Question

Container A holds 747 mL of an ideal gas at 2.20 atm


Container A holds 747 mL of an ideal gas at 2.20 atm. Container B holds 149 mL of a different ideal gas at 4.90 atm. 

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 If the gases are allowed to mix together, what is the resulting pressure?


 P= _______  atm

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Answer #1

By using the following equation we can calculate the final pressure

V1P1= V2P2

Where,

V1 = initial volume of gas in any container : Va = 747 ml

P1= initial pressure of gas in any container : Pa = 2.20 atm

V2= final volume of gas after mixing = V in container A + V in container B = Va + Vb = 747 ml + 149 ml =

P2=Final pressure of gas after mixing of gas = ?

Va x Pa = P2 x (Va + Vb)

747 ml x 2.20 atm = P2 (896 ml)

P2 = (747 ml x 2.2 atm)/896 ml

P2 = 1663.4 ml x atm /896 ml

P2 = 1.83 atm


answered by: gavin
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