Question

Container A holds 722 mL of ideal gas at 2.30 atm. Container B holds 114 mL of ideal gas at 4.80 atm. If the gases are allowed to mix together, what is the partial pressure of each gas in the total volume?

Answer 1

for gas A:

we have:

P1 = 2.30 atm

V1 = 722 mL

V2 = 836 mL (This is total volume)

we have below equation to be used:

P1*V1 = P2*V2

2.30 atm * 722.0 mL = P2 * 836.0 mL

P2 = 1.9864 atm

Answer: partial pressure of gas A = 1.99 atm

for gas B:

we have:

P1 = 4.80 atm

V1 = 114 mL

V2 = 836 mL

we have below equation to be used:

P1*V1 = P2*V2

4.80 atm * 114.0 mL = P2 * 836.0 mL

P2 = 0.6545 atm

Answer: partial pressure of gas A = 0.655 atm