/*! 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 An air bubble has a volume of \(... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 37

An air bubble has a volume 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 1.96 cm^3.

Step by step solution

01

Calculate initial pressure

First, calculate the initial pressure on the bubble when it is still 100m below the surface. This pressure comprises atmospheric pressure and pressure due to the water column above the bubble. Use the relation \(P_{initial} = P_{atmospheric} + \rho gh\), where \(P_{atmospheric} = 1.013 \times 10^{5} \, Pa\) (standard atmospheric pressure), \(\rho = 1 \times 10^{3} \, kg/m^3\) (density of water), \(g = 9.8 \, m/s^2\) (acceleration due to gravity), and \(h = 100 \, m\).
02

Calculate final pressure

The final pressure on the bubble is the atmospheric pressure at the surface of the lake. This is \(P_{final} = P_{atmospheric} = 1.013 \times 10^{5} \, Pa\).
03

Apply Boyle's law

Knowing the initial and final pressures on the bubble and its initial volume, we can use Boyle's law to find the final volume. Boyle's law states that \(P_{initial}V_{initial} = P_{final}V_{final}\). Solving for \(V_{final}\), we get \(V_{final} = \frac{P_{initial}V_{initial}}{P_{final}}\). Plug in all given values and solve for \(V_{final}\).

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Key Concepts

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

pressure in fluids
Pressure in fluids is a key concept when discussing bubbles under water, like in our submarine scenario. It refers to the force exerted by fluid molecules pressing against a surface. Underneath the surface, pressure increases with depth because not only is there atmospheric pressure, but also the pressure from the weight of the water above.
This is calculated using the formula:
  • \( P_{total} = P_{atmospheric} + \rho gh \)
Where,
  • \(P_{atmospheric}\) is the atmospheric pressure.
  • \(\rho\) is the density of the water.
  • \(g\) is the acceleration due to gravity.
  • \(h\) is the depth below the surface.
This concept is crucial to understanding the forces acting on the bubble as it travels to the surface.
volume change
The change in volume of the bubble from the depths of the lake to the surface is dictated by Boyle's Law, which states that the pressure of a gas times its volume is constant if the temperature doesn't change.
Mathematically, this principle is represented as
  • \( P_{1}V_{1} = P_{2}V_{2} \)
Where,
  • \(P_{1}\) is the initial pressure.
  • \(V_{1}\) is the initial volume.
  • \(P_{2}\) is the final pressure.
  • \(V_{2}\) is the final volume.
As the bubble rises and the pressure decreases, the volume of the bubble increases. This is because, according to Boyle's Law, if the pressure exerted on a gas decreases while the temperature remains constant, its volume must increase to fulfill the equation's equality.
density of water
The density of water is a measure of how much mass of water occupies a specific volume. It's typically referred to as 1000 kg/m鲁.
In calculations involving pressure changes within a fluid, such as water, density is a crucial variable. For our submarine bubble problem, understanding that water exerts pressure due to its density helps solve how pressure changes at different depths.
  • The higher the density, the greater the force it exerts at a given depth.
  • This is why pressure increases with depth, as more water weight sits above any given point.
Knowing this allows us to calculate how the weight of the water above the bubble adds to the atmospheric pressure, giving us the total initial pressure on the bubble.
atmospheric pressure
Atmospheric pressure is the pressure exerted by the weight of the atmosphere above us. This constant pressure is exerted on all surfaces that are in contact with the air. At sea level, standard atmospheric pressure is approximately 1.013 脳 10鈦� Pa.
It plays a vital role in our bubble example. When the bubble rises to the surface, the only pressure acting on it is this atmospheric pressure. Unlike when it was submerged, there's no additional water pressure at the surface.
  • Understanding atmospheric pressure helps in analyzing the point where the bubble reaches the surface.
  • Once at the surface, we consider only atmospheric pressure in our calculations of the bubble's volume using Boyle's Law.
Recognizing this constant helps simplify the calculations needed to determine how pressures change with depth.

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Most popular questions from this chapter

In a student experiment, a constant-volume gas thermometer is calibrated in dry ice \(\left(-78.5^{\circ} \mathrm{C}\right)\) and in boiling ethyl alcohol \(\left(78.0^{\circ} \mathrm{C}\right)\). The separate pressures are \(0.900 \mathrm{~atm}\) and \(1.635 \mathrm{~atm}\). (a) What value of absolute zero in degrees Celsius does the calibration yield? (b) What pressures would be found at (b) the freezing and (c) boiling points of water? Hint: Use the linear relationship \(P=A+B T\), where \(A\) and \(B\) are constants.

Show that the temperature \(-40^{\circ}\) is unique in that it has the same numerical value on the Celsius and Fahrenheit scales.

Before beginning a long trip on a hot day, a driver inflates an automobile tire to a gauge pressure of \(1.80 \mathrm{~atm}\) at \(300 \mathrm{~K}\). At the end of the trip, the gauge pressure has increased to \(2.20\) atm. (a) Assuming the volume has remained constant, what is the tempera- ture of the air inside the tire? (b) What percentage of the original mass of air in the tire should be released so the pressure returns to its original value? Assume the temperature remains at the value found in part (a) and the volume of the tire remains constant as air is released.

Superman leaps in front of Lois Lane to save her from a volley of bullets. In a 1-minute interval, an automatic weapon fires 150 bullets, each of mass \(8.0 \mathrm{~g}\), at \(400 \mathrm{~m} / \mathrm{s}\). The bullets strike his mighty chest, which has an area of \(0.75 \mathrm{~m}^{2}\). Find the average force exerted on Superman's chest if the bullets bounce back after an elastic, head- on collision.

One mole of oxygen gas is at a pressure of \(6.00 \mathrm{~atm}\) and a temperature of \(27.0^{\circ} \mathrm{C}\). (a) If the gas is heated at constant volume until the pressure triples, what is the final temperature? (b) If the gas is heated so that both the pressure and volume are doubled, what is the final temperature?
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