Question

Helium is mixed with oxygen gas for deep-sea divers. Calculate the percent by volume of oxygen gas in the mixture if the diver has to submerge to a depth where the total pressure is 3.8 atm. The partial pressure of oxygen is maintained at 0.20 atm at this depth.

Answer 1

Answer : The percent by volume of oxygen gas in the mixture is, 5.3 %

Explanation :

According to the Raoult's law,

$p^o_A=X_A\times p_A$

where,

$p_A$ = total partial pressure of solution = 3.8 atm

$p^o_A$ = partial pressure of oxygen = 0.20 atm

$X_A$ = mole fraction of oxygen = ?

Now put all the given values in this formula, we get:

$p^o_A=X_A\times p_A$

$0.20atm=X_A\times 3.8atm$

$X_A=0.0526$

Now we have to calculate the percent by volume of oxygen gas in the mixture.

The mole percent of oxygen gas = $0.0526\times 100=5.3\%$

As we know that, there is a direct relation between the volume of moles.

So, mole percent of oxygen gas = volume percent of oxygen gas

Volume percent of oxygen gas = $0.0526\times 100=5.3\%$

Therefore, the percent by volume of oxygen gas in the mixture is, 5.3 %