Respuesta :
Explanation:
The given data is as follows.
Moles of oxygen gas = 0.150 mol, Moles of nitrogen gas = 0.116 mol
Moles of argon gas = 0.211 mol, Volume of the flask = 0.5 L
Temperature = 298 K
So, total number of moles will be calculated as follows.
[tex]n_{total} = n_{O_{2}} + n_{N_{2}} + n_{Ar}[/tex]
= (0.150 + 0.116 + 0.211) mol
= 0.477 mol
Now, using ideal gas equation we will calculate the total pressure of given gas mixture as follows.
PV = nRT
[tex]P \times 0.5 L = 0.477 mol \times 0.0821 Latm/mol K \times 298 K[/tex]
P = 23.34 atm
Mole fraction of [tex]N_{2}[/tex] gas will be calculated as follows.
Mole fraction = [tex]\frac{\text{moles of nitrogen}}{\text{total no. of moles}}[/tex]
= [tex]\frac{0.116 mol}{0.477 mol}[/tex]
= 0.243
Using Dalton's law, [tex]P_{i} = P_{total} \times x_{i}[/tex]
where, [tex]P_{i}[/tex] = partial pressure of a gas
[tex]P_{total}[/tex] = total pressure
[tex]x_{i}[/tex] = mole fraction of gas
Therefore, putting the values into Dalton's law to calculate the partial pressure of nitrogen gas as follows.
[tex]P_{i} = P_{total} \times x_{i}[/tex]
= [tex]23.34 atm \times 0.243[/tex]
= 5.67 atm
Thus, we can conclude that the partial pressure of [tex]N_{2}[/tex] into the given mixture is 5.67 atm.
