Answer:
a)
He pressure: 15,44 atm
Nâ‚‚ pressure: 5,88 atm
Oâ‚‚ pressure: 10,59 atm
b) 33,2% Oâ‚‚
c) Partial pressure of Oâ‚‚ 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 -Where the k value depends of the temperature and it is possible to assume this as 1-
Thus, the moles of He are: 35,0Ă—300,0 = 10500 moles of He
Moles of nitrogen: 200,0*20,0 = 4000 moles of Nâ‚‚
Moles of oxygen: 180,0*40,0 = 7200 moles of Oâ‚‚
a) The pressure of each gas is n/V where V is the sum of each room volume (300,0L + 200,0L + 180,0L = 680,0L)
He pressure: 10500moles/680 = 15,44 atm
Nâ‚‚ pressure: 4000moles/680 = 5,88 atm
Oâ‚‚ pressure: 7200moles/680 = 10,59 atm
b) The mole fraction of Oâ‚‚ is:
[tex]\frac{7200 moles}{10500 moles + 4000 moles + 7200 moles}*100 =[/tex] 33,2% Oâ‚‚
c) The moles of Oâ‚‚ and Nâ‚‚ are the same, Thus, partial pressure of Oâ‚‚ is the same, 10,59 atm.
I hope it helps!