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Chapter 5: Problem 23

If a barometer were built using water \(\left(d=1.0 \mathrm{g} / \mathrm{cm}^{3}\right)\) instead of mercury \(\left(d=13.6 \mathrm{g} / \mathrm{cm}^{3}\right)\) , would the column of water be higher than, lower than, or the same as the column of mercury at 1.00 atm ? If the level is different, by what factor? Explain.

Short Answer

Expert verified
A barometer built with water instead of mercury would have a higher water column at 1.00 atm. The water column would be around 10.33 m, while the mercury column would be approximately 0.76 m. The height of the water column would be 13.6 times higher than the height of the mercury column.

Step by step solution

01

Determine the hydrostatic pressure for both columns

Hydrostatic pressure can be calculated using the formula: \(P = \rho gh \) Where \(P\) is the pressure, \(\rho\) is the density of the liquid, \(g\) is the acceleration due to gravity, and \(h\) is the height of the liquid column. For water: Density (\(\rho_w\)) = 1.00 g/cm鲁 For mercury: Density (\(\rho_m\)) = 13.6 g/cm鲁 At 1.00 atm, the pressure (\(P\)) = 1.00 脳 101325 Pa = 1013.25 hPa (1 m = 100 cm)
02

Find the height of the water column

Using the formula for hydrostatic pressure, we solve for the height of the water column (\(h_w\)): \( P = \rho_w g h_w \) \( h_w = \frac{P}{\rho_w g} \) Converting the units for \(\rho_w\) into kg/m鲁: \( \rho_w = \frac{1000}{1000} \times 1000 \, kg/m鲁 = 1000 \, kg/m鲁\) Now, we can find the height of the water column: \( h_w = \frac{101325 \, Pa}{1000 \, kg/m鲁 \times 9.81 \, m/s虏} \) \( h_w \approx 10.33\, m \)
03

Find the height of the mercury column

Using the same hydrostatic pressure formula, we solve for the height of the mercury column (\(h_m\)): \( P = \rho_m g h_m \) \( h_m = \frac{P}{\rho_m g} \) Converting the units for \(\rho_m\) into kg/m鲁: \( \rho_m = \frac{13.6}{1000} \times 1000 \times 1000 \times \frac{1}{1000} \, kg/m鲁 = 13600 \, kg/m鲁\) Now, we can find the height of the mercury column: \( h_m = \frac{101325 \, Pa}{13600 \, kg/m鲁 \times 9.81 \, m/s虏} \) \( h_m \approx 0.76\, m \)
04

Compare the heights of the two columns and find the factor

Comparing the heights of the two columns, we find: \( h_w \approx 10.33\, m \) (Water column height) \( h_m \approx 0.76\, m \) (Mercury column height) The height of the water column is higher than the height of the mercury column, so the column of water would be higher at 1.00 atm. Now, let's find the factor by which the height of the water column differs from that of the mercury column: Factor = \(\frac{h_w}{h_m}\) Factor 鈮� \(\frac{10.33 \, m}{0.76 \, m}\) Factor 鈮� 13.6
05

Conclusion

A barometer built with water instead of mercury would have a column of water that is higher than the column of mercury, at 1.00 atm. The height of the water column would be 13.6 times higher than the height of the mercury column.

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Key Concepts

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

Density of Liquids
Density is a fundamental concept that describes how much mass is contained in a given volume for a substance. It plays a key role in determining how different liquids behave under pressure.

The density of a liquid can be calculated using the formula:
  • \( \rho = \frac{m}{V} \)
where \( \rho \) is the density, \( m \) is mass, and \( V \) is volume.

Water, for example, has a density of about 1.00 g/cm鲁, which is much lower than the density of mercury at 13.6 g/cm鲁. This difference in density affects the height of liquid columns in barometers, as less dense liquids like water require a taller column to exert the same pressure as more dense liquids like mercury.
Barometer
A barometer is an instrument used to measure atmospheric pressure. By observing the height of a liquid column, typically mercury in traditional barometers, it indicates changes in atmospheric conditions.

In a barometer:
  • Higher liquid levels indicate lower atmospheric pressure.
  • Lower liquid levels suggest higher atmospheric pressure.
If a barometer used water instead of mercury, the water column would be significantly taller due to its lower density. This means the barometer would need to be much larger to accommodate the height.
Atmospheric Pressure
Atmospheric pressure is the force exerted by the weight of the atmosphere above us. It affects everything around us and can be measured using barometers.

At sea level, standard atmospheric pressure is about 101325 Pa (Pascals) or 1 atm (atmosphere).

Atmospheric pressure is responsible for holding the column of liquid in a barometer.
  • In mercury barometers, the atmospheric pressure keeps the mercury at about 0.76 m height.
  • In water barometers, the lower density results in a much taller column height, around 10.33 m, under the same pressure conditions.
Fluid Mechanics
Fluid mechanics is the study of fluids and how they move. It helps us understand concepts like hydrostatic pressure, which is the pressure exerted by a fluid at rest.

In the context of a barometer, the hydrostatic pressure equation:
  • \( P = \rho gh \)
allows us to calculate the pressure exerted by the column of liquid. Here, \( P \) is pressure, \( \rho \) is density, \( g \) is gravitational acceleration, and \( h \) is the height of the fluid column.

This equation demonstrates why less dense liquids need a taller column to exert the same pressure as denser ones. Understanding these principles is vital in many areas of engineering and science.

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Most popular questions from this chapter

Consider the unbalanced chemical equation below: $$\mathrm{CaSiO}_{3}(s)+\mathrm{HF}(g) \longrightarrow \mathrm{CaF}_{2}(a q)+\mathrm{SiF}_{4}(g)+\mathrm{H}_{2} \mathrm{O}( )$$ Suppose a 32.9 -g sample of \(\mathrm{CaSiO}_{3}\) is reacted with 31.8 \(\mathrm{L}\) of \(\mathrm{HF}\) at \(27.0^{\circ} \mathrm{C}\) and 1.00 \(\mathrm{atm}\) . Assuming the reaction goes to completion, calculate the mass of the \(\mathrm{SiF}_{4}\) and \(\mathrm{H}_{2} \mathrm{O}\) produced in the reaction.

Calculate the pressure exerted by 0.5000 mole of \(\mathrm{N}_{2}\) in a 10.000 -L container at \(25.0^{\circ} \mathrm{C}\) a. using the ideal gas law. b. using the van der Waals equation. c. Compare the results. d. Compare the results with those in Exercise 123.

Metallic molybdenum can be produced from the mineral molybdenite, \(\mathrm{MoS}_{2}\). The mineral is first oxidized in air to molybdenum trioxide and sulfur dioxide. Molybdenum trioxide is then reduced to metallic molybdenum using hydrogen gas. The balanced equations are $$\operatorname{MoS}_{2}(s)+\frac{7}{2} \mathrm{O}_{2}(g) \longrightarrow \operatorname{MoO}_{3}(s)+2 \mathrm{SO}_{2}(g)$$ $$\mathrm{MoO}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{Mo}(s)+3 \mathrm{H}_{2} \mathrm{O}(l)$$ Calculate the volumes of air and hydrogen gas at \(17^{\circ} \mathrm{C}\) and 1.00 atm that are necessary to produce \(1.00 \times 10^{3} \mathrm{kg}\) pure molybdenum from \(\mathrm{MoS}_{2}\) . Assume air contains 21\(\%\) oxygen by volume, and assume 100\(\%\) yield for each reaction.

Silicon tetrachloride \(\left(\mathrm{SiCl}_{4}\right)\) and trichlorosilane \(\left(\mathrm{SiHCl}_{3}\right)\) are both starting materials for the production of electronics-grade silicon. Calculate the densities of pure \(\mathrm{SiCl}_{4}\) and pure \(\mathrm{SiHCl}_{4}\) vapor at \(85^{\circ} \mathrm{C}\) and 635 torr.

Consider the following chemical equation. $$2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ If 25.0 \(\mathrm{mL}\) \(\mathrm{NO}_{2}\) gas is completely converted to \(\mathrm{N}_{2} \mathrm{O}_{4}\) gas under the same conditions, what volume will the \(\mathrm{N}_{2} \mathrm{O}_{4}\) occupy?
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