Question

a sample of an ideal gas at 1.00 atm and a volume of 1.49 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 55.0 atm, what was the volume of the sample? assume that the temperature was held constant.
v =

Explanation:

Step1: Recall Boyle's Law

Boyle's Law states that \(P_1V_1 = P_2V_2\) (for a gas at constant temperature), where \(P_1\) and \(V_1\) are the initial pressure and volume, and \(P_2\) and \(V_2\) are the final pressure and volume.

Step2: Rearrange the formula to solve for \(V_2\)

We can rewrite the formula as \(V_2=\frac{P_1V_1}{P_2}\).

Step3: Substitute the given values

Given \(P_1 = 1.00\space atm\), \(V_1=1.49\space L\), and \(P_2 = 55.0\space atm\). Substitute these into the formula: \(V_2=\frac{1.00\times1.49}{55.0}\).

Step4: Calculate \(V_2\)

\(V_2=\frac{1.49}{55.0}\approx0.0271\space L\)

Answer:

\(0.0271\)