Question

5. A scuba driver has a 10 L tank that is filled with Nitrox or otherwise known as "enriched air. Therefore, it contains 5.00 g of oxygen gas (O2) and 8.31 g of nitrogen gas (N2). The pressure of the tank is 179.6 atm. What is the partial pressure of both oxygen and nitrogen?

Answer 1

Answer : The partial pressure of both oxygen and nitrogen is, 61.8 atm and 117.8 atm respectively.

Explanation :

First we have to calculate the moles of
`O_2` and
`N_2`

`\text{Moles of }O_2=\frac{\text{Given mass }O_2}{\text{Molar mass }O_2}=(5.00g)/(32g/mol)=0.156mol`

and,

`\text{Moles of }N_2=\frac{\text{Given mass }N_2}{\text{Molar mass }N_2}=(8.31g)/(28g/mol)=0.297mol`

Now we have to calculate the mole fraction of
`O_2` and
`N_2`

`\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }O_2+\text{Moles of }N_2}`

`\text{Mole fraction of }O_2=(0.156)/(0.156+0.297)=0.344`

and,

`\text{Mole fraction of }N_2=\frac{\text{Moles of }N_2}{\text{Moles of }O_2+\text{Moles of }N_2}`

`\text{Mole fraction of }O_2=(0.297)/(0.156+0.297)=0.656`

Now we have to calculate the partial pressure of both oxygen and nitrogen.

According to the Raoult's law,

`p_i=X_i* p_T`

where,

`p_i` = partial pressure of gas

`p_T` = total pressure of gas = 179.6 atm

`X_i` = mole fraction of gas

`p_(O_2)=X_(O_2)* p_T`

`p_(O_2)=0.344* 179.6atm=61.8atm`

and,

`p_(N_2)=X_(N_2)* p_T`

`p_(N_2)=0.656* 179.6atm=117.8atm`

Thus, the partial pressure of both oxygen and nitrogen is, 61.8 atm and 117.8 atm respectively.