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A balloon’s internal pressure is 1.25 atm, and its volume is 2.50 L. If the balloon is taken to the bottom of the ocean that has a pressure of 95.0 atm, what would be its new volume? Use: PV = PV l

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Final answer:

Using Boyle's Law, the new volume of a balloon with an initial pressure of 1.25 atm and volume of 2.50 L when taken to a pressure of 95.0 atm at the bottom of the ocean would be approximately 0.0329 L.

Step-by-step explanation:

The student's question asks us to apply the ideal gas law concept, specifically Boyle's Law, to determine the new volume of a balloon when subjected to a significant increase in external pressure. Boyle's Law states that for a given mass of an ideal gas at constant temperature, the product of the pressure and volume of the gas is a constant (P1V1 = P2V2).

Given that the balloon’s internal pressure is 1.25 atm and its volume is 2.50 L initially, and it is taken to the bottom of the ocean with a pressure of 95.0 atm, we can find the new volume by rearranging the formula to V2 = (P1V1)/P2.

Substituting the known values, we get V2 = (1.25 atm * 2.50 L) / 95.0 atm = 0.0329 L (rounded to four significant figures).

Therefore, the new volume of the balloon when taken to the bottom of the ocean would be approximately 0.0329 liters.

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