8.31 J/mol K) can be treated as classical particles if their de Broglie wavelength L is much smaller than the average separation between the particles d. Consider monatomic helium gas (mHec^2 = 3727 MeV, molar mass 4g/mol) at 1.0 atmosphere of pressure (1.0×10^5) and the room temperature ( T = 293 K)

Question

The atoms in gas ( gas constant R=8.31 J/mol K) can be treated  as classical particles if their de Broglie wavelength    L  is much smaller than the average separation  between the particles d. Consider monatomic helium gas (mHec^2  = 3727 MeV, molar mass 4g/mol) at 1.0 atmosphere of pressure (1.0×10^5) and the room temperature ( T = 293 K)
a.	Estimate d for this gas ( d = ( # atoms/vol) ^-1/3)
b.	Find the average de Broglie wavelength L of the atoms in the gas ( KE = ( 3/2) KB T,  KB = 8.617 * 10 –E5 eV/K)
c.	Find the pressure that would make d equal to L from b
d.	Find the temperature that would make the average L equal to d from part a

I posted afew hour and saw the answer , I thank you so much, but I don’t understand why V helium is still 22,4 L at 1.0 atm, and 293 K)

Why m = 4*1.67*10-E24, but not m=4/6.02*10E23
please explain more carefully

drwls · Answer

The V of 1 mole of any ideal gas is 24.0 l at 293 K and 1 atm. That comes from the ideal gas equation V = nRT/P

m is the mass of helium atom in grams. I multiplied the proton-neutron mass by 4. Either way of computing is gives the same answer to the accuracy I was using.

There may be computational errors in what I did earlier but I beleive the reasoning is correct. I usually prefer not to follow problems through to numerical answers, but instead just show the way and let students do the rest.