Question

Container A holds 732 mL of ideal gas at 2.70 atm. Container B holds 129 mL...

Container A holds 732 mL of ideal gas at 2.70 atm. Container B holds 129 mL of ideal gas at 4.50 atm. If the gases are allowed to mix together, what is the partial pressure of each gas in the total volume?

Homework Answers

Answer #1

From the Dalton's Law of Partial Pressures, each gas will exert its pressure independently of the other.

Calculate the pressure of Gas A

Boyle's Law is

P1V1=P1V2

Given for container A,

P1=2.70 atm, V1=732 mL

P2=? and V2=732 mL + 129 mL=861 mL

Therefore, P2=(P1×V1)/V2=(2.70 atm×732mL)/861mL=2.295 atm

Calculate the pressure of Gas B

For container B,

P1=4.50 atm, V1=129mL
P2=? and V2=129 mL + 73 mL=861 mL

P2=(P1×V1)/V2=(4.50 atm×129mL)/861mL=0.6742 atm

Calculate the total pressure

Total pressure, Ptot=PA+PB

Ptot=2.295 atm + 0.6742 atm=2.969 atm ~ 2.97 atm.

Thanks and please let me know if you have any doubt.

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