A diver at a depth of \(1.0 \times 10^{2} \mathrm{m},\) where the pressure is 11.0 atm, releases a bubble with a volume of 100.0 \(\mathrm{mL}\) . What is the volume of the bubble when it reaches the surface? Assume a pressure of 1.00 atm at the surface.

Short Answer

Expert verified
The final volume \(V2\) of the bubble when it reaches the surface is 1100.0 mL.

Step by step solution

01

Write Boyle's Law Formula

Boyle's Law is represented by the equation \(P1 \times V1 = P2 \times V2\), where \(P1\) and \(V1\) represent the initial pressure and volume, and \(P2\) and \(V2\) represent the final pressure and volume.
02

Insert Given Numbers

In this case, \(P1\) is 11.0 atm, \(V1\) is 100.0 mL, and \(P2\) is 1.00 atm. Substitute these values into the Boyle's Law equation. This gives the equation \(11.0 \, \text{atm} \times 100.0 \, \text{mL} = 1.00 \, \text{atm} \times V2\).
03

Solve for \(V2\)

To solve for \(V2\), divide both sides of the equation by \(P2\) which is 1.00 atm. This gives the equation \(V2 = (11.0 \, \text{atm} \times 100.0 \, \text{mL}) / 1.00 \, \text{atm}\).
04

Compute \(V2\)

Just calculate the right side of the equation to find the value of \(V2\). The atm units will cancel each other out, leaving only the value in mL.

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