A sample of an ideal gas at 1.00 atm and a volume of 1.31L was placed in a weighted balloon and dropped into the ocean. As the pressure had increased to 45.0 atm, what was the volume of the sample? Assume that the temperature was held constant.

P1V1 = P2V2

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To solve this problem, we can use the combined gas law, which relates the initial and final conditions of a gas sample. The combined gas law equation is:

(P1 * V1) / T1 = (P2 * V2) / T2

where P1 and P2 are the initial and final pressures, V1 and V2 are the initial and final volumes, and T1 and T2 are the initial and final temperatures.

In this problem, we are told that the temperature is held constant, so T1 = T2. Moreover, we are given P1 = 1.00 atm, V1 = 1.31L, and P2 = 45.0 atm. We need to find V2, the final volume.

Substituting the given values into the formula, we get:

(1.00 atm * 1.31 L) / T1 = (45.0 atm * V2) / T2

Since T1 = T2, we can cancel out the temperature terms:

1.00 atm * 1.31 L = 45.0 atm * V2

Simplifying further:

1.31 L = 45.0 atm * V2

Now, let's solve for V2. Divide both sides by 45.0 atm:

1.31 L / 45.0 atm = V2

Calculating this, we find:

V2 ≈ 0.029 L

Therefore, the volume of the sample when the pressure increased to 45.0 atm is approximately 0.029 L.