Question

Container A holds 712 mL of an ideal gas at 2.20 atm. Container B holds 189 mL of a different ideal gas at 4.20 atm. Container A and container B are glass spheres connnected by a tube with a stopcock. Container A is larger than container B. If the gases are allowed to mix together, what is the resulting pressure?

Answer 1

We have the Boyle’s law P1V1 = P2V2

Calculation of pressure of gas A:

From Boyle’s law, P1V1 = P2V2

Here, P1 = 2.2 atm, V1 = 712 mL

P2 = ? and V2 = 712 + 189 mL = 901 mL

P2 = (2.2 x 712)/901 = 1.74 atm

Calculation of pressure of gas B:

From Boyle’s law, P1V1 = P2V2

Here, P1 = 4.2 atm, V1 = 189 mL

P2 = ? and V2 = 712 + 189 mL = 901 mL

P2 = (4.2 x 189)/901 = 0.88 atm

Resulting pressure P = Pa + Pb = 1.74 + 0.88 = 2.62 atm