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Chapter 6: Problem 91

Explain why it is necessary to include the density of \(\mathrm{Hg}(1)\) and the value of the acceleration due to gravity, \(g,\) in a precise definition of a millimeter of mercury (page 194 ).

Short Answer

Expert verified
The density of mercury and the acceleration due to gravity (g) are crucial in defining a millimeter of mercury because they determine the pressure exerted by a 1 millimeter high column of mercury, which is what a 'millimeter of mercury' is defined as.

Step by step solution

01

Understand the concept of a millimeter of mercury

A millimeter of mercury (mmHg) is a unit of pressure, defined as the additional pressure that would be generated by a column of mercury exactly 1 millimeter high. It's an important unit in various scientific fields.
02

Discuss the role of mercury density

The density of an object is mass divided by volume. Denser substances will exert a greater pressure at the same height, because they have more mass in the same volume. Hence, the density of mercury is critical to calculating the pressure it will exert in a column of a given height. If we use a different substance with a different density, the resulting pressure (and thus, the value of one 'millimeter' of this substance) would be different.
03

Discuss the role of gravity

Gravity pulls things downwards, exerting a force on them. This is what causes fluid in a column (like our mercury) to exert pressure at the bottom of the column. More gravity would lead to a higher pressure, as the fluid is pulled down harder. Hence, the acceleration due to gravity (g) is critical in calculating how much pressure the mercury will exert.
04

Conclude with the necessity of both density and gravity

The exact pressure a column of mercury will exert (and thus, the definition of a mmHg) is directly dependent on both the density of mercury and the force exerted by gravity. Without these two values, it would be impossible to precisely define what a millimeter of mercury means in terms of pressure.

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Most popular questions from this chapter

A 2.650 g sample of a gaseous compound occupies \(428 \mathrm{mL}\) at \(24.3^{\circ} \mathrm{C}\) and \(742 \mathrm{mmHg} .\) The compound consists of \(15.5 \%\) C \(, 23.0 \%\) Cl, and \(61.5 \%\) F, by mass. What is its molecular formula?

Ammonium nitrite, \(\mathrm{NH}_{4} \mathrm{NO}_{2}\), decomposes according to the chemical equation below. $$\mathrm{NH}_{4} \mathrm{NO}_{2}(\mathrm{s}) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ What is the total volume of products obtained when \(128 \mathrm{g} \mathrm{NH}_{4} \mathrm{NO}_{2}\) decomposes at \(819^{\circ} \mathrm{C}\) and \(101 \mathrm{kPa} ?\)

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The amount of ozone, \(\mathrm{O}_{3}\), in a mixture of gases can be determined by passing the mixture through a solution of excess potassium iodide, KI. Ozone reacts with the iodide ion as follows: $$\begin{aligned} \mathrm{O}_{3}(\mathrm{g})+3 \mathrm{I}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1) & \longrightarrow \\ \mathrm{O}_{2}(\mathrm{g})+\mathrm{I}_{3}^{-}(\mathrm{aq}) &+2 \mathrm{OH}^{-}(\mathrm{aq}) \end{aligned}$$ The amount of \(I_{3}^{-}\) produced is determined by titrating with thiosulfate ion, \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}:\) $$\mathrm{I}_{3}^{-}(\mathrm{aq})+2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(\mathrm{aq}) \longrightarrow 3 \mathrm{I}^{-}(\mathrm{aq})+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}(\mathrm{aq})$$ A mixture of gases occupies a volume of \(53.2 \mathrm{L}\) at \(18^{\circ} \mathrm{C}\) and \(0.993 \mathrm{atm} .\) The mixture is passed slowly through a solution containing an excess of KI to ensure that all the ozone reacts. The resulting solution requires \(26.2 \mathrm{mL}\) of \(0.1359 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\) to titrate to the end point. Calculate the mole fraction of ozone in the original mixture.

A sounding balloon is a rubber bag filled with \(\mathrm{H}_{2}(\mathrm{g})\) and carrying a set of instruments (the payload). Because this combination of bag, gas, and payload has a smaller mass than a corresponding volume of air, the balloon rises. As the balloon rises, it expands. From the table below, estimate the maximum height to which a spherical balloon can rise given the mass of balloon, \(1200 \mathrm{g} ;\) payload, \(1700 \mathrm{g}\) : quantity of \(\mathrm{H}_{2}(\mathrm{g})\) in balloon, \(120 \mathrm{ft}^{3}\) at \(0.00^{\circ} \mathrm{C}\) and \(1.00 \mathrm{atm}\); diameter of balloon at maximum height, 25 ft. Air pressure and temperature as functions of altitude are: $$\begin{array}{ccl} \hline \text { Altitude, km } & \text { Pressure, mb } & \text { Temperature, } \mathrm{K} \\ \hline 0 & 1.0 \times 10^{3} & 288 \\ 5 & 5.4 \times 10^{2} & 256 \\ 10 & 2.7 \times 10^{2} & 223 \\ 20 & 5.5 \times 10^{1} & 217 \\ 30 & 1.2 \times 10^{1} & 230 \\ 40 & 2.9 \times 10^{0} & 250 \\ 50 & 8.1 \times 10^{-1} & 250 \\ 60 & 2.3 \times 10^{-1} & 256 \\ \hline \end{array}$$
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