A 295- flask contains pure helium at a pressure of 743 torr. A second flask with a volume of 465 mL

contains pure argon at a pressure of 712 torr.If the two flasks are connected through a stopcock and the stopcock is opened, what is the partial pressure of helium?
If the two flasks are connected through a stopcock and the stopcock is opened, what is the partial pressure of argon?
If the two flasks are connected through a stopcock and the stopcock is opened, what is the total pressure?

First calculate the moles of each gas using the ideal gas law.

PV = nRT

n = PV / RT

For the He flask:

P = 743/760 atm = 0.978 atm
V = 0.295 L
R = 0.0821 L atm / K mole
T = assume 298 K (doesn't matter)

n He = PV / RT = (0.978)(0.295) / (0.0821)(298) = 0.0118 moles He

For the Ar flask:

P = 712/760 atm = 0.937 atm
V = 0.465 L
R = 0.0821 L atm / K mole
T = 298 K (same T as above)

n Ar = 0.0178 moles Ar

Total gas moles = moles He + moles Ar = 0.0118 + 0.0178 = 0.0296 moles

mole fraction He = 0.0118 / 0.0296 = 0.399; the rest is Ar (1 - 0.399 = 0.601).

When you connect the two flasks to make one flask,

(P He)(V He) + (P Ar)(V Ar) = (P new)(V new) . . .we can use torr and mL in this equation.
(743)(295) + (712)(465) = (P new)(295 + 465)
P new = 724 torr

P Ar after mixing = mole fraction Ar x P total = (0.601)(724 torr) = 435 torr