A gas in a balloon has a volume of 12.0 L at a pressure of 1.0 atm. If the balloon is taken underwater where the pressure increases to 3.0 atm, what will be the new volume of the gas, assuming temperature stays constant?

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We start from the principle that, at constant temperature, gas pressure and volume are related by Boyle's Law: P1V1 = P2V2.
Option analysis begins with the given values: P1 = 1 atm, V1 = 12.0 L, and P2 = 3.0 atm. Using Boyle's Law, V2 = (P1 × V1) / P2 = (1 atm × 12.0 L) / 3.0 atm = 12.0 / 3.0 = 4.0 L.
Now, evaluate each provided option against the calculation:
- 3.0 L: If V2 were 3.0 L, then P2 would need to be (P1V1)/V2 = (1 atm × 12.0 L) / 3.0 L = 4.0 atm. Since P2 is given as 3.0 atm, this option does not satisfy Boyle's law under the stated conditions, so it is inconsistent.
- 8.0 L: For V2 = 8.0 L, P2 would be (1 atm × 12.0 L) / 8.0 L = 1.5 atm. This is far from the stated 3.0 atm, so this volume is not correct under the constant-temperature assumption.
- 6.0 L: If V2 were 6.0 L, then P2 would be (1 atm × 12.0 L) / 6.0 L = 2.0 atm. Again, this does not match the given 3.0 atm, so this option is incorrect.
- 4.0 L: With V2 = 4.0 L, P2 would be (1 atm × 12.0 L) / 4.0 L = 3.0 atm, which exactly corresponds to the stated final pressure and the constant-temperature condition. This one is consistent with Boyle’s Law.
In summary, the volume that aligns with the given final pressure under constant temperature is 4.0 L, while the other options fail to satisfy P1V1 = P2V2 under the provided conditions.

2254 CHEM 0101 SEC2010 HW 9

A gas in a balloon has a volume of 12.0 L at a pressure of 1.0 atm. If the balloon is taken underwater where the pressure increases to 3.0 atm, what will be the new volume of the gas, assuming temperature stays constant?

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