The following question is purely hypothetical!

Question

At a party, one of the balloons containing 1,770 cm^3 of air at STP can withstand a pressure of 5.57 x 10^5 Pa. If Chris sits on the balloon and reduces its volume to 295 cm^3, what is the new pressure? Wil the balloon pop? (Convert torr to Pascals to help determine if the balloon will pop.)

the correct answer is 6.08 x 10^5 Pa and Yes. I need an explanation of this quiestion. Thank you!

Answer 1

Temperature will remain constant, so we are essentially dealing with Boyle's Law:

P1V1 = P2V2

P1 = 1 atm = 760 torr

V1 = 1770 cm3

P2 = ?

V2 = 295 cm3

Solving for P2 we have...

(760)(1770) = (P2)(295)

P2 = 4560 torr

Converting this to Pascals, we have...

4560 torr x 133.3 Pa/torr = 607,848 Pa

We can write this as 6.07848x10⁵ Pa and correcting for significant figures we have 6.08x10⁵ Pa

The balloon WILL pop because it can only handle a pressure of 5.57x10⁵ Pa.