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Chapter 11: Problem 29

A 1.00-m-tall container is filled to the brim, partway with mercury and the rest of the way with water. The container is open to the atmosphere. What must be the depth of the mercury so that the absolute pressure on the bottom of the container is twice the atmospheric pressure?

Short Answer

Expert verified
The mercury depth must be approximately 0.76 m.

Step by step solution

01

Understand the Problem

We have a container filled with mercury and water to a total height of 1 meter. The problem asks us to find the depth of mercury such that the absolute pressure at the bottom is twice the atmospheric pressure.
02

Identify Key Formulas

To find the pressure, use the formula for liquid pressure: \( P = \rho gh \), where \( \rho \) is the density, \( g \) is the acceleration due to gravity, and \( h \) is the height of the liquid.
03

Express Total Absolute Pressure

The absolute pressure at the bottom of the container is the sum of the atmospheric pressure, the pressure due to the mercury, and the pressure due to the water. Thus, \( P_\text{bottom} = P_\text{atm} + \rho_\text{Hg}g h_\text{Hg} + \rho_\text{water}g (1 - h_\text{Hg}) \).
04

Set Up the Equation for Twice the Atmospheric Pressure

We need \( P_\text{bottom} = 2P_\text{atm} \). Substitute the expression for \( P_\text{bottom} \): \[ 2P_\text{atm} = P_\text{atm} + \rho_\text{Hg}g h_\text{Hg} + \rho_\text{water}g (1 - h_\text{Hg}) \].
05

Simplify the Equation

Rearrange the equation to focus on the term involving \( h_\text{Hg} \): \[ P_\text{atm} = \rho_\text{Hg}g h_\text{Hg} + \rho_\text{water}g (1 - h_\text{Hg}) \].
06

Isolate the Mercury Depth

Solve for \( h_\text{Hg} \): \[ h_\text{Hg} = \frac{P_\text{atm} - \rho_\text{water}g}{(\rho_\text{Hg} - \rho_\text{water})g} \].
07

Substitute Known Values

Use \( \rho_\text{Hg} = 13,600 \text{ kg/m}^3 \), \( \rho_\text{water} = 1,000 \text{ kg/m}^3 \), \( g = 9.81 \text{ m/s}^2 \), and \( P_\text{atm} = 101,325 \text{ Pa} \): \[ h_\text{Hg} = \frac{101,325 - 1,000 \cdot 9.81}{(13,600 - 1,000) \cdot 9.81} \].
08

Calculate the Mercury Depth

Plug in the values and calculate: \( h_\text{Hg} \approx 0.76 \text{ m} \).

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Key Concepts

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

Density of Mercury
The density of mercury is significantly higher than that of many common liquids, like water. Specifically, mercury has a density of \(13,600 \ \text{kg/m}^3\). This means it is very heavy per unit volume. When discussing density:
  • Density is defined as mass per unit volume.
  • High density indicates that there is a lot of mass in a small volume.
Since mercury is much denser than water, it generates a higher pressure at any given depth. In the context of a liquid-filled container, this means mercury contributes significantly to the total pressure at the bottom. The heavy nature of mercury is utilized in various applications, especially in measuring devices like barometers.
Density of Water
Water is a standard reference when discussing fluid densities, with a density of \(1,000 \ \text{kg/m}^3\). This is considered relatively low, especially compared to substances like mercury. Here's why density matters:
  • Water鈥檚 lower density means it exerts less pressure at a given depth compared to a denser fluid.
  • This property of water makes it less effective in pressurizing the bottom of a container compared to denser substances.
In our exercise, the remaining space in the container that isn't filled with mercury is filled with water. This means it assists in adding to the overall pressure, though its contribution is smaller compared to mercury due to its lower density.
Hydrostatic Pressure
Hydrostatic pressure is the pressure exerted by a fluid at equilibrium due to the force of gravity. It's an essential concept for understanding scenarios involving fluids at rest:
  • The formula for hydrostatic pressure is \(P = \rho gh\), with \( \rho \) as density, \( g \) as gravitational acceleration, and \( h \) as height or depth.
  • This pressure increases with depth as more fluid weight pushes down from above.
With mercury and water in the container, each contributes to the total hydrostatic pressure at the container's bottom. Mercury, being denser, contributes more significantly to this pressure. The total hydrostatic pressure is the sum contribution from both fluids.
Atmospheric Pressure
Atmospheric pressure is the pressure exerted by the weight of the atmosphere above us. It is approximately \(101,325 \ \text{Pa}\) at sea level. When solving fluid problems such as the one given, atmospheric pressure forms part of the absolute pressure at the bottom of the container.
  • Absolute pressure is the sum of atmospheric pressure and any additional pressure exerted by the fluid(s) in the container.
  • In the problem where the aim is to make the absolute pressure twice the atmospheric pressure, atmospheric pressure provides the baseline value that must be exceeded by the pressure exerted by the contained fluids.
This concept is critical in ensuring that the calculated pressures unlike the surrounding air are accurately considered in real-world applications. Atmospheric pressure will always play a role unless a container is in a vacuum.

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Most popular questions from this chapter

A 1.3-m length of horizontal pipe has a radius of \(6.4 \times 10^{-3} \mathrm{~m}\). Water within the pipe flows with a volume flow rate of \(9.0 \times 10^{-3} \mathrm{~m}^{3} / \mathrm{s}\) out of the right end of the pipe and into the air. What is the pressure in the flowing water at the left end of the pipe if the water behaves as (a) an ideal fluid and (b) a viscous fluid \(\left(\eta=1.00 \times 10^{-3} \mathrm{~Pa} \cdot \mathrm{s}\right)\) ?

Interactive LearningWare 11.2 at reviews the approach taken in problems such as this one. A small crack occurs at the base of a 15.0 -m-high dam. The effective crack area through which water leaves is \(1.30 \times 10^{-3} \mathrm{~m}^{2}\). (a) Ignoring viscous losses, what is the speed of water flowing through the crack? (b) How many cubic meters of water per second leave the dam?

A glass bottle of soda is sealed with a screw cap. The absolute pressure of the carbon dioxide inside the bottle is \(1.80 \times 10^{5} \mathrm{~Pa}\). Assuming that the top and bottom surfaces of the cap each have an area of \(4.10 \times 10^{-4} \mathrm{~m}^{2},\) obtain the magnitude of the force that the screw thread exerts on the cap in order to keep it on the bottle. The air pressure outside the bottle is one atmosphere.

(a) The mass and radius of the sun are \(1.99 \times 10^{30} \mathrm{~kg}\) and \(6.96 \times 10^{8} \mathrm{~m}\). What is its density? (b) If a solid object is made from a material that has the same density as the sun, would it sink or float in water? Why? (c) Would a solid object sink or float in water if it were made from a material whose density was the same as that of the planet Saturn (mass \(=5.7 \times 10^{26} \mathrm{~kg},\) radius \(\left.=6.0 \times 10^{7} \mathrm{~m}\right)\) ? Provide a reason for your answer.

A hydrometer is a device used to measure the density of a liquid. It is a cylindrical tube weighted at one end, so that it floats with the heavier end downward. It is contained inside a large "medicine dropper," into which the liquid is drawn using a squeeze bulb (see the drawing). For use with your car, marks are put on the tube so that the level at which it floats indicates whether the liquid is battery acid (more dense) or antifreeze (less dense). (a) Compared to the weight \(W\) of the tube, how much buoyant force is needed to make the tube float in either battery acid or antifreeze? (b) Is a greater volume of battery acid or a greater volume of antifreeze displaced by the hydrometer to provide the necessary buoyant force? (c) Which mark is farther up from the bottom of the tube? Justify your answers.
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