/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 \(\mathrm{A} \mathrm{n}\) air bu... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 37

\(\mathrm{A} \mathrm{n}\) air bubble has a voltume of \(1.50 \mathrm{~cm}^{3}\) when it is released by a submarine \(100 \mathrm{~m}\) below the surface of a lake. What is the volume of the bubble when it reaches the surface? Assume the temperature and the number of air molecules in the bubble remain constant during its ascent.

Short Answer

Expert verified
The volume of the air bubble when it reaches the surface of the lake is 16.1 cm鲁.

Step by step solution

01

Calculate the initial and final pressures

The pressure exerted by a fluid column is the product of the depth (h), the gravitational acceleration (g) and the fluid density (蟻). For the submarine at 100 m depth, the initial pressure (P1) is atmospheric pressure plus the pressure from the water column: P1 = Patm + h1*蟻water*g. Patm is given as 101300 Pa, h1 is the depth of 100 m, 蟻water is the density of water, (1000 kg/m鲁), and g is the gravitational acceleration(9.81 m/s虏). So, P1 = 101300 Pa + 100 m * 1000 kg/m鲁 * 9.81 m/s虏 = 101300 Pa + 981000 Pa = 1082300 Pa. The final pressure (P2) is simply the atmospheric pressure at the water surface, 101300 Pa.
02

Apply Boyle's Law

According to Boyle's law, the initial volume and pressure (P1*V1) will be equal to the final volume and pressure (P2*V2). We are given that V1 is 1.50 cm鲁 (or 1.50 * 10^-6 m鲁, because 1 m鲁 = 10^6 cm鲁), and we want to find V2. So, P1*V1 = P2*V2, becomes 1082300 Pa * 1.50 * 10^-6 m鲁 = 101300 Pa * V2.
03

Solve for the final volume V2

We can solve for V2 by rearranging the equation: V2 = (P1*V1) / P2 = (1082300 Pa * 1.50 * 10^-6 m鲁) / 101300 Pa = 1.61 * 10^-5 m鲁 = 16.1 cm鲁.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Pressure Calculation
Understanding how pressure is calculated is essential in physics. Whenever an object is submerged in a fluid, it experiences pressure that can be broken into two parts: atmospheric pressure and the pressure from the fluid above it.
  • Atmospheric pressure is the pressure exerted by the weight of air in the Earth's atmosphere. At sea level, this is approximately 101300 pascals (Pa).
  • The pressure from a fluid column depends on the depth of the object within the fluid. It's calculated using the formula: Pressure = depth \(h\) \(\times\) fluid density \(\rho\) \(\times\) gravitational acceleration \(g\).
  • For water, the density \(\rho\) can be approximated as 1000 kg/m鲁, and \(g\) is around 9.81 m/s虏.
When calculating the total pressure on the submarine at 100 m depth, you add both the atmospheric pressure and the pressure from the water column. So, if you can measure or calculate the pressure at various points, it's possible to predict changes in behavior of objects like air bubbles, given changes in pressure.
Volume Change
Volume change is a crucial concept in understanding how gases behave under varying pressures. According to Boyle's Law, the volume of a gas is inversely proportional to the pressure it is under, provided the temperature and the number of particles are constant.
  • Boyle's Law equation is \( P_1 \times V_1 = P_2 \times V_2 \), where \( P_1 \) and \( V_1 \) are the initial pressure and volume, and \( P_2 \) and \( V_2 \) are the final pressure and volume.
  • If a gas is under higher pressure, its volume will be smaller, and conversely, if the pressure decreases, the volume increases.
  • In our exercise, the air bubble starts with a volume of 1.50 cm鲁 under significant pressure 100 meters underwater, and expands as it ascends to the surface, where the pressure is less. By maintaining the temperature and the number of molecules constant, we can accurately use Boyle's Law to calculate its volume changes.
This principle is vital in many areas, from understanding how divers manage changes in buoyancy, to how airbags deploy in a controlled manner.
Submarine Depth
Submarine depth plays a crucial role in understanding the behavior of objects like air bubbles within a body of water. When you release an air bubble deep underwater, it goes through a transition as it moves upwards and the surrounding pressure changes.
  • At greater depths, the pressure exerted on a body or object is significantly higher, due to the weight of the water above it.
  • This is why submarines are engineered to withstand outstanding pressures as they go deeper 鈥� every 10 meters of water depth adds approximately 1 atmospheric pressure (approx. 101300 Pa).
  • In the case of a bubble released at 100 meters, by the time it reaches the surface, the pressure impacting it decreases, causing it to expand.
Understanding how depth affects pressure and related physical properties helps in planning activities and equipment design for marine operations, enabling safe and efficient underwater ventures.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A grandfather clock is controlled by a swinging brass pendulum that is \(1.3 \mathrm{~m}\) long at a temperature of \(20^{\circ} \mathrm{C}\). (a) What is the length of the pendulum rod when the temperature drops to \(0.0^{\circ} \mathrm{C}_{7}^{\text {(b) If a pendulum's period is given }}\) by \(T=2 \pi \sqrt{L / g}\), where \(L\) is its length, does the change in length of the rod cause the clock to rum fast or slow?

A \(7.00-\mathrm{L}\) vessel contains \(3.50\) moles of ideal gas at a pressure of \(1.60 \times 10^{6} \mathrm{~Pa}\). Find (a) the temperature of the gas and (b) the average kinetic energy of a gas molecule in the vessel. (c) What additional information would you need if you were asked to find the average speed of a gas molecule?

On the scale invented by French scientist \(\mathrm{R} .\) A. \(\mathrm{F}\). \(\mathrm{de}\) R猫aumer \((1683-1757)\), the freezing point of water is \(0^{\circ}\) but the boiling point is \(80^{\circ}\), (a) Find a formula converting temperatures \(T_{F}\) in Fahrenheit to the temperatures \(T_{\text {Rif }}\) of this scale. (b) What is \(98.6 \mathrm{~F}\) on de R茅aumer's seale?

The average coefficient of volume expansion for carbon tetrachloride is \(5.81 \times 10^{-4}\left({ }^{\circ} \mathrm{C}\right)^{-1}\). If a \(50.0\) -gal steel container is filled completely with carbon tetrachloride when the temperature is \(10.0^{\circ} \mathrm{C}\), how much will spill over when the temperature rises to \(30.0^{\circ} \mathrm{C}\) ?

In a period of \(1.0 \mathrm{~s}, 5.0 \times 10^{23}\) nitrogen molecules strike a wall of area \(8.0 \mathrm{~cm}^{2}\). If the molecules move at \(300 \mathrm{~m} / \mathrm{s}\) and suike the wall head on in a perfectly elastic collision, find the pressure exerted on the wall. (The mass of one \(\mathrm{N}_{2}\) molecule is \(\left.4.68 \times 10^{-26} \mathrm{~kg} .\right)\)
See all solutions

Recommended explanations on Physics Textbooks

Electromagnetism

Read Explanation

Solid State Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Famous Physicists

Read Explanation

Physical Quantities And Units

Read Explanation

Medical Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.