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 $N_2$

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

and,

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

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

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

$\text{Mole fraction of }O_2=\frac{0.156}{0.156+0.297}=0.344$

and,

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

$\text{Mole fraction of }O_2=\frac{0.297}{0.156+0.297}=0.656$

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

According to the Raoult's law,

$p_i=X_i\times p_T$

where,

$p_i$ = partial pressure of gas

$p_T$ = total pressure of gas  = 179.6 atm

$X_i$ = mole fraction of gas

$p_{O_2}=X_{O_2}\times p_T$

$p_{O_2}=0.344\times 179.6atm=61.8atm$

and,

$p_{N_2}=X_{N_2}\times p_T$

$p_{N_2}=0.656\times 179.6atm=117.8atm$

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