After 0.600 L of #Ar# at 1.40 atm and 210 #"^o#C is mixed with 0.200L of #O_2# at 398 torr and 120 #"^o#C in a 400 -mL flask at 29 #"^o#C, what is the pressure in the flask?

1 Answer
Jul 2, 2018

Answer:

Well, FIRST off we calculate the molar quantities of each gas...by means of the Ideal Gas equation...

#n=(PV)/(RT)#...

Explanation:

#n_"Ar"=(1.40*atmxx0.600*L)/(0.0821*(L*atm)/(K*mol)xx483.15*K)=0.0212*mol#

#n_(O_2)=((398*mm)/(760*mm*Hg*atm^-1)xx0.200*L)/(0.0821*(L*atm)/(K*mol)xx393.15*K)=0.00325*mol#

And now we ADD these molar quantities, and calculate the pressure expressed under the NEW given conditions...

#P_"mixture"=(0.245*molxx0.0821*(L*atm)/(K*mol)xx302.15*K)/(400*mLxx10^-3*L*mL^-1)#

I make this approx. #15*atm# but do not trust my arithmetic...and there is a lot of rithmetick here...