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Chapter 12: Problem 37

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 volume of the bubble when it reaches the surface is 1.1 L.

Step by step solution

01

Define the initial conditions

By reading the problem, the initial conditions of the bubble rising to the surface from the depth are given: Initial volume \( V_1 = 100.0 \, mL = 0.1 \, L \) (since 1L = 1000mL), and initial pressure \( P_1 = 11.0 \, atm \). The pressure at the surface, which is our final pressure, is \( P_2 = 1.00 \, atm \). The final volume \( V_2 \) is what we need to calculate.
02

Apply Boyle's Law

According to Boyle's Law, at constant temperature the volume and pressure of a gas have an inverse relationship, expressed as \( P_1 \cdot V_1 = P_2 \cdot V_2 \).
03

Substitute the given values into Boyle's Law equation

The given values are substituted into the equation: \( (11.0 \, atm) \cdot (0.1 \, L) = (1.00 \, atm) \cdot V_2 \). Division by 1.00 atm from both sides results in the equation: \( V_2 = \frac{(11.0 \, atm) \cdot (0.1 \, L)}{(1.00 \, atm)} \).
04

Solve for the final volume

Upon solving the equation, one finds that \( V_2 = 1.1 \, L \).

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