Problem 18
Question
The compound 1 -iodododecane is a nonvolatile liquid with a density of \(1.20 \mathrm{~g} / \mathrm{mL}\). The density of mercury is \(13.6 \mathrm{~g} / \mathrm{mL}\). What do you predict for the height of a barometer column based on 1 -iodododecane, when the atmospheric pressure is 749 torr?
Step-by-Step Solution
Verified Answer
The height of a barometer column based on 1-iodododecane when the atmospheric pressure is 749 torr can be predicted to be approximately 67.0 millimeters.
1Step 1: Convert the atmospheric pressure to pascals
Since we are given the atmospheric pressure in torr, we need to convert it to pascals (Pa) to keep our units consistent. We know that 1 torr = 133.322 Pa. So we can convert 749 torr to pascals as follows:
\(pressure_{atmosphere\_Pa} = pressure_{atmosphere\_torr} \times 133.322\)
\(pressure_{atmosphere\_Pa} = 749 \times 133.322 = 99884.778\,Pa\)
2Step 2: Calculate the height of the barometer column
Now, we can use the formula mentioned in the analysis to calculate the height of the 1-iodododecane column.
\(height = \frac{density_{substance} \times atmospheric\_pressure}{density_{mercury} \times pressure_{substance}}\)
\(height = \frac{1.20 \, g/mL \times 99884.778\, Pa}{13.6 \, g/mL \times 99884.778\, Pa}\)
Since the pressure of 1-iodododecane is equal to the atmospheric pressure, we can cancel out the pressure values in the equation:
\(height = \frac{1.20 \, g/mL}{13.6 \, g/mL}\)
3Step 3: Solve for the height of the barometer column
Now, we simply solve for the height of the 1-iodododecane column:
\(height = \frac{1.20}{13.6}\)
\(height = 0.0882\)
Since the result is in a dimensionless format, we need to multiply it by the height of the mercury column at standard atmospheric pressure (760 mmHg) to get the height of the 1-iodododecane column in millimeters:
\(height_{iodododecane} = height \times 760\, mmHg\)
\(height_{iodododecane} = 0.0882 \times 760\)
\(height_{iodododecane} = 67.032\,mm\)
We can approximate the height of the 1-iodododecane column to be approximately 67.0 millimeters when the atmospheric pressure is 749 torr.
Key Concepts
Atmospheric Pressure ConversionDensity ComparisonPascals and TorrNonvolatile Liquids
Atmospheric Pressure Conversion
Atmospheric pressure is often measured in different units around the world. Two common ones are torr and pascals (Pa). Converting one unit to another ensures accuracy in calculations. To change atmospheric pressure from torr to pascals, we use the conversion factor where 1 torr equals 133.322 pascals.
A practical example is provided in the exercise. Here, 749 torr is converted using the formula:
A practical example is provided in the exercise. Here, 749 torr is converted using the formula:
- Multiply 749 by 133.322.
- Resulting in 99884.778 pascals.
Density Comparison
Density plays a key role in understanding the behavior of substances in varying pressures. It is the mass per unit volume, often expressed in grams per milliliter (g/mL).
In the exercise, the densities of two different liquids are compared:
In the exercise, the densities of two different liquids are compared:
- 1-Iodododecane has a density of 1.20 g/mL.
- Mercury has a higher density of 13.6 g/mL.
Pascals and Torr
Pascals (Pa) and torr are units of pressure, and understanding their relationship is essential for scientific calculations. Pascals are the SI unit used globally for scientific work, whereas torr often appear in meteorology and some other fields.
To compare them, remember:
To compare them, remember:
- 1 torr = 133.322 pascals.
Nonvolatile Liquids
Nonvolatile liquids, such as the 1-iodododecane mentioned in the exercise, do not easily vaporize at room temperature. This characteristic makes them ideal for certain scientific applications, including barometry.
Key points about nonvolatile liquids include:
Key points about nonvolatile liquids include:
- They maintain their volume under varying atmospheric conditions.
- They provide a stable medium for accurate pressure measurements.
Other exercises in this chapter
Problem 14
(a) A liquid and a gas are moved to larger containers. How does their behavior differ once they are in the larger containers? Explain the difference in molecula
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Suppose that a woman weighing \(130 \mathrm{lb}\) and wearing highheeled shoes momentarily places all her weight on the heel of one foot. If the area of the hee
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The typical atmospheric pressure on top of Mt. Everest \((29,028 \mathrm{ft})\) is about 265 torr. Convert this pressure to (a) atm, b) \(\mathrm{mm} \mathrm{Hg
View solution Problem 22
Perform the following conversions: (a) 0.912 atm to torr, (b) 0.685 bar to kilopascals, (c) \(655 \mathrm{~mm}\) Hg to atmospheres, (d) \(1.323 \times 10^{5}\)
View solution