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Calculate the height of a column of ethylene glycol \(\left[\mathrm{CH}_{2}(\mathrm{OH}) \mathrm{CH}_{2}(\mathrm{OH})\right]\) that would be supported by atmospheric pressure ( 1 atm). The density of ethylene glycol is \(1.12 \mathrm{~g} / \mathrm{cm}^{3}\)

Short Answer

Expert verified
The height of the ethylene glycol column is approximately 9.22 meters.

Step by step solution

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01

Understand the Problem

We need to determine the height of an ethylene glycol column that corresponds to atmospheric pressure, which is 1 atm. We have been given the density of ethylene glycol as \(1.12 \text{ g/cm}^3\). Atmospheric pressure can also be expressed as \(101,325 \text{ Pa}\).
02

Convert Given Quantities to Compatible Units

Since we are working with pressure in pascals, we convert the density to \(\text{ kg/m}^3\):\[1.12 \text{ g/cm}^3 = 1.12 \times 1000 \text{ kg/m}^3 = 1120 \text{ kg/m}^3\]
03

Use the Pressure-Height Equation

The height \(h\) of the liquid column can be calculated using the equation:\[P = \rho g h\]where \(P\) is the pressure (\(101,325 \text{ Pa}\)), \(\rho\) is the density (\(1120 \text{ kg/m}^3\)), and \(g\) is the acceleration due to gravity (\(9.81 \text{ m/s}^2\)).
04

Solve for Height

Rearrange the formula to find \(h\):\[h = \frac{P}{\rho g}\]Substitute the values:\[h = \frac{101,325}{1120 \times 9.81} \approx \frac{101,325}{10,987.2} \approx 9.22 \text{ m}\]
05

Conclusion

The height of the ethylene glycol column supported by atmospheric pressure is approximately \(9.22 \text{ meters}\).

Key Concepts

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

Understanding Ethylene Glycol
Ethylene glycol is a common organic compound widely used in antifreeze and deicing solutions. Its chemical formula is \( ext{CH}_2( ext{OH}) ext{CH}_2( ext{OH})\). Because of its chemical structure, ethylene glycol is hydrophilic, meaning it mixes well with water. This property is important for its applications in environments where liquid-based heat transfer is needed. In calculations related to physics or engineering, knowing the density of the substance plays a crucial role. For ethylene glycol, its density is given as \(1.12 ext{ g/cm}^3\). Knowing this helps us connect the intrinsic properties of ethylene glycol with external factors like pressure.
The Process of Density Conversion
In order to perform calculations that involve units from different measurement systems, such as the metric and SI systems, it's essential to convert quantities into compatible units. Density is usually given in \( ext{g/cm}^3\), especially in chemistry, but for physics calculations involving pressure and height, the SI unit \( ext{kg/m}^3\) is more suitable.
  • To convert \(1.12 ext{ g/cm}^3\) to \( ext{kg/m}^3\), multiply by 1000.
  • This is because there are 1000 grams in a kilogram and each side of a meter cube has 100 centimeters.
  • Thus, the conversion leads to \(1120 ext{ kg/m}^3\).
This conversion step is crucial to ensure all quantities relate correctly in the pressure calculation formula.
Applying the Pressure-Height Equation
The pressure-height equation is fundamental in fluid dynamics and can help determine various properties of liquids in a column. The general equation is \[ P = \rho g h \]where:
  • \(P\) is pressure, measured in pascals \(\text{Pa}\).
  • \(\rho\) is the density of the fluid, in \(\text{kg/m}^3\).
  • \(g\) is the acceleration due to gravity, approximately \(9.81 \text{ m/s}^2\).
  • \(h\) is the height of the liquid column, in meters.
To find the height of the ethylene glycol column that atmospheric pressure can support, rearrange the equation to solve for \(h\): \[ h = \frac{P}{\rho g} \]This formula allows us to see how changes in density would directly affect the height of the liquid column supported by a given pressure.
Understanding Atmospheric Pressure
Atmospheric pressure is the force exerted onto a surface by the weight of the air above it. At sea level, it is commonly measured as 1 atmosphere (atm). For scientific calculations, it is often converted into pascals (\(101,325\) Pa). Understanding atmospheric pressure is vital because it acts as a baseline pressure that counterbalances the pressure exerted by columns of liquid or other fluids.
  • It affects everything from weather patterns to boiling points of liquids.
  • In the context of the given problem, atmospheric pressure is what 'holds up' the column of ethylene glycol.
  • Knowing this baseline helps us calculate the specific column height a certain pressure could support.
Recognizing how atmospheric pressure interacts with fluids allows us to predict and measure the behavior of liquid columns under different conditions.

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Most popular questions from this chapter

A mixture of gases contains \(0.31 \mathrm{~mol} \mathrm{CH}_{4}, 0.25 \mathrm{~mol}\) \(\mathrm{C}_{2} \mathrm{H}_{6}\), and \(0.29 \mathrm{~mol} \mathrm{C}_{3} \mathrm{H}_{8}\). The total pressure is \(1.50 \mathrm{~atm} .\) Calculate the partial pressures of the gases.

An unknown gas evolved from the fermentation of glucose is found to effuse through a porous barrier in 15.0 min. Under the same conditions of temperature and pressure, it takes an equal volume of \(\mathrm{N}_{2} 12.0 \mathrm{~min}\) to effuse through the same barrier. Calculate the molar mass of the unknown gas, and suggest what the gas might be.

What are the basic assumptions of the kinetic molecular theory of gases?

The apparatus shown here can be used to measure atomic and molecular speeds. Suppose that a beam of metal atoms is directed at a rotating cylinder in a vacuum. A small opening in the cylinder allows the atoms to strike a target area. Because the cylinder is rotating, atoms traveling at different speeds will strike the target at different positions. In time, a layer of the metal will deposit on the target area, and the variation in its thickness is found to correspond to Maxwell's speed distribution. In one experiment it is found that at \(850^{\circ} \mathrm{C}\) some bismuth (Bi) atoms struck the target at a point \(2.80 \mathrm{~cm}\) from the spot directly opposite the slit. The diameter of the cylinder is \(15.0 \mathrm{~cm},\) and it is rotating at 130 revolutions per second. (a) Calculate the speed (in \(\mathrm{m} / \mathrm{s}\) ) at which the target is moving. (Hint: The circumference of a circle is given by \(2 \pi r\), where \(r\) is the radius.) (b) Calculate the time (in seconds) it takes for the target to travel \(2.80 \mathrm{~cm} .\) (c) Determine the speed of the \(\mathrm{Bi}\) atoms. Compare your result in part (c) with the \(u_{\mathrm{rms}}\) of \(\mathrm{Bi}\) at \(850^{\circ} \mathrm{C}\). Comment on the difference.

A student breaks a thermometer and spills most of the mercury (Hg) onto the floor of a laboratory that measures \(15.2 \mathrm{~m}\) long, \(6.6 \mathrm{~m}\) wide, and \(2.4 \mathrm{~m}\) high. (a) Calculate the mass of mercury vapor (in grams) in the room at \(20^{\circ} \mathrm{C}\). The vapor pressure of mercury at \(20^{\circ} \mathrm{C}\) is \(1.7 \times 10^{-6}\) atm. (b) Does the concentration of mercury vapor exceed the air quality regulation of \(0.050 \mathrm{mg} \mathrm{Hg} / \mathrm{m}^{3}\) of air? (c) One way to deal with small quantities of spilled mercury is to spray sulfur powder over the metal. Suggest a physical and a chemical reason for this action.

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