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18 July, 22:25

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?

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  1. 19 July, 00:00
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    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
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