Question

If 4.0 g of O2(g) and 4.0 g of He(g) are placed in a 5.0-L vessel at 65 °C, what will be the partial pressure of each gas and the total pressure in the vessel?

Answer 1

Gas law assumptions allow us to analyze partial pressures of different gases separately by ignoring their interactions. Assuming that O2 and He do not react:

P(O2)
Molecular weight O2 = 32 g/mol
4 g * 1 mol/32g = 0.125 mol of O2.
Find pressure using PV =nRT
T = 65 C + 273 = 338 K
R = 0.0821 L atm/(mol K)
n = 0.125 mol O2
V = 5
PV = nRT
P = nRT/V = (0.125) (0.0821) (338)/5 = 0.694 atm = partial pressure of O2

same thing for He
Molecular Weight He = 4 g/mol
4g * 1mol/4g = 1mol He
(concept! notice that since we have a larger amount of moles of He, and therefore more molecules, we have more molecules that will collide with the walls of the container so we can expect more pressure than from something that is only present in the amount of 0.125 moles (like above).
Using the converted and explicitly stated parameters above, varying only the n value to find partial pressure for He using the ideal gas law (PV=nRT)
P = nRT/V = (1)(0.0821)(338)/5 = 5.55 atm = partial pressure of He (indeed is larger than from O2!)

And the law governing partial pressures allows us to simply sum up the partial pressures with no additional fancy math techniques to discover the total pressure;:

P (O2) + P(He) = P (tot)
0.694 + 5.55 = 6.24 atm total.
If there are no silly calculator mistakes, that should be right. Hope that helps!!

Answer 2

That is a lovely explanation. I am impressed that you are a fan of Walter Lewin.