Problem 22 - Physics

Question

Multiple-Concept Example 4 deals with the concepts used in this problem. Conceptual Example 3 is also pertinent. A bubble, located \(0.200 \mathrm{~m}\) beneath the surface in a glass of beer, rises to the top. The air pressure at the top is \(1.01 \times 10^{5} \mathrm{~Pa}\). Assume that the density of beer is the same as that of fresh water. If the temperature and number of moles of \(\mathrm{CO} 2\) remain constant as the bubble rises, find the ratio of its volume at the top to that at the bottom.

Step-by-Step Solution

Answer

The volume ratio is approximately 1.0195.

Step 1: Identify the Known Values

We are given that the bubble is **0.200 m** below the surface. The air pressure at the top is given as **\(1.01 \times 10^{5} \, \text{Pa}\)**. The density of beer is assumed to be that of fresh water, which is **\(1000 \, \text{kg/m}^3\)**. We need to find the ratio of the bubble's volume at the top to that at the bottom.

Step 2: Recall the Relevant Physics Concepts

The problem involves the concepts of pressure and volume relating to gases, encapsulated by the Ideal Gas Law and Boyle's Law. Since temperature and moles of gas remain constant, we use Boyle's Law, which states that \(P_1 V_1 = P_2 V_2\), for a constant temperature process.

Step 3: Calculate Pressure at the Bottom

The pressure at the bottom includes both the atmospheric pressure and the pressure due to the liquid column. Pressure due to the liquid is given by \(\rho g h\), where \(\rho = 1000 \, \text{kg/m}^3\) (density of water), \(g = 9.8 \, \text{m/s}^2\), and \(h = 0.200 \, \text{m}\). \[ P_{\text{bottom}} = P_{\text{top}} + \rho g h = 1.01 \times 10^{5} \, \text{Pa} + (1000 \, \text{kg/m}^3)(9.8 \, \text{m/s}^2)(0.200 \, \text{m}) \] \[ P_{\text{bottom}} = 1.01 \times 10^{5} \, \text{Pa} + 1960 \, \text{Pa} = 1.0296 \times 10^{5} \, \text{Pa} \]

Step 4: Apply Boyle's Law

Using Boyle's Law \(P_1 V_1 = P_2 V_2\), where:- \(P_1 = P_{\text{bottom}} = 1.0296 \times 10^{5} \, \text{Pa}\)- \(V_1\) and \(V_2\) represent the volumes of the bubble at the bottom and top respectively- \(P_2 = P_{\text{top}} = 1.01 \times 10^{5} \, \text{Pa}\)Rearrange for \(\frac{V_2}{V_1}\) to find:\[ \frac{V_2}{V_1} = \frac{P_1}{P_2} = \frac{1.0296 \times 10^{5} \, \text{Pa}}{1.01 \times 10^{5} \, \text{Pa}} \approx 1.0195 \]

Step 5: Compute the Final Ratio

Substitute the values calculated into the ratio:\[ \frac{V_2}{V_1} = \frac{P_{\text{bottom}}}{P_{\text{top}}} = \frac{1.0296 \times 10^{5} \, \text{Pa}}{1.01 \times 10^{5} \, \text{Pa}} \approx 1.0195 \]Thus, the volume of the bubble at the top is approximately **1.0195 times** its volume at the bottom.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in physics that relates the pressure, volume, and temperature of a gas to the number of gas molecules present. The law is usually expressed as \( PV = nRT \), where:

• \( P \) is the pressure of the gas
• \( V \) is the volume
• \( n \) is the number of moles of gas
• \( R \) is the ideal gas constant
• \( T \) is the temperature in Kelvin

This equation assumes that gases behave ideally, meaning the gas particles have no volume and no interaction between them. For many gases at high temperatures and low pressures, this approximation holds quite well.

In the context of buoyancy and gas bubbles, such as the rising bubble in a glass of beer, the ideal gas law helps us understand how changes in pressure and volume affect the behavior of the gas inside the bubble. However, since temperature and the number of moles remain constant as the bubble rises, we specifically use Boyle's Law, a derivative concept, to analyze the pressure-volume relationship in this scenario.

Pressure in Fluids

Pressure in fluids is an essential concept to understand how different forces are exerted within a liquid and gas. In fluids, pressure can be described as the force exerted by the fluid per unit area. This pressure acts equally in all directions within the fluid.

For submerged objects, like a bubble underwater, the pressure increases with depth due to the weight of the fluid above the object. This pressure increase can be calculated using the formula:\[ P = P_0 + \rho gh \]where:

• \( P_0 \) is the atmospheric pressure on the fluid's surface
• \( \rho \) is the fluid density
• \( g \) is acceleration due to gravity
• \( h \) is the depth below the surface

As a result, a bubble submerged in a liquid experiences greater pressure at the bottom due to the added pressure from the column of liquid above, leading to changes in its volume as it rises to the surface.

Density of Water

Density is a measure of how much mass is contained in a given volume and is a crucial factor in understanding fluid behaviors such as buoyancy. The density of water is typically given as \(1000 \text{ kg/m}^3\) at standard temperature and pressure (STP). In many real-world problems, other liquids, like beer in our scenario, are often approximated by the density of fresh water for simplicity.

This approximation allows us to easily calculate forces acting on objects submerged in such liquids, like pressure due to the liquid's weight, which is a critical component in fluid mechanics. By using water's density, we simplify the mathematics while maintaining sufficient accuracy for practical purposes.

Understanding density helps explain why some objects float or sink and, in the context of our example, why a bubble changes its volume as it moves through the liquid.

Atmospheric Pressure

Atmospheric pressure is the force exerted onto a surface by the weight of the air above that surface in the Earth's atmosphere. At sea level, this is generally about \(1.01 \times 10^5 \text{ Pa}\). It acts as one of the baseline pressures when calculating pressure in fluids and significantly impacts how gases behave within fluid systems.

In our example with the bubble rising through the beer, the atmospheric pressure is the pressure at the liquid’s surface. It's critical to calculate how much additional pressure from the liquid itself influences the bubble as it travels upwards.

Changes in atmospheric pressure can be observed with weather patterns (like high and low-pressure systems), and they affect how objects, including bubbles, behave when transitioning from one environment to another, especially from water to air.