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Chapter 10: Problem 17

(a) How high in meters must a column of glycerol be to exert a pressure equal to that of a \(760-\mathrm{mm}\) column of mercury? The density of glycerol is 1.26 \(\mathrm{g} / \mathrm{mL}\) , whereas that of mercury is 13.6 \(\mathrm{g} / \mathrm{mL}\) . (b) What pressure, in atmospheres, is exerted on the body of a diver if she is 15 ft below the surface of the water when the atmospheric pressure is 750 torr? Assume that the density of the water is \(1.00 \mathrm{g} / \mathrm{cm}^{3}=1.00 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3} .\) The gravitational constant is \(9.81 \mathrm{m} / \mathrm{s}^{2},\) and \(1 \mathrm{Pa}=1 \mathrm{kg} / \mathrm{m}-\mathrm{s}^{2} .\)

Short Answer

Expert verified
(a) To find the height of the glycerol column exerting the same pressure as a 760-mm column of mercury, we use the formula \( h_{glycerol} = \frac{Pressure_{Hg}}{Density_{glycerol} \times g} \). After calculation, we get \( h_{glycerol} \approx 8.17 \mathrm{m} \). (b) The total pressure exerted on the diver is the sum of the atmospheric pressure and the pressure due to the water column. Calculating the total pressure, we get \(Total Pressure_{atm} \approx 1.62 \) atmospheres.

Step by step solution

01

(a) Calculate the pressure of a 760-mm column of mercury

To find the pressure of a 760-mm column of mercury, we can use the formula, \( Pressure = Density \times g \times h \), where h is the height of the column in meters. Convert 760 mm to meters: \( h_{Hg} = 760 \frac{mm}{1} \times \frac{1m}{1000mm} = 0.76 \mathrm{m} \). So the pressure of the mercury column can be calculated as \( Pressure_{Hg} = 13.6 \frac{g}{mL} \times \frac{1000g}{1kg} \times 9.81 \frac{m}{s^{2}} \times 0.76 m \).
02

(a) Calculate the height of the glycerol column

We need to set the pressure of the glycerol column equal to the pressure of the mercury column. The equation becomes, \(Pressure_{Hg} = Density_{glycerol} \times g \times h_{glycerol}\). Rearrange the equation to solve for \(h_{glycerol}\): \( h_{glycerol} = \frac{Pressure_{Hg}}{Density_{glycerol} \times g} \). Substitute the known values and solve for \(h_{glycerol}\).
03

(b) Convert 15 ft to meters

To find the pressure exerted on the body of the diver, we first need to convert the given height from feet to meters. \(height_{water} = 15 \frac{ft}{1} \times \frac{0.3048m}{1ft} = 4.572 \mathrm{m}\).
04

(b) Convert 750 torr to Pascals

To add the atmospheric pressure to the pressure due to the water column, we need to convert 750 torr into Pascals. \(Pressure_{atm} = 750 \frac{torr}{1} \times \frac{101325 Pa}{760 torr} \).
05

(b) Calculate the pressure due to the water column

Using the formula for pressure, \( Pressure_{water} = Density_{water} \times g \times height_{water} \), we can calculate the pressure exerted by the water column. \( Pressure_{water} = 1000 \frac{kg}{m^3} \times 9.81 \frac{m}{s^2} \times 4.572 m \).
06

(b) Find the total pressure on the diver

To find the total pressure exerted on the diver, we need to add the atmospheric pressure to the pressure due to the water column: \(Total Pressure = Pressure_{atm} + Pressure_{water}\).
07

(b) Convert total pressure to atmospheres

Finally, we want the total pressure in atmospheres. Convert the total pressure from Pascals to atmospheres using the conversion factor: \( Total Pressure_{atm} = \frac{Total Pressure}{101325 Pa} \).

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Most popular questions from this chapter

Nitrogen and hydrogen gases react to form ammonia gas as follows: $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)$$ At a certain temperature and pressure, 1.2 \(\mathrm{L}\) of \(\mathrm{N}_{2}\) reacts with 3.6 \(\mathrm{Lof} \mathrm{H}_{2} .\) If all the \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) are consumed, what volume of \(\mathrm{NH}_{3},\) at the same temperature and pressure, will be produced?

A scuba diver's tank contains 0.29 \(\mathrm{kg}\) of \(\mathrm{O}_{2}\) compressed into a volume of 2.3 \(\mathrm{L}\) . (a) Calculate the gas pressure inside the tank at \(9^{\circ} \mathrm{C} (\mathbf{b})\) What volume would this oxygen occupy at \(26^{\circ} \mathrm{C}\) and 0.95 atm?

(a) What conditions are represented by the abbreviation STP? (b) What is the molar volume of an ideal gas at STP? (c) Room temperature is often assumed to be \(25^{\circ} \mathrm{C}\) . Calculate the molar volume of an ideal gas at \(25^{\circ} \mathrm{C}\) and 1 atm pressure. (d) If you measure pressure in bars instead of atmospheres, calculate the corresponding value of \(R\) in L-bar/mol-K.

The density of a gas of unknown molar mass was measured as a function of pressure at \(0^{\circ} \mathrm{C},\) as in the table that follows. (a) Determine a precise molar mass for the gas. [Hint: Graph \(d / P\) versus \(P . ](\mathbf{b})\) Why is \(d / P\) not a constant as a function of pressure? $$\begin{array}{lllll}{\text { Pressure (atm) }} & {1.00} & {0.666} & {0.500} & {0.333} & {0.250} \\ \hline \text { Density (g/L) } & {2.3074} & {1.5263} & {1.1401} & {0.7571} & {0.5660}\end{array}$$

A sample of 5.00 \(\mathrm{mL}\) of diethylether \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OC}_{2} \mathrm{H}_{5},\right.\) density \(=0.7134 \mathrm{g} / \mathrm{mL}\) ) is introduced into a 6.00 -L vessel that already contains a mixture of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2},\) whose partial pressures are \(P_{\mathrm{N}_{2}}=0.751 \mathrm{atm}\) and \(P_{\mathrm{O}_{2}}=0.208\) atm. The temperature is held at \(35.0^{\circ} \mathrm{C},\) and the diethylether totally evaporates. (a) Calculate the partial pressure of the diethylether. (b) Calculate the total pressure in the container.
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