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A bubble of gas released at the bottom of a lake increases to four times its original volume when it reaches the surface. Assuming that atmospheric pressure is equivalent to the pressure exerted by a column of water \(10 \mathrm{~m}\) high, what is the depth of the lake? (a) \(20 \mathrm{~m}\) (b) \(10 \mathrm{~m}\) (c) \(30 \mathrm{~m}\) (d) \(40 \mathrm{~m}\)

Short Answer

Expert verified
The depth of the lake is (c) 30 meters.

Step by step solution

01

Understand Boyle's Law

The problem describes a scenario based on Boyle's Law, which states that for a given amount of ideal gas kept at a constant temperature, the product of the pressure and volume is constant. Mathematically, Boyle's Law is expressed as P1 * V1 = P2 * V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume, respectively.
02

Determine Initial and Final Volumes

According to the problem, the volume of the gas bubble increases fourfold when it reaches the surface. Therefore, if the initial volume is V1, the final volume is 4 * V1.
03

Determine Initial and Final Pressures

When the bubble is at the bottom of the lake, the pressure it experiences is the atmospheric pressure plus the pressure due to the water column above it. At the surface, the pressure on the bubble is just the atmospheric pressure, which is equal to the pressure exerted by a 10 m column of water.
04

Apply Boyle's Law

Using Boyle's Law, we can set up the equation (P_atm + P_water) * V1 = P_atm * 4V1, where P_atm is the atmospheric pressure and P_water is the pressure due to the water column. Since P_atm is equal to a 10 m column of water, P_water must be equal to the difference in pressure between what the gas bubble experiences at the bottom and the surface.
05

Simplify and Solve for the Depth of the Lake

Simplify the Boyle's Law equation: (P_atm + P_water) * V1 = 4 * P_atm * V1. Cancel out V1 from both sides: P_atm + P_water = 4 * P_atm. Subtract P_atm from both sides to get P_water: 3 * P_atm = P_water. Since P_atm is the pressure of a 10 m water column, then P_water is the pressure of a 30 m water column. So, the depth of the lake, which corresponds to P_water, is 30 meters.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Physical Chemistry
Physical Chemistry is the branch of chemistry that studies the physical properties and behavior of matter, especially when matter is subject to changes in temperature, energy, force, and time. It broadly covers thermodynamics, kinetics, quantum mechanics, and statistical mechanics, providing a fundamental basis for understanding chemical reactions and the behavior of gases, liquids, and solids.

In the context of Boyle's Law problems, Physical Chemistry comes into play as it concerns the study of gases and their interactions with pressure and volume. Boyle's Law, itself a fundamental principle in Physical Chemistry, dictates how the pressure and volume of a gas relate to each other when the temperature remains constant.

Understanding the Physical Chemistry behind gases is crucial when solving problems related to variations in pressure and volume. In an educational setting, it is important to emphasize the practical application of these principles to natural occurrences—such as the behavior of a gas bubble rising through water—making abstract concepts more tangible for students.
Pressure-Volume Relationship
The Pressure-Volume Relationship is a core concept in thermodynamics which describes how the pressure of a gas tends to decrease as the volume increases when temperature is constant, and vice versa. This relationship is most commonly referred to as Boyle's Law, a principle that highlights the inverse proportion between the pressure and volume of a confined gas.

In the textbook exercise provided, students observe this concept in action as a bubble of gas rises to the surface of a lake, increasing in volume due to the decreased pressure it encounters as it ascends. Here's a simplified version:
  • At the bottom of the lake, the bubble faces high pressure due to both the water above and the atmospheric pressure.
  • As it rises, the pressure exerted on the bubble decreases, allowing it to expand, illustrating the inverse relationship between pressure and volume.

It is essential for students to grasp the concept of Pressure-Volume Relationship to understand how changes in either property affect the other. This knowledge is not only pivotal in solving Boyle's Law problems but also in various real-world applications, including understanding how breathing works in the human body and predicting the behavior of weather balloons.
Ideal Gas Law
The Ideal Gas Law is a fundamental equation in chemistry and physics that relates the pressure, volume, temperature, and number of moles of an ideal gas. The equation is expressed as PV = nRT, where P is pressure, V is volume, n is number of moles, R is the ideal gas constant, and T is temperature.

Although the original exercise revolves around Boyle's Law, the Ideal Gas Law is a more general equation that encompasses Boyle's Law when the temperature (T) and the amount of gas (n) remain constant. It enables chemists and physicists to predict the behavior of gases under various conditions.

The assumption that a gas behaves ideally means the gas particles are considered to have perfectly elastic collisions and no intermolecular forces—conditions that real gases approximate well under low pressure and high temperature. When applying the Ideal Gas Law to problems, like the one in our exercise, it's helpful to combine it with the knowledge that atmospheric pressure can be equated to the pressure exerted by a column of water, providing a tangible way to measure and apply pressure in calculations.

The comprehension of the Ideal Gas Law is key for students to solve a variety of problems in gas dynamics and to understand the principles that govern the behavior of gases in our atmosphere and in chemical reactions.

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Most popular questions from this chapter

Calculate relative rate of effusion of \(\mathrm{SO}_{2}\) to \(\mathrm{CH}_{4}\), if the mixture obtained by effusing out a : mixture with molar ratio \(\frac{n_{\mathrm{SO}_{2}}}{n_{\mathrm{CH}_{4}}}=\frac{8}{1}\) for three effusing steps: (a) \(2: 1\) (b) \(1: 4\) (c) \(1: 2\) (d) none of these

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The compressibility factor for nitrogen at \(330 \mathrm{~K}\) and \(800 \mathrm{~atm}\) is \(1.90\) and at \(570 \mathrm{~K}\) and 200 atm is \(1.10 .\) A certain mass of \(\mathrm{N}_{2}\) occupies a volume of \(1 \mathrm{dm}^{3}\) at \(330 \mathrm{~K}\) and \(800 \mathrm{~atm} .\) Calculate volume occupied by same quantity of \(\mathrm{N}_{2}\) gas at \(570 \mathrm{~K}\) and \(200 \mathrm{~atm}\) : (a) \(1 \mathrm{~L}\) (b) \(2 \mathrm{~L}\) (c) \(3 \mathrm{~L}\) (d) \(4 \mathrm{~L}\)

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