Question

A 4.0 L flask containing N2 at 15 atm is connected to a 4.0 L flask containing H2 at 7.0 atm and the gases are allowed to mix. What is the mole fraction of N2

Answer 1

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:

$X_{N_{2}} = \frac{n_{N_{2}}}{n_{t}} = \frac{n_{N_{2}}}{n_{N_{2}} + n_{H_{2}}}$   (1)

The number of moles of N₂ and H₂ can be found with the ideal gas law:

$PV = nRT$

Where:

P: is the pressure

R: is the gas constant

T: is the temperature

V: is the volume

For nitrogen gas we have:

$n_{N_{2}} = \frac{P_{N_{2}}V_{N_{2}}}{RT}$   (2)

And for hydrogen:

$n_{H_{2}} = \frac{P_{H_{2}}V_{H_{2}}}{RT}$   (3)

After entering equations (2) and (3) into (1), we get:

$X_{N_{2}} = \frac{\frac{P_{N_{2}}V_{N_{2}}}{RT}}{\frac{P_{N_{2}}V_{N_{2}}}{RT} + \frac{P_{H_{2}}V_{H_{2}}}{RT}}$  

Since RT are constants, we have:

$X_{N_{2}} = \frac{P_{N_{2}}V_{N_{2}}}{P_{N_{2}}V_{N_{2}} + P_{H_{2}}V_{H_{2}}}$                

We know that:

$P_{N_{2}} = 15 atm$                

$V_{N_{2}} = 4.0 L$                

$P_{H_{2}} = 7.0 atm$                

$V_{H_{2}} = 4.0 L$          

so:

$X_{N_{2}} = \frac{15 atm*4.0 L}{15 atm*4.0 L + 7.0 amt*4.0 L} = 0.68$                

Therefore, the mole fraction of N₂ is 0.68.

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