904. Use of Diffusion for Nuclear Energy Applications: Uranium Enrichment - 904.1. The Kinetic-Molecular Theory
LEARNING OBJECTIVES

- State the postulates of the kinetic-molecular theory
- Use this theory's postulates to explain the gas laws

The gas laws that we have seen to this point, as well as the ideal gas equation, are empirical, that is, they have been derived from experimental observations. The mathematical forms of these laws closely describe the macroscopic behavior of most gases at pressures less than about 1 or 2 atm . Although the gas laws describe relationships that have been verified by many experiments, they do not tell us why gases follow these relationships.

The kinetic molecular theory (KMT) is a simple microscopic model that effectively explains the gas laws described in previous modules of this chapter. This theory is based on the following five postulates described here. (Note: The term "molecule" will be used to refer to the individual chemical species that compose the gas, although some gases are composed of atomic species, for example, the noble gases.)

1. Gases are composed of molecules that are in continuous motion, travelling in straight lines and changing direction only when they collide with other molecules or with the walls of a container.
2. The molecules composing the gas are negligibly small compared to the distances between them.
3. The pressure exerted by a gas in a container results from collisions between the gas molecules and the container walls.
4. Gas molecules exert no attractive or repulsive forces on each other or the container walls; therefore, their collisions are elastic (do not involve a loss of energy).
5. The average kinetic energy of the gas molecules is proportional to the kelvin temperature of the gas.

The test of the KMT and its postulates is its ability to explain and describe the behavior of a gas. The various gas laws can be derived from the assumptions of the KMT, which have led chemists to believe that the assumptions of the theory accurately represent the properties of gas molecules. We will first look at the individual gas laws (Boyle's, Charles's, Amontons's, Avogadro's, and Dalton's laws) conceptually to see how the KMT explains them. Then, we will more carefully consider the relationships between molecular masses, speeds, and kinetic energies with temperature, and explain Graham's law.


905. The Kinetic-Molecular Theory Explains the Behavior of Gases, Part I -
Recalling that gas pressure is exerted by rapidly moving gas molecules and depends directly on the number of molecules hitting a unit area of the wall per unit of time, we see that the KMT conceptually explains the behavior of a gas as follows:

- Amontons's law. If the temperature is increased, the average speed and kinetic energy of the gas molecules increase. If the volume is held constant, the increased speed of the gas molecules results in more frequent and more forceful collisions with the walls of the container, therefore increasing the pressure (Figure 8.31).
- Charles's law. If the temperature of a gas is increased, a constant pressure may be maintained only if the volume occupied by the gas increases. This will result in greater average distances traveled by the molecules to reach the container walls, as well as increased wall surface area. These conditions will decrease the both the frequency of molecule-wall collisions and the number of collisions per unit area, the combined effects of which balance the effect of increased collision forces due to the greater kinetic energy at the higher temperature.
- Boyle's law. If the gas volume volume of a given amount of gas at a given temperature is decreased (that is, if the gas is compressed), the molecules will be exposed to a decreased container wall area. Collisions with
the container wall will therefore occur more frequently and the pressure exerted by the gas will increase (Figure 8.31).
- Avogadro's law. At constant pressure and temperature, the frequency and force of molecule-wall collisions are constant. Under such conditions, increasing the number of gaseous molecules will require a proportional increase in the container volume in order to yield a decrease in the number of collisions per unit area to compensate for the increased frequency of collisions (Figure 8.31).
- Dalton's Law. Because of the large distances between them, the molecules of one gas in a mixture bombard the container walls with the same frequency whether other gases are present or not, and the total pressure of a gas mixture equals the sum of the (partial) pressures of the individual gases.

FIGURE 8.31 (a) When gas temperature increases, gas pressure increases due to increased force and frequency of molecular collisions. (b) When volume decreases, gas pressure increases due to increased frequency of molecular collisions. (c) When the amount of gas increases at a constant pressure, volume increases to yield a constant number of collisions per unit wall area per unit time.


906. molecular speeds and Kinetic Energy -
The previous discussion showed that the KMT qualitatively explains the behaviors described by the various gas laws. The postulates of this theory may be applied in a more quantitative fashion to derive these individual laws. To do this, we must first look at speeds and kinetic energies of gas molecules, and the temperature of a gas sample.

In a gas sample, individual molecules have widely varying speeds; however, because of the vast number of molecules and collisions involved, the molecular speed distribution and average speed are constant. This molecular speed distribution is known as a Maxwell-Boltzmann distribution, and it depicts the relative numbers of molecules in a bulk sample of gas that possesses a given speed (Figure 8.32).

FIGURE 8.32 The molecular speed distribution for oxygen gas at 300 K is shown here. Very few molecules move at either very low or very high speeds. The number of molecules with intermediate speeds increases rapidly up to a maximum, which is the most probable speed, then drops off rapidly. Note that the most probable speed, $v_{p}$, is a little less than $400 \mathrm{~m} / \mathrm{s}$, while the root mean square speed, $u_{\mathrm{rms}}$, is closer to $500 \mathrm{~m} / \mathrm{s}$.

The kinetic energy (KE) of a particle of mass ( $m$ ) and speed $(u)$ is given by:

$$
\mathrm{KE}=\frac{1}{2} m u^{2}
$$

Expressing mass in kilograms and speed in meters per second will yield energy values in units of joules ( $\mathrm{J}=\mathrm{kg}$ $\mathrm{m}^{2} \mathrm{~s}^{-2}$ ). To deal with a large number of gas molecules, we use averages for both speed and kinetic energy. In the KMT, the root mean square speed of a particle, $\mathbf{u}_{\mathbf{r m s}}$, is defined as the square root of the average of the squares of the speeds with $n=$ the number of particles:


906. molecular speeds and Kinetic Energy -
$$
u_{\mathrm{rms}}=\sqrt{\overline{u^{2}}}=\sqrt{\frac{u_{1}^{2}+u_{2}^{2}+u_{3}^{2}+u_{4}^{2}+\ldots}{n}}
$$

The average kinetic energy for a mole of particles, $\mathrm{KE}_{\text {avg }}$, is then equal to:

$$
\mathrm{KE}_{\mathrm{avg}}=\frac{1}{2} M u_{\mathrm{rms}}^{2}
$$

where $M$ is the molar mass expressed in units of $\mathrm{kg} / \mathrm{mol}$. The $\mathrm{KE}_{\text {avg }}$ of a mole of gas molecules is also directly proportional to the temperature of the gas and may be described by the equation:

$$
\mathrm{KE}_{\mathrm{avg}}=\frac{3}{2} R T
$$

where $R$ is the gas constant and T is the kelvin temperature. When used in this equation, the appropriate form of the gas constant is $8.314 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}\left(8.314 \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)$. These two separate equations for $\mathrm{KE}_{\text {avg }}$ may be combined and rearranged to yield a relation between molecular speed and temperature:

$$
\begin{gathered}
\frac{1}{2} M u_{\mathrm{rms}}^{2}=\frac{3}{2} R T \\
u_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}}
\end{gathered}
$$


908. Calculation of $u_{\text {rms }}$ -
Calculate the root-mean-square speed for a nitrogen molecule at $30^{\circ} \mathrm{C}$.


909. Solution -
Convert the temperature into Kelvin:

$$
30^{\circ} \mathrm{C}+273=303 \mathrm{~K}
$$

Determine the molar mass of nitrogen in kilograms:

$$
\frac{28.0 \frac{\mathrm{~g}}{\mathrm{o}}}{1 \mathrm{~mol}} \times \frac{1 \mathrm{~kg}}{1000 \frac{\mathrm{~g}}{\mathrm{o}}}=0.028 \mathrm{~kg} / \mathrm{mol}
$$

Replace the variables and constants in the root-mean-square speed equation, replacing Joules with the equivalent $\mathrm{kg} \mathrm{m}{ }^{2} \mathrm{~s}^{-2}$ :

$$
\begin{gathered}
u_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}} \\
u_{\mathrm{rms}}=\sqrt{\frac{3(8.314 \mathrm{~J} / \mathrm{mol} \mathrm{~K})(303 \mathrm{~K})}{(0.028 \mathrm{~kg} / \mathrm{mol})}}=\sqrt{2.70 \times 10^{5} \mathrm{~m}^{2} \mathrm{~s}^{-2}}=519 \mathrm{~m} / \mathrm{s}
\end{gathered}
$$


910. Check Your Learning -
Calculate the root-mean-square speed for a mole of oxygen molecules at $-23^{\circ} \mathrm{C}$.


911. Answer: -
441 m/s

If the temperature of a gas increases, its $\mathrm{KE}_{\text {avg }}$ increases, more molecules have higher speeds and fewer molecules have lower speeds, and the distribution shifts toward higher speeds overall, that is, to the right. If temperature decreases, $\mathrm{KE}_{\text {avg }}$ decreases, more molecules have lower speeds and fewer molecules have higher speeds, and the distribution shifts toward lower speeds overall, that is, to the left. This behavior is illustrated for nitrogen gas in Figure 8.33.

FIGURE 8.33 The molecular speed distribution for nitrogen gas $\left(N_{2}\right)$ shifts to the right and flattens as the temperature increases; it shifts to the left and heightens as the temperature decreases.

At a given temperature, all gases have the same $\mathrm{KE}_{\text {avg }}$ for their molecules. Gases composed of lighter molecules have more high-speed particles and a higher $u_{r m s}$, with a speed distribution that peaks at relatively higher speeds. Gases consisting of heavier molecules have more low-speed particles, a lower $u_{r m s}$, and a speed distribution that peaks at relatively lower speeds. This trend is demonstrated by the data for a series of noble gases shown in Figure 8.34.

FIGURE 8.34 molecular speed is directly related to molecular mass. At a given temperature, lighter molecules move faster on average than heavier molecules.


912. LINK TO LEARNING -
The gas simulator (http://openstax.org/l/16MolecVelocity) may be used to examine the effect of temperature on molecular speeds. Examine the simulator's "energy histograms" (molecular speed distributions) and "species information" (which gives average speed values) for molecules of different masses at various temperatures.

The Kinetic-Molecular Theory Explains the Behavior of Gases, Part II
According to Graham's law, the molecules of a gas are in rapid motion and the molecules themselves are small. The average distance between the molecules of a gas is large compared to the size of the molecules. As a consequence, gas molecules can move past each other easily and diffuse at relatively fast rates.

The rate of effusion of a gas depends directly on the (average) speed of its molecules:
effusion rate $\propto u_{\text {rms }}$
Using this relation, and the equation relating molecular speed to mass, Graham's law may be easily derived as shown here:

$$
\begin{gathered}
u_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}} \\
M=\frac{3 R T}{u_{\mathrm{rms}}^{2}}=\frac{3 R T}{\bar{u}^{2}} \\
\frac{\text { effusion rate } \mathrm{A}}{\text { effusion rate } \mathrm{B}}=\frac{u_{\mathrm{rms} \mathrm{~A}}}{u_{\mathrm{rms} \mathrm{~B}}}=\frac{\sqrt{\frac{3 R T}{M_{\mathrm{A}}}}}{\sqrt{\frac{3 R T}{M_{\mathrm{B}}}}}=\sqrt{\frac{M_{\mathrm{B}}}{M_{\mathrm{A}}}}
\end{gathered}
$$

The ratio of the rates of effusion is thus derived to be inversely proportional to the ratio of the square roots of their masses. This is the same relation observed experimentally and expressed as Graham's law.


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

- Describe the physical factors that lead to deviations from ideal gas behavior
- Explain how these factors are represented in the van der Waals equation
- Define compressibility (Z) and describe how its variation with pressure reflects non-ideal behavior
- Quantify non-ideal behavior by comparing computations of gas properties using the ideal gas law and the van der Waals equation

Thus far, the ideal gas law, $P V=n R T$, has been applied to a variety of different types of problems, ranging from
reaction stoichiometry and empirical and molecular formula problems to determining the density and molar mass of a gas. As mentioned in the previous modules of this chapter, however, the behavior of a gas is often non-ideal, meaning that the observed relationships between its pressure, volume, and temperature are not accurately described by the gas laws. In this section, the reasons for these deviations from ideal gas behavior are considered.

One way in which the accuracy of $P V=n R T$ can be judged is by comparing the actual volume of 1 mole of gas (its molar volume, $V_{\mathrm{m}}$ ) to the molar volume of an ideal gas at the same temperature and pressure. This ratio is called the compressibility factor ( $\mathbf{Z}$ ) with:

$$
\mathrm{Z}=\frac{\text { molar volume of gas at same } T \text { and } P}{\text { molar volume of ideal gas at same } T \text { and } P}=\left(\frac{P V_{m}}{R T}\right)_{\text {measured }}
$$

Ideal gas behavior is therefore indicated when this ratio is equal to 1 , and any deviation from 1 is an indication of non-ideal behavior. Figure 8.35 shows plots of Z over a large pressure range for several common gases.

FIGURE 8.35 A graph of the compressibility factor $(Z)$ vs. pressure shows that gases can exhibit significant deviations from the behavior predicted by the ideal gas law.

As is apparent from Figure 8.35, the ideal gas law does not describe gas behavior well at relatively high pressures. To determine why this is, consider the differences between real gas properties and what is expected of a hypothetical ideal gas.


913. LEARNING OBJECTIVES -
As is apparent from Figure 8.35, the ideal gas law does not describe gas behavior well at relatively high pressures. To determine why this is, consider the differences between real gas properties and what is expected of a hypothetical ideal gas.

Particles of a hypothetical ideal gas have no significant volume and do not attract or repel each other. In general, real gases approximate this behavior at relatively low pressures and high temperatures. However, at high pressures, the molecules of a gas are crowded closer together, and the amount of empty space between the molecules is reduced. At these higher pressures, the volume of the gas molecules themselves becomes appreciable relative to the total volume occupied by the gas. The gas therefore becomes less compressible at these high pressures, and although its volume continues to decrease with increasing pressure, this decrease is not proportional as predicted by Boyle's law.

At relatively low pressures, gas molecules have practically no attraction for one another because they are (on average) so far apart, and they behave almost like particles of an ideal gas. At higher pressures, however, the force of attraction is also no longer insignificant. This force pulls the molecules a little closer together, slightly decreasing the pressure (if the volume is constant) or decreasing the volume (at constant pressure) (Figure 8.36). This change is more pronounced at low temperatures because the molecules have lower KE relative to the attractive forces, and so they are less effective in overcoming these attractions after colliding with one another.

FIGURE 8.36 (a) Attractions between gas molecules serve to decrease the gas volume at constant pressure compared to an ideal gas whose molecules experience no attractive forces. (b) These attractive forces will decrease the force of collisions between the molecules and container walls, therefore reducing the pressure exerted at constant volume compared to an ideal gas.

There are several different equations that better approximate gas behavior than does the ideal gas law. The first, and simplest, of these was developed by the Dutch scientist Johannes van der Waals in 1879. The van der Waals equation improves upon the ideal gas law by adding two terms: one to account for the volume of the gas molecules and another for the attractive forces between them.

$$
P V=n R T \longrightarrow\left(P+\frac{a n^{2}}{V^{2}}\right)(V-n b)=n R T
$$


913. LEARNING OBJECTIVES -
$$
P V=n R T \longrightarrow\left(P+\frac{a n^{2}}{V^{2}}\right)(V-n b)=n R T
$$

The constant a corresponds to the strength of the attraction between molecules of a particular gas, and the constant $b$ corresponds to the size of the molecules of a particular gas. The "correction" to the pressure term in the ideal gas law is $\frac{n^{2} a}{V^{2}}$, and the "correction" to the volume is $n b$. Note that when $V$ is relatively large and $n$ is relatively small, both of these correction terms become negligible, and the van der Waals equation reduces to the ideal gas law, $P V=n R T$. Such a condition corresponds to a gas in which a relatively low number of molecules is occupying a relatively large volume, that is, a gas at a relatively low pressure. Experimental values for the van der Waals constants of some common gases are given in Table 8.3.

Values of van der Waals Constants for Some Common Gases

| Gas | $\boldsymbol{a}\left(\mathrm{L}^{2} \mathrm{~atm} / \mathrm{mol}^{2}\right)$ | $\boldsymbol{b}(\mathrm{L} / \mathrm{mol})$ |
| :--- | :--- | :--- |
| $\mathrm{N}_{2}$ | 1.39 | 0.0391 |
| $\mathrm{O}_{2}$ | 1.36 | 0.0318 |
| $\mathrm{CO}_{2}$ | 3.59 | 0.0427 |
| $\mathrm{H}_{2} \mathrm{O}$ | 5.46 | 0.0305 |
| He | 0.0342 | 0.0237 |
| $\mathrm{CCl}_{4}$ | 20.4 | 0.1383 |

TABLE 8.3


913. LEARNING OBJECTIVES -
TABLE 8.3

At low pressures, the correction for intermolecular attraction, $a$, is more important than the one for molecular volume, $b$. At high pressures and small volumes, the correction for the volume of the molecules becomes important because the molecules themselves are incompressible and constitute an appreciable fraction of the total volume. At some intermediate pressure, the two corrections have opposing influences and the gas appears to follow the relationship given by $P V=n R T$ over a small range of pressures. This behavior is reflected by the "dips" in several of the compressibility curves shown in Figure 8.35. The attractive force between molecules initially makes the gas more compressible than an ideal gas, as pressure is raised ( Z decreases with increasing $P$ ). At very high pressures, the gas becomes less compressible ( Z increases with $P$ ), as the gas molecules begin to occupy an increasingly significant fraction of the total gas volume.

Strictly speaking, the ideal gas equation functions well when intermolecular attractions between gas molecules are negligible and the gas molecules themselves do not occupy an appreciable part of the whole volume. These criteria are satisfied under conditions of low pressure and high temperature. Under such conditions, the gas is said to behave ideally, and deviations from the gas laws are small enough that they may be disregarded-this is, however, very often not the case.


915. Comparison of Ideal Gas Law and van der Waals Equation -
A 4.25-L flask contains $3.46 \mathrm{~mol} \mathrm{CO}_{2}$ at $229^{\circ} \mathrm{C}$. Calculate the pressure of this sample of $\mathrm{CO}_{2}$ :
(a) from the ideal gas law
(b) from the van der Waals equation
(c) Explain the reason(s) for the difference.


916. Solution -
(a) From the ideal gas law:

$$
P=\frac{n R T}{V}=\frac{3.46 \mathrm{mot} \times 0.08206 \mathrm{t} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times 502 \mathrm{~K}}{4.25 \mathrm{t}}=33.5 \mathrm{~atm}
$$

(b) From the van der Waals equation:

$$
\begin{gathered}
\left(P+\frac{n^{2} a}{V^{2}}\right) \times(V-n b)=n R T \rightarrow P=\frac{n R T}{(V-n b)}-\frac{n^{2} a}{V^{2}} \\
P=\frac{3.46 \mathrm{~mol} \times 0.08206 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}}{}=1 \mathrm{~K}^{-1} \times 502 \mathrm{~K} \\
\left(4.25 \mathrm{~L}-3.46 \mathrm{~mol} \times 0.0427 \mathrm{~L} \mathrm{~mol}^{-1}\right)
\end{gathered} \frac{(3.46 \mathrm{~mol})^{2} \times 3.59 \mathrm{~L}^{2} \mathrm{~atm} \mathrm{~mol}^{2}}{(4.25 \mathrm{~L})^{2}} .
$$

This finally yields $P=32.4 \mathrm{~atm}$.
(c) This is not very different from the value from the ideal gas law because the pressure is not very high and the temperature is not very low. The value is somewhat different because $\mathrm{CO}_{2}$ molecules do have some volume and attractions between molecules, and the ideal gas law assumes they do not have volume or attractions.


917. Check your Learning -
A $560-\mathrm{mL}$ flask contains $21.3 \mathrm{~g} \mathrm{~N} \mathrm{~N}_{2}$ at $145^{\circ} \mathrm{C}$. Calculate the pressure of $\mathrm{N}_{2}$ :
(a) from the ideal gas law
(b) from the van der Waals equation
(c) Explain the reason(s) for the difference.


918. Answer: -
(a) 46.562 atm ; (b) 46.594 atm ; (c) The van der Waals equation takes into account the volume of the gas molecules themselves as well as intermolecular attractions.


919. Key Terms -
absolute zero temperature at which the volume of a gas would be zero according to Charles's law.
Amontons's law (also, Gay-Lussac's law) pressure of a given number of moles of gas is directly proportional to its kelvin temperature when the volume is held constant
atmosphere (atm) unit of pressure; 1 atm = $101,325 \mathrm{~Pa}$
Avogadro's law volume of a gas at constant temperature and pressure is proportional to the number of gas molecules
bar (bar or b) unit of pressure; $1 \mathrm{bar}=100,000 \mathrm{~Pa}$
barometer device used to measure atmospheric pressure
Boyle's law volume of a given number of moles of gas held at constant temperature is inversely proportional to the pressure under which it is measured
Charles's law volume of a given number of moles of gas is directly proportional to its kelvin temperature when the pressure is held constant
compressibility factor ( $Z$ ) ratio of the experimentally measured molar volume for a gas to its molar volume as computed from the ideal gas equation
Dalton's law of partial pressures total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases.
diffusion movement of an atom or molecule from a region of relatively high concentration to one of relatively low concentration (discussed in this chapter with regard to gaseous species, but applicable to species in any phase)
effusion transfer of gaseous atoms or molecules from a container to a vacuum through very small openings
Graham's law of effusion rates of diffusion and effusion of gases are inversely proportional to the square roots of their molecular masses
hydrostatic pressure pressure exerted by a fluid due to gravity
ideal gas hypothetical gas whose physical properties are perfectly described by the gas laws
ideal gas constant ( $\boldsymbol{R}$ ) constant derived from the
$8.314 \mathrm{~L} \mathrm{kPa} \mathrm{mol}^{-1} \mathrm{~K}^{-1}$
ideal gas law relation between the pressure, volume, amount, and temperature of a gas under conditions derived by combination of the simple gas laws
kinetic molecular theory theory based on simple principles and assumptions that effectively explains ideal gas behavior
manometer device used to measure the pressure of a gas trapped in a container


919. Key Terms -
kinetic molecular theory theory based on simple principles and assumptions that effectively explains ideal gas behavior
manometer device used to measure the pressure of a gas trapped in a container
mean free path average distance a molecule travels between collisions
mole fraction ( $\boldsymbol{X}$ ) concentration unit defined as the ratio of the molar amount of a mixture component to the total number of moles of all mixture components
partial pressure pressure exerted by an individual gas in a mixture
pascal (Pa) SI unit of pressure; $1 \mathrm{~Pa}=1 \mathrm{~N} / \mathrm{m}^{2}$
pounds per square inch (psi) unit of pressure common in the US
pressure force exerted per unit area
rate of diffusion amount of gas diffusing through a given area over a given time
root mean square speed ( $\boldsymbol{u}_{\text {rms }}$ ) measure of average speed for a group of particles calculated as the square root of the average squared speed
standard conditions of temperature and pressure (STP) $273.15 \mathrm{~K}\left(0^{\circ} \mathrm{C}\right)$ and $1 \mathrm{~atm}(101.325 \mathrm{kPa})$
standard molar volume volume of 1 mole of gas at STP, approximately 22.4 L for gases behaving ideally
torr unit of pressure; 1 torr $=\frac{1}{760} \mathrm{~atm}$
van der Waals equation modified version of the ideal gas equation containing additional terms to account for non-ideal gas behavior
vapor pressure of water pressure exerted by water vapor in equilibrium with liquid water in a closed container at a specific temperature


920. Key Equations -
$$
P=\frac{F}{A}
$$

$p=h \rho g$
$P V=n R T$
$P_{\text {Total }}=P_{A}+P_{B}+P_{C}+\ldots=\Sigma_{\mathrm{i}} P_{\mathrm{i}}$
$P_{A}=X_{A} P_{\text {Total }}$
$X_{A}=\frac{n_{A}}{n_{\text {Total }}}$
rate of diffusion $=\frac{\text { amount of gas passing through an area }}{\text { unit of time }}$
$\frac{\text { rate of effusion of gas } \mathrm{A}}{\text { rate of effusion of gas } \mathrm{B}}=\frac{\sqrt{m_{B}}}{\sqrt{m_{A}}}=\frac{\sqrt{\mathcal{N}_{B}}}{\sqrt{\mathcal{N}_{A}}}$
$u_{\mathrm{rms}}=\sqrt{\overline{u^{2}}}=\sqrt{\frac{u_{1}^{2}+u_{2}^{2}+u_{3}^{2}+u_{4}^{2}+\ldots}{n}}$
$\mathrm{KE}_{\mathrm{avg}}=\frac{3}{2} R T$
$u_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}}$
$\mathrm{Z}=\frac{\text { molar volume of gas at same } T \text { and } P}{\text { molar volume of ideal gas at same } T \text { and } P}=\left(\frac{P \times V_{m}}{R \times T}\right)_{\text {measured }}$
$\left(P+\frac{n^{2} a}{V^{2}}\right) \times(V-n b)=n R T$


921. Summary - 921.1. Gas Pressure
Gases exert pressure, which is force per unit area. The pressure of a gas may be expressed in the SI unit of pascal or kilopascal, as well as in many other units including torr, atmosphere, and bar. Atmospheric pressure is measured using a barometer; other gas pressures can be measured using one of several types of manometers.


921. Summary - 921.2. Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law
The behavior of gases can be described by several laws based on experimental observations of their properties. The pressure of a given amount of gas is directly proportional to its absolute temperature, provided that the volume does not change (Amontons's law). The volume of a given gas sample is directly proportional to its absolute temperature at constant pressure (Charles's law). The volume of a given amount of gas is inversely proportional to its pressure when temperature is held constant (Boyle's law). Under the same conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules (Avogadro's law).

The equations describing these laws are special cases of the ideal gas law, $P V=n R T$, where $P$ is the pressure of the gas, $V$ is its volume, $n$ is the number of moles of the gas, $T$ is its kelvin temperature, and $R$ is the ideal (universal) gas constant.


921. Summary - 921.3. Stoichiometry of Gaseous Substances, Mixtures, and Reactions
The ideal gas law can be used to derive a number of convenient equations relating directly measured
quantities to properties of interest for gaseous substances and mixtures. Appropriate rearrangement of the ideal gas equation may be made to permit the calculation of gas densities and molar masses. Dalton's law of partial pressures may be used to relate measured gas pressures for gaseous mixtures to their compositions. Avogadro's law may be used in stoichiometric computations for chemical reactions involving gaseous reactants or products.


921. Summary - 921.4. Effusion and Diffusion of Gases
Gaseous atoms and molecules move freely and randomly through space. Diffusion is the process whereby gaseous atoms and molecules are transferred from regions of relatively high concentration to regions of relatively low concentration. Effusion is a similar process in which gaseous species pass from a container to a vacuum through very small orifices. The rates of effusion of gases are inversely proportional to the square roots of their densities or to the square roots of their atoms/molecules' masses (Graham's law).


921. Summary - 921.5. The Kinetic-Molecular Theory
The kinetic molecular theory is a simple but very effective model that effectively explains ideal gas behavior. The theory assumes that gases consist of widely separated molecules of negligible volume that are in constant motion, colliding elastically with one another and the walls of their container with average speeds determined by their absolute temperatures. The individual molecules of a gas exhibit a range of speeds, the distribution of these speeds being dependent on the temperature of the
gas and the mass of its molecules.


921. Summary - 921.6. Non-Ideal Gas Behavior
Gas molecules possess a finite volume and experience forces of attraction for one another. Consequently, gas behavior is not necessarily described well by the ideal gas law. Under conditions of low pressure and high temperature, these factors are negligible, the ideal gas equation is an accurate
description of gas behavior, and the gas is said to exhibit ideal behavior. However, at lower temperatures and higher pressures, corrections for molecular volume and molecular attractions are required to account for finite molecular size and attractive forces. The van der Waals equation is a modified version of the ideal gas law that can be used to account for the non-ideal behavior of gases under these conditions.


922. Exercises - 922.1. Gas Pressure
1. Why are sharp knives more effective than dull knives? (Hint: Think about the definition of pressure.)
2. Why do some small bridges have weight limits that depend on how many wheels or axles the crossing vehicle has?
3. Why should you roll or belly-crawl rather than walk across a thinly-frozen pond?
4. A typical barometric pressure in Redding, California, is about 750 mm Hg . Calculate this pressure in atm and kPa.
5. A typical barometric pressure in Denver, Colorado, is 615 mm Hg . What is this pressure in atmospheres and kilopascals?
6. A typical barometric pressure in Kansas City is 740 torr. What is this pressure in atmospheres, in millimeters of mercury, and in kilopascals?
7. Canadian tire pressure gauges are marked in units of kilopascals. What reading on such a gauge corresponds to 32 psi?
8. During the Viking landings on Mars, the atmospheric pressure was determined to be on the average about 6.50 millibars ( $1 \mathrm{bar}=0.987 \mathrm{~atm}$ ). What is that pressure in torr and kPa ?
9. The pressure of the atmosphere on the surface of the planet Venus is about 88.8 atm. Compare that pressure in psi to the normal pressure on earth at sea level in psi.
10. A medical laboratory catalog describes the pressure in a cylinder of a gas as 14.82 MPa . What is the pressure of this gas in atmospheres and torr?
11. Consider this scenario and answer the following questions: On a mid-August day in the northeastern United States, the following information appeared in the local newspaper: atmospheric pressure at sea level 29.97 in. Hg, 1013.9 mbar.
(a) What was the pressure in kPa ?
(b) The pressure near the seacoast in the northeastern United States is usually reported near $30.0 \mathrm{in} . \mathrm{Hg}$. During a hurricane, the pressure may fall to near 28.0 in . Hg. Calculate the drop in pressure in torr.
12. Why is it necessary to use a nonvolatile liquid in a barometer or manometer?


922. Exercises - 922.1. Gas Pressure
12. Why is it necessary to use a nonvolatile liquid in a barometer or manometer?
13. The pressure of a sample of gas is measured at sea level 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


922. Exercises - 922.1. Gas Pressure
14. The pressure of a sample of gas is measured with an open-end manometer, partially shown to the right. The liquid in the manometer is mercury. Assuming atmospheric pressure is 29.92 in . Hg, determine the pressure of the gas in:
(a) torr
(b) Pa
(c) bar

15. The pressure of a sample of gas is measured at sea level with an open-end mercury manometer. Assuming atmospheric pressure is 760.0 mm Hg , determine the pressure of the gas in:
(a) mm Hg
(b) atm
(c) kPa

16. The pressure of a sample of gas is measured at sea level with an open-end mercury manometer. Assuming atmospheric pressure is 760 mm Hg , determine the pressure of the gas in:
(a) mm Hg
(b) atm
(c) kPa

17. How would the use of a volatile liquid affect the measurement of a gas using open-ended manometers vs. closed-end manometers?


922. Exercises - 922.2. Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law
18. Sometimes leaving a bicycle in the sun on a hot day will cause a blowout. Why?
19. Explain how the volume of the bubbles exhausted by a scuba diver (Figure 8.16) change as they rise to the surface, assuming that they remain intact.
20. One way to state Boyle's law is "All other things being equal, the pressure of a gas is inversely proportional to its volume." (a) What is the meaning of the term "inversely proportional?" (b) What are the "other things" that must be equal?
21. An alternate way to state Avogadro's law is "All other things being equal, the number of molecules in a gas is directly proportional to the volume of the gas." (a) What is the meaning of the term "directly proportional?" (b) What are the "other things" that must be equal?
22. How would the graph in Figure 8.12 change if the number of moles of gas in the sample used to determine the curve were doubled?
23. How would the graph in Figure 8.13 change if the number of moles of gas in the sample used to determine the curve were doubled?
24. In addition to the data found in Figure 8.13, what other information do we need to find the mass of the sample of air used to determine the graph?
25. Determine the volume of 1 mol of $\mathrm{CH}_{4}$ gas at 150 K and 1 atm, using Figure 8.12.
26. Determine the pressure of the gas in the syringe shown in Figure 8.13 when its volume is 12.5 mL , using:
(a) the appropriate graph
(b) Boyle's law
27. A spray can is used until it is empty except for the propellant gas, which has a pressure of 1344 torr at 23 ${ }^{\circ} \mathrm{C}$. If the can is thrown into a fire $\left(\mathrm{T}=475^{\circ} \mathrm{C}\right)$, what will be the pressure in the hot can?


922. Exercises - 922.2. Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law
28. What is the temperature of an 11.2 - L sample of carbon monoxide, CO , at 744 torr if it occupies 13.3 L at 55 ${ }^{\circ} \mathrm{C}$ and 744 torr?
29. A $2.50-\mathrm{L}$ volume of hydrogen measured at $-196^{\circ} \mathrm{C}$ is warmed to $100^{\circ} \mathrm{C}$. Calculate the volume of the gas at the higher temperature, assuming no change in pressure.
30. A balloon inflated with three breaths of air has a volume of 1.7 L . At the same temperature and pressure, what is the volume of the balloon if five more same-sized breaths are added to the balloon?
31. A weather balloon contains 8.80 moles of helium at a pressure of 0.992 atm and a temperature of $25^{\circ} \mathrm{C}$ at ground level. What is the volume of the balloon under these conditions?


922. Exercises - 922.2. Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law
32. The volume of an automobile air bag was 66.8 L when inflated at $25^{\circ} \mathrm{C}$ with 77.8 g of nitrogen gas. What was the pressure in the bag in kPa ?
33. How many moles of gaseous boron trifluoride, $\mathrm{BF}_{3}$, are contained in a $4.3410-\mathrm{L}$ bulb at 788.0 K if the pressure is 1.220 atm ? How many grams of $\mathrm{BF}_{3}$ ?
34. Iodine, $\mathrm{I}_{2}$, is a solid at room temperature but sublimes (converts from a solid into a gas) when warmed. What is the temperature in a $73.3-\mathrm{mL}$ bulb that contains 0.292 g of $\mathrm{I}_{2}$ vapor at a pressure of 0.462 atm ?
35. How many grams of gas are present in each of the following cases?
(a) 0.100 L of $\mathrm{CO}_{2}$ at 307 torr and $26^{\circ} \mathrm{C}$
(b) $8.75 \mathrm{~L}^{\text {of }} \mathrm{C}_{2} \mathrm{H}_{4}$, at 378.3 kPa and 483 K
(c) 221 mL of Ar at 0.23 torr and $-54^{\circ} \mathrm{C}$
36. A high altitude balloon is filled with $1.41 \times 10^{4} \mathrm{~L}$ of hydrogen at a temperature of $21^{\circ} \mathrm{C}$ and a pressure of 745 torr. What is the volume of the balloon at a height of 20 km , where the temperature is $-48^{\circ} \mathrm{C}$ and the pressure is 63.1 torr?


922. Exercises - 922.2. Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law
37. A cylinder of medical oxygen has a volume of 35.4 L , and contains $\mathrm{O}_{2}$ at a pressure of 151 atm and a temperature of $25^{\circ} \mathrm{C}$. What volume of $\mathrm{O}_{2}$ does this correspond to at normal body conditions, that is, 1 atm and $37^{\circ} \mathrm{C}$ ?
38. A large scuba tank (Figure 8.16) with a volume of 18 L is rated for a pressure of 220 bar. The tank is filled at $20^{\circ} \mathrm{C}$ and contains enough air to supply 1860 L of air to a diver at a pressure of 2.37 atm (a depth of 45 feet). Was the tank filled to capacity at $20^{\circ} \mathrm{C}$ ?
39. A 20.0-L cylinder containing 11.34 kg of butane, $\mathrm{C}_{4} \mathrm{H}_{10}$, was opened to the atmosphere. Calculate the mass of the gas remaining in the cylinder if it were opened and the gas escaped until the pressure in the cylinder was equal to the atmospheric pressure, 0.983 atm , and a temperature of $27^{\circ} \mathrm{C}$.
40. While resting, the average $70-\mathrm{kg}$ human male consumes 14 L of pure $\mathrm{O}_{2}$ per hour at $25^{\circ} \mathrm{C}$ and 100 kPa . How many moles of $\mathrm{O}_{2}$ are consumed by a 70 kg man while resting for 1.0 h ?
41. For a given amount of gas showing ideal behavior, draw labeled graphs of:
(a) the variation of $P$ with $V$
(b) the variation of $V$ with $T$
(c) the variation of $P$ with $T$
(d) the variation of $\frac{1}{P}$ with $V$
42. A liter of methane gas, $\mathrm{CH}_{4}$, at STP contains more atoms of hydrogen than does a liter of pure hydrogen gas, $\mathrm{H}_{2}$, at STP. Using Avogadro's law as a starting point, explain why.


922. Exercises - 922.2. Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law
43. The effect of chlorofluorocarbons (such as $\mathrm{CCl}_{2} \mathrm{~F}_{2}$ ) on the depletion of the ozone layer is well known. The use of substitutes, such as $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{~F}(g)$, for the chlorofluorocarbons, has largely corrected the problem. Calculate the volume occupied by 10.0 g of each of these compounds at STP:
(a) $\mathrm{CCl}_{2} \mathrm{~F}_{2}(g)$
(b) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{~F}(g)$
44. As 1 g of the radioactive element radium decays over 1 year, it produces $1.16 \times 10^{18}$ alpha particles (helium nuclei). Each alpha particle becomes an atom of helium gas. What is the pressure in pascal of the helium gas produced if it occupies a volume of 125 mL at a temperature of $25^{\circ} \mathrm{C}$ ?
45. A balloon with a volume of 100.21 L at $21^{\circ} \mathrm{C}$ and 0.981 atm is released and just barely clears the top of Mount Crumpit in British Columbia. If the final volume of the balloon is 144.53 L at a temperature of 5.24 ${ }^{\circ} \mathrm{C}$, what is the pressure experienced by the balloon as it clears Mount Crumpet?
46. If the temperature of a fixed amount of a gas is doubled at constant volume, what happens to the pressure?
47. If the volume of a fixed amount of a gas is tripled at constant temperature, what happens to the pressure?


922. Exercises - 922.3. Stoichiometry of Gaseous Substances, Mixtures, and Reactions
48. What is the density of laughing gas, dinitrogen monoxide, $\mathrm{N}_{2} \mathrm{O}$, at a temperature of 325 K and a pressure of 113.0 kPa ?
49. Calculate the density of Freon $12, \mathrm{CF}_{2} \mathrm{Cl}_{2}$, at $30.0^{\circ} \mathrm{C}$ and 0.954 atm.
50. Which is denser at the same temperature and pressure, dry air or air saturated with water vapor? Explain.
51. A cylinder of $\mathrm{O}_{2}(\mathrm{~g})$ used in breathing by patients with emphysema has a volume of 3.00 L at a pressure of 10.0 atm . If the temperature of the cylinder is $28.0^{\circ} \mathrm{C}$, what mass of oxygen is in the cylinder?
52. What is the molar mass of a gas if 0.0494 g of the gas occupies a volume of 0.100 L at a temperature $26^{\circ} \mathrm{C}$ and a pressure of 307 torr?
53. What is the molar mass of a gas if 0.281 g of the gas occupies a volume of 125 mL at a temperature $126^{\circ} \mathrm{C}$ and a pressure of 777 torr?
54. How could you show experimentally that the molecular formula of propene is $\mathrm{C}_{3} \mathrm{H}_{6}$, not $\mathrm{CH}_{2}$ ?
55. The density of a certain gaseous fluoride of phosphorus is $3.93 \mathrm{~g} / \mathrm{L}$ at STP. Calculate the molar mass of this fluoride and determine its molecular formula.
56. Consider this question: What is the molecular formula of a compound that contains $39 \% \mathrm{C}, 45 \% \mathrm{~N}$, and $16 \% \mathrm{H}$ if 0.157 g of the compound occupies 125 mL with a pressure of 99.5 kPa at $22^{\circ} \mathrm{C}$ ?
(a) Outline the steps necessary to answer the question.
(b) Answer the question.


922. Exercises - 922.3. Stoichiometry of Gaseous Substances, Mixtures, and Reactions
(a) Outline the steps necessary to answer the question.
(b) Answer the question.
57. A 36.0-L cylinder of a gas used for calibration of blood gas analyzers in medical laboratories contains 350 $\mathrm{g} \mathrm{CO}_{2}, 805 \mathrm{~g} \mathrm{O}_{2}$, and $4,880 \mathrm{~g} \mathrm{~N}_{2}$. At 25 degrees C , what is the pressure in the cylinder in atmospheres?
58. A cylinder of a gas mixture used for calibration of blood gas analyzers in medical laboratories contains $5.0 \% \mathrm{CO}_{2}, 12.0 \% \mathrm{O}_{2}$, and the remainder $\mathrm{N}_{2}$ at a total pressure of 146 atm . What is the partial pressure of each component of this gas? (The percentages given indicate the percent of the total pressure that is due to each component.)
59. A sample of gas isolated from unrefined petroleum contains $90.0 \% \mathrm{CH}_{4}, 8.9 \% \mathrm{C}_{2} \mathrm{H}_{6}$, and $1.1 \% \mathrm{C}_{3} \mathrm{H}_{8}$ at a total pressure of 307.2 kPa . What is the partial pressure of each component of this gas? (The percentages given indicate the percent of the total pressure that is due to each component.)
60. A mixture of 0.200 g of $\mathrm{H}_{2}, 1.00 \mathrm{~g}$ of $\mathrm{N}_{2}$, and 0.820 g of Ar is stored in a closed container at STP. Find the volume of the container, assuming that the gases exhibit ideal behavior.
61. Most mixtures of hydrogen gas with oxygen gas are explosive. However, a mixture that contains less than $3.0 \% \mathrm{O}_{2}$ is not. If enough $\mathrm{O}_{2}$ is added to a cylinder of $\mathrm{H}_{2}$ at 33.2 atm to bring the total pressure to 34.5 atm , is the mixture explosive?


922. Exercises - 922.3. Stoichiometry of Gaseous Substances, Mixtures, and Reactions
62. A commercial mercury vapor analyzer can detect, in air, concentrations of gaseous Hg atoms (which are poisonous) as low as $2 \times 10^{-6} \mathrm{mg} / \mathrm{L}$ of air. At this concentration, what is the partial pressure of gaseous mercury if the atmospheric pressure is 733 torr at $26^{\circ} \mathrm{C}$ ?
63. A sample of carbon monoxide was collected over water at a total pressure of 756 torr and a temperature of $18{ }^{\circ} \mathrm{C}$. What is the pressure of the carbon monoxide? (See Table 8.2 for the vapor pressure of water.)
64. In an experiment in a general chemistry laboratory, a student collected a sample of a gas over water. The volume of the gas was 265 mL at a pressure of 753 torr and a temperature of $27^{\circ} \mathrm{C}$. The mass of the gas was 0.472 g . What was the molar mass of the gas?
65. Joseph Priestley first prepared pure oxygen by heating mercuric oxide, HgO :
$2 \mathrm{HgO}(s) \longrightarrow 2 \mathrm{Hg}(l)+\mathrm{O}_{2}(g)$
(a) Outline the steps necessary to answer the following question: What volume of $\mathrm{O}_{2}$ at $23{ }^{\circ} \mathrm{C}$ and 0.975 atm is produced by the decomposition of 5.36 g of HgO ?
(b) Answer the question.
66. Cavendish prepared hydrogen in 1766 by the novel method of passing steam through a red-hot gun barrel:
$4 \mathrm{H}_{2} \mathrm{O}(g)+3 \mathrm{Fe}(s) \longrightarrow \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+4 \mathrm{H}_{2}(g)$


922. Exercises - 922.3. Stoichiometry of Gaseous Substances, Mixtures, and Reactions
(a) Outline the steps necessary to answer the following question: What volume of $\mathrm{H}_{2}$ at a pressure of 745 torr and a temperature of $20^{\circ} \mathrm{C}$ can be prepared from the reaction of 15.0 g of $\mathrm{H}_{2} \mathrm{O}$ ?
(b) Answer the question.
67. The chlorofluorocarbon $\mathrm{CCl}_{2} \mathrm{~F}_{2}$ can be recycled into a different compound by reaction with hydrogen to produce $\mathrm{CH}_{2} \mathrm{~F}_{2}(\mathrm{~g})$, a compound useful in chemical manufacturing:
$\mathrm{CCl}_{2} \mathrm{~F}_{2}(g)+4 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{2} \mathrm{~F}_{2}(g)+2 \mathrm{HCl}(g)$
(a) Outline the steps necessary to answer the following question: What volume of hydrogen at 225 atm and $35.5^{\circ} \mathrm{C}$ would be required to react with 1 ton $\left(1.000 \times 10^{3} \mathrm{~kg}\right)$ of $\mathrm{CCl}_{2} \mathrm{~F}_{2}$ ?
(b) Answer the question.
68. Automobile air bags are inflated with nitrogen gas, which is formed by the decomposition of solid sodium azide $\left(\mathrm{NaN}_{3}\right)$. The other product is sodium metal. Calculate the volume of nitrogen gas at $27{ }^{\circ} \mathrm{C}$ and 756 torr formed by the decomposition of 125 g of sodium azide.
69. Lime, CaO , is produced by heating calcium carbonate, $\mathrm{CaCO}_{3}$; carbon dioxide is the other product.
(a) Outline the steps necessary to answer the following question: What volume of carbon dioxide at 875 K and 0.966 atm is produced by the decomposition of 1 ton $\left(1.000 \times 10^{3} \mathrm{~kg}\right)$ of calcium carbonate?
(b) Answer the question.


922. Exercises - 922.3. Stoichiometry of Gaseous Substances, Mixtures, and Reactions
(b) Answer the question.
70. Before small batteries were available, carbide lamps were used for bicycle lights. Acetylene gas, $\mathrm{C}_{2} \mathrm{H}_{2}$, and solid calcium hydroxide were formed by the reaction of calcium carbide, $\mathrm{CaC}_{2}$, with water. The ignition of the acetylene gas provided the light. Currently, the same lamps are used by some cavers, and calcium carbide is used to produce acetylene for carbide cannons.
(a) Outline the steps necessary to answer the following question: What volume of $\mathrm{C}_{2} \mathrm{H}_{2}$ at 1.005 atm and $12.2{ }^{\circ} \mathrm{C}$ is formed by the reaction of 15.48 g of $\mathrm{CaC}_{2}$ with water?
(b) Answer the question.
71. Calculate the volume of oxygen required to burn 12.00 L of ethane gas, $\mathrm{C}_{2} \mathrm{H}_{6}$, to produce carbon dioxide and water, if the volumes of $\mathrm{C}_{2} \mathrm{H}_{6}$ and $\mathrm{O}_{2}$ are measured under the same conditions of temperature and pressure.
72. What volume of $\mathrm{O}_{2}$ at STP is required to oxidize 8.0 L of NO at STP to $\mathrm{NO}_{2}$ ? What volume of $\mathrm{NO}_{2}$ is produced at STP?
73. Consider the following questions:
(a) What is the total volume of the $\mathrm{CO}_{2}(\mathrm{~g})$ and $\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$ at $600^{\circ} \mathrm{C}$ and 0.888 atm produced by the combustion of 1.00 L of $\mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g})$ measured at STP?
(b) What is the partial pressure of $\mathrm{H}_{2} \mathrm{O}$ in the product gases?
74. Methanol, $\mathrm{CH}_{3} \mathrm{OH}$, is produced industrially by the following reaction:


922. Exercises - 922.3. Stoichiometry of Gaseous Substances, Mixtures, and Reactions
74. Methanol, $\mathrm{CH}_{3} \mathrm{OH}$, is produced industrially by the following reaction:
$\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \xrightarrow{\text { copper catalyst } 300^{\circ} \mathrm{C}, 300 \mathrm{~atm}} \mathrm{CH}_{3} \mathrm{OH}(g)$
Assuming that the gases behave as ideal gases, find the ratio of the total volume of the reactants to the final volume.
75. What volume of oxygen at 423.0 K and a pressure of 127.4 kPa is produced by the decomposition of 129.7 g of $\mathrm{BaO}_{2}$ to BaO and $\mathrm{O}_{2}$ ?
76. A 2.50-L sample of a colorless gas at STP decomposed to give 2.50 L of $\mathrm{N}_{2}$ and 1.25 L of $\mathrm{O}_{2}$ at STP. What is the colorless gas?
77. Ethanol, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$, is produced industrially from ethylene, $\mathrm{C}_{2} \mathrm{H}_{4}$, by the following sequence of reactions:
$3 \mathrm{C}_{2} \mathrm{H}_{4}+2 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{HSO}_{4}+\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{SO}_{4}$
$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{HSO}_{4}+\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{SO}_{4}+3 \mathrm{H}_{2} \mathrm{O} \longrightarrow 3 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+2 \mathrm{H}_{2} \mathrm{SO}_{4}$


922. Exercises - 922.3. Stoichiometry of Gaseous Substances, Mixtures, and Reactions
What volume of ethylene at STP is required to produce 1.000 metric ton ( 1000 kg ) of ethanol if the overall yield of ethanol is $90.1 \%$ ?
78. One molecule of hemoglobin will combine with four molecules of oxygen. If 1.0 g of hemoglobin combines with 1.53 mL of oxygen at body temperature $\left(37^{\circ} \mathrm{C}\right)$ and a pressure of 743 torr, what is the molar mass of hemoglobin?
79. A sample of a compound of xenon and fluorine was confined in a bulb with a pressure of 18 torr. Hydrogen was added to the bulb until the pressure was 72 torr. Passage of an electric spark through the mixture produced Xe and HF. After the HF was removed by reaction with solid KOH , the final pressure of xenon and unreacted hydrogen in the bulb was 36 torr. What is the empirical formula of the xenon fluoride in the original sample? (Note: Xenon fluorides contain only one xenon atom per molecule.)
80. One method of analyzing amino acids is the van Slyke method. The characteristic amino groups $\left(-\mathrm{NH}_{2}\right)$ in protein material are allowed to react with nitrous acid, $\mathrm{HNO}_{2}$, to form $\mathrm{N}_{2}$ gas. From the volume of the gas, the amount of amino acid can be determined. A $0.0604-\mathrm{g}$ sample of a biological sample containing glycine, $\mathrm{CH}_{2}\left(\mathrm{NH}_{2}\right) \mathrm{COOH}$, was analyzed by the van Slyke method and yielded 3.70 mL of $\mathrm{N}_{2}$ collected over water at a pressure of 735 torr and $29^{\circ} \mathrm{C}$. What was the percentage of glycine in the sample?
$\mathrm{CH}_{2}\left(\mathrm{NH}_{2}\right) \mathrm{CO}_{2} \mathrm{H}+\mathrm{HNO}_{2} \longrightarrow \mathrm{CH}_{2}(\mathrm{OH}) \mathrm{CO}_{2} \mathrm{H}+\mathrm{H}_{2} \mathrm{O}+\mathrm{N}_{2}$


922. Exercises - 922.4. Effusion and Diffusion of Gases
81. A balloon filled with helium gas takes 6 hours to deflate to $50 \%$ of its original volume. How long will it take for an identical balloon filled with the same volume of hydrogen gas (instead of helium) to decrease its volume by $50 \%$ ?
82. Explain why the numbers of molecules are not identical in the left- and right-hand bulbs shown in the center illustration of Figure 8.27.
83. Starting with the definition of rate of effusion and Graham's finding relating rate and molar mass, show how to derive the Graham's law equation, relating the relative rates of effusion for two gases to their molecular masses.
84. Heavy water, $\mathrm{D}_{2} \mathrm{O}$ (molar mass $=20.03 \mathrm{~g} \mathrm{~mol}^{-1}$ ), can be separated from ordinary water, $\mathrm{H}_{2} \mathrm{O}$ (molar mass $=$ 18.01), as a result of the difference in the relative rates of diffusion of the molecules in the gas phase. Calculate the relative rates of diffusion of $\mathrm{H}_{2} \mathrm{O}$ and $\mathrm{D}_{2} \mathrm{O}$.
85. Which of the following gases diffuse more slowly than oxygen? $\mathrm{F}_{2}, \mathrm{Ne}, \mathrm{N}_{2} \mathrm{O}, \mathrm{C}_{2} \mathrm{H}_{2}, \mathrm{NO}, \mathrm{Cl}_{2}, \mathrm{H}_{2} \mathrm{~S}$


922. Exercises - 922.4. Effusion and Diffusion of Gases
86. During the discussion of gaseous diffusion for enriching uranium, it was claimed that ${ }^{235} \mathrm{UF}_{6}$ diffuses $0.4 \%$ faster than ${ }^{238} \mathrm{UF}_{6}$. Show the calculation that supports this value. The molar mass of ${ }^{235} \mathrm{UF}_{6}=$ $235.043930+6 \times 18.998403=349.034348 \mathrm{~g} / \mathrm{mol}$, and the molar mass of ${ }^{238} \mathrm{UF}_{6}=238.050788+6 \times$ $18.998403=352.041206 \mathrm{~g} / \mathrm{mol}$.
87. Calculate the relative rate of diffusion of ${ }^{1} \mathrm{H}_{2}$ (molar mass $2.0 \mathrm{~g} / \mathrm{mol}$ ) compared with ${ }^{2} \mathrm{H}_{2}$ (molar mass 4.0 $\mathrm{g} / \mathrm{mol}$ ) and the relative rate of diffusion of $\mathrm{O}_{2}$ (molar mass $32 \mathrm{~g} / \mathrm{mol}$ ) compared with $\mathrm{O}_{3}$ (molar mass $48 \mathrm{~g} /$ mol ).
88. A gas of unknown identity diffuses at a rate of $83.3 \mathrm{~mL} / \mathrm{s}$ in a diffusion apparatus in which carbon dioxide diffuses at the rate of $102 \mathrm{~mL} / \mathrm{s}$. Calculate the molecular mass of the unknown gas.


922. Exercises - 922.4. Effusion and Diffusion of Gases
89. When two cotton plugs, one moistened with ammonia and the other with hydrochloric acid, are simultaneously inserted into opposite ends of a glass tube that is 87.0 cm long, a white ring of $\mathrm{NH}_{4} \mathrm{Cl}$ forms where gaseous $\mathrm{NH}_{3}$ and gaseous HCl first come into contact. $\mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)$ At approximately what distance from the ammonia moistened plug does this occur? (Hint: Calculate the rates of diffusion for both $\mathrm{NH}_{3}$ and HCl , and find out how much faster $\mathrm{NH}_{3}$ diffuses than HCl .)


922. Exercises - 922.5. The Kinetic-Molecular Theory
90. Using the postulates of the kinetic molecular theory, explain why a gas uniformly fills a container of any shape.
91. Can the speed of a given molecule in a gas double at constant temperature? Explain your answer.
92. Describe what happens to the average kinetic energy of ideal gas molecules when the conditions are changed as follows:
(a) The pressure of the gas is increased by reducing the volume at constant temperature.
(b) The pressure of the gas is increased by increasing the temperature at constant volume.
(c) The average speed of the molecules is increased by a factor of 2.
93. The distribution of molecular speeds in a sample of helium is shown in Figure 8.34. If the sample is cooled, will the distribution of speeds look more like that of $\mathrm{H}_{2}$ or of $\mathrm{H}_{2} \mathrm{O}$ ? Explain your answer.
94. What is the ratio of the average kinetic energy of a $\mathrm{SO}_{2}$ molecule to that of an $\mathrm{O}_{2}$ molecule in a mixture of two gases? What is the ratio of the root mean square speeds, $u_{\text {rms }}$, of the two gases?
95. A 1-L sample of CO initially at STP is heated to 546 K , and its volume is increased to 2 L .
(a) What effect do these changes have on the number of collisions of the molecules of the gas per unit area of the container wall?
(b) What is the effect on the average kinetic energy of the molecules?
(c) What is the effect on the root mean square speed of the molecules?
96. The root mean square speed of $\mathrm{H}_{2}$ molecules at $25^{\circ} \mathrm{C}$ is about $1.6 \mathrm{~km} / \mathrm{s}$. What is the root mean square speed of a $\mathrm{N}_{2}$ molecule at $25^{\circ} \mathrm{C}$ ?
97. Answer the following questions:
(a) Is the pressure of the gas in the hot-air balloon shown at the opening of this chapter greater than, less than, or equal to that of the atmosphere outside the balloon?


922. Exercises - 922.5. The Kinetic-Molecular Theory
97. Answer the following questions:
(a) Is the pressure of the gas in the hot-air balloon shown at the opening of this chapter greater than, less than, or equal to that of the atmosphere outside the balloon?
(b) Is the density of the gas in the hot-air balloon shown at the opening of this chapter greater than, less than, or equal to that of the atmosphere outside the balloon?
(c) At a pressure of 1 atm and a temperature of $20^{\circ} \mathrm{C}$, dry air has a density of $1.2256 \mathrm{~g} / \mathrm{L}$. What is the (average) molar mass of dry air?
(d) The average temperature of the gas in a hot-air balloon is $1.30 \times 10^{2}{ }^{\circ} \mathrm{F}$. Calculate its density, assuming the molar mass equals that of dry air.
(e) The lifting capacity of a hot-air balloon is equal to the difference in the mass of the cool air displaced by the balloon and the mass of the gas in the balloon. What is the difference in the mass of 1.00 L of the cool air in part (c) and the hot air in part (d)?
(f) An average balloon has a diameter of 60 feet and a volume of $1.1 \times 10^{5} \mathrm{ft}^{3}$. What is the lifting power of such a balloon? If the weight of the balloon and its rigging is 500 pounds, what is its capacity for carrying passengers and cargo?
(g) A balloon carries 40.0 gallons of liquid propane (density $0.5005 \mathrm{~g} / \mathrm{L}$ ). What volume of $\mathrm{CO}_{2}$ and $\mathrm{H}_{2} \mathrm{O}$ gas is produced by the combustion of this propane?
(h) A balloon flight can last about 90 minutes. If all of the fuel is burned during this time, what is the approximate rate of heat loss (in $\mathrm{kJ} / \mathrm{min}$ ) from the hot air in the bag during the flight?


922. Exercises - 922.5. The Kinetic-Molecular Theory
98. Show that the ratio of the rate of diffusion of Gas 1 to the rate of diffusion of Gas $2, \frac{R_{1}}{R_{2}}$, is the same at $0^{\circ} \mathrm{C}$ and $100^{\circ} \mathrm{C}$.


922. Exercises - 922.6. Non-Ideal Gas Behavior
99. Graphs showing the behavior of several different gases follow. Which of these gases exhibit behavior significantly different from that expected for ideal gases?

Temperature

PV/RT

Temperature

Temperature
100. Explain why the plot of $P V$ for $\mathrm{CO}_{2}$ differs from that of an ideal gas.

101. Under which of the following sets of conditions does a real gas behave most like an ideal gas, and for which conditions is a real gas expected to deviate from ideal behavior? Explain.
(a) high pressure, small volume
(b) high temperature, low pressure
(c) low temperature, high pressure
102. Describe the factors responsible for the deviation of the behavior of real gases from that of an ideal gas.
103. For which of the following gases should the correction for the molecular volume be largest: $\mathrm{CO}, \mathrm{CO}_{2}, \mathrm{H}_{2}, \mathrm{He}, \mathrm{NH}_{3}, \mathrm{SF}_{6}$ ?
104. A 0.245 -L flask contains $0.467 \mathrm{~mol} \mathrm{CO}_{2}$ at $159{ }^{\circ} \mathrm{C}$. Calculate the pressure:
(a) using the ideal gas law
(b) using the van der Waals equation
(c) Explain the reason for the difference.
(d) Identify which correction (that for P or V ) is dominant and why.
105. Answer the following questions:
(a) If XX behaved as an ideal gas, what would its graph of Z vs. P look like?
(b) For most of this chapter, we performed calculations treating gases as ideal. Was this justified?
(c) What is the effect of the volume of gas molecules on Z? Under what conditions is this effect small? When is it large? Explain using an appropriate diagram.
(d) What is the effect of intermolecular attractions on the value of Z? Under what conditions is this effect small? When is it large? Explain using an appropriate diagram.
(e) In general, under what temperature conditions would you expect Z to have the largest deviations from the Z for an ideal gas?

418 8•Exercises


923. CHAPTER 9 Thermochemistry -
Figure 9.1 Sliding a match head along a rough surface initiates a combustion reaction that produces energy in the form of heat and light. (credit: modification of work by Laszlo Ilyes)


924. CHAPTER OUTLINE - 924.3. Enthalpy
9.4 Strengths of Ionic and Covalent Bonds

INTRODUCTION Chemical reactions, such as those that occur when you light a match, involve changes in energy as well as matter. Societies at all levels of development could not function without the energy released by chemical reactions. In 2012, about $85 \%$ of US energy consumption came from the combustion of petroleum products, coal, wood, and garbage. We use this energy to produce electricity ( $38 \%$ ); to transport food, raw materials, manufactured goods, and people (27\%); for industrial production (21\%); and to heat and power our homes and businesses (10\%). ${ }^{\frac{1}{2}}$ While these combustion reactions help us meet our essential energy needs, they are also recognized by the majority of the scientific community as a major contributor to global climate change.

Useful forms of energy are also available from a variety of chemical reactions other than combustion. For example, the energy produced by the batteries in a cell phone, car, or flashlight results from chemical reactions. This chapter introduces many of the basic ideas necessary to explore the relationships between chemical changes and energy, with a focus on thermal energy.


924. CHAPTER OUTLINE - 924.3. Enthalpy
[^6]The chemical bond is simply another form of energy, and its strength is indicated in exactly the same units as any other process involving energy. That is, in joules (or kilojoules) - or, when taken on a molar basis, in J/mol or $\mathrm{kJ} / \mathrm{mol}$. All of the principles involving energy we have examined up to now apply to bond energies exactly as they do to all other forms of energy: that is, energy may change form or be absorbed or released, but it cannot be created or destroyed. The energy of a chemical bond is indicated by the bond enthalpy, taken by convention to be the energy required to break a chemical bond. The same energy is released when a chemical bond is formed. Given that a chemical reaction involves breaking some bonds and making others, the enthalpy change for a chemical reaction can be estimated by analyzing the bonds broken and formed during the reaction. While this procedure gives an estimate of the overall enthalpy, it should be noted that a far more precise method involves using enthalpies of formation; the fact that there can be significant differences between, for example, an $\mathrm{O}-\mathrm{H}$ bond in water and that in acetic acid accounts for the discrepancy in calculating $\Delta H$ using bond enthalpies versus the use of enthalpies of formation.


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

- Define energy, distinguish types of energy, and describe the nature of energy changes that accompany chemical and physical changes
- Distinguish the related properties of heat, thermal energy, and temperature
- Define and distinguish specific heat and heat capacity, and describe the physical implications of both
- Perform calculations involving heat, specific heat, and temperature change

Chemical changes and their accompanying changes in energy are important parts of our everyday world (Figure 9.2). The macronutrients in food (proteins, fats, and carbohydrates) undergo metabolic reactions that provide the energy to keep our bodies functioning. We burn a variety of fuels (gasoline, natural gas, coal) to produce energy for transportation, heating, and the generation of electricity. Industrial chemical reactions use enormous amounts of energy to produce raw materials (such as iron and aluminum). Energy is then used to manufacture those raw materials into useful products, such as cars, skyscrapers, and bridges.

FIGURE 9.2 The energy involved in chemical changes is important to our daily lives: (a) A cheeseburger for lunch provides the energy you need to get through the rest of the day; (b) the combustion of gasoline provides the energy that moves your car (and you) between home, work, and school; and (c) coke, a processed form of coal, provides the energy needed to convert iron ore into iron, which is essential for making many of the products we use daily. (credit a: modification of work by "Pink Sherbet Photography"/Flickr; credit b: modification of work by Jeffery Turner)


925. LEARNING OBJECTIVES -
Over $90 \%$ of the energy we use comes originally from the sun. Every day, the sun provides the earth with almost 10,000 times the amount of energy necessary to meet all of the world's energy needs for that day. Our challenge is to find ways to convert and store incoming solar energy so that it can be used in reactions or chemical processes that are both convenient and nonpolluting. Plants and many bacteria capture solar energy through photosynthesis. We release the energy stored in plants when we burn wood or plant products such as ethanol. We also use this energy to fuel our bodies by eating food that comes directly from plants or from animals that got their energy by eating plants. Burning coal and petroleum also releases stored solar energy: These fuels are fossilized plant and animal matter.

This chapter will introduce the basic ideas of an important area of science concerned with the amount of heat absorbed or released during chemical and physical changes-an area called thermochemistry. The concepts introduced in this chapter are widely used in almost all scientific and technical fields. Food scientists use them to determine the energy content of foods. Biologists study the energetics of living organisms, such as the metabolic combustion of sugar into carbon dioxide and water. The oil, gas, and transportation industries, renewable energy providers, and many others endeavor to find better methods to produce energy for our commercial and personal needs. Engineers strive to improve energy efficiency, find better ways to heat and cool our homes, refrigerate our food and drinks, and meet the energy and cooling needs of computers and electronics, among other applications. Understanding thermochemical principles is essential for chemists, physicists, biologists, geologists, every type of engineer, and just about anyone who studies or does any kind of science.


926. Energy -
Energy can be defined as the capacity to supply heat or do work. One type of work ( $\boldsymbol{w}$ ) is the process of causing matter to move against an opposing force. For example, we do work when we inflate a bicycle tire-we move matter (the air in the pump) against the opposing force of the air already in the tire.

Like matter, energy comes in different types. One scheme classifies energy into two types: potential energy, the energy an object has because of its relative position, composition, or condition, and kinetic energy, the energy that an object possesses because of its motion. Water at the top of a waterfall or dam has potential energy because of its position; when it flows downward through generators, it has kinetic energy that can be used to do work and produce electricity in a hydroelectric plant (Figure 9.3). A battery has potential energy because the chemicals within it can produce electricity that can do work.

(a)

(b)

FIGURE 9.3 (a) Water at a higher elevation, for example, at the top of Victoria Falls, has a higher potential energy than water at a lower elevation. As the water falls, some of its potential energy is converted into kinetic energy. (b) If the water flows through generators at the bottom of a dam, such as the Hoover Dam shown here, its kinetic energy is converted into electrical energy. (credit a: modification of work by Steve Jurvetson; credit b: modification of work by "curimedia"/Wikimedia commons)

Energy can be converted from one form into another, but all of the energy present before a change occurs always exists in some form after the change is completed. This observation is expressed in the law of conservation of energy: during a chemical or physical change, energy can be neither created nor destroyed, although it can be changed in form. (This is also one version of the first law of thermodynamics, as you will learn later.)


926. Energy -
When one substance is converted into another, there is always an associated conversion of one form of energy into another. Heat is usually released or absorbed, but sometimes the conversion involves light, electrical energy, or some other form of energy. For example, chemical energy (a type of potential energy) is stored in the molecules that compose gasoline. When gasoline is combusted within the cylinders of a car's engine, the rapidly expanding gaseous products of this chemical reaction generate mechanical energy (a type of kinetic
energy) when they move the cylinders' pistons.
According to the law of conservation of matter (seen in an earlier chapter), there is no detectable change in the total amount of matter during a chemical change. When chemical reactions occur, the energy changes are relatively modest and the mass changes are too small to measure, so the laws of conservation of matter and energy hold well. However, in nuclear reactions, the energy changes are much larger (by factors of a million or so), the mass changes are measurable, and matter-energy conversions are significant. This will be examined in more detail in a later chapter on nuclear chemistry.


927. Thermal Energy, Temperature, and Heat -
Thermal energy is kinetic energy associated with the random motion of atoms and molecules. Temperature is a quantitative measure of "hot" or "cold." When the atoms and molecules in an object are moving or vibrating quickly, they have a higher average kinetic energy (KE), and we say that the object is "hot." When the atoms and molecules are moving slowly, they have lower average KE, and we say that the object is "cold" (Figure 9.4). Assuming that no chemical reaction or phase change (such as melting or vaporizing) occurs, increasing the amount of thermal energy in a sample of matter will cause its temperature to increase. And, assuming that no chemical reaction or phase change (such as condensation or freezing) occurs, decreasing the amount of thermal energy in a sample of matter will cause its temperature to decrease.

Hot water
(a)

Cold water
(b)

FIGURE 9.4 (a) The molecules in a sample of hot water move more rapidly than (b) those in a sample of cold water.


928. LINK TO LEARNING -
Click on this interactive simulation (http://openstax.org/l/16PHETtempFX) to view the effects of temperature on molecular motion.

Most substances expand as their temperature increases and contract as their temperature decreases. This property can be used to measure temperature changes, as shown in Figure 9.5. The operation of many thermometers depends on the expansion and contraction of substances in response to temperature changes.

FIGURE 9.5 (a) In an alcohol or mercury thermometer, the liquid (dyed red for visibility) expands when heated and contracts when cooled, much more so than the glass tube that contains the liquid. (b) In a bimetallic thermometer, two different metals (such as brass and steel) form a two-layered strip. When heated or cooled, one of the metals (brass) expands or contracts more than the other metal (steel), causing the strip to coil or uncoil. Both types of thermometers have a calibrated scale that indicates the temperature. (credit a: modification of work by "dwstucke"/Flickr)


929. LINK TO LEARNING -
The following demonstration (http://openstax.org/l/16Bimetallic) allows one to view the effects of heating and cooling a coiled bimetallic strip.

Heat ( $\boldsymbol{q}$ ) is the transfer of thermal energy between two bodies at different temperatures. Heat flow (a redundant term, but one commonly used) increases the thermal energy of one body and decreases the thermal energy of the other. Suppose we initially have a high temperature (and high thermal energy) substance (H) and a low temperature (and low thermal energy) substance ( L ). The atoms and molecules in H have a higher average KE than those in L . If we place substance H in contact with substance L , the thermal energy will flow spontaneously from substance H to substance L . The temperature of substance H will decrease, as will the average KE of its molecules; the temperature of substance $L$ will increase, along with the average KE of its molecules. Heat flow will continue until the two substances are at the same temperature (Figure 9.6).

FIGURE 9.6 (a) Substances H and L are initially at different temperatures, and their atoms have different average kinetic energies. (b) When they contact each other, collisions between the molecules result in the transfer of kinetic (thermal) energy from the hotter to the cooler matter. (c) The two objects reach "thermal equilibrium" when both substances are at the same temperature and their molecules have the same average kinetic energy.


930. LINK TO LEARNING -
Click on the PhET simulation (http://openstax.org/l/16PHETenergy) to explore energy forms and changes. Visit
the Energy Systems tab to create combinations of energy sources, transformation methods, and outputs. Click on Energy Symbols to visualize the transfer of energy.

Matter undergoing chemical reactions and physical changes can release or absorb heat. A change that releases heat is called an exothermic process. For example, the combustion reaction that occurs when using an oxyacetylene torch is an exothermic process-this process also releases energy in the form of light as evidenced by the torch's flame (Figure 9.7). A reaction or change that absorbs heat is an endothermic process. A cold pack used to treat muscle strains provides an example of an endothermic process. When the substances in the cold pack (water and a salt like ammonium nitrate) are brought together, the resulting process absorbs heat, leading to the sensation of cold.

(a)

(b)

FIGURE 9.7 (a) An oxyacetylene torch produces heat by the combustion of acetylene in oxygen. The energy released by this exothermic reaction heats and then melts the metal being cut. The sparks are tiny bits of the molten metal flying away. (b) A cold pack uses an endothermic process to create the sensation of cold. (credit a: modification of work by "Skatebiker"/Wikimedia commons)


930. LINK TO LEARNING -
Historically, energy was measured in units of calories (cal). A calorie is the amount of energy required to raise one gram of water by 1 degree C (1 kelvin). However, this quantity depends on the atmospheric pressure and the starting temperature of the water. The ease of measurement of energy changes in calories has meant that the calorie is still frequently used. The Calorie (with a capital C), or large calorie, commonly used in quantifying food energy content, is a kilocalorie. The SI unit of heat, work, and energy is the joule. A joule (J) is defined as the amount of energy used when a force of 1 newton moves an object 1 meter. It is named in honor of the English physicist James Prescott Joule. One joule is equivalent to $1 \mathrm{~kg} \mathrm{~m}{ }^{2} / \mathrm{s}^{2}$, which is also called 1 newton-meter. A kilojoule (kJ) is 1000 joules. To standardize its definition, 1 calorie has been set to equal 4.184 joules.

We now introduce two concepts useful in describing heat flow and temperature change. The heat capacity (C) of a body of matter is the quantity of heat ( $q$ ) it absorbs or releases when it experiences a temperature change $(\Delta T)$ of 1 degree Celsius (or equivalently, 1 kelvin):

$$
C=\frac{q}{\Delta T}
$$

Heat capacity is determined by both the type and amount of substance that absorbs or releases heat. It is therefore an extensive property-its value is proportional to the amount of the substance.

For example, consider the heat capacities of two cast iron frying pans. The heat capacity of the large pan is five times greater than that of the small pan because, although both are made of the same material, the mass of the large pan is five times greater than the mass of the small pan. More mass means more atoms are present in the larger pan, so it takes more energy to make all of those atoms vibrate faster. The heat capacity of the small cast
iron frying pan is found by observing that it takes $18,150 \mathrm{~J}$ of energy to raise the temperature of the pan by $50.0^{\circ} \mathrm{C}$ :


930. LINK TO LEARNING -
$$
C_{\text {small pan }}=\frac{18,140 \mathrm{~J}}{50.0^{\circ} \mathrm{C}}=363 \mathrm{~J} /{ }^{\circ} \mathrm{C}
$$

The larger cast iron frying pan, while made of the same substance, requires $90,700 \mathrm{~J}$ of energy to raise its temperature by $50.0^{\circ} \mathrm{C}$. The larger pan has a (proportionally) larger heat capacity because the larger amount of material requires a (proportionally) larger amount of energy to yield the same temperature change:

$$
C_{\text {large pan }}=\frac{90,700 \mathrm{~J}}{50.0^{\circ} \mathrm{C}}=1814 \mathrm{~J} /{ }^{\circ} \mathrm{C}
$$

The specific heat capacity (c) of a substance, commonly called its "specific heat," is the quantity of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius (or 1 kelvin):

$$
c=\frac{q}{\mathrm{~m} \Delta \mathrm{~T}}
$$

Specific heat capacity depends only on the kind of substance absorbing or releasing heat. It is an intensive property-the type, but not the amount, of the substance is all that matters. For example, the small cast iron frying pan has a mass of 808 g . The specific heat of iron (the material used to make the pan) is therefore:

$$
c_{\text {iron }}=\frac{18,140 \mathrm{~J}}{(808 \mathrm{~g})\left(50.0^{\circ} \mathrm{C}\right)}=0.449 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C}
$$

The large frying pan has a mass of 4040 g . Using the data for this pan, we can also calculate the specific heat of iron:


930. LINK TO LEARNING -
The large frying pan has a mass of 4040 g . Using the data for this pan, we can also calculate the specific heat of iron:

$$
c_{\text {iron }}=\frac{90,700 \mathrm{~J}}{(4040 \mathrm{~g})\left(50.0^{\circ} \mathrm{C}\right)}=0.449 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C}
$$

Although the large pan is more massive than the small pan, since both are made of the same material, they both yield the same value for specific heat (for the material of construction, iron). Note that specific heat is measured in units of energy per temperature per mass and is an intensive property, being derived from a ratio of two extensive properties (heat and mass). The molar heat capacity, also an intensive property, is the heat capacity per mole of a particular substance and has units of $\mathrm{J} / \mathrm{mol}{ }^{\circ} \mathrm{C}$ (Figure 9.8).

FIGURE 9.8 Because of its larger mass, a large frying pan has a larger heat capacity than a small frying pan. Because they are made of the same material, both frying pans have the same specific heat. (credit: Mark Blaser)
Water has a relatively high specific heat (about $4.2 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C}$ for the liquid and $2.09 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C}$ for the solid); most metals have much lower specific heats (usually less than $1 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}$ ). The specific heat of a substance varies somewhat with temperature. However, this variation is usually small enough that we will treat specific heat as constant over the range of temperatures that will be considered in this chapter. Specific heats of some common substances are listed in Table 9.1.

Specific Heats of Common Substances at $25^{\circ} \mathrm{C}$ and 1 bar


930. LINK TO LEARNING -
Specific Heats of Common Substances at $25^{\circ} \mathrm{C}$ and 1 bar

| Substance | Symbol (state) | Specific Heat ( $\mathrm{J} / \mathrm{g}^{\circ} \mathrm{C}$ ) |
| :---: | :---: | :---: |
| helium | $\mathrm{He}(\mathrm{g})$ | 5.193 |
| water | $\mathrm{H}_{2} \mathrm{O}(\mathrm{I})$ | 4.184 |
| ethanol | $\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}(\mathrm{I})$ | 2.376 |
| ice | $\mathrm{H}_{2} \mathrm{O}(s)$ | 2.093 (at $\left.-10^{\circ} \mathrm{C}\right)$ |
| water vapor | $\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$ | 1.864 |
| nitrogen | $\mathrm{N}_{2}(\mathrm{~g})$ | 1.040 |
| air |  | 1.007 |
| oxygen | $\mathrm{O}_{2}(\mathrm{~g})$ | 0.918 |
| aluminum | $\mathrm{Al}(\mathrm{s})$ | 0.897 |
| carbon dioxide | $\mathrm{CO}_{2}(\mathrm{~g})$ | 0.853 |
| argon | $\operatorname{Ar}(\mathrm{g})$ | 0.522 |
| iron | $\mathrm{Fe}(s)$ | 0.449 |
| copper | $\mathrm{Cu}(s)$ | 0.385 |
| lead | $\mathrm{Pb}(s)$ | 0.130 |
| gold | $\mathrm{Au}(s)$ | 0.129 |
| silicon | Si(s) | 0.712 |

TABLE 9.1

If we know the mass of a substance and its specific heat, we can determine the amount of heat, $q$, entering or leaving the substance by measuring the temperature change before and after the heat is gained or lost:


930. LINK TO LEARNING -
If we know the mass of a substance and its specific heat, we can determine the amount of heat, $q$, entering or leaving the substance by measuring the temperature change before and after the heat is gained or lost:

$$
\begin{aligned}
& q=(\text { specific heat }) \times(\text { mass of substance }) \times(\text { temperature change }) \\
& q=c \times m \times \Delta T=c \times m \times\left(T_{\text {final }}-T_{\text {initial }}\right)
\end{aligned}
$$

In this equation, $c$ is the specific heat of the substance, $m$ is its mass, and $\Delta T$ (which is read "delta $T$ ") is the temperature change, $T_{\text {final }}-T_{\text {initial }}$. If a substance gains thermal energy, its temperature increases, its final temperature is higher than its initial temperature, $T_{\text {final }}-T_{\text {initial }}$ has a positive value, and the value of $q$ is positive. If a substance loses thermal energy, its temperature decreases, the final temperature is lower than the initial temperature, $T_{\text {final }}-T_{\text {initial }}$ has a negative value, and the value of $q$ is negative.


932. Measuring Heat -
A flask containing $8.0 \times 10^{2} \mathrm{~g}$ of water is heated, and the temperature of the water increases from $21^{\circ} \mathrm{C}$ to 85 ${ }^{\circ} \mathrm{C}$. How much heat did the water absorb?


933. Solution -
To answer this question, consider these factors:

- the specific heat of the substance being heated (in this case, water)
- the amount of substance being heated (in this case, $8.0 \times 10^{2} \mathrm{~g}$ )
- the magnitude of the temperature change (in this case, from $21^{\circ} \mathrm{C}$ to $85^{\circ} \mathrm{C}$ ).

The specific heat of water is $4.184 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C}$, so to heat 1 g of water by $1^{\circ} \mathrm{C}$ requires 4.184 J . We note that since 4.184 J is required to heat 1 g of water by $1^{\circ} \mathrm{C}$, we will need 800 times as much to heat $8.0 \times 10^{2} \mathrm{~g}$ of water by 1 ${ }^{\circ} \mathrm{C}$. Finally, we observe that since 4.184 J are required to heat 1 g of water by $1^{\circ} \mathrm{C}$, we will need 64 times as much to heat it by $64^{\circ} \mathrm{C}$ (that is, from $21^{\circ} \mathrm{C}$ to $85^{\circ} \mathrm{C}$ ).

This can be summarized using the equation:


933. Solution -
This can be summarized using the equation:

$$
\begin{aligned}
& q=c \times m \times \Delta T=c \times m \times\left(T_{\text {final }}-T_{\text {initial }}\right) \\
= & \left(4.184 \mathrm{~J} / \frac{\mathrm{g}}{}{ }^{\circ} \mathrm{C}\right) \times\left(8.0 \times 10^{2} \frac{\mathrm{~g}}{\mathrm{~g}}\right) \times(85-21)^{\circ} \mathrm{C} \\
= & \left(4.184 \mathrm{~J} / \frac{\mathrm{g}}{}{ }^{\circ} \mathrm{C}\right) \times\left(8.0 \times 10^{2} \frac{\mathrm{~g}}{\mathrm{C}}\right) \times(64)^{\circ} \mathrm{C} \\
= & 210,000 \mathrm{~J}\left(=2.1 \times 10^{2} \mathrm{~kJ}\right)
\end{aligned}
$$

Because the temperature increased, the water absorbed heat and $q$ is positive.


934. Check Your Learning -
How much heat, in joules, must be added to a 502 g iron skillet to increase its temperature from $25^{\circ} \mathrm{C}$ to 250 ${ }^{\circ} \mathrm{C}$ ? The specific heat of iron is $0.449 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C}$.


935. Answer: -
$5.07 \times 10^{4} \mathrm{~J}$

Note that the relationship between heat, specific heat, mass, and temperature change can be used to determine any of these quantities (not just heat) if the other three are known or can be deduced.


937. Determining Other Quantities -
A piece of unknown metal weighs 348 g . When the metal piece absorbs 6.64 kJ of heat, its temperature increases from $22.4^{\circ} \mathrm{C}$ to $43.6^{\circ} \mathrm{C}$. Determine the specific heat of this metal (which might provide a clue to its identity).


938. Solution -
Since mass, heat, and temperature change are known for this metal, we can determine its specific heat using the relationship:

$$
q=c \times m \times \Delta T=c \times m \times\left(T_{\text {final }}-T_{\text {initial }}\right)
$$

Substituting the known values:

$$
6640 \mathrm{~J}=c \times(348 \mathrm{~g}) \times(43.6-22.4)^{\circ} \mathrm{C}
$$

Solving:

$$
c=\frac{6640 \mathrm{~J}}{(348 \mathrm{~g}) \times\left(21.2^{\circ} \mathrm{C}\right)}=0.900 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C}
$$

Comparing this value with the values in Table 9.1, this value matches the specific heat of aluminum, which suggests that the unknown metal may be aluminum.


939. Check Your Learning -
A piece of unknown metal weighs 217 g . When the metal piece absorbs 1.43 kJ of heat, its temperature increases from $24.5^{\circ} \mathrm{C}$ to $39.1^{\circ} \mathrm{C}$. Determine the specific heat of this metal, and predict its identity.


940. Answer: -
$c=0.451 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}$; the metal is likely to be iron


942. Solar Thermal Energy Power Plants -
The sunlight that reaches the earth contains thousands of times more energy than we presently capture. Solar thermal systems provide one possible solution to the problem of converting energy from the sun into energy we can use. Large-scale solar thermal plants have different design specifics, but all concentrate sunlight to heat some substance; the heat "stored" in that substance is then converted into electricity.

The Solana Generating Station in Arizona's Sonora Desert produces 280 megawatts of electrical power. It uses parabolic mirrors that focus sunlight on pipes filled with a heat transfer fluid (HTF) (Figure 9.9). The HTF then does two things: It turns water into steam, which spins turbines, which in turn produces electricity, and it melts and heats a mixture of salts, which functions as a thermal energy storage system. After the sun goes down, the molten salt mixture can then release enough of its stored heat to produce steam to run the turbines for 6 hours. Molten salts are used because they possess a number of beneficial properties, including high heat capacities and thermal conductivities.

FIGURE 9.9 This solar thermal plant uses parabolic trough mirrors to concentrate sunlight. (credit a: modification of work by Bureau of Land Management)

The 377-megawatt Ivanpah Solar Generating System, located in the Mojave Desert in California, is the largest solar thermal power plant in the world (Figure 9.10). Its 170,000 mirrors focus huge amounts of sunlight on three water-filled towers, producing steam at over $538^{\circ} \mathrm{C}$ that drives electricity-producing turbines. It produces enough energy to power 140,000 homes. Water is used as the working fluid because of its large heat capacity and heat of vaporization.

(a)

(b)

FIGURE 9.10 (a) The Ivanpah solar thermal plant uses 170,000 mirrors to concentrate sunlight on water-filled towers. (b) It covers 4000 acres of public land near the Mojave Desert and the California-Nevada border. (credit a: modification of work by Craig Dietrich; credit b: modification of work by "USFWS Pacific Southwest Region"/Flickr)


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

- Explain the technique of calorimetry
- Calculate and interpret heat and related properties using typical calorimetry data

One technique we can use to measure the amount of heat involved in a chemical or physical process is known as calorimetry. Calorimetry is used to measure amounts of heat transferred to or from a substance. To do so, the heat is exchanged with a calibrated object (calorimeter). The temperature change measured by the calorimeter is used to derive the amount of heat transferred by the process under study. The measurement of heat transfer using this approach requires the definition of a system (the substance or substances undergoing the chemical or physical change) and its surroundings (all other matter, including components of the measurement apparatus, that serve to either provide heat to the system or absorb heat from the system).

A calorimeter is a device used to measure the amount of heat involved in a chemical or physical process. For example, when an exothermic reaction occurs in solution in a calorimeter, the heat produced by the reaction is absorbed by the solution, which increases its temperature. When an endothermic reaction occurs, the heat required is absorbed from the thermal energy of the solution, which decreases its temperature (Figure 9.11). The temperature change, along with the specific heat and mass of the solution, can then be used to calculate the amount of heat involved in either case.

FIGURE 9.11 In a calorimetric determination, either (a) an exothermic process occurs and heat, $q$, is negative, indicating that thermal energy is transferred from the system to its surroundings, or (b) an endothermic process occurs and heat, $q$, is positive, indicating that thermal energy is transferred from the surroundings to the system.

Calorimetry measurements are important in understanding the heat transferred in reactions involving everything from microscopic proteins to massive machines. During her time at the National Bureau of Standards, research chemist Reatha Clark King performed calorimetric experiments to understand the precise heats of various flourine compounds. Her work was important to NASA in their quest for better rocket fuels.


943. LEARNING OBJECTIVES -
Scientists use well-insulated calorimeters that all but prevent the transfer of heat between the calorimeter and its environment, which effectively limits the "surroundings" to the nonsystem components with the calorimeter (and the calorimeter itself). This enables the accurate determination of the heat involved in chemical processes, the energy content of foods, and so on. General chemistry students often use simple calorimeters constructed from polystyrene cups (Figure 9.12). These easy-to-use "coffee cup" calorimeters allow more heat exchange with the outside environment, and therefore produce less accurate energy values.

FIGURE 9.12 A simple calorimeter can be constructed from two polystyrene cups. A thermometer and stirrer extend through the cover into the reaction mixture.

Commercial solution calorimeters are also available. Relatively inexpensive calorimeters often consist of two thin-walled cups that are nested in a way that minimizes thermal contact during use, along with an insulated cover, handheld stirrer, and simple thermometer. More expensive calorimeters used for industry and research typically have a well-insulated, fully enclosed reaction vessel, motorized stirring mechanism, and a more accurate temperature sensor (Figure 9.13).

FIGURE 9.13 Commercial solution calorimeters range from (a) simple, inexpensive models for student use to (b) expensive, more accurate models for industry and research.

Before discussing the calorimetry of chemical reactions, consider a simpler example that illustrates the core idea behind calorimetry. Suppose we initially have a high-temperature substance, such as a hot piece of metal $(M)$, and a low-temperature substance, such as cool water (W). If we place the metal in the water, heat will flow from M to W . The temperature of M will decrease, and the temperature of W will increase, until the two substances have the same temperature-that is, when they reach thermal equilibrium (Figure 9.14). If this occurs in a calorimeter, ideally all of this heat transfer occurs between the two substances, with no heat gained or lost by either its external environment. Under these ideal circumstances, the net heat change is zero:

$$
q_{\text {substance } \mathrm{M}}+q_{\text {substance } \mathrm{W}}=0
$$


943. LEARNING OBJECTIVES -
$$
q_{\text {substance } \mathrm{M}}+q_{\text {substance } \mathrm{W}}=0
$$

This relationship can be rearranged to show that the heat gained by substance M is equal to the heat lost by substance W:

$$
q_{\text {substance } \mathrm{M}}=-q_{\text {substance } \mathrm{W}}
$$

The magnitude of the heat (change) is therefore the same for both substances, and the negative sign merely shows that $q_{\text {substance } \mathrm{M}}$ and $q_{\text {substance W }}$ are opposite in direction of heat flow (gain or loss) but does not indicate the arithmetic sign of either $q$ value (that is determined by whether the matter in question gains or loses heat, per definition). In the specific situation described, $q_{\text {substance } M}$ is a negative value and $q_{\text {substance } \mathrm{W}}$ is positive, since heat is transferred from M to W .

FIGURE 9.14 In a simple calorimetry process, (a) heat, $q$, is transferred from the hot metal, $M$, to the cool water, W, until (b) both are at the same temperature.


945. Heat Transfer between Substances at Different Temperatures -
A 360.0-g piece of rebar (a steel rod used for reinforcing concrete) is dropped into 425 mL of water at $24.0^{\circ} \mathrm{C}$. The final temperature of the water was measured as $42.7^{\circ} \mathrm{C}$. Calculate the initial temperature of the piece of rebar. Assume the specific heat of steel is approximately the same as that for iron (Table 9.1), and that all heat transfer occurs between the rebar and the water (there is no heat exchange with the surroundings).


946. Solution -
The temperature of the water increases from $24.0^{\circ} \mathrm{C}$ to $42.7^{\circ} \mathrm{C}$, so the water absorbs heat. That heat came from the piece of rebar, which initially was at a higher temperature. Assuming that all heat transfer was between the rebar and the water, with no heat "lost" to the outside environment, then heat given off by rebar = -heat taken in by water, or:

$$
q_{\text {rebar }}=-q_{\text {water }}
$$

Since we know how heat is related to other measurable quantities, we have:

$$
(c \times m \times \Delta T)_{\mathrm{rebar}}=-(c \times m \times \Delta T)_{\mathrm{water}}
$$

Letting $\mathrm{f}=$ final and $\mathrm{i}=$ initial, in expanded form, this becomes:

$$
c_{\text {rebar }} \times m_{\text {rebar }} \times\left(T_{\mathrm{f}, \text { rebar }}-T_{\mathrm{i}, \text { rebar }}\right)=-c_{\text {water }} \times m_{\text {water }} \times\left(T_{\mathrm{f}, \text { water }}-T_{\mathrm{i}, \text { water }}\right)
$$

The density of water is $1.0 \mathrm{~g} / \mathrm{mL}$, so 425 mL of water $=425 \mathrm{~g}$. Noting that the final temperature of both the rebar and water is $42.7^{\circ} \mathrm{C}$, substituting known values yields:


946. Solution -
$$
\begin{gathered}
\left(0.449 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C}\right)(360.0 \mathrm{~g})\left(42.7{ }^{\circ} \mathrm{C}-T_{\mathrm{i}, \text { rebar }}\right)=-\left(4.184 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C}\right)(425 \mathrm{~g})\left(42.7^{\circ} \mathrm{C}-24.0^{\circ} \mathrm{C}\right) \\
T_{i, r e b a r}=\frac{\left(4.184 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C}\right)(425 \mathrm{~g})\left(42.7^{\circ} \mathrm{C}-24.0^{\circ} \mathrm{C}\right)}{\left(0.449 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C}\right)(360.0 \mathrm{~g})}+42.7^{\circ} \mathrm{C}
\end{gathered}
$$

Solving this gives $T_{i, \text { rebar }}=248^{\circ} \mathrm{C}$, so the initial temperature of the rebar was $248{ }^{\circ} \mathrm{C}$.


947. Check Your Learning -
A $248-\mathrm{g}$ piece of copper is dropped into 390 mL of water at $22.6^{\circ} \mathrm{C}$. The final temperature of the water was
measured as $39.9^{\circ} \mathrm{C}$. Calculate the initial temperature of the piece of copper. Assume that all heat transfer occurs between the copper and the water.


948. Answer: -
The initial temperature of the copper was $335.6^{\circ} \mathrm{C}$.


949. Check Your Learning -
A 248 -g piece of copper initially at $314^{\circ} \mathrm{C}$ is dropped into 390 mL of water initially at $22.6^{\circ} \mathrm{C}$. Assuming that all heat transfer occurs between the copper and the water, calculate the final temperature.


950. Answer: -
The final temperature (reached by both copper and water) is $38.7^{\circ} \mathrm{C}$.

This method can also be used to determine other quantities, such as the specific heat of an unknown metal.


952. Identifying a Metal by Measuring Specific Heat -
A 59.7 g piece of metal that had been submerged in boiling water was quickly transferred into 60.0 mL of water initially at $22.0^{\circ} \mathrm{C}$. The final temperature is $28.5^{\circ} \mathrm{C}$. Use these data to determine the specific heat of the metal. Use this result to identify the metal.


953. Solution -
Assuming perfect heat transfer, heat given off by metal $=-$ heat taken in by water, or:

$$
q_{\text {metal }}=-q_{\text {water }}
$$

In expanded form, this is:

$$
c_{\text {metal }} \times m_{\text {metal }} \times\left(T_{\mathrm{f}, \text { metal }}-T_{\mathrm{i}, \text { metal }}\right)=-c_{\text {water }} \times m_{\text {water }} \times\left(T_{\mathrm{f}, \text { water }}-T_{\mathrm{i}, \text { water }}\right)
$$

Noting that since the metal was submerged in boiling water, its initial temperature was $100.0^{\circ} \mathrm{C}$; and that for water, $60.0 \mathrm{~mL}=60.0 \mathrm{~g}$; we have:

$$
\left(c_{\text {metal }}\right)(59.7 \mathrm{~g})\left(28.5^{\circ} \mathrm{C}-100.0^{\circ} \mathrm{C}\right)=-\left(4.18 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C}\right)(60.0 \mathrm{~g})\left(28.5^{\circ} \mathrm{C}-22.0^{\circ} \mathrm{C}\right)
$$

Solving this:

$$
c_{\text {metal }}=\frac{-\left(4.184 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C}\right)(60.0 \mathrm{~g})\left(6.5^{\circ} \mathrm{C}\right)}{(59.7 \mathrm{~g})\left(-71.5^{\circ} \mathrm{C}\right)}=0.38 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C}
$$


953. Solution -
Comparing this with values in Table 9.1, our experimental specific heat is closest to the value for copper (0.39 $\mathrm{J} / \mathrm{g}{ }^{\circ} \mathrm{C}$ ), so we identify the metal as copper.