chem help pls?
2016-10-18 5:46 pm
Container A holds 712 mL of ideal gas at 2.10 atm. Container B holds 174 mL of ideal gas at 4.80 atm. If the gases are allowed to mix together, what is the resulting pressure?
回答 (1)
2016-10-18 6:13 pm
P₁V₁ = P₂V₂
P₂ = P₁ × (V₁/V₂)
Volume of the gas mixture = (712 + 174) mL = 886 mL
In the gas mixture :
Partial pressure due to the ideal gas originally in Container A = (2.10 atm) × (712/886)
Partial pressure due to the ideal gas originally in Container B = (4.80 atm) × (174/886)
Resulting pressure = (2.10 atm) × (712/886) + (4.80 atm) × (174/886) = 2.63 atm
P₂ = P₁ × (V₁/V₂)
Volume of the gas mixture = (712 + 174) mL = 886 mL
In the gas mixture :
Partial pressure due to the ideal gas originally in Container A = (2.10 atm) × (712/886)
Partial pressure due to the ideal gas originally in Container B = (4.80 atm) × (174/886)
Resulting pressure = (2.10 atm) × (712/886) + (4.80 atm) × (174/886) = 2.63 atm
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