A sample of an ideal gas at 1.00 atm and a volume of 1.61 L was placed in a weighted balloon and dropped into the ocean. As the sample descended, the water pressure compressed the balloon and reduced its volume. When the pressure had increased to 45.0 atm, what was the volume of the sample? Assume that the temperature was held constant.
To solve this problem, we can use Boyle's Law, which states that the pressure of a gas is inversely proportional to its volume, as long as the temperature and the amount of gas remain constant. The formula for Boyle's Law is:
P1 * V1 = P2 * V2
Where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.
In this case, the initial pressure (P1) is 1.00 atm, the initial volume (V1) is 1.61 L, and the final pressure (P2) is 45.0 atm. We want to find the final volume (V2).
Substituting the known values into the formula, we have:
(1.00 atm) * (1.61 L) = (45.0 atm) * V2
Now we can solve for V2:
V2 = (1.00 atm) * (1.61 L) / (45.0 atm)
V2 ≈ 0.0356 L
Therefore, when the pressure increased to 45.0 atm, the volume of the sample was approximately 0.0356 L.