Answer:
2.89 atm is the partial pressure of helium.
Explanation:
Number of atoms of neon gas = x
Number of atoms of helium gas = 3x
Moles of neon gas = [tex]n_1=N_A\times x=xN_A[/tex]
Moles of helium gas = [tex]n_2=N_A\times 3x=3xN_A[/tex]
Concentration of mixture ,c= [tex]\frac{n(moles)}{V(volume(L))}= 0.150 mol/L[/tex]
Temperature of the mixture = T = 313 K
Total pressure of the mixture = P
[tex]PV=nRT[/tex]
[tex]P=\frac{n}{V}RT[/tex]
[tex]P=cRT=0.150 mol/L\times 0.0821 atm L/mol K\times 313 K[/tex]
P = 3.8546 atm
Mole fraction of neon = [tex]\chi_1=\frac{n_1}{n_1+n_2}=\frac{xN_A}{xN_A+3xN_A}=0.25[/tex]
Mole fraction of helium= [tex]\chi_2=\frac{n_2}{n_1+n_2}=\frac{3xN_A}{xN_A+3xN_A}=0.75[/tex]
Partial pressure of helium gas can be determined by the help of Dalton's:
[tex]p_{He}=P\times \chi_2=3.8546 atm\times 0.75=2.89 atm [/tex]
2.89 atm is the partial pressure of helium.
The partial pressure of helium is mathematically given as
p{He}=2.89 atm
Question Parameter(s):
The mixture of helium and neon gases at 313 K contains 3 times the number of helium atoms as neon atoms.
Generally, the equation for the ideal gas is mathematically given as
PV=nRT
Therefore
P=cRT
P=0.150 * 0.0821 313 K
P = 3.8546 atm
In conclusion, mole fraction of neon and helium
[tex]x_1=\frac{n_1}{n_1+n_2}\\\\x_1=\frac{xN_A}{xN_A+3xN_A}\\[/tex]
x_1=0.25
For helium
[tex]x_2=\frac{3xN_A}{xN_A+3xN_A}[/tex]
x_2=0.75
Hence, Partial pressure of helium gas
p{He}=P*x_2
p{He}=3.8546 * 0.75
p{He}=2.89 atm
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