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Chapter 10: Problem 18

(a) 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? \((\mathbf{b})\) What is the pressure, in atmospheres, on the body of a diver if he is $21 \mathrm{ft}$ below the surface of the water when the atmospheric pressure is 742 torr?

Short Answer

Expert verified
The height of a 1-iodododecane barometer column at atmospheric pressure of 749 torr is approximately \(8.48 \: \text{m}\). The pressure on a diver at 21 ft below the surface of the water when the atmospheric pressure is 742 torr is approximately \(1.596 \: \text{atm}\).

Step by step solution

01

(Part A: Convert atmospheric pressure to pascal)

: First, convert the given atmospheric pressure from torr to pascal using the conversion factor. \( P_\text{atm} = 749 \: \text{torr} \cdot \frac{133.33 \: \text{Pa}}{1 \: \text{torr}} = 99892.17 \: \text{Pa} \)
02

(Part A: Calculate the height of the 1-iodododecane column)

: Next, use the formula \(P = \rho g h\) to find the height (h) of the 1-iodododecane column. Rearranging for h, we get \( h = \frac{P_\text{atm}}{\rho_\text{ido} g} \) Density of 1-iodododecane, \(\rho_\text{ido} = 1.20 \: \text{g/mL} = 1200 \: \text{kg/m}^3\) Substitute the values: \(h = \frac{99892.17 \: \text{Pa}}{(1200 \: \text{kg/m}^3)(9.81 \: \text{m/s}^2)} = 8.48 \: \text{m}\) So, the height of the 1-iodododecane barometer column is \(8.48 \: \text{m}\).
03

(Part B: Convert feet to meters and torr to pascal)

: First, convert 21 feet to meters and atmospheric pressure to pascal. Depth of the diver, \(h = 21 \: \text{ft} \cdot \frac{1 \: \text{m}}{3.2808 \: \text{ft}} = 6.400 \: \text{m}\) Atmospheric pressure, \(P_\text{atm} = 742 \: \text{torr} \cdot \frac{133.33 \: \text{Pa}}{1 \: \text{torr}} = 98910.26 \: \text{Pa}\)
04

(Part B: Calculate the pressure exerted by the water column)

: Use the formula \(P = \rho_\text{w} g h\) to find the additional pressure exerted by the water column on the diver. Density of water, \(\rho_\text{w} = 1000 \: \text{kg/m}^3\) Pressure due to water, \(P_\text{w} = (1000 \: \text{kg/m}^3)(9.81 \: \text{m/s}^2)(6.400 \: \text{m}) = 62844 \: \text{Pa}\)
05

(Part B: Calculate the total pressure on the diver in atmospheres)

: Add the atmospheric and water pressure, then convert the total pressure to atmospheres. Total pressure, \(P_\text{total} = P_\text{atm} + P_\text{w} = 98910.26 \: \text{Pa} + 62844 \: \text{Pa} = 161754.26 \: \text{Pa}\) Convert to atmospheres, \(\frac{1 \: \text{atm}}{101325 \: \text{Pa}} \cdot P_\text{total} = \frac{161754.26 \: \text{Pa}}{101325 \: \text{Pa}} = 1.596 \: \text{atm}\) The pressure on the diver is \(1.596 \: \text{atm}\).

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Most popular questions from this chapter

A sample of \(3.00 \mathrm{~g}\) of \(\mathrm{SO}_{2}(g)\) originally in a \(5.00-\mathrm{L}\) vessel at \(21{ }^{\circ} \mathrm{C}\) is transferred to a \(10.0-\mathrm{L}\) vessel at \(26^{\circ} \mathrm{C}\). A sample of $2.35 \mathrm{~g}\( of \)\mathrm{N}_{2}(g)\( originally in a \)2.50-\mathrm{L}$ vessel at \(20^{\circ} \mathrm{C}\) is transferred to this same \(10.0-\mathrm{L}\) vessel. \((\mathbf{a})\) What is the partial pressure of \(\mathrm{SO}_{2}(g)\) in the larger container? (b) What is the partial pressure of \(\mathrm{N}_{2}(g)\) in this vessel? (c) What is the total pressure in the vessel?

In the contact process, sulfur dioxide and oxygen gas react to form sulfur trioxide as follows: $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)$$ At a certain temperature and pressure, \(50 \mathrm{~L}\) of \(\mathrm{SO}_{2}\) reacts with \(25 \mathrm{~L}\) of \(\mathrm{O}_{2}\). If all the \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) are consumed, what volume of \(\mathrm{SO}_{3}\), at the same temperature and pressure, will be produced?

If \(5.15 \mathrm{~g}\) of \(\mathrm{Ag}_{2} \mathrm{O}\) is sealed in a \(75.0-\mathrm{mL}\) tube filled with 101.3 \(\mathrm{kPa}\) of \(\mathrm{N}_{2}\) gas at \(32{ }^{\circ} \mathrm{C},\) and the tube is heated to $320^{\circ} \mathrm{C}\(, the \)\mathrm{Ag}_{2} \mathrm{O}$ decomposes to form oxygen and silver. What is the total pressure inside the tube assuming the volume of the tube remains constant?

Cyclopropane, a gas used with oxygen as a general anesthetic, is composed of \(85.7 \% \mathrm{C}\) and \(14.3 \% \mathrm{H}\) by mass. \((\mathbf{a})\) If $1.56 \mathrm{~g}\( of cyclopropane has a volume of \)1.00 \mathrm{~L}$ at 99.7 \(\mathrm{kPa}\) and \(50.0^{\circ} \mathrm{C}\), what is the molecular formula of cyclopropane? (b) Judging from its molecular formula, would you expect cyclopropane to deviate more or less than Ar from ideal-gas behavior at moderately high pressures and room temperature? Explain. (c) Would cyclopropane effuse through a pinhole faster or more slowly than methane, \(\mathrm{CH}_{4} ?\)

At constant pressure, the mean free path \((\lambda)\) of a gas molecule is directly proportional to temperature. At constant temperature, \(\lambda\) is inversely proportional to pressure. If you compare two different gas molecules at the same temperature and pressure, \(\lambda\) is inversely proportional to the square of the diameter of the gas molecules. Put these facts together to create a formula for the mean free path of a gas molecule with a proportionality constant (call it \(R_{\mathrm{mfp}}\), like the ideal-gas constant) and define units for \(R_{\mathrm{mfp}^{-}}\)
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