Question

Two gases are mixed in a 10.0 L fixed volume flask: 2.0L of O2 at 2.00atm and 8.0L of N2 at 3.00 atm. A. Calculate the partial pressure for each gas and the total pressure. B. How much (volume in liters) argon at 5.0 atm must be added to the 10.0L flask to lower the mole fraction of nitrogen to .5? C. What is the total pressure of the mixture from Part B?

Answer 1

first we find moles of each using PV = nRT equation

for O2 ,    2 x 2 = n x 0.08206 x 298 ,   nO2 moles = 0.1636

for N2 , 3 x 8 = n x 0.08206 x 298 , nN2 = 0.9814

total moles = 0.1636+0.9814 = 1.145

Total Pressure = ( n x R x T/V) = ( 1.145 x 0.08206 x 298 /10) = 2.8 atm

Mol fraction of O2= ( moles of O2/total gas moles) = ( 0.1636 / 1.145) = 0.143 ,

mol fraction of N2 = 1-0.143 = 0.857

Partial Pressure of O2 = mol fraction of O2 x total pressure = ( 0.143 x 2.8) =0.4 atm

Partial Pressure of N2 = 2.8-0.4 = 2.4 atm

B) Let Argon moles be   n

then mol fraction of N2 = 0.5 = ( N2 moles/ total gas moles) = 0.9814 / ( 1.145+n)

0.5 ( 1.145+n) = ( 0.9814)

0.5725 +0.5n = 0.9814

n = 0.8178 , P = 5 atm   we find volume of Argon

( 5 x V) = ( 0.8178 x 0.08206 x 298)

V = 4 L   is volume fo ARgon

Total gas moles = 0.8178+1.145 = 1.9628

( P x 10) = ( 1.9628 x 0.08206x298)

P total = 4.8 atm