A swimmer has 8.20 x 10^-4 m^3 of air in his lungs when he dives into a lake. Assuming the pressure of the air is 95 percent of the external pressure at all times, what is the volume of the air at a depth of 10.0 m? Assume that the atmospheric pressure at the surface is 1.013 x 10^5 Pa, the density of the lake water is 1.00 x 10^3 kg/m^3, and the temperature is constant.
To find the volume of air at a depth of 10.0 m, we can use Boyle's Law, which states that the volume of a gas is inversely proportional to its pressure when temperature is constant.
Boyle's Law formula: P1 * V1 = P2 * V2
Let's break down the given information step by step:
Step 1: Determine the starting pressure and volume of the air.
The starting pressure is given as 95% of the external pressure at all times.
External pressure = atmospheric pressure = 1.013 x 10^5 Pa
Starting pressure = 0.95 * 1.013 x 10^5 Pa
The starting volume of air is given as 8.20 x 10^-4 m^3.
Starting volume = 8.20 x 10^-4 m^3
Step 2: Determine the pressure at a depth of 10.0 m.
The pressure at a depth in a fluid can be calculated using the hydrostatic pressure formula.
Hydrostatic pressure = density * gravity * height
Given:
Density of lake water = 1.00 x 10^3 kg/m^3
Height = 10.0 m
Gravity = 9.81 m/s^2
Pressure at a depth of 10.0 m = (1.00 x 10^3 kg/m^3) * (9.81 m/s^2) * (10.0 m)
Step 3: Calculate the final volume of air at a depth of 10.0 m using Boyle's Law.
P1 * V1 = P2 * V2
(0.95 * 1.013 x 10^5 Pa) * (8.20 x 10^-4 m^3) = (Pressure at a depth of 10.0 m) * V2
Substitute the values and solve for V2:
V2 = [(0.95 * 1.013 x 10^5 Pa) * (8.20 x 10^-4 m^3)] / (Pressure at a depth of 10.0 m)
Perform the calculations to find the final volume of air at a depth of 10.0 m.