What is the volume in this handball?

Question

a handball with the internal volume of 60. cm3 was filled with air to a pressure of 1.35 atm. a player filled a syringe to the 25-cm3 mark with air at 1.00 atm and injected it into the handball. calculate the pressure inside the tampered handball, assuming no volume change.

Answer 1

The initial pressure was 1.35 atm, so there must have been 60 cm3 x 1.35 = 81cm3 of air, had the ball been allowed to expand to 1 atm pressure.  Then 25 cm3 were added, so we now have what would be 106 cm3 of air.

So we are forcing a mass of gas that would normally occupy 106cm3, into a 60cm3 space.  So the pressure is thus 106 / 60 = 1.77 atm.

Answer 2

34.13

Answer 3

The Ideal Gas Laws should tell you that, since you are only adding mass...

(What does your text say?...have you studied them?)

At constant vol and temp, pressure is directly proportional to mols (mass in this case) in a closed system...so

P = 1.35 x (60 + 25) / 60 = 1.91 atm