A mixture of oxygen and argon gases in a 704 L flask at 56 C

A mixture of oxygen and argon gases, in a 7.04 L flask at 56 C, contains 1.67 grams of oxygen and 831 in the flask is atm and the total pressure in the flask is

Solution

no of moles of O2 = W/G.M.Wt
                   = 1.67/32 = 0.052moles
V = 7.04L
T = 56+273 = 329K
PV = nRT
P = nRT/V
   = 0.052*0.0821*329/7.04   = 0.2atm
The partial pressure of O2 PO2 = 0.2atm
no of moles of Ar = W/G.A.Wt
                   = 8.31/40 = 0.21moles
Total no of moles of oxygen and argon = 0.052+0.21 = 0.262moles
V = 7.04L
T = 56+273 = 329K
PV = nRT
P = nRT/V
   = 0.262*0.0821*329/7.04   = 1 atm >>>>>answer
The total pressure of the mixture = 1atm

2. no of moles of Xe = W/G.A.Wt
                      = 87.6/131   = 0.6687moles
V = 9.5L
P = 2.36atm
T = 28+273 = 301K
PV = nRT
n   = PV/RT
     = 2.36*9.5/0.0821*301   = 0.9moles
total no of moles of Xe and helium = 0.9 moles
n Xe + n He = 0.9 moles
0.6687 + n He = 0.9
n He           = 0.9-0.6687   = 0.2313 moles
mass of He = no of moles * gram atomic mass
            = 0.2313*4 = 0.9252g
3

N2                                  CO2                     mixture
P1 = 2.2atm                         P2 = 2.05atm            P
V1 = 1.28L                         V2 = 2.04L             V = 1.28+2.04 = 3.32L

   P    = P1V1 + P2V2/V
         = 2.2*1.28+2.05*2.04/3.32
         = 6.998/3.32   = 2.1atm