[Energyemerald]

Hello everybody!

I have two questions that I don't even know where to begin on. The first one is multiple choice and I have all of the possible answers. The second one is write-out. They both regard the gas laws. If anyone knows how to do these, I would be most appreciative.

1) A 14.0 L cylinder contains 5.60 g N₂, 79.9 g Ar, and 6.40 g O₂. What is the total pressure in ATM at 300K? ( R = Idea Gas Constant) (Hint: No calculator Required)
The options are:  A) 120R     B) 20R     C) 30R     D) 26R     E) 60R

2) Two cylinders at 300K are connected by a closed stopcock valve. The right-hand cylinder contains 2.4 L of hydrogen at .600 atm. The left cylinder is larger and contains 6.8 L of helium at 1.40 atm.

- How many moles of each gas?
I can solve this: Since 22.4 L = 1 mole, 2.4/22.4 = .107 mol H₂
Also, since 22.4 L = 1 mole, 6.8/22.4 = .304 mole He

- Find total pressure when the valve is open.
No idea. This is like the other question when it asked the total pressure.

-Partial pressure of both gasses at 300K when stopcock is opened.
I need to know the total pressure to find the partial pressures, I believe.

Can anybody help me solve these questions?

THANKS!

[Energyemerald]

Edit:

I have figured these out!

In question one, PV=nRT. Substitute the TOTAL values in for P and N.

So, using those masses, one can find the number of moles from the molar masses: .2 mol N2, 1 mol Ar, and .2 mol O2. That was easy enough. It gets tricky when you have to plug it into the equation.
P(total) = R * 1.4 mol * 300K / 14.0 L = 30 R

In question two,

Moles can be found by plugging into PV=nRT again.
(.600 atm * 2.40 L)  / (R*300K) = n = .0585 mol H2
(1.4 atm * 6.8 L) / (R * 300 K ) = n = .386 mol He
Total can be found by adding these two values together.
Total pressure = (.456 mol * R * 300K) (9.2 L) = 1.22 atm
Use ratios to find partial pressures!