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Chapter 5: Problem 7

A long glass tube, sealed at one end, has an inner diameter of 10.0 \(\mathrm{mm}\). The tube is filled with water and inverted into a pail of water. If the atmospheric pressure is \(755 \mathrm{mmHg}\), how high (in \(\mathrm{mmH}_{2} \mathrm{O}\) ) is the column of water in the tube \(\left(d\right.\) of \(\mathrm{Hg}=13.5 \mathrm{~g} / \mathrm{mL} ; d\) of \(\left.\mathrm{H}_{2} \mathrm{O}=1.00 \mathrm{~g} / \mathrm{mL}\right) ?\)

Short Answer

Expert verified
10192.5 mmH\textsubscript{2}O

Step by step solution

01

Understand the Relationship Between Pressure and Height

The pressure exerted by the column of water inside the tube will be equal to the atmospheric pressure outside. We need to express the heights in terms of the respective fluid densities and pressures.
02

Relate Mercury and Water Pressures

Use the given densities and the atmospheric pressure to find the height of the water column. The pressure exerted by the mercury column can be expressed as: \[ P_{atm} = h_{Hg} * d_{Hg} \]Where:- \(P_{atm}\) = atmospheric pressure in mmHg- \(h_{Hg}\) = height of mercury column- \(d_{Hg}\) = density of mercury
03

Convert Atmospheric Pressure to mmH\textsubscript{2}O

Next, we convert this atmospheric pressure to the equivalent height of a water column using the density of water. Rearrange the pressure equation in terms of water: \[ P_{atm} = h_{H_2O} * d_{H_2O} \]Where:- \( h_{H_2O} \) = height of the water column- \(d_{H_2O}\) = density of water
04

Solve for Height of the Water Column

Equating the two expressions for \(P_{atm}\):\[ h_{Hg} * d_{Hg} = h_{H_2O} * d_{H_2O} \]Solve for \(h_{H_2O}\):\[ h_{H_2O} = \frac{h_{Hg} * d_{Hg}}{d_{H_2O}} \]Substitute the given values:\[ h_{H_2O} = \frac{755 \ \mathrm{mmHg} * 13.5 \ \mathrm{\ g / mL}}{1.00 \ \mathrm{\ g / mL}} \]
05

Calculate

Calculate the value:\[ h_{H_2O} = \frac{755 \ \mathrm{mmHg} * 13.5 \ \mathrm{\ g / mL}}{1.00 \ \mathrm{\ g / mL}} = 10192.5 \ \mathrm{mmH_2O} \]

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Key Concepts

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

pressure and fluid column height
Understanding the relationship between pressure and fluid column height is essential in this exercise. Pressure is a measure of force exerted on an area. It can be exerted by liquids and gases, and it is proportional to the height of the fluid column above the point where the pressure is being measured.
For example, the pressure at the bottom of a water column is greater if the column is taller.
This is because more water weight sits above it. For this exercise, we use this principle to link atmospheric pressure to the height of a water column in the tube.
fluid column
A fluid column in this context refers to the vertical height of water or mercury inside the glass tube.
When we invert a water-filled tube into a pail of water, the fluid column inside the tube stays elevated. This occurs due to the atmospheric pressure acting on the open side of the tube, supporting the water column.
The height of the water column can be calculated using the density of the fluid and the atmospheric pressure, as demonstrated in the solution.
density
Density is a crucial concept in understanding how the height of the fluid column affects pressure. It is defined as mass per unit volume, and different substances have different densities.
The denser the fluid, the higher the pressure it exerts at a given height.
In this problem, mercury has a density of 13.5 g/mL, much higher than water, which has a density of 1.00 g/mL. Using the relationship between density and pressure, we can determine that the same atmospheric pressure supports a much shorter column of mercury than water.
atmospheric pressure
Atmospheric pressure is the pressure exerted by the weight of the atmosphere above the surface of the Earth. It varies with altitude, weather, and temperature.
In this problem, the atmospheric pressure is given as 755 mmHg. This is the force per unit area exerted by the atmosphere around us.
By relating it to the height and density of the fluid column, we can find the height of the water column in the tube. This calculation involves equalizing the pressures exerted by the mercury and water columns, which relies on their densities to solve for the desired height.

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Most popular questions from this chapter

Does \(\mathrm{SF}_{6}\) (boiling point \(=16^{\circ} \mathrm{C}\) at 1 atm) behave more ideally at \(150^{\circ} \mathrm{C}\) or at \(20^{\circ} \mathrm{C}\) ? Explain.

Allotropes are different molecular forms of an element, such as dioxygen \(\left(\mathrm{O}_{2}\right)\) and ozone \(\left(\mathrm{O}_{3}\right) .\) (a) What is the density of each oxygen allotrope at \(0^{\circ} \mathrm{C}\) and 760 torr? (b) Calculate the ratio of densities, \(d_{\mathrm{O}_{3}} / d_{\mathrm{O}_{2}}\) and explain the significance of this number.

A large portion of metabolic energy arises from the biological combustion of glucose: $$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ (a) If this reaction is carried out in an expandable container at \(37^{\circ} \mathrm{C}\) and \(780 .\) torr, what volume of \(\mathrm{CO}_{2}\) is produced from \(20.0 \mathrm{~g}\) of glucose and excess \(\mathrm{O}_{2} ?\) (b) If the reaction is carried out at the same conditions with the stoichiometric amount of \(\mathrm{O}_{2}\), what is the partial pressure of each gas when the reaction is \(50 \%\) complete \((10.0 \mathrm{~g}\) of glucose remains)?

Hemoglobin is the protein that transports \(\mathrm{O}_{2}\) through the blood from the lungs to the rest of the body. To do so, each molecule of hemoglobin combines with four molecules of \(\mathrm{O}_{2}\). If \(1.00 \mathrm{~g}\) of hemoglobin combines with \(1.53 \mathrm{~mL}\) of \(\mathrm{O}_{2}\) at \(37^{\circ} \mathrm{C}\) and 743 torr, what is the molar mass of hemoglobin?

A mixture of \(\mathrm{CO}_{2}\) and \(\mathrm{Kr}\) weighs \(35.0 \mathrm{~g}\) and exerts a pressure of 0.708 atm in its container. since \(\mathrm{Kr}\) is expensive, you wish to recover it from the mixture. After the \(\mathrm{CO}_{2}\) is completely removed by absorption with \(\mathrm{NaOH}(s),\) the pressure in the container is 0.250 atm. How many grams of \(\mathrm{CO}_{2}\) were originally present? How many grams of Kr can you recover?
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