You are in a submarine, the hallway in front of you is connected to two rooms. The hallway has a capacity of 300.0 L and is filled with 35.0 atm of helium gas. Room A has a capacity of 200.0 L and holds 20.0 atm of nitrogen gas. Room B has 40.0 atm of oxygen gas in 180.0 L. a.If you open both rooms to the hallway, what would be pressure of each gas in the combined space of the hallway and two rooms? b.What would be the mol fraction of oxygen in that gas? c.After mixing the gases, if you let gas escape until the total pressure in the hallway was 1.2 atm, what would be the final partial pressure of oxygen?

Respuesta :

Answer:

a)

He pressure: 15,44 atm

Nβ‚‚ pressure: 5,88 atm

Oβ‚‚ pressure: 10,59 atm

b) 33,2% Oβ‚‚

c) Partial pressure of Oβ‚‚ 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: 35,0Γ—300,0 = 10500 moles of He

Moles of nitrogen: 200,0*20,0 = 4000 moles of Nβ‚‚

Moles of oxygen: 180,0*40,0 = 7200 moles of Oβ‚‚

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

Nβ‚‚ pressure: 4000moles/680 = 5,88 atm

Oβ‚‚ pressure: 7200moles/680 = 10,59 atm

b) The mole fraction of Oβ‚‚ is:

[tex]\frac{7200 moles}{10500 moles + 4000 moles + 7200 moles}*100 =[/tex] 33,2% Oβ‚‚

c) The moles of Oβ‚‚ and Nβ‚‚ are the same, Thus, partial pressure of Oβ‚‚ is the same, 10,59 atm.

I hope it helps!