Answer:
<em>a)</em>
He pressure: <em>15,44 atm</em>
N₂ pressure: <em>5,88 atm</em>
O₂ pressure: <em>10,59 atm</em>
<em>b) </em>33,2% O₂
<em>c) </em>Partial pressure of O₂ is 10,59 atm
Explanation:
To solve this problem you can use gas law:
PV = nRT
You can assume temperature is constant and define "k" as RT to obtain:
PV = nk -<em>Where the k value depends of the temperature and it is possible to assume this as 1-</em>
Thus, the moles of He are: 35,0×300,0 =<em> 10500 moles of He</em>
Moles of nitrogen: 200,0*20,0 = <em>4000 moles of N₂</em>
Moles of oxygen: 180,0*40,0 = <em>7200 moles of O₂</em>
a) The pressure of each gas is n/V where V is the sum of each room volume (300,0L + 200,0L + 180,0L = <em>680,0L</em>)
He pressure: 10500moles/680 = <em>15,44 atm</em>
N₂ pressure: 4000moles/680 = <em>5,88 atm</em>
O₂ pressure: 7200moles/680 = <em>10,59 atm</em>
<em>b) </em>The mole fraction of O₂ is:
<em>33,2% O₂</em>
<em>c) </em>The moles of O₂ and N₂ are the same, Thus, partial pressure of O₂ is the same, 10,59 atm.
I hope it helps!