A balloon filled with 2.00 L of helium initially at 1.25 atm of pressure rises into the atmosphere. When the surrounding pressure reaches 410. mmHg, the balloon will burst. If 1 atm = 760. mmHg, what volume will the balloon occupy in the instant before it bursts?

final volume the balloon occupy in instant before it burst is 4.634 L

Explanation:

given data

helium filled V1 = 2 L

initially pressure P1 = 1.25 atm

finally pressure P2 = 410 mmHg

1 atm = 760 mmHg

to find out

at what volume the balloon occupy in instant before it burst

solution

we have given

760 mmHg = 1atm

and 1 mmHg = 1/760

so 410mmHg = 410/760 atm

so that final pressure is = 410/760 atm

we use here Boyle's Law that is

Boyle's law states that "the volume of an ideal gas is inversely proportional to the pressure of the gas at constant temperatures"

so P ∝ 1/V

P1V1 = P2V2

V2 = P1V1 / P2

put all value

V2 = 1.25 (2) / (410/760)

V2 = 2.5 / 0.53947 = 4.634

so final volume the balloon occupy in instant before it burst is 4.634 L