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You have a balloon filled with 5 mL of air at a pressure of 5 atm. You squeeze the ballon changing the volume to 2.5mL how will the pressure cuange? What will the new pressure be?

User Bgschiller
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Final answer:

If a balloon filled with 5 mL of air at a pressure of 5 atm is squeezed to a volume of 2.5 mL, the new pressure will increase to 10 atm under Boyle's Law, which states that volume and pressure are inversely proportional at a constant temperature.

Step-by-step explanation:

The question is concerning the relationship between the volume and pressure of a gas contained in a balloon, which is described by Boyle's Law. According to Boyle's Law, the volume of a given amount of gas held at a constant temperature varies inversely with the applied pressure when the temperature is held constant. If we start with a balloon filled with 5 mL of air at a pressure of 5 atm and then squeeze the balloon to reduce the volume to 2.5 mL, the new pressure can be calculated using the formula P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.

In this case: (5 atm) * (5 mL) = (P2) * (2.5 mL); solving for P2 gives us: P2 = (5 atm * 5 mL) / 2.5 mL = 10 atm. Thus, if you reduce the volume of the balloon to 2.5 mL, the air pressure inside the balloon will increase to 10 atm, assuming temperature remains constant.

User Nunoarruda
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