762. Exercises - 762.4. Reaction Yields
76. The phosphorus pentoxide used to produce phosphoric acid for cola soft drinks is prepared by burning phosphorus in oxygen.
(a) What is the limiting reactant when 0.200 mol of $\mathrm{P}_{4}$ and 0.200 mol of $\mathrm{O}_{2}$ react according to
$\mathrm{P}_{4}+5 \mathrm{O}_{2} \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}$
(b) Calculate the percent yield if 10.0 g of $\mathrm{P}_{4} \mathrm{O}_{10}$ is isolated from the reaction.
77. Would you agree to buy 1 trillion $(1,000,000,000,000)$ gold atoms for $\$ 5$ ? Explain why or why not. Find the current price of gold at http://money.cnn.com/data/commodities/ ( 1 troy ounce $=31.1 \mathrm{~g}$ )


762. Exercises - 762.5. Quantitative Chemical Analysis
78. What volume of $0.0105-\mathrm{M} \mathrm{HBr}$ solution is required to titrate 125 mL of a $0.0100-\mathrm{MCa}(\mathrm{OH})_{2}$ solution? $\mathrm{Ca}(\mathrm{OH})_{2}(a q)+2 \mathrm{HBr}(a q) \longrightarrow \mathrm{CaBr}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)$
79. Titration of a $20.0-\mathrm{mL}$ sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidity of the rain is due to the presence of sulfuric acid, what was the concentration of sulfuric acid in this sample of rain?
80. What is the concentration of NaCl in a solution if titration of 15.00 mL of the solution with $0.2503 \mathrm{M} \mathrm{AgNO}_{3}$ requires 20.22 mL of the $\mathrm{AgNO}_{3}$ solution to reach the end point?
$\mathrm{AgNO}_{3}(a q)+\mathrm{NaCl}(a q) \longrightarrow \mathrm{AgCl}(s)+\mathrm{NaNO}_{3}(a q)$
81. In a common medical laboratory determination of the concentration of free chloride ion in blood serum, a serum sample is titrated with a $\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}$ solution.
$2 \mathrm{Cl}^{-}(a q)+\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}(a q) \longrightarrow 2 \mathrm{NO}_{3}{ }^{-}(a q)+\mathrm{HgCl}_{2}(s)$
What is the $\mathrm{Cl}^{-}$concentration in a $0.25-\mathrm{mL}$ sample of normal serum that requires 1.46 mL of $8.25 \times 10^{-4}$ $M \mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})$ to reach the end point?


762. Exercises - 762.5. Quantitative Chemical Analysis
82. Potatoes can be peeled commercially by soaking them in a $3-M$ to $6-M$ solution of sodium hydroxide, then removing the loosened skins by spraying them with water. Does a sodium hydroxide solution have a suitable concentration if titration of 12.00 mL of the solution requires 30.6 mL of 1.65 M HCI to reach the end point?
83. A sample of gallium bromide, $\mathrm{GaBr}_{3}$, weighing 0.165 g was dissolved in water and treated with silver nitrate, $\mathrm{AgNO}_{3}$, resulting in the precipitation of 0.299 g AgBr . Use these data to compute the \%Ga (by mass) $\mathrm{GaBr}_{3}$.
84. The principal component of mothballs is naphthalene, a compound with a molecular mass of about 130 amu, containing only carbon and hydrogen. A $3.000-\mathrm{mg}$ sample of naphthalene burns to give 10.3 mg of $\mathrm{CO}_{2}$. Determine its empirical and molecular formulas.
85. A $0.025-\mathrm{g}$ sample of a compound composed of boron and hydrogen, with a molecular mass of $\sim 28 \mathrm{amu}$, burns spontaneously when exposed to air, producing 0.063 g of $\mathrm{B}_{2} \mathrm{O}_{3}$. What are the empirical and molecular formulas of the compound?


762. Exercises - 762.5. Quantitative Chemical Analysis
86. Sodium bicarbonate (baking soda), $\mathrm{NaHCO}_{3}$, can be purified by dissolving it in hot water $\left(60^{\circ} \mathrm{C}\right)$, filtering to remove insoluble impurities, cooling to $0^{\circ} \mathrm{C}$ to precipitate solid $\mathrm{NaHCO}_{3}$, and then filtering to remove the solid, leaving soluble impurities in solution. Any $\mathrm{NaHCO}_{3}$ that remains in solution is not recovered. The solubility of $\mathrm{NaHCO}_{3}$ in hot water of $60^{\circ} \mathrm{C}$ is $164 \mathrm{~g} / \mathrm{L}$. Its solubility in cold water of $0{ }^{\circ} \mathrm{C}$ is $69 \mathrm{~g} / \mathrm{L}$. What is the percent yield of $\mathrm{NaHCO}_{3}$ when it is purified by this method?
87. What volume of 0.600 MHCl is required to react completely with 2.50 g of sodium hydrogen carbonate? $\mathrm{NaHCO}_{3}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{NaCl}(a q)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$
88. What volume of $0.08892 \mathrm{MHNO}_{3}$ is required to react completely with 0.2352 g of potassium hydrogen phosphate?
$2 \mathrm{HNO}_{3}(a q)+\mathrm{K}_{2} \mathrm{HPO}_{4}(a q) \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(a q)+2 \mathrm{KNO}_{3}(a q)$
89. What volume of a $0.3300-M$ solution of sodium hydroxide would be required to titrate 15.00 mL of 0.1500 $M$ oxalic acid?


762. Exercises - 762.5. Quantitative Chemical Analysis
89. What volume of a $0.3300-M$ solution of sodium hydroxide would be required to titrate 15.00 mL of 0.1500 $M$ oxalic acid?
$\mathrm{C}_{2} \mathrm{O}_{4} \mathrm{H}_{2}(a q)+2 \mathrm{NaOH}(a q) \longrightarrow \mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)$
90. What volume of a $0.00945-M$ solution of potassium hydroxide would be required to titrate 50.00 mL of a sample of acid rain with a $\mathrm{H}_{2} \mathrm{SO}_{4}$ concentration of $1.23 \times 10^{-4} \mathrm{M}$.
$\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{KOH}(a q) \longrightarrow \mathrm{K}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)$
91. A sample of solid calcium hydroxide, $\mathrm{Ca}(\mathrm{OH})_{2}$, is allowed to stand in water until a saturated solution is formed. A titration of 75.00 mL of this solution with $5.00 \times 10^{-2} \mathrm{M} \mathrm{HCl}$ requires 36.6 mL of the acid to reach the end point.
$\mathrm{Ca}(\mathrm{OH})_{2}(a q)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{CaCl}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)$
What is the molarity?
92. What mass of $\mathrm{Ca}(\mathrm{OH})_{2}$ will react with 25.0 g of butanoic to form the preservative calcium butanoate according to the equation?


762. Exercises - 762.5. Quantitative Chemical Analysis
93. How many milliliters of a $0.1500-M$ solution of KOH will be required to titrate 40.00 mL of a $0.0656-M$ solution of $\mathrm{H}_{3} \mathrm{PO}_{4}$ ?
$\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+2 \mathrm{KOH}(a q) \longrightarrow \mathrm{K}_{2} \mathrm{HPO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)$
94. Potassium hydrogen phthalate, $\mathrm{KHC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}$, or KHP , is used in many laboratories, including general chemistry laboratories, to standardize solutions of base. KHP is one of only a few stable solid acids that can be dried by warming and weighed. A $0.3420-\mathrm{g}$ sample of $\mathrm{KHC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}$ reacts with 35.73 mL of a NaOH solution in a titration. What is the molar concentration of the NaOH ?
$\mathrm{KHC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}(a q)+\mathrm{NaOH}(a q) \longrightarrow \mathrm{KNaC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(a q)$
95. The reaction of $\mathrm{WCl}_{6}$ with Al at $\sim 400^{\circ} \mathrm{C}$ gives black crystals of a compound containing only tungsten and chlorine. A sample of this compound, when reduced with hydrogen, gives 0.2232 g of tungsten metal and hydrogen chloride, which is absorbed in water. Titration of the hydrochloric acid thus produced requires 46.2 mL of 0.1051 M NaOH to reach the end point. What is the empirical formula of the black tungsten chloride?


763. CHAPTER 8 <br> Gases -
Figure 8.1 The hot air inside these balloons is less dense than the surrounding cool air. This results in a buoyant force that causes the balloons to rise when their guy lines are untied. (credit: modification of work by Anthony Quintano)


764. CHAPTER OUTLINE - 764.1. Gas Pressure
8.2 Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law
8.3 Stoichiometry of Gaseous Substances, Mixtures, and Reactions


764. CHAPTER OUTLINE - 764.2. Effusion and Diffusion of Gases
8.5 The Kinetic-Molecular Theory
8.6 Non-Ideal Gas Behavior

INTRODUCTION We are surrounded by an ocean of gas-the atmosphere-and many of the properties of gases are familiar to us from our daily activities. Heated gases expand, which can make a hot air balloon rise (Figure 8.1) or cause a blowout in a bicycle tire left in the sun on a hot day.

Gases have played an important part in the development of chemistry. In the seventeenth and eighteenth centuries, many scientists investigated gas behavior, providing the first mathematical descriptions of the behavior of matter.

In this chapter, we will examine the relationships between gas temperature, pressure, amount, and volume. We will study a simple theoretical model and use it to analyze the experimental behavior of gases. The results of these analyses will show us the limitations of the theory and how to improve on it.


765. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Define the property of pressure
- Define and convert among the units of pressure measurements
- Describe the operation of common tools for measuring gas pressure
- Calculate pressure from manometer data

The earth's atmosphere exerts a pressure, as does any other gas. Although we do not normally notice atmospheric pressure, we are sensitive to pressure changes-for example, when your ears "pop" during takeoff and landing while flying, or when you dive underwater. Gas pressure is caused by the force exerted by gas molecules colliding with the surfaces of objects (Figure 8.2). Although the force of each collision is very small, any surface of appreciable area experiences a large number of collisions in a short time, which can result in a high pressure. In fact, normal air pressure is strong enough to crush a metal container when not balanced by equal pressure from inside the container.

FIGURE 8.2 The atmosphere above us exerts a large pressure on objects at the surface of the earth, roughly equal to the weight of a bowling ball pressing on an area the size of a human thumbnail.


766. LINK TO LEARNING -
A dramatic illustration (http://openstax.org/l/16atmospressur1) of atmospheric pressure is provided in this brief video, which shows a railway tanker car imploding when its internal pressure is decreased.

A smaller scale demonstration (http://openstax.org/l/16atmospressur2) of this phenomenon is briefly explained.

Atmospheric pressure is caused by the weight of the column of air molecules in the atmosphere above an object, such as the tanker car. At sea level, this pressure is roughly the same as that exerted by a full-grown African elephant standing on a doormat, or a typical bowling ball resting on your thumbnail. These may seem like huge amounts, and they are, but life on earth has evolved under such atmospheric pressure. If you actually perch a bowling ball on your thumbnail, the pressure experienced is twice the usual pressure, and the sensation is unpleasant.

In general, pressure is defined as the force exerted on a given area: $P=\frac{F}{A}$. Note that pressure is directly proportional to force and inversely proportional to area. Thus, pressure can be increased either by increasing the amount of force or by decreasing the area over which it is applied; pressure can be decreased by decreasing the force or increasing the area.

Let's apply this concept to determine which exerts a greater pressure in Figure 8.3-the elephant or the figure skater? A large African elephant can weigh 7 tons, supported on four feet, each with a diameter of about 1.5 ft
(footprint area of $250 \mathrm{in}^{2}$ ), so the pressure exerted by each foot is about $14 \mathrm{lb} / \mathrm{in}^{2}$ :

$$
\text { pressure per elephant foot }=14,000 \frac{\mathrm{lb}}{\text { elephant }} \times \frac{1 \text { elephant }}{4 \text { feet }} \times \frac{1 \text { foot }}{250 \mathrm{in}^{2}}=14 \mathrm{lb} / \mathrm{in}^{2}
$$


766. LINK TO LEARNING -
The figure skater weighs about 120 lbs , supported on two skate blades, each with an area of about $2 \mathrm{in}^{2}$, so the pressure exerted by each blade is about $30 \mathrm{lb} / \mathrm{in}^{2}$ :

$$
\text { pressure per skate blade }=120 \frac{\mathrm{lb}}{\text { skater }} \times \frac{1 \text { skater }}{2 \text { blades }} \times \frac{1 \text { blade }}{2 \mathrm{in}^{2}}=30 \mathrm{lb} / \mathrm{in}^{2}
$$

Even though the elephant is more than one hundred-times heavier than the skater, it exerts less than one-half of the pressure. On the other hand, if the skater removes their skates and stands with bare feet (or regular footwear) on the ice, the larger area over which their weight is applied greatly reduces the pressure exerted:

$$
\text { pressure per human foot }=120 \frac{\mathrm{lb}}{\text { skater }} \times \frac{1 \text { skater }}{2 \text { feet }} \times \frac{1 \text { foot }}{30 \mathrm{in}^{2}}=2 \mathrm{lb} / \mathrm{in}^{2}
$$

FIGURE 8.3 Although (a) an elephant's weight is large, creating a very large force on the ground, (b) the figure skater exerts a much higher pressure on the ice due to the small surface area of the skates. (credit a: modification of work by Guido da Rozze; credit b: modification of work by Ryosuke Yagi)


766. LINK TO LEARNING -
The SI unit of pressure is the pascal ( $\mathbf{P a}$ ), with $1 \mathrm{~Pa}=1 \mathrm{~N} / \mathrm{m}^{2}$, where N is the newton, a unit of force defined as 1 $\mathrm{kg} \mathrm{m} / \mathrm{s}^{2}$. One pascal is a small pressure; in many cases, it is more convenient to use units of kilopascal ( $1 \mathrm{kPa}=$ 1000 Pa ) or bar ( $1 \mathrm{bar}=100,000 \mathrm{~Pa}$ ). In the United States, pressure is often measured in pounds of force on an area of one square inch-pounds per square inch (psi)-for example, in car tires. Pressure can also be measured using the unit atmosphere (atm), which originally represented the average sea level air pressure at the approximate latitude of Paris $\left(45^{\circ}\right)$. Table 8.1 provides some information on these and a few other common units for pressure measurements

Pressure Units

| Unit Name and Abbreviation | Definition or Relation to Other Unit |
| :--- | :--- |
| pascal (Pa) | $1 \mathrm{~Pa}=1 \mathrm{~N} / \mathrm{m}^{2}$
 recommended IUPAC unit |
| kilopascal (kPa) | $1 \mathrm{kPa}=1000 \mathrm{~Pa}$ |
| pounds per square inch (psi) | air pressure at sea level is ~14.7 psi |
| atmosphere (atm) | 1 atm $=101,325 \mathrm{~Pa}=760$ torr
 air pressure at sea level is $\sim 1 \mathrm{~atm}$ |

TABLE 8.1


766. LINK TO LEARNING -
TABLE 8.1

| Unit Name and Abbreviation | Definition or Relation to Other Unit |
| :--- | :--- |
| bar (bar, or b) | 1 bar $=100,000 \mathrm{~Pa}$ (exactly)
 commonly used in meteorology |
| millibar (mbar, or mb) | $1000 \mathrm{mbar}=1 \mathrm{bar}$ |
| inches of mercury (in. Hg) | $1 \mathrm{in} . \mathrm{Hg}=3386 \mathrm{~Pa}$
 used by aviation industry, also some weather reports |
| torr | 1 torr $=\frac{1}{760}$ atm
 named after Evangelista Torricelli, inventor of the barometer |
| millimeters of mercury (mm Hg) | $1 \mathrm{~mm} \mathrm{Hg} \mathrm{\sim 1} \mathrm{torr}$ |

TABLE 8.1


768. Conversion of Pressure Units -
The United States National Weather Service reports pressure in both inches of Hg and millibars. Convert a pressure of 29.2 in . Hg into:
(a) torr
(b) atm
(c) kPa
(d) mbar


769. Solution -
This is a unit conversion problem. The relationships between the various pressure units are given in Table 8.1.
(a) $29.2 \mathrm{im} \mathrm{Hg} \times \frac{25.4 \mathrm{~mm}}{1 \mathrm{im}} \times \frac{1 \text { torr }}{1 \mathrm{mmHg}}=742$ torr
(b) $742 \mathrm{tOrf} \times \frac{1 \mathrm{~atm}}{760 \text { tort }}=0.976 \mathrm{~atm}$
(c) 742 torf $\times \frac{101.325 \mathrm{kPa}}{760 \text { tort }}=98.9 \mathrm{kPa}$
(d) $98.9 \mathrm{kPa} \times \frac{1000 \mathrm{~Pa}}{1 \mathrm{kPa}} \times \frac{1 \text { bar }}{100,000 \mathrm{~Pa}} \times \frac{1000 \mathrm{mbar}}{1 \text { bar }}=989 \mathrm{mbar}$


770. Check Your Learning -
A typical barometric pressure in Kansas City is 740 torr. What is this pressure in atmospheres, in millimeters of mercury, in kilopascals, and in bar?


771. Answer: -
0.974 atm; $740 \mathrm{~mm} \mathrm{Hg} ; 98.7 \mathrm{kPa} ; 0.987$ bar

We can measure atmospheric pressure, the force exerted by the atmosphere on the earth's surface, with a barometer (Figure 8.4). A barometer is a glass tube that is closed at one end, filled with a nonvolatile liquid such as mercury, and then inverted and immersed in a container of that liquid. The atmosphere exerts pressure on the liquid outside the tube, the column of liquid exerts pressure inside the tube, and the pressure at the liquid surface is the same inside and outside the tube. The height of the liquid in the tube is therefore
proportional to the pressure exerted by the atmosphere.

FIGURE 8.4 In a barometer, the height, $h$, of the column of liquid is used as a measurement of the air pressure. Using very dense liquid mercury (left) permits the construction of reasonably sized barometers, whereas using water (right) would require a barometer more than 30 feet tall.

If the liquid is water, normal atmospheric pressure will support a column of water over 10 meters high, which is rather inconvenient for making (and reading) a barometer. Because mercury ( Hg ) is about 13.6-times denser than water, a mercury barometer only needs to be $\frac{1}{13.6}$ as tall as a water barometer-a more suitable size. Standard atmospheric pressure of 1 atm at sea level ( $101,325 \mathrm{~Pa}$ ) corresponds to a column of mercury that is about 760 mm (29.92 in.) high. The torr was originally intended to be a unit equal to one millimeter of mercury, but it no longer corresponds exactly. The pressure exerted by a fluid due to gravity is known as hydrostatic pressure, $p$ :

$$
p=h \rho g
$$

where $h$ is the height of the fluid, $\rho$ is the density of the fluid, and $g$ is acceleration due to gravity.


773. Calculation of Barometric Pressure -
Show the calculation supporting the claim that atmospheric pressure near sea level corresponds to the pressure exerted by a column of mercury that is about 760 mm high. The density of mercury $=13.6 \mathrm{~g} / \mathrm{cm}^{3}$.


774. Solution -
The hydrostatic pressure is given by $p=h \rho g$, with $h=760 \mathrm{~mm}, \rho=13.6 \mathrm{~g} / \mathrm{cm}^{3}$, and $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$. Plugging these values into the equation and doing the necessary unit conversions will give us the value we seek. (Note: We are expecting to find a pressure of $\sim 101,325 \mathrm{~Pa}$.)

$$
\begin{gathered}
101,325 \mathrm{~N} / \mathrm{m}^{2}=101,325 \frac{\mathrm{~kg} \cdot \mathrm{~m} / \mathrm{s}^{2}}{\mathrm{~m}^{2}}=101,325 \frac{\mathrm{~kg}}{\mathrm{~m} \cdot \mathrm{~s}^{2}} \\
p=\left(760 \mathrm{~mm} \times \frac{1 \mathrm{~m}}{1000 \mathrm{~mm}}\right) \times\left(\frac{13.6 \mathrm{~g}}{1 \mathrm{~cm}^{3}} \times \frac{1 \mathrm{~kg}}{1000 \mathrm{~g}} \times \frac{(100 \mathrm{~cm})^{3}}{(1 \mathrm{~m})^{3}}\right) \times\left(\frac{9.81 \mathrm{~m}}{1 \mathrm{~s}^{2}}\right) \\
=(0.760 \mathrm{~m})\left(13,600 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=1.01 \times 10^{5} \mathrm{~kg} / \mathrm{ms}^{2}=1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}
\end{gathered}
$$

$$
=1.01 \times 10^{5} \mathrm{~Pa}
$$


775. Check Your Learning -
Calculate the height of a column of water at $25^{\circ} \mathrm{C}$ that corresponds to normal atmospheric pressure. The density of water at this temperature is $1.0 \mathrm{~g} / \mathrm{cm}^{3}$.


776. Answer: -
10.3 m

A manometer is a device similar to a barometer that can be used to measure the pressure of a gas trapped in a container. A closed-end manometer is a U-shaped tube with one closed arm, one arm that connects to the gas to be measured, and a nonvolatile liquid (usually mercury) in between. As with a barometer, the distance between the liquid levels in the two arms of the tube ( $h$ in the diagram) is proportional to the pressure of the gas in the container. An open-end manometer (Figure 8.5) is the same as a closed-end manometer, but one of its arms is open to the atmosphere. In this case, the distance between the liquid levels corresponds to the difference in pressure between the gas in the container and the atmosphere.

FIGURE 8.5 A manometer can be used to measure the pressure of a gas. The (difference in) height between the liquid levels ( $h$ ) is a measure of the pressure. Mercury is usually used because of its large density.


778. Calculation of Pressure Using a Closed-End Manometer -
The pressure of a sample of gas is measured with a closed-end manometer, as shown to the right. The liquid in the manometer is mercury. Determine the pressure of the gas in:
(a) torr
(b) Pa
(c) bar


779. Solution -
The pressure of the gas is equal to a column of mercury of height 26.4 cm . (The pressure at the bottom horizontal line is equal on both sides of the tube. The pressure on the left is due to the gas and the pressure on the right is due to 26.4 cm Hg , or mercury.) We could use the equation $p=h \rho g$ as in Example 8.2, but it is simpler to just convert between units using Table 8.1.
(a) $26.4 \mathrm{cmHg} \times \frac{10 \mathrm{mmHz}}{1 \mathrm{cmHg}} \times \frac{1 \text { torr }}{1 \mathrm{mmHg}}=264$ torr
(b) 264 tori $\times \frac{1 \mathrm{Am}}{760 \text { torr }} \times \frac{101,325 \mathrm{~Pa}}{1+35,200 \mathrm{~Pa}}$
(c) $35,200 \mathrm{~Pa} \times \frac{1 \mathrm{bar}}{100,000 \mathrm{~Pa}}=0.352 \mathrm{bar}$


780. Check Your Learning -
The pressure of a sample of gas is measured with a closed-end manometer. The liquid in the manometer is mercury. Determine the pressure of the gas in:
(a) torr
(b) Pa
(c) bar


781. Answer: -
(a) $\sim 150$ torr; (b) $\sim 20,000 \mathrm{~Pa}$; (c) $\sim 0.20$ bar


783. Calculation of Pressure Using an Open-End Manometer -
The pressure of a sample of gas is measured at sea level with an open-end Hg (mercury) manometer, as shown to the right. Determine the pressure of the gas in:
(a) mm Hg
(b) atm
(c) kPa


784. Solution -
The pressure of the gas equals the hydrostatic pressure due to a column of mercury of height 13.7 cm plus the pressure of the atmosphere at sea level. (The pressure at the bottom horizontal line is equal on both sides of the tube. The pressure on the left is due to the gas and the pressure on the right is due to 13.7 cm of Hg plus atmospheric pressure.)
(a) In mm Hg , this is: $137 \mathrm{~mm} \mathrm{Hg}+760 \mathrm{~mm} \mathrm{Hg}=897 \mathrm{~mm} \mathrm{Hg}$
(b) $897 \mathrm{mmHg} \times \frac{1 \mathrm{~atm}}{760 \mathrm{~mm} \mathrm{Hg}}=1.18 \mathrm{~atm}$
(c) $1.18 \mathrm{~atm} \times \frac{101.325 \mathrm{kPa}}{1 \mathrm{~atm}}=1.20 \times 10^{2} \mathrm{kPa}$


785. Check Your Learning -
The pressure of a sample of gas is measured at sea level with an open-end Hg manometer, as shown to the right. Determine the pressure of the gas in:
(a) mm Hg
(b) atm
(c) kPa

Answer:
(a) 642 mm Hg ; (b) 0.845 atm ; (c) 85.6 kPa


787. Measuring Blood Pressure -
Blood pressure is measured using a device called a sphygmomanometer (Greek sphygmos = "pulse"). It consists of an inflatable cuff to restrict blood flow, a manometer to measure the pressure, and a method of determining when blood flow begins and when it becomes impeded (Figure 8.6). Since its invention in 1881, it has been an essential medical device. There are many types of sphygmomanometers: manual ones that require a stethoscope and are used by medical professionals; mercury ones, used when the most accuracy is required; less accurate mechanical ones; and digital ones that can be used with little training but that have limitations. When using a sphygmomanometer, the cuff is placed around the upper arm and inflated until blood flow is completely blocked, then slowly released. As the heart beats, blood forced through the arteries causes a rise in pressure. This rise in pressure at which blood flow begins is the systolic pressure-the peak pressure in the cardiac cycle. When the cuff's pressure equals the arterial systolic pressure, blood flows past the cuff, creating audible sounds that can be heard using a stethoscope. This is followed by a decrease in pressure as the heart's ventricles prepare for another beat. As cuff pressure continues to decrease, eventually sound is no longer heard; this is the diastolic pressure-the lowest pressure (resting phase) in the cardiac cycle. Blood pressure units from a sphygmomanometer are in terms of millimeters of mercury ( mm Hg ).

(a)

(b)

FIGURE 8.6 (a) A medical technician prepares to measure a patient's blood pressure with a sphygmomanometer. (b) A typical sphygmomanometer uses a valved rubber bulb to inflate the cuff and a diaphragm gauge to measure pressure. (credit a: modification of work by Master Sgt. Jeffrey Allen)


789. Meteorology, Climatology, and Atmospheric Science -
Throughout the ages, people have observed clouds, winds, and precipitation, trying to discern patterns and make predictions: when it is best to plant and harvest; whether it is safe to set out on a sea voyage; and much more. We now face complex weather and atmosphere-related challenges that will have a major impact on our civilization and the ecosystem. Several different scientific disciplines use chemical principles to help us better understand weather, the atmosphere, and climate. These are meteorology, climatology, and atmospheric science. Meteorology is the study of the atmosphere, atmospheric phenomena, and atmospheric effects on earth's weather. Meteorologists seek to understand and predict the weather in the short term, which can save lives and benefit the economy. Weather forecasts (Figure 8.7) are the result of thousands of measurements of air pressure, temperature, and the like, which are compiled, modeled, and analyzed in weather centers worldwide.

FIGURE 8.7 Meteorologists use weather maps to describe and predict weather. Regions of high (H) and low (L) pressure have large effects on weather conditions. The gray lines represent locations of constant pressure known as isobars. (credit: modification of work by National Oceanic and Atmospheric Administration)

In terms of weather, low-pressure systems occur when the earth's surface atmospheric pressure is lower than the surrounding environment: Moist air rises and condenses, producing clouds. Movement of moisture and air within various weather fronts instigates most weather events.

The atmosphere is the gaseous layer that surrounds a planet. Earth's atmosphere, which is roughly 100-125 km thick, consists of roughly $78.1 \%$ nitrogen and $21.0 \%$ oxygen, and can be subdivided further into the regions shown in Figure 8.8: the exosphere (furthest from earth, $>700 \mathrm{~km}$ above sea level), the thermosphere ( $80-700 \mathrm{~km}$ ), the mesosphere ( $50-80 \mathrm{~km}$ ), the stratosphere (second lowest level of our atmosphere, $12-50 \mathrm{~km}$ above sea level), and the troposphere (up to 12 km above sea level, roughly $80 \%$ of the earth's atmosphere by mass and the layer where most weather events originate). As you go higher in the troposphere, air density and temperature both decrease.


789. Meteorology, Climatology, and Atmospheric Science -
FIGURE 8.8 Earth's atmosphere has five layers: the troposphere, the stratosphere, the mesosphere, the thermosphere, and the exosphere.

Climatology is the study of the climate, averaged weather conditions over long time periods, using atmospheric data. However, climatologists study patterns and effects that occur over decades, centuries, and millennia, rather than shorter time frames of hours, days, and weeks like meteorologists. Atmospheric science is an even broader field, combining meteorology, climatology, and other scientific disciplines that study the atmosphere.


790. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Identify the mathematical relationships between the various properties of gases
- Use the ideal gas law, and related gas laws, to compute the values of various gas properties under specified conditions

During the seventeenth and especially eighteenth centuries, driven both by a desire to understand nature and a quest to make balloons in which they could fly (Figure 8.9), a number of scientists established the relationships between the macroscopic physical properties of gases, that is, pressure, volume, temperature, and amount of gas. Although their measurements were not precise by today's standards, they were able to determine the mathematical relationships between pairs of these variables (e.g., pressure and temperature, pressure and volume) that hold for an ideal gas-a hypothetical construct that real gases approximate under certain conditions. Eventually, these individual laws were combined into a single equation-the ideal gas law-that relates gas quantities for gases and is quite accurate for low pressures and moderate temperatures. We will consider the key developments in individual relationships (for pedagogical reasons not quite in historical order), then put them together in the ideal gas law.

(a)

(b)

(c)

FIGURE 8.9 In 1783, the first (a) hydrogen-filled balloon flight, (b) manned hot air balloon flight, and (c) manned hydrogen-filled balloon flight occurred. When the hydrogen-filled balloon depicted in (a) landed, the frightened villagers of Gonesse reportedly destroyed it with pitchforks and knives. The launch of the latter was reportedly viewed by 400,000 people in Paris.


791. Pressure and Temperature: Amontons's Law -
Imagine filling a rigid container attached to a pressure gauge with gas and then sealing the container so that no gas may escape. If the container is cooled, the gas inside likewise gets colder and its pressure is observed to decrease. Since the container is rigid and tightly sealed, both the volume and number of moles of gas remain constant. If we heat the sphere, the gas inside gets hotter (Figure 8.10) and the pressure increases.

FIGURE 8.10 The effect of temperature on gas pressure: When the hot plate is off, the pressure of the gas in the sphere is relatively low. As the gas is heated, the pressure of the gas in the sphere increases.

This relationship between temperature and pressure is observed for any sample of gas confined to a constant volume. An example of experimental pressure-temperature data is shown for a sample of air under these conditions in Figure 8.11. We find that temperature and pressure are linearly related, and if the temperature is on the kelvin scale, then $P$ and $T$ are directly proportional (again, when volume and moles of gas are held constant); if the temperature on the kelvin scale increases by a certain factor, the gas pressure increases by the same factor.

| Temperature
 $\left({ }^{\circ} \mathbf{C}\right)$ | Temperature
 $\mathbf{( K )}$ | Pressure
 $\mathbf{( k P a )}$ |
| :---: | :---: | :---: |
| -100 | 173 | 36.0 |
| -50 | 223 | 46.4 |
| 0 | 273 | 56.7 |
| 50 | 323 | 67.1 |
| 100 | 373 | 77.5 |
| 150 | 423 | 88.0 |

FIGURE 8.11 For a constant volume and amount of air, the pressure and temperature are directly proportional, provided the temperature is in kelvin. (Measurements cannot be made at lower temperatures because of the condensation of the gas.) When this line is extrapolated to lower pressures, it reaches a pressure of 0 at $-273^{\circ} \mathrm{C}$, which is 0 on the kelvin scale and the lowest possible temperature, called absolute zero.


791. Pressure and Temperature: Amontons's Law -
Guillaume Amontons was the first to empirically establish the relationship between the pressure and the temperature of a gas ( $\sim 1700$ ), and Joseph Louis Gay-Lussac determined the relationship more precisely (~1800). Because of this, the $P$ - $T$ relationship for gases is known as either Amontons's law or Gay-Lussac's law. Under either name, it states that the pressure of a given amount of gas is directly proportional to its temperature on the kelvin scale when the volume is held constant. Mathematically, this can be written:

$$
P \propto T \text { or } P=\text { constant } \times T \text { or } P=k \times T
$$

where $\alpha$ means "is proportional to," and $k$ is a proportionality constant that depends on the identity, amount, and volume of the gas.

For a confined, constant volume of gas, the ratio $\frac{P}{T}$ is therefore constant (i.e., $\frac{P}{T}=k$ ). If the gas is initially in "Condition 1" (with $P=P_{1}$ and $T=T_{1}$ ), and then changes to "Condition 2" (with $P=P_{2}$ and $T=T_{2}$ ), we have that $\frac{P_{1}}{T_{1}}=k$ and $\frac{P_{2}}{T_{2}}=k$, which reduces to $\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}$. This equation is useful for pressure-temperature calculations for a confined gas at constant volume. Note that temperatures must be on the kelvin scale for any gas law calculations ( 0 on the kelvin scale and the lowest possible temperature is called absolute zero). (Also note that there are at least three ways we can describe how the pressure of a gas changes as its temperature changes: We can use a table of values, a graph, or a mathematical equation.)


793. Predicting Change in Pressure with Temperature -
A can of hair spray is used until it is empty except for the propellant, isobutane gas.
(a) On the can is the warning "Store only at temperatures below $120^{\circ} \mathrm{F}\left(48.8^{\circ} \mathrm{C}\right)$. Do not incinerate." Why?
(b) The gas in the can is initially at $24^{\circ} \mathrm{C}$ and 360 kPa , and the can has a volume of 350 mL . If the can is left in a car that reaches $50^{\circ} \mathrm{C}$ on a hot day, what is the new pressure in the can?


794. Solution -
(a) The can contains an amount of isobutane gas at a constant volume, so if the temperature is increased by heating, the pressure will increase proportionately. High temperature could lead to high pressure, causing the can to burst. (Also, isobutane is combustible, so incineration could cause the can to explode.)
(b) We are looking for a pressure change due to a temperature change at constant volume, so we will use Amontons's/Gay-Lussac's law. Taking $P_{1}$ and $T_{1}$ as the initial values, $T_{2}$ as the temperature where the pressure is unknown and $P_{2}$ as the unknown pressure, and converting ${ }^{\circ} \mathrm{C}$ to K , we have:

$$
\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}} \text { which means that } \frac{360 \mathrm{kPa}}{297 \mathrm{~K}}=\frac{P_{2}}{323 \mathrm{~K}}
$$

Rearranging and solving gives: $P_{2}=\frac{360 \mathrm{kPa} \times 323 \mathrm{~K}}{297 \mathrm{~K}}=390 \mathrm{kPa}$


795. Check Your Learning -
A sample of nitrogen, $\mathrm{N}_{2}$, occupies 45.0 mL at $27^{\circ} \mathrm{C}$ and 600 torr. What pressure will it have if cooled to $-73^{\circ} \mathrm{C}$ while the volume remains constant?


796. Answer: -
400 torr


797. Volume and Temperature: Charles's Law -
If we fill a balloon with air and seal it, the balloon contains a specific amount of air at atmospheric pressure, let's say 1 atm. If we put the balloon in a refrigerator, the gas inside gets cold and the balloon shrinks (although both the amount of gas and its pressure remain constant). If we make the balloon very cold, it will shrink a great deal, and it expands again when it warms up.


798. LINK TO LEARNING -
This video (http://openstax.org/l/16CharlesLaw) shows how cooling and heating a gas causes its volume to decrease or increase, respectively.

These examples of the effect of temperature on the volume of a given amount of a confined gas at constant pressure are true in general: The volume increases as the temperature increases, and decreases as the temperature decreases. Volume-temperature data for a 1-mole sample of methane gas at 1 atm are listed and graphed in Figure 8.12.

| Temperature $\left({ }^{\circ} \mathrm{C}\right)$ | Temperature (K) | Volume (L) |
| :---: | :---: | :---: |
| -3 | 270 | 22 |
| -23 | 250 | 21 |
| -53 | 220 | 18 |
| -162 | 111 | 9 |

FIGURE 8.12 The volume and temperature are linearly related for 1 mole of methane gas at a constant pressure of 1 atm. If the temperature is in kelvin, volume and temperature are directly proportional. The line stops at 111 K because methane liquefies at this temperature; when extrapolated, it intersects the graph's origin, representing a temperature of absolute zero.

The relationship between the volume and temperature of a given amount of gas at constant pressure is known as Charles's law in recognition of the French scientist and balloon flight pioneer Jacques Alexandre César Charles. Charles's law states that the volume of a given amount of gas is directly proportional to its temperature on the kelvin scale when the pressure is held constant.

Mathematically, this can be written as:

$$
V \alpha T \text { or } V=\text { constant } \cdot T \text { or } V=k \cdot T \text { or } V_{1} / T_{1}=V_{2} / T_{2}
$$


798. LINK TO LEARNING -
with $k$ being a proportionality constant that depends on the amount and pressure of the gas.
For a confined, constant pressure gas sample, $\frac{V}{T}$ is constant (i.e., the ratio $=k$ ), and as seen with the $P-T$ relationship, this leads to another form of Charles's law: $\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$.


800. Predicting Change in Volume with Temperature -
A sample of carbon dioxide, $\mathrm{CO}_{2}$, occupies 0.300 L at $10^{\circ} \mathrm{C}$ and 750 torr. What volume will the gas have at $30^{\circ} \mathrm{C}$ and 750 torr?


801. Solution -
Because we are looking for the volume change caused by a temperature change at constant pressure, this is a job for Charles's law. Taking $V_{1}$ and $T_{1}$ as the initial values, $T_{2}$ as the temperature at which the volume is unknown and $V_{2}$ as the unknown volume, and converting ${ }^{\circ} \mathrm{C}$ into K we have:

$$
\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}} \text { which means that } \frac{0.300 \mathrm{~L}}{283 \mathrm{~K}}=\frac{V_{2}}{303 \mathrm{~K}}
$$

Rearranging and solving gives: $V_{2}=\frac{0.300 \mathrm{~L} \times 303 \mathrm{~K}}{283 \mathrm{~K}}=0.321 \mathrm{~L}$
This answer supports our expectation from Charles's law, namely, that raising the gas temperature (from 283 K to 303 K ) at a constant pressure will yield an increase in its volume (from 0.300 L to 0.321 L ).


802. Check Your Learning -
A sample of oxygen, $\mathrm{O}_{2}$, occupies 32.2 mL at $30^{\circ} \mathrm{C}$ and 452 torr. What volume will it occupy at $-70^{\circ} \mathrm{C}$ and the same pressure?


803. Answer: -
21.6 mL


805. Measuring Temperature with a Volume Change -
Temperature is sometimes measured with a gas thermometer by observing the change in the volume of the gas as the temperature changes at constant pressure. The hydrogen in a particular hydrogen gas thermometer has a volume of $150.0 \mathrm{~cm}^{3}$ when immersed in a mixture of ice and water $\left(0.00^{\circ} \mathrm{C}\right)$. When immersed in boiling liquid ammonia, the volume of the hydrogen, at the same pressure, is $131.7 \mathrm{~cm}^{3}$. Find the temperature of boiling ammonia on the kelvin and Celsius scales.


806. Solution -
A volume change caused by a temperature change at constant pressure means we should use Charles's law. Taking $V_{1}$ and $T_{1}$ as the initial values, $T_{2}$ as the temperature at which the volume is unknown and $V_{2}$ as the unknown volume, and converting ${ }^{\circ} \mathrm{C}$ into K we have:

$$
\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}} \text { which means that } \frac{150.0 \mathrm{~cm}^{3}}{273.15 \mathrm{~K}}=\frac{131.7 \mathrm{~cm}^{3}}{T_{2}}
$$

Rearrangement gives $T_{2}=\frac{131.7 \mathrm{~cm}^{3} \times 273.15 \mathrm{~K}}{150.0 \mathrm{~cm}^{3}}=239.8 \mathrm{~K}$
Subtracting 273.15 from 239.8 K , we find that the temperature of the boiling ammonia on the Celsius scale is $-33.4^{\circ} \mathrm{C}$.


807. Check Your Learning -
What is the volume of a sample of ethane at 467 K and 1.1 atm if it occupies 405 mL at 298 K and 1.1 atm?


808. Answer: -
635 mL


809. Volume and Pressure: Boyle's Law -
If we partially fill an airtight syringe with air, the syringe contains a specific amount of air at constant temperature, say $25^{\circ} \mathrm{C}$. If we slowly push in the plunger while keeping temperature constant, the gas in the syringe is compressed into a smaller volume and its pressure increases; if we pull out the plunger, the volume increases and the pressure decreases. This example of the effect of volume on the pressure of a given amount of a confined gas is true in general. Decreasing the volume of a contained gas will increase its pressure, and increasing its volume will decrease its pressure. In fact, if the volume increases by a certain factor, the pressure decreases by the same factor, and vice versa. Volume-pressure data for an air sample at room temperature are graphed in Figure 8.13.

FIGURE 8.13 When a gas occupies a smaller volume, it exerts a higher pressure; when it occupies a larger volume, it exerts a lower pressure (assuming the amount of gas and the temperature do not change). Since $P$ and $V$ are inversely proportional, a graph of $\frac{1}{P}$ vs. $V$ is linear.

Unlike the $P-T$ and $V-T$ relationships, pressure and volume are not directly proportional to each other. Instead, $P$ and $V$ exhibit inverse proportionality: Increasing the pressure results in a decrease of the volume of the gas. Mathematically this can be written:

$$
P \propto 1 / V \text { or } P=k \cdot 1 / V \text { or } P \cdot V=k \text { or } P_{1} V_{1}=P_{2} V_{2}
$$


809. Volume and Pressure: Boyle's Law -
with $k$ being a constant. Graphically, this relationship is shown by the straight line that results when plotting the inverse of the pressure ( $\frac{1}{P}$ ) versus the volume $(V)$, or the inverse of volume $\left(\frac{1}{V}\right)$ versus the pressure $(P)$. Graphs with curved lines are difficult to read accurately at low or high values of the variables, and they are more difficult to use in fitting theoretical equations and parameters to experimental data. For those reasons, scientists often try to find a way to "linearize" their data. If we plot $P$ versus $V$, we obtain a hyperbola (see Figure 8.14).

FIGURE 8.14 The relationship between pressure and volume is inversely proportional. (a) The graph of $P$ vs. $V$ is a hyperbola, whereas (b) the graph of $\left(\frac{1}{P}\right)$ vs. $V$ is linear.

The relationship between the volume and pressure of a given amount of gas at constant temperature was first published by the English natural philosopher Robert Boyle over 300 years ago. It is summarized in the statement now known as Boyle's law: The volume of a given amount of gas held at constant temperature is inversely proportional to the pressure under which it is measured.


811. Volume of a Gas Sample -
The sample of gas in Figure 8.13 has a volume of 15.0 mL at a pressure of 13.0 psi. Determine the pressure of the gas at a volume of 7.5 mL , using:
(a) the $P$ - $V$ graph in Figure 8.13
(b) the $\frac{1}{P}$ vs. $V$ graph in Figure 8.13
(c) the Boyle's law equation

Comment on the likely accuracy of each method.


812. Solution -
(a) Estimating from the $P$ - $V$ graph gives a value for $P$ somewhere around 27 psi .
(b) Estimating from the $\frac{1}{P}$ versus $V$ graph give a value of about 26 psi .
(c) From Boyle's law, we know that the product of pressure and volume ( $P V$ ) for a given sample of gas at a constant temperature is always equal to the same value. Therefore we have $P_{1} V_{1}=k$ and $P_{2} V_{2}=k$ which means that $P_{1} V_{1}=P_{2} V_{2}$.

Using $P_{1}$ and $V_{1}$ as the known values 13.0 psi and $15.0 \mathrm{~mL}, P_{2}$ as the pressure at which the volume is unknown, and $V_{2}$ as the unknown volume, we have:

$$
P_{1} V_{1}=P_{2} V_{2} \text { or } 13.0 \mathrm{psi} \times 15.0 \mathrm{~mL}=P_{2} \times 7.5 \mathrm{~mL}
$$

Solving:

$$
P_{2}=\frac{13.0 \mathrm{psi} \times 15.0 \mathrm{~mL}}{7.5 \mathrm{~mL}}=26 \mathrm{psi}
$$

It was more difficult to estimate well from the $P$ - $V$ graph, so (a) is likely more inaccurate than (b) or (c). The calculation will be as accurate as the equation and measurements allow.


813. Check Your Learning -
The sample of gas in Figure 8.13 has a volume of 30.0 mL at a pressure of 6.5 psi . Determine the volume of the gas at a pressure of 11.0 psi , using:
(a) the $P$ - $V$ graph in Figure 8.13
(b) the $\frac{1}{P}$ vs. $V$ graph in Figure 8.13
(c) the Boyle's law equation

Comment on the likely accuracy of each method.


814. Answer: -
(a) about $17-18 \mathrm{~mL}$; (b) $\sim 18 \mathrm{~mL}$; (c) 17.7 mL ; it was more difficult to estimate well from the $P$ - $V$ graph, so (a) is likely more inaccurate than (b); the calculation will be as accurate as the equation and measurements allow


816. Breathing and Boyle's Law -
What do you do about 20 times per minute for your whole life, without break, and often without even being aware of it? The answer, of course, is respiration, or breathing. How does it work? It turns out that the gas laws apply here. Your lungs take in gas that your body needs (oxygen) and get rid of waste gas (carbon dioxide). Lungs are made of spongy, stretchy tissue that expands and contracts while you breathe. When you inhale, your diaphragm and intercostal muscles (the muscles between your ribs) contract, expanding your chest cavity and making your lung volume larger. The increase in volume leads to a decrease in pressure (Boyle's law). This causes air to flow into the lungs (from high pressure to low pressure). When you exhale, the process reverses: Your diaphragm and rib muscles relax, your chest cavity contracts, and your lung volume decreases, causing the pressure to increase (Boyle's law again), and air flows out of the lungs (from high pressure to low pressure). You then breathe in and out again, and again, repeating this Boyle's law cycle for the rest of your life (Figure 8.15).

FIGURE 8.15 Breathing occurs because expanding and contracting lung volume creates small pressure differences between your lungs and your surroundings, causing air to be drawn into and forced out of your lungs.


817. Moles of Gas and Volume: Avogadro's Law -
The Italian scientist Amedeo Avogadro advanced a hypothesis in 1811 to account for the behavior of gases, stating that equal volumes of all gases, measured under the same conditions of temperature and pressure, contain the same number of molecules. Over time, this relationship was supported by many experimental observations as expressed by Avogadro's law: For a confined gas, the volume ( $V$ ) and number of moles (n) are directly proportional if the pressure and temperature both remain constant.

In equation form, this is written as:

$$
V \propto n \text { or } V=k \times n \text { or } \frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}
$$

Mathematical relationships can also be determined for the other variable pairs, such as $P$ versus $n$, and $n$ versus $T$.


818. LINK TO LEARNING -
Visit this interactive PhET simulation (http://openstax.org/l/16IdealGasLaw) to investigate the relationships between pressure, volume, temperature, and amount of gas. Use the simulation to examine the effect of changing one parameter on another while holding the other parameters constant (as described in the preceding sections on the various gas laws).


819. The Ideal Gas Law -
To this point, four separate laws have been discussed that relate pressure, volume, temperature, and the number of moles of the gas:

- Boyle's law: $P V=$ constant at constant $T$ and $n$
- Amontons's law: $\frac{P}{T}=$ constant at constant $V$ and $n$
- Charles's law: $\frac{V}{T}=$ constant at constant $P$ and $n$
- Avogadro's law: $\frac{V}{n}=$ constant at constant $P$ and $T$

Combining these four laws yields the ideal gas law, a relation between the pressure, volume, temperature, and number of moles of a gas:

$$
P V=n R T
$$

where $P$ is the pressure of a gas, $V$ is its volume, $n$ is the number of moles of the gas, $T$ is its temperature on the kelvin scale, and $R$ is a constant called the ideal gas constant or the universal gas constant. The units used to express pressure, volume, and temperature will determine the proper form of the gas constant as required by dimensional analysis, the most commonly encountered values being $0.08206 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ and 8.314 kPa L $\mathrm{mol}^{-1} \mathrm{~K}^{-1}$.

Gases whose properties of $P, V$, and $T$ are accurately described by the ideal gas law (or the other gas laws) are said to exhibit ideal behavior or to approximate the traits of an ideal gas. An ideal gas is a hypothetical construct that may be used along with kinetic molecular theory to effectively explain the gas laws as will be described in a later module of this chapter. Although all the calculations presented in this module assume ideal behavior, this assumption is only reasonable for gases under conditions of relatively low pressure and high temperature. In the final module of this chapter, a modified gas law will be introduced that accounts for the non-ideal behavior observed for many gases at relatively high pressures and low temperatures.


819. The Ideal Gas Law -
The ideal gas equation contains five terms, the gas constant $R$ and the variable properties $P, V, n$, and $T$. Specifying any four of these terms will permit use of the ideal gas law to calculate the fifth term as demonstrated in the following example exercises.


821. Using the Ideal Gas Law -
Methane, $\mathrm{CH}_{4}$, is being considered for use as an alternative automotive fuel to replace gasoline. One gallon of gasoline could be replaced by 655 g of $\mathrm{CH}_{4}$. What is the volume of this much methane at $25^{\circ} \mathrm{C}$ and 745 torr?


822. Solution -
We must rearrange $P V=n R T$ to solve for $V: V=\frac{n R T}{P}$
If we choose to use $R=0.08206 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$, then the amount must be in moles, temperature must be in kelvin, and pressure must be in atm.

Converting into the "right" units:

$$
\begin{gathered}
n=655 \mathrm{gCH}_{4} \times \frac{1 \mathrm{~mol}}{16.043-\mathrm{g}_{4} \mathrm{CH}_{4}}=40.8 \mathrm{~mol} \\
T=25^{\circ} \mathrm{C}+273=298 \mathrm{~K} \\
P=745 \mathrm{tmm} \times \frac{1 \mathrm{~atm}}{760 \mathrm{tmf}}=0.980 \mathrm{~atm} \\
V=\frac{n R T}{P}=\frac{(40.8 \mathrm{mor})\left(0.08206 \mathrm{~L} \mathrm{atmol}^{-1} \mathrm{~K}^{-1}\right)(298 \mathrm{~K})}{0.980 \mathrm{~atm}}=1.02 \times 10^{3} \mathrm{~L}
\end{gathered}
$$

It would require 1020 L ( 269 gal ) of gaseous methane at about 1 atm of pressure to replace 1 gal of gasoline. It requires a large container to hold enough methane at 1 atm to replace several gallons of gasoline.


823. Check Your Learning -
Calculate the pressure in bar of 2520 moles of hydrogen gas stored at $27^{\circ} \mathrm{C}$ in the $180-\mathrm{L}$ storage tank of a modern hydrogen-powered car.


824. Answer: -
350 bar

If the number of moles of an ideal gas are kept constant under two different sets of conditions, a useful mathematical relationship called the combined gas law is obtained: $\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}$ using units of atm, L, and K. Both sets of conditions are equal to the product of $n \times R$ (where $n=$ the number of moles of the gas and $R$ is the ideal gas law constant).


826. Using the Combined Gas Law -
When filled with air, a typical scuba tank with a volume of 13.2 L has a pressure of 153 atm (Figure 8.16). If the water temperature is $27^{\circ} \mathrm{C}$, how many liters of air will such a tank provide to a diver's lungs at a depth of approximately 70 feet in the ocean where the pressure is 3.13 atm ?

FIGURE 8.16 Scuba divers use compressed air to breathe while underwater. (credit: modification of work by Mark Goodchild)

Letting 1 represent the air in the scuba tank and 2 represent the air in the lungs, and noting that body temperature (the temperature the air will be in the lungs) is $37^{\circ} \mathrm{C}$, we have:

$$
\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}} \rightarrow \frac{(153 \mathrm{~atm})(13.2 \mathrm{~L})}{(300 \mathrm{~K})}=\frac{(3.13 \mathrm{~atm})\left(V_{2}\right)}{(310 \mathrm{~K})}
$$

Solving for $V_{2}$ :

$$
V_{2}=\frac{(153 \mathrm{~atm})(13.2 \mathrm{~L})(310 \mathrm{~K})}{(300 \mathrm{~K})(3.13 \mathrm{~atm})}=667 \mathrm{~L}
$$

(Note: Be advised that this particular example is one in which the assumption of ideal gas behavior is not very reasonable, since it involves gases at relatively high pressures and low temperatures. Despite this limitation, the calculated volume can be viewed as a good "ballpark" estimate.)


827. Check Your Learning -
A sample of ammonia is found to occupy 0.250 L under laboratory conditions of $27^{\circ} \mathrm{C}$ and 0.850 atm . Find the volume of this sample at $0^{\circ} \mathrm{C}$ and 1.00 atm .


828. Answer: -
0.193 L


830. The Interdependence between Ocean Depth and Pressure in Scuba Diving -
Whether scuba diving at the Great Barrier Reef in Australia (shown in Figure 8.17) or in the Caribbean, divers must understand how pressure affects a number of issues related to their comfort and safety.

FIGURE 8.17 Scuba divers, whether at the Great Barrier Reef or in the Caribbean, must be aware of buoyancy, pressure equalization, and the amount of time they spend underwater, to avoid the risks associated with pressurized gases in the body. (credit: Kyle Taylor)


830. The Interdependence between Ocean Depth and Pressure in Scuba Diving -
Pressure increases with ocean depth, and the pressure changes most rapidly as divers reach the surface. The pressure a diver experiences is the sum of all pressures above the diver (from the water and the air). Most pressure measurements are given in units of atmospheres, expressed as "atmospheres absolute" or ATA in the diving community: Every 33 feet of salt water represents 1 ATA of pressure in addition to 1 ATA of pressure from the atmosphere at sea level. As a diver descends, the increase in pressure causes the body's air pockets in the ears and lungs to compress; on the ascent, the decrease in pressure causes these air pockets to expand, potentially rupturing eardrums or bursting the lungs. Divers must therefore undergo equalization by adding air to body airspaces on the descent by breathing normally and adding air to the mask by breathing out of the nose or adding air to the ears and sinuses by equalization techniques; the corollary is also true on ascent, divers must release air from the body to maintain equalization. Buoyancy, or the ability to control whether a diver sinks or floats, is controlled by the buoyancy compensator (BCD). If a diver is ascending, the air in their BCD expands because of lower pressure according to Boyle's law (decreasing the pressure of gases increases the volume). The expanding air increases the buoyancy of the diver, and they begin to ascend. The diver must vent air from the BCD or risk an uncontrolled ascent that could rupture the lungs. In descending, the increased pressure causes the air in the BCD to compress and the diver sinks much more quickly; the diver must add air to the BCD or risk an uncontrolled descent, facing much higher pressures near the ocean floor. The pressure also impacts how long a diver can stay underwater before ascending. The deeper a diver dives, the more compressed the air that is breathed because of increased pressure: If a diver dives 33 feet, the pressure is 2 ATA and the air would be compressed to one-half of its original volume. The diver uses up available air twice as fast as at the surface.


831. Standard Conditions of Temperature and Pressure -
We have seen that the volume of a given quantity of gas and the number of molecules (moles) in a given volume of gas vary with changes in pressure and temperature. Chemists sometimes make comparisons against a standard temperature and pressure (STP) for reporting properties of gases: 273.15 K and 1 atm (101.325 $\mathrm{kPa}) .{ }^{\frac{1}{2}}$ At STP, one mole of an ideal gas has a volume of about $22.4 \mathrm{~L}-$ this is referred to as the standard molar volume (Figure 8.18).

[^4]

FIGURE 8.18 Regardless of its chemical identity, one mole of gas behaving ideally occupies a volume of $\sim 22.4 \mathrm{~L}$ at STP.


831. Standard Conditions of Temperature and Pressure - 831.1. Stoichiometry of Gaseous Substances, Mixtures, and Reactions LEARNING OBJECTIVES
By the end of this section, you will be able to:

- Use the ideal gas law to compute gas densities and molar masses
- Perform stoichiometric calculations involving gaseous substances
- State Dalton's law of partial pressures and use it in calculations involving gaseous mixtures

The study of the chemical behavior of gases was part of the basis of perhaps the most fundamental chemical revolution in history. French nobleman Antoine Lavoisier, widely regarded as the "father of modern chemistry," changed chemistry from a qualitative to a quantitative science through his work with gases. He discovered the law of conservation of matter, discovered the role of oxygen in combustion reactions, determined the composition of air, explained respiration in terms of chemical reactions, and more. He was a casualty of the French Revolution, guillotined in 1794. Of his death, mathematician and astronomer JosephLouis Lagrange said, "It took the mob only a moment to remove his head; a century will not suffice to reproduce it." ${ }^{2}$ Much of the knowledge we do have about Lavoisier's contributions is due to his wife, MarieAnne Paulze Lavoisier, who worked with him in his lab. A trained artist fluent in several languages, she created detailed illustrations of the equipment in his lab, and translated texts from foreign scientists to complement his knowledge. After his execution, she was instrumental in publishing Lavoisier's major treatise, which unified many concepts of chemistry and laid the groundwork for significant further study.

As described in an earlier chapter of this text, we can turn to chemical stoichiometry for answers to many of the questions that ask "How much?" The essential property involved in such use of stoichiometry is the amount of substance, typically measured in moles ( $n$ ). For gases, molar amount can be derived from convenient experimental measurements of pressure, temperature, and volume. Therefore, these measurements are useful in assessing the stoichiometry of pure gases, gas mixtures, and chemical reactions involving gases. This section will not introduce any new material or ideas, but will provide examples of applications and ways to integrate concepts we have already discussed.

[^5]


832. Gas Density and Molar Mass -
The ideal gas law described previously in this chapter relates the properties of pressure $P$, volume $V$, temperature $T$, and molar amount $n$. This law is universal, relating these properties in identical fashion regardless of the chemical identity of the gas:

$$
P V=n R T
$$

The density $d$ of a gas, on the other hand, is determined by its identity. As described in another chapter of this text, the density of a substance is a characteristic property that may be used to identify the substance.

$$
d=\frac{m}{V}
$$

Rearranging the ideal gas equation to isolate $V$ and substituting into the density equation yields

$$
d=\frac{m P}{n R T}=\left(\frac{m}{n}\right) \frac{P}{R T}
$$

The ratio $m / n$ is the definition of molar mass, $\mathcal{M}$ :

$$
\mathscr{M}=\frac{m}{n}
$$

The density equation can then be written

$$
d=\frac{\mathscr{M} P}{R T}
$$

This relation may be used for calculating the densities of gases of known identities at specified values of pressure and temperature as demonstrated in Example 8.11.


834. Measuring Gas Density -
What is the density of molecular nitrogen gas at STP?


835. Solution -
The molar mass of molecular nitrogen, $\mathrm{N}_{2}$, is $28.01 \mathrm{~g} / \mathrm{mol}$. Substituting this value along with standard temperature and pressure into the gas density equation yields

$$
d=\frac{\mathscr{M} P}{R T}=\frac{(28.01 \mathrm{~g} / \mathrm{mol})(1.00 \mathrm{~atm})}{\left(0.0821 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)(273 \mathrm{~K})}=1.25 \mathrm{~g} / \mathrm{L}
$$


836. Check Your Learning -
What is the density of molecular hydrogen gas at $17.0^{\circ} \mathrm{C}$ and a pressure of 760 torr?


837. Answer: -
$\mathrm{d}=0.0847 \mathrm{~g} / \mathrm{L}$

When the identity of a gas is unknown, measurements of the mass, pressure, volume, and temperature of a sample can be used to calculate the molar mass of the gas (a useful property for identification purposes). Combining the ideal gas equation

$$
P V=n R T
$$

and the definition of molar mass

$$
\mathscr{M}=\frac{m}{n}
$$

yields the following equation:

$$
\mathscr{M}=\frac{m R T}{P V}
$$

Determining the molar mass of a gas via this approach is demonstrated in Example 8.12.


839. Determining the Molecular Formula of a Gas from its Molar Mass and Empirical Formula -
Cyclopropane, a gas once used with oxygen as a general anesthetic, is composed of $85.7 \%$ carbon and $14.3 \%$ hydrogen by mass. Find the empirical formula. If 1.56 g of cyclopropane occupies a volume of 1.00 L at 0.984 atm and $50^{\circ} \mathrm{C}$, what is the molecular formula for cyclopropane?


840. Solution -
First determine the empirical formula of the gas. Assume 100 g and convert the percentage of each element into grams. Determine the number of moles of carbon and hydrogen in the $100-\mathrm{g}$ sample of cyclopropane. Divide by the smallest number of moles to relate the number of moles of carbon to the number of moles of hydrogen. In the last step, realize that the smallest whole number ratio is the empirical formula:

$$
\begin{array}{lll}
85.7 \mathrm{~g} \mathrm{C} \times \frac{1 \mathrm{~mol} \mathrm{C}}{12.01 \mathrm{~g} \mathrm{C}}=7.136 \mathrm{~mol} \mathrm{C} & \frac{7.136}{7.136}=1.00 \mathrm{~mol} \mathrm{C} \\
14.3 \mathrm{~g} \mathrm{H} \times \frac{1 \mathrm{~mol} \mathrm{H}}{1.01 \mathrm{~g} \mathrm{H}}=14.158 \mathrm{~mol} \mathrm{H} & \frac{14.158}{7.136}=1.98 \mathrm{~mol} \mathrm{H}
\end{array}
$$

Empirical formula is $\mathrm{CH}_{2}$ [empirical mass (EM) of $14.03 \mathrm{~g} / \mathrm{emp}$ irical unit].
Next, use the provided values for mass, pressure, temperature and volume to compute the molar mass of the gas:

$$
\mathscr{M}=\frac{m R T}{P V}=\frac{(1.56 \mathrm{~g})\left(0.0821 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)(323 \mathrm{~K})}{(0.984 \mathrm{~atm})(1.00 \mathrm{~L})}=42.0 \mathrm{~g} / \mathrm{mol}
$$

Comparing the molar mass to the empirical formula mass shows how many empirical formula units make up a molecule:


840. Solution -
Comparing the molar mass to the empirical formula mass shows how many empirical formula units make up a molecule:

$$
\frac{M}{E M}=\frac{42.0 \mathrm{~g} / \mathrm{mol}}{14.0 \mathrm{~g} / \mathrm{mol}}=3
$$

The molecular formula is thus derived from the empirical formula by multiplying each of its subscripts by three:

$$
\left(\mathrm{CH}_{2}\right)_{3}=\mathrm{C}_{3} \mathrm{H}_{6}
$$


841. Check Your Learning -
Acetylene, a fuel used welding torches, is composed of $92.3 \% \mathrm{C}$ and $7.7 \% \mathrm{H}$ by mass. Find the empirical formula. If 1.10 g of acetylene occupies of volume of 1.00 L at 1.15 atm and $59.5^{\circ} \mathrm{C}$, what is the molecular formula for acetylene?


842. Answer: -
Empirical formula, CH; Molecular formula, $\mathrm{C}_{2} \mathrm{H}_{2}$


844. Determining the Molar Mass of a Volatile Liquid -
The approximate molar mass of a volatile liquid can be determined by:

1. Heating a sample of the liquid in a flask with a tiny hole at the top, which converts the liquid into gas that
may escape through the hole
2. Removing the flask from heat at the instant when the last bit of liquid becomes gas, at which time the flask will be filled with only gaseous sample at ambient pressure
3. Sealing the flask and permitting the gaseous sample to condense to liquid, and then weighing the flask to determine the sample's mass (see Figure 8.19)

FIGURE 8.19 When the volatile liquid in the flask is heated past its boiling point, it becomes gas and drives air out of the flask. At $t_{l \longrightarrow g}$, the flask is filled with volatile liquid gas at the same pressure as the atmosphere. If the flask is then cooled to room temperature, the gas condenses and the mass of the gas that filled the flask, and is now liquid, can be measured. (credit: modification of work by Mark Ott)

Using this procedure, a sample of chloroform gas weighing 0.494 g is collected in a flask with a volume of 129 $\mathrm{cm}^{3}$ at $99.6^{\circ} \mathrm{C}$ when the atmospheric pressure is 742.1 mm Hg . What is the approximate molar mass of chloroform?


845. Solution -
Since $\mathcal{M}=\frac{m}{n}$ and $n=\frac{P V}{R T}$, substituting and rearranging gives $\mathcal{M}=\frac{m R T}{P V}$,
then

$$
\mathcal{M}=\frac{m R T}{P V}=\frac{(0.494 \mathrm{~g}) \times 0.08206 \mathrm{~L} \cdot \mathrm{~atm} / \mathrm{mol} \mathrm{~K} \times 372.8 \mathrm{~K}}{0.976 \mathrm{~atm} \times 0.129 \mathrm{~L}}=120 \mathrm{~g} / \mathrm{mol} .
$$


846. Check Your Learning -
A sample of phosphorus that weighs $3.243 \times 10^{-2} \mathrm{~g}$ exerts a pressure of 31.89 kPa in a $56.0-\mathrm{mL}$ bulb at $550^{\circ} \mathrm{C}$. What are the molar mass and molecular formula of phosphorus vapor?


847. Answer: -
$124 \mathrm{~g} / \mathrm{mol} \mathrm{P}_{4}$


848. The Pressure of a Mixture of Gases: Dalton's Law -
Unless they chemically react with each other, the individual gases in a mixture of gases do not affect each other's pressure. Each individual gas in a mixture exerts the same pressure that it would exert if it were present alone in the container (Figure 8.20). The pressure exerted by each individual gas in a mixture is called its partial pressure. This observation is summarized by Dalton's law of partial pressures: The total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases:

$$
P_{\text {Total }}=P_{A}+P_{B}+P_{C}+\ldots=\Sigma_{\mathrm{i}} P_{\mathrm{i}}
$$

In the equation $P_{\text {Total }}$ is the total pressure of a mixture of gases, $P_{A}$ is the partial pressure of gas $A ; P_{B}$ is the partial pressure of gas $B ; P_{C}$ is the partial pressure of gas $C$; and so on.

FIGURE 8.20 If equal-volume cylinders containing gasses at pressures of $300 \mathrm{kPa}, 450 \mathrm{kPa}$, and 600 kPa are all combined in the same-size cylinder, the total pressure of the gas mixture is 1350 kPa .

The partial pressure of gas A is related to the total pressure of the gas mixture via its mole fraction ( $\boldsymbol{X}$ ), a unit of concentration defined as the number of moles of a component of a solution divided by the total number of moles of all components:

$$
P_{A}=X_{A} \times P_{\text {Total }} \quad \text { where } \quad X_{A}=\frac{n_{A}}{n_{\text {Total }}}
$$

where $P_{A}, X_{A}$, and $n_{A}$ are the partial pressure, mole fraction, and number of moles of gas A, respectively, and $n_{\text {Total }}$ is the number of moles of all components in the mixture.


850. The Pressure of a Mixture of Gases -
A 10.0-L vessel contains $2.50 \times 10^{-3} \mathrm{~mol}$ of $\mathrm{H}_{2}, 1.00 \times 10^{-3} \mathrm{~mol}$ of He , and $3.00 \times 10^{-4} \mathrm{~mol}$ of Ne at $35^{\circ} \mathrm{C}$.
(a) What are the partial pressures of each of the gases?
(b) What is the total pressure in atmospheres?


851. Solution -
The gases behave independently, so the partial pressure of each gas can be determined from the ideal gas equation, using $P=\frac{n R T}{V}$ :

$$
\begin{aligned}
& P_{\mathrm{H}_{2}}=\frac{\left(2.50 \times 10^{-3} \mathrm{mot}\right)\left(0.08206 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)(308 \mathrm{~K})}{10.0 \mathrm{t}}=6.32 \times 10^{-3} \mathrm{~atm} \\
& P_{\mathrm{He}}=\frac{\left(1.00 \times 10^{-3} \mathrm{mot}\right)\left(0.08206 \mathrm{t} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)(308 \mathrm{~K})}{10.0 \mathrm{t}}=2.53 \times 10^{-3} \mathrm{~atm} \\
& \left.P_{\mathrm{Ne}}=\frac{\left(3.00 \times 10^{-4} \mathrm{mot}\right)\left(0.08206 \mathrm{t} \mathrm{~atm} \mathrm{mmol}^{-1} \mathrm{~K}^{-1}\right)}{10.0 \mathrm{t}}(308 \mathrm{~K})\right)=7.58 \times 10^{-4} \mathrm{~atm}
\end{aligned}
$$

The total pressure is given by the sum of the partial pressures:

$$
P_{\mathrm{T}}=P_{\mathrm{H}_{2}}+P_{\mathrm{He}}+P_{\mathrm{Ne}}=(0.00632+0.00253+0.00076) \mathrm{atm}=9.61 \times 10^{-3} \mathrm{~atm}
$$


852. Check Your Learning -
A 5.73-L flask at $25^{\circ} \mathrm{C}$ contains 0.0388 mol of $\mathrm{N}_{2}, 0.147 \mathrm{~mol}$ of CO , and $0.0803 \mathrm{~mol} \mathrm{of} \mathrm{H}_{2}$. What is the total pressure in the flask in atmospheres?


853. Answer: -
1.137 atm

Here is another example of this concept, but dealing with mole fraction calculations.


855. The Pressure of a Mixture of Gases -
A gas mixture used for anesthesia contains 2.83 mol oxygen, $\mathrm{O}_{2}$, and 8.41 mol nitrous oxide, $\mathrm{N}_{2} \mathrm{O}$. The total pressure of the mixture is 192 kPa .
(a) What are the mole fractions of $\mathrm{O}_{2}$ and $\mathrm{N}_{2} \mathrm{O}$ ?
(b) What are the partial pressures of $\mathrm{O}_{2}$ and $\mathrm{N}_{2} \mathrm{O}$ ?


856. Solution -
The mole fraction is given by $X_{A}=\frac{n_{A}}{n_{\text {Total }}}$ and the partial pressure is $P_{A}=X_{A} \times P_{\text {Total }}$.
For $\mathrm{O}_{2}$,

$$
X_{O_{2}}=\frac{n_{O_{2}}}{n_{\text {Total }}}=\frac{2.83 \mathrm{~mol}}{(2.83+8.41) \mathrm{mol}}=0.252
$$

and $P_{O_{2}}=X_{O_{2}} \times P_{\text {Total }}=0.252 \times 192 \mathrm{kPa}=48.4 \mathrm{kPa}$
For $\mathrm{N}_{2} \mathrm{O}$,

$$
X_{N_{2} O}=\frac{n_{N_{2} O}}{n_{\text {Total }}}=\frac{8.41 \mathrm{~mol}}{(2.83+8.41) \mathrm{mol}}=0.748
$$

and
$P_{N_{2} O}=X_{N_{2} O} \times P_{\text {Total }}=0.748 \times 192 \mathrm{kPa}=144 \mathrm{kPa}$


857. Check Your Learning -
What is the pressure of a mixture of 0.200 g of $\mathrm{H}_{2}, 1.00 \mathrm{~g}$ of $\mathrm{N}_{2}$, and 0.820 g of Ar in a container with a volume of 2.00 L at $20^{\circ} \mathrm{C}$ ?


858. Answer: -
1.87 atm


859. Collection of Gases over Water -
A simple way to collect gases that do not react with water is to capture them in a bottle that has been filled with water and inverted into a dish filled with water. The pressure of the gas inside the bottle can be made equal to the air pressure outside by raising or lowering the bottle. When the water level is the same both inside and outside the bottle (Figure 8.21), the pressure of the gas is equal to the atmospheric pressure, which can be measured with a barometer.

FIGURE 8.21 When a reaction produces a gas that is collected above water, the trapped gas is a mixture of the gas produced by the reaction and water vapor. If the collection flask is appropriately positioned to equalize the water levels both within and outside the flask, the pressure of the trapped gas mixture will equal the atmospheric pressure outside the flask (see the earlier discussion of manometers).

However, there is another factor we must consider when we measure the pressure of the gas by this method. Water evaporates and there is always gaseous water (water vapor) above a sample of liquid water. As a gas is collected over water, it becomes saturated with water vapor and the total pressure of the mixture equals the partial pressure of the gas plus the partial pressure of the water vapor. The pressure of the pure gas is therefore equal to the total pressure minus the pressure of the water vapor-this is referred to as the "dry" gas pressure, that is, the pressure of the gas only, without water vapor. The vapor pressure of water, which is the pressure exerted by water vapor in equilibrium with liquid water in a closed container, depends on the temperature (Figure 8.22); more detailed information on the temperature dependence of water vapor can be found in Table 8.2, and vapor pressure will be discussed in more detail in the chapter on liquids.

FIGURE 8.22 This graph shows the vapor pressure of water at sea level as a function of temperature.

Vapor Pressure of Ice and Water in Various Temperatures at Sea Level


859. Collection of Gases over Water -
FIGURE 8.22 This graph shows the vapor pressure of water at sea level as a function of temperature.

Vapor Pressure of Ice and Water in Various Temperatures at Sea Level

| Temperature $\left({ }^{\circ} \mathrm{C}\right)$ | Pressure (torr) | Temperature ( ${ }^{\circ} \mathrm{C}$ ) | Pressure (torr) | Temperature ( ${ }^{\circ} \mathrm{C}$ ) | Pressure (torr) |
| :---: | :---: | :---: | :---: | :---: | :---: |
| -10 | 1.95 | 18 | 15.5 | 30 | 31.8 |
| -5 | 3.0 | 19 | 16.5 | 35 | 42.2 |
| -2 | 3.9 | 20 | 17.5 | 40 | 55.3 |
| 0 | 4.6 | 21 | 18.7 | 50 | 92.5 |
| 2 | 5.3 | 22 | 19.8 | 60 | 149.4 |
| 4 | 6.1 | 23 | 21.1 | 70 | 233.7 |
| 6 | 7.0 | 24 | 22.4 | 80 | 355.1 |
| 8 | 8.0 | 25 | 23.8 | 90 | 525.8 |
| 10 | 9.2 | 26 | 25.2 | 95 | 633.9 |
| 12 | 10.5 | 27 | 26.7 | 99 | 733.2 |
| 14 | 12.0 | 28 | 28.3 | 100.0 | 760.0 |
| 16 | 13.6 | 29 | 30.0 | 101.0 | 787.6 |

TABLE 8.2


861. Pressure of a Gas Collected Over Water -
If 0.200 L of argon is collected over water at a temperature of $26^{\circ} \mathrm{C}$ and a pressure of 750 torr in a system like that shown in Figure 8.21, what is the partial pressure of argon?


862. Solution -
According to Dalton's law, the total pressure in the bottle ( 750 torr) is the sum of the partial pressure of argon and the partial pressure of gaseous water:

$$
P_{\mathrm{T}}=P_{\mathrm{Ar}}+P_{\mathrm{H}_{2} \mathrm{O}}
$$

Rearranging this equation to solve for the pressure of argon gives:

$$
P_{\mathrm{Ar}}=P_{\mathrm{T}}-P_{\mathrm{H}_{2} \mathrm{O}}
$$

The pressure of water vapor above a sample of liquid water at $26^{\circ} \mathrm{C}$ is 25.2 torr (Appendix E), so:

$$
P_{\mathrm{Ar}}=750 \text { torr }-25.2 \text { torr }=725 \text { torr }
$$


863. Check Your Learning -
A sample of oxygen collected over water at a temperature of $29.0^{\circ} \mathrm{C}$ and a pressure of 764 torr has a volume of 0.560 L . What volume would the dry oxygen have under the same conditions of temperature and pressure?


864. Answer: -
0.583 L


865. Chemical Stoichiometry and Gases -
Chemical stoichiometry describes the quantitative relationships between reactants and products in chemical reactions.

We have previously measured quantities of reactants and products using masses for solids and volumes in conjunction with the molarity for solutions; now we can also use gas volumes to indicate quantities. If we know the volume, pressure, and temperature of a gas, we can use the ideal gas equation to calculate how many moles of the gas are present. If we know how many moles of a gas are involved, we can calculate the volume of a gas at any temperature and pressure.


866. Avogadro's Law Revisited -
Sometimes we can take advantage of a simplifying feature of the stoichiometry of gases that solids and solutions do not exhibit: All gases that show ideal behavior contain the same number of molecules in the same volume (at the same temperature and pressure). Thus, the ratios of volumes of gases involved in a chemical reaction are given by the coefficients in the equation for the reaction, provided that the gas volumes are measured at the same temperature and pressure.

We can extend Avogadro's law (that the volume of a gas is directly proportional to the number of moles of the gas) to chemical reactions with gases: Gases combine, or react, in definite and simple proportions by volume, provided that all gas volumes are measured at the same temperature and pressure. For example, since nitrogen and hydrogen gases react to produce ammonia gas according to $\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)$, a given volume of nitrogen gas reacts with three times that volume of hydrogen gas to produce two times that volume of ammonia gas, if pressure and temperature remain constant.

The explanation for this is illustrated in Figure 8.23. According to Avogadro's law, equal volumes of gaseous $\mathrm{N}_{2}$, $\mathrm{H}_{2}$, and $\mathrm{NH}_{3}$, at the same temperature and pressure, contain the same number of molecules. Because one molecule of $\mathrm{N}_{2}$ reacts with three molecules of $\mathrm{H}_{2}$ to produce two molecules of $\mathrm{NH}_{3}$, the volume of $\mathrm{H}_{2}$ required is three times the volume of $\mathrm{N}_{2}$, and the volume of $\mathrm{NH}_{3}$ produced is two times the volume of $\mathrm{N}_{2}$.

FIGURE 8.23 One volume of $\mathrm{N}_{2}$ combines with three volumes of $\mathrm{H}_{2}$ to form two volumes of $\mathrm{NH}_{3}$.


868. Reaction of Gases -
Propane, $\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})$, is used in gas grills to provide the heat for cooking. What volume of $\mathrm{O}_{2}(\mathrm{~g})$ measured at $25{ }^{\circ} \mathrm{C}$ and 760 torr is required to react with 2.7 L of propane measured under the same conditions of temperature and pressure? Assume that the propane undergoes complete combustion.


869. Solution -
The ratio of the volumes of $\mathrm{C}_{3} \mathrm{H}_{8}$ and $\mathrm{O}_{2}$ will be equal to the ratio of their coefficients in the balanced equation for the reaction:

$$
\begin{array}{lll}
\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) & \longrightarrow & 3 \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\
1 \text { volume }+5 \text { volumes } & & 3 \text { volumes }+4 \text { volumes }
\end{array}
$$

From the equation, we see that one volume of $\mathrm{C}_{3} \mathrm{H}_{8}$ will react with five volumes of $\mathrm{O}_{2}$ :

$$
2.7 \mathrm{LC}_{3} \mathrm{H}_{8}-\frac{5 \mathrm{LO}_{2}}{1 \mathrm{LC}_{3} \mathrm{H}_{8}}=13.5 \mathrm{~L} \mathrm{O}_{2}
$$

A volume of 13.5 L of $\mathrm{O}_{2}$ will be required to react with $2.7 \mathrm{~L}^{\text {of }} \mathrm{C}_{3} \mathrm{H}_{8}$.


870. Check Your Learning -
An acetylene tank for an oxyacetylene welding torch provides 9340 L of acetylene gas, $\mathrm{C}_{2} \mathrm{H}_{2}$, at $0^{\circ} \mathrm{C}$ and 1 atm. How many tanks of oxygen, each providing $7.00 \times 10^{3} \mathrm{~L}$ of $\mathrm{O}_{2}$ at $0^{\circ} \mathrm{C}$ and 1 atm , will be required to burn the acetylene?

$$
2 \mathrm{C}_{2} \mathrm{H}_{2}+5 \mathrm{O}_{2} \longrightarrow 4 \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}
$$


871. Answer: -
3.34 tanks ( $2.34 \times 10^{4} \mathrm{~L}$ )


873. Volumes of Reacting Gases -
Ammonia is an important fertilizer and industrial chemical. Suppose that a volume of 683 billion cubic feet of gaseous ammonia, measured at $25^{\circ} \mathrm{C}$ and 1 atm , was manufactured. What volume of $\mathrm{H}_{2}(\mathrm{~g})$, measured under the same conditions, was required to prepare this amount of ammonia by reaction with $\mathrm{N}_{2}$ ?

$$
\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)
$$


874. Solution -
Because equal volumes of $\mathrm{H}_{2}$ and $\mathrm{NH}_{3}$ contain equal numbers of molecules and each three molecules of $\mathrm{H}_{2}$ that react produce two molecules of $\mathrm{NH}_{3}$, the ratio of the volumes of $\mathrm{H}_{2}$ and $\mathrm{NH}_{3}$ will be equal to $3: 2$. Two volumes of $\mathrm{NH}_{3}$, in this case in units of billion $\mathrm{ft}^{3}$, will be formed from three volumes of $\mathrm{H}_{2}$ :

$$
683 \text { billion } \mathrm{ft}^{3} \mathrm{NH}_{3} \times \frac{3 \text { billion } \mathrm{ft}^{3} \mathrm{H}_{2}}{2 \text { bitlion } \mathrm{ft}^{3} \mathrm{NH}_{3}}=1.02 \times 10^{3} \text { billion } \mathrm{ft}^{3} \mathrm{H}_{2}
$$

The manufacture of 683 billion $\mathrm{ft}^{3}$ of $\mathrm{NH}_{3}$ required 1020 billion $\mathrm{ft}^{3}$ of $\mathrm{H}_{2}$. (At $25^{\circ} \mathrm{C}$ and 1 atm, this is the volume of a cube with an edge length of approximately 1.9 miles.)


875. Check Your Learning -
What volume of $\mathrm{O}_{2}(\mathrm{~g})$ measured at $25{ }^{\circ} \mathrm{C}$ and 760 torr is required to react with 17.0 L of ethylene, $\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})$, measured under the same conditions of temperature and pressure? The products are $\mathrm{CO}_{2}$ and water vapor.


876. Answer: -
51.0 L


878. Volume of Gaseous Product -
What volume of hydrogen at $27^{\circ} \mathrm{C}$ and 723 torr may be prepared by the reaction of 8.88 g of gallium with an excess of hydrochloric acid?

$$
2 \mathrm{Ga}(s)+6 \mathrm{HCl}(a q) \longrightarrow 2 \mathrm{GaCl}_{3}(a q)+3 \mathrm{H}_{2}(g)
$$


879. Solution -
Convert the provided mass of the limiting reactant, Ga , to moles of hydrogen produced:

$$
8.88 \mathrm{~g} \mathrm{Ga} \times \frac{1 \mathrm{molGa}}{69.723 \mathrm{~g} \mathrm{ga}} \times \frac{3 \mathrm{~mol} \mathrm{H}_{2}}{2 \mathrm{molGa}}=0.191 \mathrm{~mol} \mathrm{H}_{2}
$$

Convert the provided temperature and pressure values to appropriate units ( K and atm, respectively), and then use the molar amount of hydrogen gas and the ideal gas equation to calculate the volume of gas:

$$
V=\left(\frac{n R T}{P}\right)=\frac{0.191 \mathrm{mot} \times 0.08206 \mathrm{~L} \mathrm{atmmol}^{-1} \mathrm{~K}^{-1} \times 300 \mathrm{~K}}{0.951 \mathrm{~atm}}=4.94 \mathrm{~L}
$$


880. Check Your Learning -
Sulfur dioxide is an intermediate in the preparation of sulfuric acid. What volume of $\mathrm{SO}_{2}$ at $343{ }^{\circ} \mathrm{C}$ and 1.21 atm is produced by burning 1.00 kg of sulfur in excess oxygen?


881. Answer: -
$1.30 \times 10^{3} \mathrm{~L}$


883. Greenhouse Gases and Climate Change -
The thin skin of our atmosphere keeps the earth from being an ice planet and makes it habitable. In fact, this is due to less than $0.5 \%$ of the air molecules. Of the energy from the sun that reaches the earth, almost $\frac{1}{3}$ is reflected back into space, with the rest absorbed by the atmosphere and the surface of the earth. Some of the energy that the earth absorbs is re-emitted as infrared (IR) radiation, a portion of which passes back out through the atmosphere into space. Most of this IR radiation, however, is absorbed by certain atmospheric gases, effectively trapping heat within the atmosphere in a phenomenon known as the greenhouse effect. This effect maintains global temperatures within the range needed to sustain life on earth. Without our atmosphere, the earth's average temperature would be lower by more than $30^{\circ} \mathrm{C}$ (nearly $60^{\circ} \mathrm{F}$ ). The major greenhouse gases (GHGs) are water vapor, carbon dioxide, methane, and ozone. Since the Industrial Revolution, human activity has been increasing the concentrations of GHGs, which have changed the energy balance and are significantly altering the earth's climate (Figure 8.24).

FIGURE 8.24 Greenhouse gases trap enough of the sun's energy to make the planet habitable-this is known as the greenhouse effect. Human activities are increasing greenhouse gas levels, warming the planet and causing more extreme weather events.

There is strong evidence from multiple sources that higher atmospheric levels of $\mathrm{CO}_{2}$ are caused by human activity, with fossil fuel burning accounting for about $\frac{3}{4}$ of the recent increase in $\mathrm{CO}_{2}$. Reliable data from ice cores reveals that $\mathrm{CO}_{2}$ concentration in the atmosphere is at the highest level in the past 800,000 years; other evidence indicates that it may be at its highest level in 20 million years. In recent years, the $\mathrm{CO}_{2}$ concentration has increased from preindustrial levels of $\sim 280 \mathrm{ppm}$ to more than 400 ppm today (Figure 8.25).

Carbon Dioxide in the Atmosphere


883. Greenhouse Gases and Climate Change -
Carbon Dioxide in the Atmosphere

FIGURE $8.25 \mathrm{CO}_{2}$ levels over the past 700,000 years were typically from 200-300 ppm, with a steep, unprecedented increase over the past 50 years.


884. LINK TO LEARNING -
Click here (http://openstax.org/l/16GlobalWarming) to see a 2-minute video explaining greenhouse gases and global warming.


886. Susan Solomon -
Atmospheric and climate scientist Susan Solomon (Figure 8.26) is the author of one of The New York Times books of the year (The Coldest March, 2001), one of Time magazine's 100 most influential people in the world (2008), and a working group leader of the Intergovernmental Panel on Climate Change (IPCC), which was the recipient of the 2007 Nobel Peace Prize. She helped determine and explain the cause of the formation of the ozone hole over Antarctica, and has authored many important papers on climate change.

She has been awarded the top scientific honors in the US and France (the National Medal of Science and the Grande Medaille, respectively), and is a member of the National Academy of Sciences, the Royal Society, the French Academy of Sciences, and the European Academy of Sciences. Formerly a professor at the University of Colorado, she is now at MIT, and continues to work at NOAA.

For more information, watch this video (http://openstax.org/l/16SusanSolomon) about Susan Solomon.

FIGURE 8.26 Susan Solomon's research focuses on climate change and has been instrumental in determining the cause of the ozone hole over Antarctica. (credit: National Oceanic and Atmospheric Administration)


887. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Define and explain effusion and diffusion
- State Graham's law and use it to compute relevant gas properties

If you have ever been in a room when a piping hot pizza was delivered, you have been made aware of the fact that gaseous molecules can quickly spread throughout a room, as evidenced by the pleasant aroma that soon reaches your nose. Although gaseous molecules travel at tremendous speeds (hundreds of meters per second), they collide with other gaseous molecules and travel in many different directions before reaching the desired target. At room temperature, a gaseous molecule will experience billions of collisions per second. The mean free path is the average distance a molecule travels between collisions. The mean free path increases with decreasing pressure; in general, the mean free path for a gaseous molecule will be hundreds of times the diameter of the molecule

In general, we know that when a sample of gas is introduced to one part of a closed container, its molecules very quickly disperse throughout the container; this process by which molecules disperse in space in response to differences in concentration is called diffusion (shown in Figure 8.27). The gaseous atoms or molecules are, of course, unaware of any concentration gradient, they simply move randomly-regions of higher concentration have more particles than regions of lower concentrations, and so a net movement of species from high to low concentration areas takes place. In a closed environment, diffusion will ultimately result in equal concentrations of gas throughout, as depicted in Figure 8.27. The gaseous atoms and molecules continue to move, but since their concentrations are the same in both bulbs, the rates of transfer between the bulbs are equal (no net transfer of molecules occurs).

FIGURE 8.27 (a) Two gases, $\mathrm{H}_{2}$ and $\mathrm{O}_{2}$, are initially separated. (b) When the stopcock is opened, they mix together.


887. LEARNING OBJECTIVES -
FIGURE 8.27 (a) Two gases, $\mathrm{H}_{2}$ and $\mathrm{O}_{2}$, are initially separated. (b) When the stopcock is opened, they mix together.

The lighter gas, $\mathrm{H}_{2}$, passes through the opening faster than $\mathrm{O}_{2}$, so just after the stopcock is opened, more $\mathrm{H}_{2}$ molecules move to the $\mathrm{O}_{2}$ side than $\mathrm{O}_{2}$ molecules move to the $\mathrm{H}_{2}$ side. (c) After a short time, both the slowermoving $\mathrm{O}_{2}$ molecules and the faster-moving $\mathrm{H}_{2}$ molecules have distributed themselves evenly on both sides of the vessel.

We are often interested in the rate of diffusion, the amount of gas passing through some area per unit time:

$$
\text { rate of diffusion }=\frac{\text { amount of gas passing through an area }}{\text { unit of time }}
$$

The diffusion rate depends on several factors: the concentration gradient (the increase or decrease in concentration from one point to another); the amount of surface area available for diffusion; and the distance the gas particles must travel. Note also that the time required for diffusion to occur is inversely proportional to the rate of diffusion, as shown in the rate of diffusion equation.

A process involving movement of gaseous species similar to diffusion is effusion, the escape of gas molecules through a tiny hole such as a pinhole in a balloon into a vacuum (Figure 8.28). Although diffusion and effusion rates both depend on the molar mass of the gas involved, their rates are not equal; however, the ratios of their rates are the same.

Diffusion

Effusion

FIGURE 8.28 Diffusion involves the unrestricted dispersal of molecules throughout space due to their random motion. When this process is restricted to passage of molecules through very small openings in a physical barrier, the process is called effusion.


887. LEARNING OBJECTIVES -
FIGURE 8.28 Diffusion involves the unrestricted dispersal of molecules throughout space due to their random motion. When this process is restricted to passage of molecules through very small openings in a physical barrier, the process is called effusion.

If a mixture of gases is placed in a container with porous walls, the gases effuse through the small openings in the walls. The lighter gases pass through the small openings more rapidly (at a higher rate) than the heavier ones (Figure 8.29). In 1832, Thomas Graham studied the rates of effusion of different gases and formulated Graham's law of effusion: The rate of effusion of a gas is inversely proportional to the square root of the mass of its particles:

$$
\text { rate of effusion } \propto \frac{1}{\sqrt{\mathcal{M}}}
$$

This means that if two gases A and B are at the same temperature and pressure, the ratio of their effusion rates is inversely proportional to the ratio of the square roots of the masses of their particles:

$$
\frac{\text { rate of effusion of } A}{\text { rate of effusion of } B}=\frac{\sqrt{\mathcal{N}_{B}}}{\sqrt{\mathcal{N}_{A}}}
$$

FIGURE 8.29 The left photograph shows two balloons inflated with different gases, helium (orange) and argon (blue).The right-side photograph shows the balloons approximately 12 hours after being filled, at which time the helium balloon has become noticeably more deflated than the argon balloon, due to the greater effusion rate of the lighter helium gas. (credit: modification of work by Paul Flowers)


889. Applying Graham's Law to Rates of Effusion -
Calculate the ratio of the rate of effusion of hydrogen to the rate of effusion of oxygen.


890. Solution -
From Graham's law, we have:

$$
\frac{\text { rate of effusion of hydrogen }}{\text { rate of effusion of oxygen }}=\frac{\sqrt{32 \mathrm{gmol}^{-1}}}{\sqrt{2 \mathrm{gmol}^{-1}}}=\frac{\sqrt{16}}{\sqrt{1}}=\frac{4}{1}
$$

Hydrogen effuses four times as rapidly as oxygen.


891. Check Your Learning -
At a particular pressure and temperature, nitrogen gas effuses at the rate of $79 \mathrm{~mL} / \mathrm{s}$. Under the same conditions, at what rate will sulfur dioxide effuse?


892. Answer: -
$52 \mathrm{~mL} / \mathrm{s}$


894. Effusion Time Calculations -
It takes 243 s for $4.46 \times 10^{-5} \mathrm{~mol}$ Xe to effuse through a tiny hole. Under the same conditions, how long will it take $4.46 \times 10^{-5} \mathrm{~mol} \mathrm{Ne}$ to effuse?


895. Solution -
It is important to resist the temptation to use the times directly, and to remember how rate relates to time as well as how it relates to mass. Recall the definition of rate of effusion:

$$
\text { rate of effusion }=\frac{\text { amount of gas transferred }}{\text { time }}
$$

and combine it with Graham's law:

$$
\frac{\text { rate of effusion of gas } \mathrm{Xe}}{\text { rate of effusion of gas Ne }}=\frac{\sqrt{\mathcal{M}_{\mathrm{Ne}}}}{\sqrt{\mathcal{M}_{\mathrm{Xe}}}}
$$

To get:

$$
\frac{\frac{\text { amount of Xe transferred }}{\text { time for Xe }}}{\frac{\text { amount of Ne transferred }}{\text { time for Ne }}}=\frac{\sqrt{\mathcal{M}_{\mathrm{Ne}}}}{\sqrt{\mathcal{M}_{\mathrm{Xe}}}}
$$

Noting that amount of $A=$ amount of $B$, and solving for time for $N e$ :

$$
\frac{\frac{\text { amount of } \mathrm{Xe}}{\text { time for } \mathrm{Xe}}}{\frac{\text { amount of Ne }}{\text { time for } \mathrm{Ne}}}=\frac{\text { time for } \mathrm{Ne}}{\text { time for } \mathrm{Xe}}=\frac{\sqrt{\mathcal{M}_{\mathrm{Ne}}}}{\sqrt{\mathcal{M}_{\mathrm{Xe}}}}=\frac{\sqrt{\mathcal{M}_{\mathrm{Ne}}}}{\sqrt{\mathcal{M}_{\mathrm{Xe}}}}
$$

and substitute values:

$$
\frac{\text { time for } \mathrm{Ne}}{243 \mathrm{~s}}=\sqrt{\frac{20.2 \mathrm{~g} \mathrm{~mol}}{131.3 \mathrm{~m} \mathrm{~mol}}}=0.392
$$

Finally, solve for the desired quantity:

$$
\text { time for } \mathrm{Ne}=0.392 \times 243 \mathrm{~s}=95.3 \mathrm{~s}
$$


895. Solution -
Finally, solve for the desired quantity:

$$
\text { time for } \mathrm{Ne}=0.392 \times 243 \mathrm{~s}=95.3 \mathrm{~s}
$$

Note that this answer is reasonable: Since Ne is lighter than Xe , the effusion rate for Ne will be larger than that for Xe , which means the time of effusion for Ne will be smaller than that for Xe .


896. Check Your Learning -
A party balloon filled with helium deflates to $\frac{2}{3}$ of its original volume in 8.0 hours. How long will it take an identical balloon filled with the same number of moles of air ( $\mathcal{M}=28.2 \mathrm{~g} / \mathrm{mol}$ ) to deflate to $\frac{1}{2}$ of its original volume?


897. Answer: -
32 h


899. Determining Molar Mass Using Graham's Law -
An unknown gas effuses 1.66 times more rapidly than $\mathrm{CO}_{2}$. What is the molar mass of the unknown gas? Can you make a reasonable guess as to its identity?


900. Solution -
From Graham's law, we have:

$$
\frac{\text { rate of effusion of Unknown }}{\text { rate of effusion of } \mathrm{CO}_{2}}=\frac{\sqrt{\mathcal{M}_{\mathrm{CO}_{2}}}}{\sqrt{\mathcal{M}_{\text {Unknown }}}}
$$

Plug in known data:

$$
\frac{1.66}{1}=\frac{\sqrt{44.0 \mathrm{~g} / \mathrm{mol}}}{\sqrt{\mathcal{M}_{\text {Unknown }}}}
$$

Solve:

$$
\mathcal{M}_{\text {Unknown }}=\frac{44.0 \mathrm{~g} / \mathrm{mol}}{(1.66)^{2}}=16.0 \mathrm{~g} / \mathrm{mol}
$$

The gas could well be $\mathrm{CH}_{4}$, the only gas with this molar mass.


901. Check Your Learning -
Hydrogen gas effuses through a porous container 8.97-times faster than an unknown gas. Estimate the molar mass of the unknown gas.


902. Answer: -
$163 \mathrm{~g} / \mathrm{mol}$


904. Use of Diffusion for Nuclear Energy Applications: Uranium Enrichment -
Gaseous diffusion has been used to produce enriched uranium for use in nuclear power plants and weapons. Naturally occurring uranium contains only $0.72 \%$ of ${ }^{235} \mathrm{U}$, the kind of uranium that is "fissile," that is, capable of sustaining a nuclear fission chain reaction. Nuclear reactors require fuel that is $2-5 \%{ }^{235} \mathrm{U}$, and nuclear bombs need even higher concentrations. One way to enrich uranium to the desired levels is to take advantage of Graham's law. In a gaseous diffusion enrichment plant, uranium hexafluoride ( $\mathrm{UF}_{6}$, the only uranium compound that is volatile enough to work) is slowly pumped through large cylindrical vessels called diffusers, which contain porous barriers with microscopic openings. The process is one of diffusion because the other side of the barrier is not evacuated. The ${ }^{235} \mathrm{UF}_{6}$ molecules have a higher average speed and diffuse through the barrier a little faster than the heavier ${ }^{238} \mathrm{UF}_{6}$ molecules. The gas that has passed through the barrier is slightly enriched in ${ }^{235} \mathrm{UF}_{6}$ and the residual gas is slightly depleted. The small difference in molecular weights between ${ }^{235} \mathrm{UF}_{6}$ and ${ }^{238} \mathrm{UF}_{6}$ only about $0.4 \%$ enrichment, is achieved in one diffuser (Figure 8.30). But by connecting many diffusers in a sequence of stages (called a cascade), the desired level of enrichment can be attained.

FIGURE 8.30 In a diffuser, gaseous $U F_{6}$ is pumped through a porous barrier, which partially separates ${ }^{235} \mathrm{UF}_{6}$ from ${ }^{238} \mathrm{UF}_{6}$ The $\mathrm{UF}_{6}$ must pass through many large diffuser units to achieve sufficient enrichment in ${ }^{235} \mathrm{U}$.


904. Use of Diffusion for Nuclear Energy Applications: Uranium Enrichment -
The large scale separation of gaseous ${ }^{235} \mathrm{UF}_{6}$ from ${ }^{238} \mathrm{UF}_{6}$ was first done during the World War II, at the atomic energy installation in Oak Ridge, Tennessee, as part of the Manhattan Project (the development of the first atomic bomb). Although the theory is simple, this required surmounting many daunting technical challenges to make it work in practice. The barrier must have tiny, uniform holes (about $10^{-6} \mathrm{~cm}$ in diameter) and be porous enough to produce high flow rates. All materials (the barrier, tubing, surface coatings, lubricants, and gaskets) need to be able to contain, but not react with, the highly reactive and corrosive $\mathrm{UF}_{6}$.

Because gaseous diffusion plants require very large amounts of energy (to compress the gas to the high
pressures required and drive it through the diffuser cascade, to remove the heat produced during compression, and so on), it is now being replaced by gas centrifuge technology, which requires far less energy. A current hot political issue is how to deny this technology to Iran, to prevent it from producing enough enriched uranium for them to use to make nuclear weapons.