Question

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?

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.