The mole fraction of N₂ after the mixture of 4.0 L of N₂ at 15 atm with 4.0 L of H₂ at 7.0 atm is 0.68.
We can calculate the mole fraction of N₂ with the following equation:
[tex] X_{N_{2}} = \frac{n_{N_{2}}}{n_{t}} = \frac{n_{N_{2}}}{n_{N_{2}} + n_{H_{2}}} [/tex] (1)
The number of moles of N₂ and H₂ can be found with the ideal gas law:
[tex] PV = nRT [/tex]
Where:
P: is the pressure
R: is the gas constant
T: is the temperature
V: is the volume
For nitrogen gas we have:
[tex] n_{N_{2}} = \frac{P_{N_{2}}V_{N_{2}}}{RT} [/tex] (2)
And for hydrogen:
[tex] n_{H_{2}} = \frac{P_{H_{2}}V_{H_{2}}}{RT} [/tex] (3)
After entering equations (2) and (3) into (1), we get:
[tex] X_{N_{2}} = \frac{\frac{P_{N_{2}}V_{N_{2}}}{RT}}{\frac{P_{N_{2}}V_{N_{2}}}{RT} + \frac{P_{H_{2}}V_{H_{2}}}{RT}} [/tex]
Since RT are constants, we have:
[tex] X_{N_{2}} = \frac{P_{N_{2}}V_{N_{2}}}{P_{N_{2}}V_{N_{2}} + P_{H_{2}}V_{H_{2}}} [/tex]
We know that:
[tex] P_{N_{2}} = 15 atm[/tex]
[tex] V_{N_{2}} = 4.0 L[/tex]
[tex] P_{H_{2}} = 7.0 atm[/tex]
[tex] V_{H_{2}} = 4.0 L[/tex]
so:
[tex] X_{N_{2}} = \frac{15 atm*4.0 L}{15 atm*4.0 L + 7.0 amt*4.0 L} = 0.68 [/tex]
Therefore, the mole fraction of N₂ is 0.68.
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