A scuba diver creates a spherical bubble with a radius of 2.5 cm at a depth of 30.0 m where th

step 1 For this question based on ideal gas law, PV equals to NRT. So the pressure in the deep sea and

A scuba diver creates a spherical bubble with a radius of...

A scuba diver creates a spherical bubble with a radius of $2.5 \mathrm{~cm}$ at a depth of $30.0 \mathrm{~m}$ where the total pressure (including atmospheric pressure) is $4.00 \mathrm{~atm}$. What is the radius of the bubble when it reaches the surface of the water? (Assume that the atmospheric pressure is $1.00 \mathrm{~atm}$ and the temperature is $298 \mathrm{~K}$.)

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Key Concepts

Spherical Volume Calculation

The volume of a sphere is given by the formula V = (4/3)?r³, where r is the radius of the sphere. This concept is crucial in this problem because it allows the conversion between the bubble's radius and its volume. Knowledge of the sphere's geometry helps determine the new radius after applying the changes in pressure.

Boyle's Law

Boyle's Law is a fundamental principle in thermodynamics which states that for a fixed amount of gas at a constant temperature, the pressure and volume are inversely proportional. This means that if the pressure decreases, the volume increases, and vice versa. In this problem, Boyle's Law is used to relate the bubble's volume at depth to its volume at the surface, where the pressure is different.

Pressure-Volume Relationship in Fluids

In fluids, the external pressure can significantly affect the volume of a gas bubble. As a diver ascends and external pressure decreases, the bubble expands according to the pressure-volume relationship defined by Boyle's Law. Understanding this relationship is key to analyzing how a bubble's physical dimensions change with variations in depth.

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