Question

Consider a lake that is about \(40 \mathrm{~m}\) deep. A gas bubble with a diameter of \(1.0 \mathrm{~mm}\) originates at the bottom of a lake where the pressure is \(405.3 \mathrm{kPa}\). Calculate its volume when the bubble reaches the surface of the lake where the pressure is 98 \(\mathrm{kPa}\), assuming that the temperature does not change.

Short answer

The volume of the gas bubble when it reaches the surface of the lake is approximately \(2.15 \times 10^{-3} \mathrm{cm^3}\).

Step-by-step solution

Identify the Ideal Gas Law

The Ideal Gas Law can be expressed as: \(PV = nRT\) where P is the pressure, V is the volume, n is the amount of gas (in moles), R is the universal gas constant, and T is the temperature. Since the temperature and the amount of gas do not change in this problem, the Ideal Gas Law can be simplified to: \(\frac{P_1V_1}{P_2V_2}=1\) where the subscripts 1 and 2 refer to the bottom and surface of the lake, respectively.

Calculate the initial volume of the bubble

We are given the diameter of the bubble at the bottom of the lake as 1.0 mm. We will first calculate the radius: Radius = Diameter / 2 = \(1.0 \mathrm{mm} / 2 = 0.5 \mathrm{mm}\) Now, convert the radius to meters: Radius = \(0.5 \times 10^{-3} \mathrm{m}\) The volume of a sphere can be calculated using the formula: Volume = \(\frac{4}{3}\pi r^3\) Substituting the radius value, we get: \(V_1 = \frac{4}{3}\pi (0.5 \times 10^{-3})^3 = 5.24 \times 10^{-10} \mathrm{m^3}\)

Calculate the volume of the bubble at the surface

We will now apply the simplified Ideal Gas Law equation to find the volume at the surface \(V_2\): \(\frac{P_1V_1}{P_2V_2}=1\) Rearrange the formula to isolate \(V_2\): \(V_2 = \frac{P_1V_1}{P_2}\) Substitute the given pressures and initial volume: \(V_2 = \frac{405.3 \times 10^3 \mathrm{Pa} \times 5.24 \times 10^{-10} \mathrm{m^3}}{98 \times 10^3 \mathrm{Pa}}\) Now, calculate the value of \(V_2\): \(V_2 = 2.15 \times 10^{-9} \mathrm{m^3}\)

Convert the volume to cubic centimeters

The volume at the surface is given in cubic meters, but it is often more convenient to express it in cubic centimeters. To convert the volume from cubic meters to cubic centimeters, use the conversion factor: 1 \(\mathrm{m^3} = 10^6 \mathrm{cm^3}\) \(V_2 = 2.15 \times 10^{-9} \mathrm{m^3} \times 10^6 \mathrm{cm^3/m^3} = 2.15 \times 10^{-3} \mathrm{cm^3}\) The volume of the gas bubble when it reaches the surface of the lake is approximately 2.15 x 10^{-3} cubic centimeters.

Key concepts

Understanding Volume Calculation for a Bubble

In the problem, we start with calculating the initial volume of a gas bubble. The bubble is spherical, and its volume is determined using the formula for the volume of a sphere:
\[ V = \frac{4}{3}\pi r^3 \]where \( r \) is the radius. Initially, the given diameter of the bubble is \( 1.0 \text{ mm} \). To find the radius, we divide by 2, giving us \( 0.5 \text{ mm} \) or \( 0.5 \times 10^{-3} \text{ m} \) when converted to meters.

• Calculate the radius: Diameter/2
• Convert radius to meters for standard units
• Use sphere volume formula to find initial volume

Substitute the radius into the volume formula to calculate the initial volume, \( V_1 \), which yields \( 5.24 \times 10^{-10} \text{ m}^3 \). Volume calculations are key, especially when dealing with spherical shapes like bubbles, because they tell us how much space an object occupies.

Exploring Pressure Changes for the Bubble

Pressure changes have a significant impact on the volume of gases. In this context, we're working with a gas bubble that rises from the bottom of a lake to the surface. At the bottom, the pressure is higher at \( 405.3 \text{ kPa} \), while at the surface pressure decreases to \( 98 \text{ kPa} \). As per the Ideal Gas Law,
\[ PV = nRT \]We can simplify it under constant temperature and gas quantity conditions to:
\[ \frac{P_1V_1}{P_2V_2}=1 \]where \( P_1 \) and \( P_2 \) are pressures at the bottom and surface, respectively, and \( V_1 \) and \( V_2 \) are the corresponding volumes.

• Recognize pressure decrease from bottom to surface
• Utilize Ideal Gas Law with simplified assumptions
• Correlate pressure decrease with volume increase of bubble

As pressure decreases, the volume of the bubble increases in response. This is a classic illustration of Boyle's Law which states that the volume of a gas is inversely proportional to its pressure under constant temperature.

The Journey and Expansion of the Gas Bubble

A gas bubble's life underwater is influenced by pressure and its state of expansion. When the bubble is at the lake's depth, it experiences high pressure leading to its smaller volume.
As the bubble ascends, reduced pressure allows the bubble to expand increasing its volume. This is calculated using the simplified Ideal Gas Law. Initially, the bubble's volume is \( 5.24 \times 10^{-10} \text{ m}^3 \) at the bottom of the lake, and it reaches approximately \( 2.15 \times 10^{-9} \text{ m}^3 \) by the time it surfaces.

• Bubble's volume increase aligns with pressure decrease
• Calculating volume at different points highlights gas expansion
• Convert final volume to more convenient units

This calculated journey, dictated by volume and pressure dynamics, reinforces fundamental gas law principles and serves as a simple yet powerful example of gas behavior in variable pressure environments such as underwater.