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?

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Answer 1 · 2018-07-02T06:41:14

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

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

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

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

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...

After 0.600 L of ArAr at 1.40 atm and 210 "^o"^oC is mixed with 0.200L of O_2O_2 at 398 torr and 120 "^o"^oC in a 400 -mL flask at 29 "^o"^oC, what is the pressure in the flask?