A balloon filled with 2.00L of helium initially at 1.15 atm of pressure rises into the atmosphere. When the surrounding pressure reaches 370.mmHg, the balloon will burst. What volume will the ballon occupy in the instant before it bursts?
From the problem we can identify that:
Vi = 2.00 L Pi = 1.15 atm
Vf = ? Pf = 370 mmHg
Before we calculate the final volume we have to take the pressures to the same unit. We can convert the final pressure from mmHg to atm. We know that 760 mmHg is equal to 1 atm, so:
760 mmHg = 1 atm
Pf = 370 mmHg * 1 atm/(760 mmHg) = 0.487 atm
Pf = 0.487 atm
Since the process is at constant temperature, we can apply this formula to calculate the final volume:
Pi * Vi = Pf * Vf
Vf = Pi * Vi / Pf
Vf = 1.15 atm * 2.00 L / 0.487 atm
Vf = 4.72 L
Answer: the volume that the balloon will occupy is 4.72 L.