Question

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?

Answer 1

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...