Question

A sample of an ideal gas at 1.00 atm

and a volume of 1.38 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 15.0 atm,
what was the volume of the sample? Assume that the temperature was held constant.

Answer 1

We can use Boyle's Law, which states that the pressure and volume of a fixed amount of gas at a constant temperature are inversely proportional, to solve this problem. Mathematically, Boyle's Law can be expressed as:

P1V1 = P2V2

where P1 and V1 are the initial pressure and volume of the gas, respectively, and P2 and V2 are the final pressure and volume of the gas, respectively.

In this problem, we are given that the initial pressure and volume of the gas are 1.00 atm and 1.38 L, respectively, and that the final pressure is 15.0 atm. We can rearrange Boyle's Law to solve for V2:

V2 = (P1V1) / P2

Substituting the given values, we get:

V2 = (1.00 atm x 1.38 L) / 15.0 atm

Simplifying, we get:

V2 = 0.092 L

Therefore, the volume of the gas sample when the pressure had increased to 15.0 atm is 0.092 L.