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
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.
To answer your first question, the volume of helium gas is considered to be 22.4 L at standard temperature and pressure (STP). This is a standard condition used for gases, where the temperature is 273 K (0 degrees Celsius) and the pressure is 1 atm. However, in the given question, the gas is not at STP, it is at room temperature (293 K) and a pressure of 1.0 atm. Therefore, the volume of the gas is not necessarily 22.4 L.
Regarding your second question, the value of m in this case is not the number of helium atoms, but rather the molar mass of helium (given as 4 g/mol). The molar mass represents the mass of one mole of a substance, which contains Avogadro's number (6.02 × 10^23) of atoms or molecules. So, when we want to calculate the mass of a certain number of moles, we use the molar mass.
To find the mass of one helium atom, we divide the molar mass by Avogadro's number:
m = 4 g/mol / (6.02 × 10^23 atoms/mol)
m ≈ 4/6.02 × 10^(-23 + 23) g
m ≈ (4/6.02) × 10^(-23 + 23) g
m ≈ 6.64 × 10^(-24) g
Please note that this calculation assumes the mass of one helium atom is evenly distributed among the 6.02 × 10^23 atoms in one mole.