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?

a)

He pressure: 15,44 atm

N₂ pressure: 5,88 atm

O₂ pressure: 10,59 atm

b) 33,2% O₂

c) 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 -Where the k value depends of the temperature and it is possible to assume this as 1-

Thus, the moles of He are: 35,0×300,0 = 10500 moles of He

Moles of nitrogen: 200,0*20,0 = 4000 moles of N₂

Moles of oxygen: 180,0*40,0 = 7200 moles of O₂

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

N₂ pressure: 4000moles/680 = 5,88 atm

O₂ pressure: 7200moles/680 = 10,59 atm

b) The mole fraction of O₂ is:

33,2% O₂

c) The moles of O₂ and N₂ are the same, Thus, partial pressure of O₂ is the same, 10,59 atm.

I hope it helps!

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.