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Answer on Question #83513
Container A holds 782 mL of ideal gas at 2.80 atm. Container B holds 174 mL of ideal gas at 4.40 atm. If the gases are allowed to mix together, what is the resulting pressure?
Solution
We should use Dalton's Law of Partial Pressures to calculate the resulting pressure:
Ptotal = PA + PB
Find pressure (PA) of ideal gas that was in the container A when two gases were mixed together. We should use Boyle's Law:
P1V1 = P2V2
where P1 = 2.80 atm, V1 = 782 mL, P2 = PA , V2 = 782 mL + 174 mL = 956 mL
2.80*782 = PA*956
PA = 2.29 atm
Find pressure (PB) of ideal gas that was in the container B when two gases were mixed together. We should use Boyle's Law:
P1V1 = P2V2
where P1 = 4.40 atm, V1 = 174 mL, P2 = PB , V2 = 782 mL + 174 mL = 956 mL
4.40*174 = PB*956
PB = 0.8 atm
Ptotal = PA + PB= 2.29 atm + 0.8 atm = 3.09 atm
Answer: 3.09 atm