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A balloon filled with 2.00 L of helium initially at 1.85 atm of pressure rises into the atmosphere. When the surrounding pressure reaches 470. mmHg, the balloon will burst. If 1 atm = 760. mmHg, what volume will the balloon occupy in the instant before it bursts?

Sagot :

Neetooidn

The volume will the balloon occupy in the instant before it bursts will be 5.98 L.

Initial volume of helium (V1) = 2.00 L

Initial pressure of helium (P1) = 1.85 atm

Final pressure of helium (P2) = 470 mmHg or 0.61842 atm

Final volume of helium (V2) = ?

We will calculate the final volume by using the following equation;

    P1V1 = P2V2

Rearrange it for V2

    V2 = P1V1 / P2

    V2 = (1.85 atm × 2.00 L) / 0.61842 atm

    V2 = 5.98 L

So 5.98 L volume will the balloon occupy in the instant before it bursts.

You can also learn about pressure from the following question:

https://brainly.com/question/12971272

#SPJ4

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