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Example 1.2.2 | 2688 | 2688 | A certain food service gives the following choices for dinner: $E_1$, soup or tomato 1.2-2 juice; $E_2$, steak or shrimp; $E_3$, French fried potatoes, mashed potatoes, or a baked potato; $E_4$, corn or peas; $E_5$, jello, tossed salad, cottage cheese, or coleslaw; $E_6$, cake, cookies, pudding, brownie, vanilla ice cream, chocolate ice cream, or orange sherbet; $E_7$, coffee, tea, milk, or punch. How many different dinner selections are possible if one of the listed choices is made for each of $E_1, E_2, \ldots$, and $E_7$ ? | By the multiplication principle, there are
$(2)(2)(3)(2)(4)(7)(4)=2688$
different combinations.
| stat |
||
Example 1.4.5 | 0.9966 | 0.9966 | A rocket has a built-in redundant system. In this system, if component $K_1$ fails, it is bypassed and component $K_2$ is used. If component $K_2$ fails, it is bypassed and component $K_3$ is used. (An example of a system with these kinds of components is three computer systems.) Suppose that the probability of failure of any one component is 0.15 , and assume that the failures of these components are mutually independent events. Let $A_i$ denote the event that component $K_i$ fails for $i=1,2,3$. What is the probability that the system fails? |
Because the system fails if $K_1$ fails and $K_2$ fails and $K_3$ fails, the probability that the system does not fail is given by
$$
\begin{aligned}
P\left[\left(A_1 \cap A_2 \cap A_3\right)^{\prime}\right] & =1-P\left(A_1 \cap A_2 \cap A_3\right) \\
& =1-P\left(A_1\right) P\left(A_2\right) P\left(A_3\right) \\
& =1-(0.15)^3 \\
& =0.9966 .
\end{aligned}
$$ | stat |
||
Example 1.3.3 | \frac{2}{3} | 0.66666666666 | Suppose that $P(A)=0.7, P(B)=0.3$, and $P(A \cap B)=0.2$. Given that the outcome of the experiment belongs to $B$, what then is the probability of $A$ ? | We are effectively restricting the sample space to $B$; of the probability $P(B)=0.3,0.2$ corresponds to $P(A \cap B)$ and hence to $A$. That is, $0.2 / 0.3=2 / 3$ of the probability of $B$ corresponds to $A$. Of course, by the formal definition, we also obtain
$$
P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{0.2}{0.3}=\frac{2}{3}
$$ | stat |
||
Example 1.2.12 | 210 | 210 | A coin is flipped 10 times and the sequence of heads and tails is observed. What is the number of possible 10-tuplets that result in four heads and six tails? | A coin is flipped 10 times and the sequence of heads and tails is observed. The number of possible 10-tuplets that result in four heads and six tails is
$$
\left(\begin{array}{c}
10 \\
4
\end{array}\right)=\frac{10 !}{4 ! 6 !}=\frac{10 !}{6 ! 4 !}=\left(\begin{array}{c}
10 \\
6
\end{array}\right)=210 .
$$
| stat |
||
Example 1.2.14 | 1260 | 1260 | Among nine orchids for a line of orchids along one wall, three are white, four lavender, and two yellow. How many color displays are there? | The number of different color displays is
$$
\left(\begin{array}{c}
9 \\
3,4,2
\end{array}\right)=\frac{9 !}{3 ! 4 ! 2 !}=1260
$$
| stat |
||
Example 1.1.5 | 0.59 | 0.59 | A survey was taken of a group's viewing habits of sporting events on TV during I.I-5 the last year. Let $A=\{$ watched football $\}, B=\{$ watched basketball $\}, C=\{$ watched baseball $\}$. The results indicate that if a person is selected at random from the surveyed group, then $P(A)=0.43, P(B)=0.40, P(C)=0.32, P(A \cap B)=0.29$, $P(A \cap C)=0.22, P(B \cap C)=0.20$, and $P(A \cap B \cap C)=0.15$. Find $P(A \cup B \cup C)$. | $$
\begin{aligned}
P(A \cup B \cup C)= & P(A)+P(B)+P(C)-P(A \cap B)-P(A \cap C) \\
& -P(B \cap C)+P(A \cap B \cap C) \\
= & 0.43+0.40+0.32-0.29-0.22-0.20+0.15 \\
= & 0.59
\end{aligned}
$$ | stat |
||
Example 1.3.10 | $\frac{41}{90}$ | 0.444444444444444 | A grade school boy has five blue and four white marbles in his left pocket and four blue and five white marbles in his right pocket. If he transfers one marble at random from his left to his right pocket, what is the probability of his then drawing a blue marble from his right pocket? | For notation, let $B L, B R$, and $W L$ denote drawing blue from left pocket, blue from right pocket, and white from left pocket, respectively. Then
$$
\begin{aligned}
P(B R) & =P(B L \cap B R)+P(W L \cap B R) \\
& =P(B L) P(B R \mid B L)+P(W L) P(B R \mid W L) \\
& =\frac{5}{9} \cdot \frac{5}{10}+\frac{4}{9} \cdot \frac{4}{10}=\frac{41}{90}
\end{aligned}
$$
is the desired probability. | stat |
||
Example 1.1.4 | 0.95 | 0.95 | A faculty leader was meeting two students in Paris, one arriving by train from Amsterdam and the other arriving by train from Brussels at approximately the same time. Let $A$ and $B$ be the events that the respective trains are on time. Suppose we know from past experience that $P(A)=0.93, P(B)=0.89$, and $P(A \cap B)=0.87$. Find $P(A \cup B)$. | $$P(A \cup B) =P(A)+P(B)-P(A \cap B)=0.93+0.89-0.87=0.95$$ | stat |
||
Example 1.2.4 | 358800 | 358800 | What is the number of possible four-letter code words, selecting from the 26 letters in the alphabet? | The number of possible four-letter code words, selecting from the 26 letters in the alphabet, in which all four letters are different is
$$
{ }_{26} P_4=(26)(25)(24)(23)=\frac{26 !}{22 !}=358,800 .
$$ | stat |
||
$\mathrm{~kJ} \mathrm{~mol}^{-1}$ | 11.9 | -131.1 | -131.1 | Consider the half-cell reaction $\operatorname{AgCl}(s)+\mathrm{e}^{-} \rightarrow$ $\operatorname{Ag}(s)+\mathrm{Cl}^{-}(a q)$. If $\mu^{\circ}(\mathrm{AgCl}, s)=-109.71 \mathrm{~kJ} \mathrm{~mol}^{-1}$, and if $E^{\circ}=+0.222 \mathrm{~V}$ for this half-cell, calculate the standard Gibbs energy of formation of $\mathrm{Cl}^{-}(a q)$. | thermo |
||
$\mathrm{~kJ} \mathrm{~mol}^{-1}$ | 6.37 | 28 | 28 | $\mathrm{N}_2 \mathrm{O}_3$ dissociates according to the equilibrium $\mathrm{N}_2 \mathrm{O}_3(\mathrm{~g}) \rightleftharpoons \mathrm{NO}_2(\mathrm{~g})+\mathrm{NO}(\mathrm{g})$. At $298 \mathrm{~K}$ and one bar pressure, the degree of dissociation defined as the ratio of moles of $\mathrm{NO}_2(g)$ or $\mathrm{NO}(g)$ to the moles of the reactant assuming no dissociation occurs is $3.5 \times 10^{-3}$. Calculate $\Delta G_R^{\circ}$ for this reaction. | thermo |
||
$10^6$ | 1.1 | 1.27 | 1.27 | Approximately how many oxygen molecules arrive each second at the mitochondrion of an active person with a mass of $84 \mathrm{~kg}$ ? The following data are available: Oxygen consumption is about $40 . \mathrm{mL}$ of $\mathrm{O}_2$ per minute per kilogram of body weight, measured at $T=300 . \mathrm{K}$ and $P=1.00 \mathrm{~atm}$. In an adult there are about $1.6 \times 10^{10}$ cells per kg body mass. Each cell contains about 800 . mitochondria. | thermo |
||
$Å$ | 19.46 | Angstrom | 12 | 12 | In a FRET experiment designed to monitor conformational changes in T4 lysozyme, the fluorescence intensity fluctuates between 5000 and 10,000 counts per second.
Assuming that 7500 counts represents a FRET efficiency of 0.5 , what is the change in FRET pair separation distance during the reaction? For the tetramethylrhodamine/texas red FRET pair employed $r_0=50 . Å$. | thermo |
|
$10^8 \mathrm{~J}$ | 5.4 | 4.85 | 4.85 | An air conditioner is a refrigerator with the inside of the house acting as the cold reservoir and the outside atmosphere acting as the hot reservoir. Assume that an air conditioner consumes $1.70 \times 10^3 \mathrm{~W}$ of electrical power, and that it can be idealized as a reversible Carnot refrigerator. If the coefficient of performance of this device is 3.30 , how much heat can be extracted from the house in a day? | thermo |
||
$\mathrm{~Pa}$ | 8.14 | 425 | 425 | You have collected a tissue specimen that you would like to preserve by freeze drying. To ensure the integrity of the specimen, the temperature should not exceed $-5.00{ }^{\circ} \mathrm{C}$. The vapor pressure of ice at $273.16 \mathrm{~K}$ is $624 \mathrm{~Pa}$. What is the maximum pressure at which the freeze drying can be carried out? | thermo |
||
$\mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ | 15.22 | 7.82 | 7.82 | The molar constant volume heat capacity for $\mathrm{I}_2(\mathrm{~g})$ is $28.6 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$. What is the vibrational contribution to the heat capacity? You can assume that the contribution from the electronic degrees of freedom is negligible. | thermo |
||
$\mathrm{~nm}^2$ | 17.1 | 0.318 | 0.318 | The diffusion coefficient for $\mathrm{CO}_2$ at $273 \mathrm{~K}$ and $1 \mathrm{~atm}$ is $1.00 \times 10^{-5} \mathrm{~m}^2 \mathrm{~s}^{-1}$. Estimate the collisional cross section of $\mathrm{CO}_2$ given this diffusion coefficient. | thermo |
||
$10^3 \mathrm{~J}^{\circ} \mathrm{C}^{-1}$
| 4.15 | 6.64 | 6.64 | Benzoic acid, $1.35 \mathrm{~g}$, is reacted with oxygen in a constant volume calorimeter to form $\mathrm{H}_2 \mathrm{O}(l)$ and $\mathrm{CO}_2(g)$ at $298 \mathrm{~K}$. The mass of the water in the inner bath is $1.55 \times$ $10^3 \mathrm{~g}$. The temperature of the calorimeter and its contents rises $2.76 \mathrm{~K}$ as a result of this reaction. Calculate the calorimeter constant. | thermo |
||
18.37 | 0.15 | 0.15 | The activation energy for a reaction is $50 . \mathrm{J} \mathrm{mol}^{-1}$. Determine the effect on the rate constant for this reaction with a change in temperature from $273 \mathrm{~K}$ to $298 \mathrm{~K}$. | thermo |
|||
$\mathrm{~s}$ | 17.21 | 22 | 22 | How long will it take to pass $200 . \mathrm{mL}$ of $\mathrm{H}_2$ at $273 \mathrm{~K}$ through a $10 . \mathrm{cm}$-long capillary tube of $0.25 \mathrm{~mm}$ if the gas input and output pressures are 1.05 and $1.00 \mathrm{~atm}$, respectively? | thermo |
||
$\mathrm{~nm}$ | 10.23 | 1.4 | 1.4 | Calculate the Debye-Hückel screening length $1 / \kappa$ at $298 \mathrm{~K}$ in a $0.0075 \mathrm{~m}$ solution of $\mathrm{K}_3 \mathrm{PO}_4$. | thermo |
||
$\mathrm{~K}$ | 2.13 | 322 | 322 | A system consisting of $82.5 \mathrm{~g}$ of liquid water at $300 . \mathrm{K}$ is heated using an immersion heater at a constant pressure of 1.00 bar. If a current of $1.75 \mathrm{~A}$ passes through the $25.0 \mathrm{ohm}$ resistor for 100 .s, what is the final temperature of the water? | thermo |
||
$\mathrm{~K}$ | 15.4 | 1310 | 1310 | For an ensemble consisting of a mole of particles having two energy levels separated by $1000 . \mathrm{cm}^{-1}$, at what temperature will the internal energy equal $3.00 \mathrm{~kJ}$ ? | thermo |
||
$\mathrm{~J}$ | 2.10 | 0.46 | 0.46 | A muscle fiber contracts by $3.5 \mathrm{~cm}$ and in doing so lifts a weight. Calculate the work performed by the fiber. Assume the muscle fiber obeys Hooke's law $F=-k x$ with a force constant $k$ of $750 . \mathrm{N} \mathrm{m}^{-1}$. | thermo |
||
% | 1.5 | 32 | 32 | A gas sample is known to be a mixture of ethane and butane. A bulb having a $230.0 \mathrm{~cm}^3$ capacity is filled with the gas to a pressure of $97.5 \times 10^3 \mathrm{~Pa}$ at $23.1^{\circ} \mathrm{C}$. If the mass of the gas in the bulb is $0.3554 \mathrm{~g}$, what is the mole percent of butane in the mixture?
| thermo |
||
$10^{21}$ | 1.6 | 1.11 | 1.11 | One liter of fully oxygenated blood can carry 0.18 liters of $\mathrm{O}_2$ measured at $T=298 \mathrm{~K}$ and $P=1.00 \mathrm{~atm}$. Calculate the number of moles of $\mathrm{O}_2$ carried per liter of blood. Hemoglobin, the oxygen transport protein in blood has four oxygen binding sites. How many hemoglobin molecules are required to transport the $\mathrm{O}_2$ in $1.0 \mathrm{~L}$ of fully oxygenated blood? | thermo |
||
$\mathrm{~K}$ | 13.15 | 4152 | 4152 | Consider a collection of molecules where each molecule has two nondegenerate energy levels that are separated by $6000 . \mathrm{cm}^{-1}$. Measurement of the level populations demonstrates that there are exactly 8 times more molecules in the ground state than in the upper state. What is the temperature of the collection? | thermo |
||
$\mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ | 5.14 | -191.2 | -191.2 | Calculate $\Delta S^{\circ}$ for the reaction $3 \mathrm{H}_2(g)+\mathrm{N}_2(g) \rightarrow$ $2 \mathrm{NH}_3(g)$ at $725 \mathrm{~K}$. Omit terms in the temperature-dependent heat capacities higher than $T^2 / \mathrm{K}^2$. | thermo |
||
17.15 | 1.33 | 1.33 | The thermal conductivities of acetylene $\left(\mathrm{C}_2 \mathrm{H}_2\right)$ and $\mathrm{N}_2$ at $273 \mathrm{~K}$ and $1 \mathrm{~atm}$ are 0.01866 and $0.0240 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~m}^{-1} \mathrm{~s}^{-1}$, respectively. Based on these data, what is the ratio of the collisional cross section of acetylene relative to $\mathrm{N}_2$ ? | thermo |
|||
$\mathrm{~s}$ | 18.39 | 269 | 269 | Consider the gas phase thermal decomposition of 1.0 atm of $\left(\mathrm{CH}_3\right)_3 \mathrm{COOC}\left(\mathrm{CH}_3\right)_3(\mathrm{~g})$ to acetone $\left(\mathrm{CH}_3\right)_2 \mathrm{CO}(\mathrm{g})$ and ethane $\left(\mathrm{C}_2 \mathrm{H}_6\right)(\mathrm{g})$, which occurs with a rate constant of $0.0019 \mathrm{~s}^{-1}$. After initiation of the reaction, at what time would you expect the pressure to be $1.8 \mathrm{~atm}$ ? | thermo |
||
$\mathrm{~atm}$ | 8.13 | 1.95 | 1.95 | Autoclaves that are used to sterilize surgical tools require a temperature of $120 .{ }^{\circ} \mathrm{C}$ to kill some bacteria. If water is used for this purpose, at what pressure must the autoclave operate? | thermo |
||
$10^{17}$ | 14.6 | 3.9 | 3.9 | Imagine gaseous $\mathrm{Ar}$ at $298 \mathrm{~K}$ confined to move in a two-dimensional plane of area $1.00 \mathrm{~cm}^2$. What is the value of the translational partition function? | thermo |
||
$10^{-9}$ | 15.47 | 2.25 | 2.25 | Determine the equilibrium constant for the dissociation of sodium at $298 \mathrm{~K}: \mathrm{Na}_2(g) \rightleftharpoons 2 \mathrm{Na}(g)$. For $\mathrm{Na}_2$, $B=0.155 \mathrm{~cm}^{-1}, \widetilde{\nu}=159 \mathrm{~cm}^{-1}$, the dissociation energy is $70.4 \mathrm{~kJ} / \mathrm{mol}$, and the ground-state electronic degeneracy for $\mathrm{Na}$ is 2 . | thermo |
||
$10^3 \mathrm{~Pa}$ | 8.39 | 1.51 | 1.51 | At $298.15 \mathrm{~K}, \Delta G_f^{\circ}(\mathrm{HCOOH}, g)=-351.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\Delta G_f^{\circ}(\mathrm{HCOOH}, l)=-361.4 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Calculate the vapor pressure of formic acid at this temperature. | thermo |
||
$10^{-5} \mathrm{~m}^2 \mathrm{~s}^{-1}$ | 17.2 | 1.06 | 1.06 | The collisional cross section of $\mathrm{N}_2$ is $0.43 \mathrm{~nm}^2$. What is the diffusion coefficient of $\mathrm{N}_2$ at a pressure of $1 \mathrm{~atm}$ and a temperature of $298 \mathrm{~K}$ ? | thermo |
||
$\mathrm{~L}$ | 1.8 | 37.9 | 37.9 | A vessel contains $1.15 \mathrm{~g}$ liq $\mathrm{H}_2 \mathrm{O}$ in equilibrium with water vapor at $30 .{ }^{\circ} \mathrm{C}$. At this temperature, the vapor pressure of $\mathrm{H}_2 \mathrm{O}$ is 31.82 torr. What volume increase is necessary for all the water to evaporate?
| thermo |
||
$10^{-9} \mathrm{~J}$ | 8.7 | 2.89 | 2.89 | A cell is roughly spherical with a radius of $20.0 \times 10^{-6} \mathrm{~m}$. Calculate the work required to expand the cell surface against the surface tension of the surroundings if the radius increases by a factor of three. Assume the cell is surrounded by pure water and that $T=298.15 \mathrm{~K}$. | thermo |
||
$\mathrm{~bar}$ | 3.6 | 93.4 | 93.4 | A vessel is filled completely with liquid water and sealed at $13.56^{\circ} \mathrm{C}$ and a pressure of 1.00 bar. What is the pressure if the temperature of the system is raised to $82.0^{\circ} \mathrm{C}$ ? Under these conditions, $\beta_{\text {water }}=2.04 \times 10^{-4} \mathrm{~K}^{-1}$, $\beta_{\text {vessel }}=1.42 \times 10^{-4} \mathrm{~K}^{-1}$, and $\kappa_{\text {water }}=4.59 \times 10^{-5} \mathrm{bar}^{-1}$. | thermo |
||
$10^4 \mathrm{~m}$ | 12.32 | 1.6 | 1.6 | A crude model for the molecular distribution of atmospheric gases above Earth's surface (denoted by height $h$ ) can be obtained by considering the potential energy due to gravity:
$$
P(h)=e^{-m g h / k T}
$$
In this expression $m$ is the per-particle mass of the gas, $g$ is the acceleration due to gravity, $k$ is a constant equal to $1.38 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}$, and $T$ is temperature. Determine $\langle h\rangle$ for methane $\left(\mathrm{CH}_4\right)$ using this distribution function. | thermo |
||
$\mathrm{~g}$ | 4.33 | 49 | 49 | A camper stranded in snowy weather loses heat by wind convection. The camper is packing emergency rations consisting of $58 \%$ sucrose, $31 \%$ fat, and $11 \%$ protein by weight. Using the data provided in Problem P4.32 and assuming the fat content of the rations can be treated with palmitic acid data and the protein content similarly by the protein data in Problem P4.32, how much emergency rations must the camper consume in order to compensate for a reduction in body temperature of $3.5 \mathrm{~K}$ ? Assume the heat capacity of the body equals that of water. Assume the camper weighs $67 \mathrm{~kg}$. | thermo |
||
$\mathrm{Torr}$ | 9.8 | 170 | 170 | At 303 . K, the vapor pressure of benzene is 120 . Torr and that of hexane is 189 Torr. Calculate the vapor pressure of a solution for which $x_{\text {benzene }}=0.28$ assuming ideal behavior. | thermo |
||
$\mathrm{~kJ} \mathrm{~mol}^{-1}$ | 15.45 | -57.2 | -57.2 | Determine the molar standard Gibbs energy for ${ }^{35} \mathrm{Cl}^{35} \mathrm{Cl}$ where $\widetilde{\nu}=560 . \mathrm{cm}^{-1}, B=0.244 \mathrm{~cm}^{-1}$, and the ground electronic state is nondegenerate. | thermo |
||
$\mathrm{~kJ} \mathrm{~mol}^{-1}$ | 6.12 | 132.9 | 132.9 | For the reaction $\mathrm{C}($ graphite $)+\mathrm{H}_2 \mathrm{O}(g) \rightleftharpoons$ $\mathrm{CO}(g)+\mathrm{H}_2(g), \Delta H_R^{\circ}=131.28 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $298.15 \mathrm{~K}$. Use the values of $C_{P, m}^{\circ}$ at $298.15 \mathrm{~K}$ in the data tables to calculate $\Delta H_R^{\circ}$ at $125.0^{\circ} \mathrm{C}$. | thermo |
||
10.6 | 0.0547 | 0.0547 | Calculate the mean ionic activity of a $0.0350 \mathrm{~m} \mathrm{Na}_3 \mathrm{PO}_4$ solution for which the mean activity coefficient is 0.685 . | thermo |
|||
$^{\circ} \mathrm{C}$ | 8.43 | -3.5 | -3.5 | Consider the transition between two forms of solid tin, $\mathrm{Sn}(s$, gray $) \rightarrow \mathrm{Sn}(s$, white $)$. The two phases are in equilibrium at 1 bar and $18^{\circ} \mathrm{C}$. The densities for gray and white tin are 5750 and $7280 \mathrm{~kg} \mathrm{~m}^{-3}$, respectively, and the molar entropies for gray and white tin are 44.14 and $51.18 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$, respectively. Calculate the temperature at which the two phases are in equilibrium at 350. bar. | thermo |
||
$\mathrm{~L}$ | 9.22 | -0.10 | -0.10 | The densities of pure water and ethanol are 997 and $789 \mathrm{~kg} \mathrm{~m}^{-3}$, respectively. For $x_{\text {ethanol }}=0.35$, the partial molar volumes of ethanol and water are 55.2 and $17.8 \times 10^{-3} \mathrm{~L} \mathrm{~mol}^{-1}$, respectively. Calculate the change in volume relative to the pure components when $2.50 \mathrm{~L}$ of a solution with $x_{\text {ethanol }}=0.35$ is prepared. | thermo |
||
16.15 | 0.132 | 0.132 | For $\mathrm{N}_2$ at $298 \mathrm{~K}$, what fraction of molecules has a speed between 200. and $300 . \mathrm{m} / \mathrm{s}$ ? | thermo |
|||
$\mathrm{~bar}$ | 1.3 | 26.9 | 26.9 | Calculate the pressure exerted by Ar for a molar volume of $1.31 \mathrm{~L} \mathrm{~mol}^{-1}$ at $426 \mathrm{~K}$ using the van der Waals equation of state. The van der Waals parameters $a$ and $b$ for Ar are 1.355 bar dm ${ }^6 \mathrm{~mol}^{-2}$ and $0.0320 \mathrm{dm}^3 \mathrm{~mol}^{-1}$, respectively. Is the attractive or repulsive portion of the potential dominant under these conditions?
| thermo |
||
$\mathrm{~K}$ | 8.25 | 344 | 344 | For water, $\Delta H_{\text {vaporization }}$ is $40.656 \mathrm{~kJ} \mathrm{~mol}^{-1}$, and the normal boiling point is $373.12 \mathrm{~K}$. Calculate the boiling point for water on the top of Mt. Everest (elevation $8848 \mathrm{~m}$ ), where the barometric pressure is 253 Torr. | thermo |
||
9.24 | 0.466 | 0.466 | An ideal solution is formed by mixing liquids $\mathrm{A}$ and $B$ at $298 \mathrm{~K}$. The vapor pressure of pure A is 151 Torr and that of pure B is 84.3 Torr. If the mole fraction of $\mathrm{A}$ in the vapor is 0.610 , what is the mole fraction of $\mathrm{A}$ in the solution? | thermo |
|||
$\mathrm{~m}^2$ | 5.42 | 19.4 | 19.4 | The mean solar flux at Earth's surface is $\sim 2.00 \mathrm{~J}$ $\mathrm{cm}^{-2} \mathrm{~min}^{-1}$. In a nonfocusing solar collector, the temperature reaches a value of $79.5^{\circ} \mathrm{C}$. A heat engine is operated using the collector as the hot reservoir and a cold reservoir at $298 \mathrm{~K}$. Calculate the area of the collector needed to produce 1000. W. Assume that the engine operates at the maximum Carnot efficiency. | thermo |
||
$\mathrm{~g}$ | 2.4 | 15 | 15 | A hiker caught in a thunderstorm loses heat when her clothing becomes wet. She is packing emergency rations that if completely metabolized will release $35 \mathrm{~kJ}$ of heat per gram of rations consumed. How much rations must the hiker consume to avoid a reduction in body temperature of $2.5 \mathrm{~K}$ as a result of heat loss? Assume the heat capacity of the body equals that of water and that the hiker weighs $51 \mathrm{~kg}$. | thermo |
||
6.30 | 0.241 | 0.241 | Calculate the degree of dissociation of $\mathrm{N}_2 \mathrm{O}_4$ in the reaction $\mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g)$ at 300 . $\mathrm{K}$ and a total pressure of 1.50 bar. Do you expect the degree of dissociation to increase or decrease as the temperature is increased to 550 . K? Assume that $\Delta H_R^{\circ}$ is independent of temperature. | thermo |
|||
$10^3 \mathrm{~J}$ | 6.20 | -9.54 | -9.54 | Calculate $\Delta G$ for the isothermal expansion of $2.25 \mathrm{~mol}$ of an ideal gas at $325 \mathrm{~K}$ from an initial pressure of 12.0 bar to a final pressure of 2.5 bar. | thermo |
||
$10^{34} \mathrm{~m}^{-3} \mathrm{~s}^{-1}$ | 16.30 | 8.44 | 8.44 | Determine the total collisional frequency for $\mathrm{CO}_2$ at $1 \mathrm{~atm}$ and $298 \mathrm{~K}$. | thermo |
||
9.9 | 0.312 | 0.312 | The volatile liquids $A$ and $\mathrm{B}$, for which $P_A^*=165$ Torr and $P_B^*=85.1$ Torr are confined to a piston and cylinder assembly. Initially, only the liquid phase is present. As the pressure is reduced, the first vapor is observed at a total pressure of 110 . Torr. Calculate $x_{\mathrm{A}}$ | thermo |
|||
$10^3 \mathrm{~g} \mathrm{~mol}^{-1}$ | 9.7 | 1.45 | 1.45 | The osmotic pressure of an unknown substance is measured at $298 \mathrm{~K}$. Determine the molecular weight if the concentration of this substance is $31.2 \mathrm{~kg} \mathrm{~m}^{-3}$ and the osmotic pressure is $5.30 \times 10^4 \mathrm{~Pa}$. The density of the solution is $997 \mathrm{~kg} \mathrm{~m}^{-3}$. | thermo |
||
$\mathrm{~K}$ | 7.4 | 310 | 310 | One mole of Ar initially at 310 . K undergoes an adiabatic expansion against a pressure $P_{\text {external }}=0$ from a volume of $8.5 \mathrm{~L}$ to a volume of $82.0 \mathrm{~L}$. Calculate the final temperature using the ideal gas | thermo |
||
$\mathrm{~J} \mathrm{~s}^{-1}$ | 5.33 | 773 | 773 | A refrigerator is operated by a $0.25-\mathrm{hp}(1 \mathrm{hp}=$ 746 watts) motor. If the interior is to be maintained at $4.50^{\circ} \mathrm{C}$ and the room temperature on a hot day is $38^{\circ} \mathrm{C}$, what is the maximum heat leak (in watts) that can be tolerated? Assume that the coefficient of performance is $50 . \%$ of the maximum theoretical value. | thermo |
||
4.34 | 1.8 | 1.8 | In order to get in shape for mountain climbing, an avid hiker with a mass of $60 . \mathrm{kg}$ ascends the stairs in the world's tallest structure, the $828 \mathrm{~m}$ tall Burj Khalifa in Dubai, United Arab Emirates. Assume that she eats energy bars on the way up and that her body is $25 \%$ efficient in converting the energy content of the bars into the work of climbing. How many energy bars does she have to eat if a single bar produces $1.08 \times 10^3 \mathrm{~kJ}$ of energy upon metabolizing? | thermo |
|||
$10^{24}$ | 18.14 | 1.43 | 1.43 | The half-life of ${ }^{238} \mathrm{U}$ is $4.5 \times 10^9$ years. How many disintegrations occur in $1 \mathrm{~min}$ for a $10 \mathrm{mg}$ sample of this element? | thermo |
||
$\mathrm{~mol} \mathrm{~kg}^{-1}$ | 10.13 | 0.321 | 0.321 | Calculate the ionic strength in a solution that is 0.0750 $m$ in $\mathrm{K}_2 \mathrm{SO}_4, 0.0085 \mathrm{~m}$ in $\mathrm{Na}_3 \mathrm{PO}_4$, and $0.0150 \mathrm{~m}$ in $\mathrm{MgCl}_2$. | thermo |
||
5.17 | 2.4 | 2.4 | The interior of a refrigerator is typically held at $36^{\circ} \mathrm{F}$ and the interior of a freezer is typically held at $0.00^{\circ} \mathrm{F}$. If the room temperature is $65^{\circ} \mathrm{F}$, by what factor is it more expensive to extract the same amount of heat from the freezer than from the refrigerator? Assume that the theoretical limit for the performance of a reversible refrigerator is valid in this case. | thermo |
|||
14.16 | 5840 | 5840 | Calculate the rotational partition function for $\mathrm{SO}_2$ at $298 \mathrm{~K}$ where $B_A=2.03 \mathrm{~cm}^{-1}, B_B=0.344 \mathrm{~cm}^{-1}$, and $B_C=0.293 \mathrm{~cm}^{-1}$ | thermo |
|||
$\mathrm{~K}$ | 15.2 | 655 | 655 | For a two-level system where $v=1.50 \times 10^{13} \mathrm{~s}^{-1}$, determine the temperature at which the internal energy is equal to $0.25 \mathrm{Nhv}$, or $1 / 2$ the limiting value of $0.50 \mathrm{Nhv}$. | thermo |
||
$10^6$ | 6.10 | 4.76 | 4.76 | Calculate $K_P$ at $600 . \mathrm{K}$ for the reaction $\mathrm{N}_2 \mathrm{O}_4(l) \rightleftharpoons 2 \mathrm{NO}_2(g)$ assuming that $\Delta H_R^{\circ}$ is constant over the interval 298-725 K. | thermo |
||
$\mathrm{~m}$ | 2.5 | 30 | 30 | Count Rumford observed that using cannon boring machinery a single horse could heat $11.6 \mathrm{~kg}$ of ice water $(T=273 \mathrm{~K})$ to $T=355 \mathrm{~K}$ in 2.5 hours. Assuming the same rate of work, how high could a horse raise a $225 \mathrm{~kg}$ weight in 2.5 minutes? Assume the heat capacity of water is $4.18 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~g}^{-1}$. | thermo |
||
$\mathrm{~K}$ | 13.22 | 432 | 432 | The vibrational frequency of $I_2$ is $208 \mathrm{~cm}^{-1}$. At what temperature will the population in the first excited state be half that of the ground state? | thermo |
||
$\mathrm{~K}^{-1}$ | 5.5 | 57.2 | 57.2 | One mole of $\mathrm{H}_2 \mathrm{O}(l)$ is compressed from a state described by $P=1.00$ bar and $T=350$. K to a state described by $P=590$. bar and $T=750$. K. In addition, $\beta=2.07 \times 10^{-4} \mathrm{~K}^{-1}$ and the density can be assumed to be constant at the value $997 \mathrm{~kg} \mathrm{~m}^{-3}$. Calculate $\Delta S$ for this transformation, assuming that $\kappa=0$.
| thermo |
||
$\mathrm{~K}$ | 3.5 | 292 | 292 | A mass of $34.05 \mathrm{~g}$ of $\mathrm{H}_2 \mathrm{O}(s)$ at $273 \mathrm{~K}$ is dropped into $185 \mathrm{~g}$ of $\mathrm{H}_2 \mathrm{O}(l)$ at $310 . \mathrm{K}$ in an insulated container at 1 bar of pressure. Calculate the temperature of the system once equilibrium has been reached. Assume that $C_{P, m}$ for $\mathrm{H}_2 \mathrm{O}(l)$ is constant at its values for $298 \mathrm{~K}$ throughout the temperature range of interest. | thermo |
||
$\mathrm{~kJ} \mathrm{~mol}^{-1}$ | 4.4 | 91.7 | 91.7 | Calculate $\Delta H_f^{\circ}$ for $N O(g)$ at $975 \mathrm{~K}$, assuming that the heat capacities of reactants and products are constant over the temperature interval at their values at $298.15 \mathrm{~K}$. | thermo |
||
$10^4$ | 13.27 | 5.85 | 5.85 | A two-level system is characterized by an energy separation of $1.30 \times 10^{-18} \mathrm{~J}$. At what temperature will the population of the ground state be 5 times greater than that of the excited state? | thermo |
||
$\mathrm{~K}$ | 14.5 | 0.068 | 0.068 | At what temperature are there Avogadro's number of translational states available for $\mathrm{O}_2$ confined to a volume of 1000. $\mathrm{cm}^3$ ? | thermo |
||
$10^{-4}$ | 11.25 | 4.16 | 4.16 | The half-cell potential for the reaction $\mathrm{O}_2(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l)$ is $+1.03 \mathrm{~V}$ at $298.15 \mathrm{~K}$ when $a_{\mathrm{O}_2}=1.00$. Determine $a_{\mathrm{H}^{+}}$ | thermo |
||
$\mathrm{~cm}^3$ | 9.5 | -8 | -8 | The partial molar volumes of water and ethanol in a solution with $x_{\mathrm{H}_2 \mathrm{O}}=0.45$ at $25^{\circ} \mathrm{C}$ are 17.0 and $57.5 \mathrm{~cm}^3 \mathrm{~mol}^{-1}$, respectively. Calculate the volume change upon mixing sufficient ethanol with $3.75 \mathrm{~mol}$ of water to give this concentration. The densities of water and ethanol are 0.997 and $0.7893 \mathrm{~g} \mathrm{~cm}^{-3}$, respectively, at this temperature. | thermo |
||
$\mathrm{~J}\mathrm{~K}^{-1}$ | 5.6 | 58.2 | 58.2 | In this problem, 3.00 mol of liquid mercury is transformed from an initial state characterized by $T_i=300 . \mathrm{K}$ and $P_i=1.00$ bar to a final state characterized by $T_f=600 . \mathrm{K}$ and $P_f=3.00$ bar.
Calculate $\Delta S$ for this process; $\beta=1.81 \times 10^{-4} \mathrm{~K}^{-1}, \rho=13.54 \mathrm{~g} \mathrm{~cm}^{-3}$, and $C_{P, m}$ for $\mathrm{Hg}(l)=27.98 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$. |
Because the volume changes only slightly with temperature and pressure over the range indicated,
$$
\begin{aligned}
& \Delta S=\int_{T_i}^{T_f} \frac{C_P}{T} d T-\int_{P_i}^{P_f} V \beta d P \approx n C_{P, m} \ln \frac{T_f}{T_i}-n V_{m, i} \beta\left(P_f-P_i\right) \\
& =3.00 \mathrm{~mol} \times 27.98 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times \ln \frac{600 . \mathrm{K}}{300 . \mathrm{K}} \\
& -3.00 \mathrm{~mol} \times \frac{200.59 \mathrm{~g} \mathrm{~mol}^{-1}}{13.54 \mathrm{~g} \mathrm{~cm}^{-3} \times \frac{10^6 \mathrm{~cm}^3}{\mathrm{~m}^3}} \times 1.81 \times 10^{-4} \mathrm{~K}^{-1} \times 2.00 \mathrm{bar} \\
& \times 10^5 \mathrm{Pabar}^{-1} \\
& =58.2 \mathrm{~J} \mathrm{~K}^{-1}-1.61 \times 10^{-3} \mathrm{~J} \mathrm{~K}^{-1}=58.2 \mathrm{~J} \mathrm{~K}^{-1} \\
&
\end{aligned}
$$
| thermo |
|
$\mathrm{~K}$ | 15.2 | 449 | 449 | For an ensemble consisting of 1.00 moles of particles having two energy levels separated by $h v=1.00 \times 10^{-20} \mathrm{~J}$, at what temperature will the internal energy of this system equal $1.00 \mathrm{~kJ}$ ? | Using the expression for total energy and recognizing that $N=n N_A$,
$$
U=-\left(\frac{\partial \ln Q}{\partial \beta}\right)_V=-n N_A\left(\frac{\partial \ln q}{\partial \beta}\right)_V
$$
Evaluating the preceding expression and paying particular attention to units, we get
$$
\begin{aligned}
& U=-n N_A\left(\frac{\partial}{\partial \beta} \ln q\right)_V=-\frac{n N_A}{q}\left(\frac{\partial q}{\partial \beta}\right)_V \\
& \frac{U}{n N_A}=\frac{-1}{\left(1+e^{-\beta h \nu}\right)}\left(\frac{\partial}{\partial \beta}\left(1+e^{-\beta h \nu}\right)\right)_V \\
&=\frac{h \nu e^{-\beta h \nu}}{1+e^{-\beta h \nu}}=\frac{h \nu}{e^{\beta h \nu}+1} \\
& \frac{n N_A h \nu}{U}-1=e^{\beta h \nu} \\
& \ln \left(\frac{n N_A h \nu}{U}-1\right)=\beta h \nu=\frac{h \nu}{k T}
\end{aligned}
$$
$$
\begin{aligned}
T & =\frac{h \nu}{k \ln \left(\frac{n N_A h \nu}{U}-1\right)} \\
= & \frac{1.00 \times 10^{-20} \mathrm{~J}}{\left(1.38 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}\right) \ln \left(\frac{(1.00 \mathrm{~mol})\left(6.022 \times 10^{23} \mathrm{~mol}^{-1}\right)\left(1.00 \times 10^{-20} \mathrm{~J}\right)}{\left(1.00 \times 10^3 \mathrm{~J}\right)}-1\right)} \\
& =449 \mathrm{~K}
\end{aligned}
$$ | thermo |
|
$\mathrm{~nm}$ | 17.9 | 1.94 | 1.94 | The sedimentation coefficient of lysozyme $\left(\mathrm{M}=14,100 \mathrm{~g} \mathrm{~mol}^{-1}\right)$ in water at $20^{\circ} \mathrm{C}$ is $1.91 \times 10^{-13} \mathrm{~s}$ and the specific volume is $0.703 \mathrm{~cm}^3 \mathrm{~g}^{-1}$. The density of water at this temperature is $0.998 \mathrm{~g} \mathrm{~cm}^{-3}$ and $\eta=1.002 \mathrm{cP}$. Assuming lysozyme is spherical, what is the radius of this protein? |
The frictional coefficient is dependent on molecular radius. Solving Equation for $f$, we obtain
$$
\begin{aligned}
& f=\frac{m(1-\overline{\mathrm{V}} \rho)}{\overline{\mathrm{s}}}=\frac{\frac{\left(14,100 \mathrm{~g} \mathrm{~mol}^{-1}\right)}{\left(6.022 \times 10^{23} \mathrm{~mol}^{-1}\right)}\left(1-(0.703 \mathrm{~mL} \mathrm{~g})\left(0.998 \mathrm{~g} \mathrm{~mL}^{-1}\right)\right)}{1.91 \times 10^{-13} \mathrm{~s}} \\
& =3.66 \times 10^{-8} \mathrm{~g} \mathrm{~s}^{-1} \\
&
\end{aligned}
$$
The frictional coefficient is related to the radius for a spherical particle by Equation such that
$$
r=\frac{f}{6 \pi \eta}=\frac{3.66 \times 10^{-8} \mathrm{~g} \mathrm{~s}^{-1}}{6 \pi\left(1.002 \mathrm{~g} \mathrm{~m}^{-1} \mathrm{~s}^{-1}\right)}=1.94 \times 10^{-9} \mathrm{~m}=1.94 \mathrm{~nm}
$$
| thermo |
|
$ \mathrm{Jmol}^{-1} \mathrm{~K}^{-1}$ | 15.5 | 164 | 146 | Determine the standard molar entropy of $\mathrm{Ne}$ under standard thermodynamic conditions. |
Beginning with the expression for entropy:
$$
\begin{aligned}
S & =\frac{5}{2} R+R \ln \left(\frac{V}{\Lambda^3}\right)-R \ln N_A \\
& =\frac{5}{2} R+R \ln \left(\frac{V}{\Lambda^3}\right)-54.75 R \\
& =R \ln \left(\frac{V}{\Lambda^3}\right)-52.25 R
\end{aligned}
$$
The conventional standard state is defined by $T=298 \mathrm{~K}$ and $V_m=24.4 \mathrm{~L}\left(0.0244 \mathrm{~m}^3\right)$. The thermal wavelength for $\mathrm{Ne}$ is
$$
\begin{aligned}
\Lambda & =\left(\frac{h^2}{2 \pi m k T}\right)^{1 / 2} \\
& =\left(\frac{\left(6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)^2}{2 \pi\left(\frac{0.02018 \mathrm{~kg} \mathrm{~mol}^{-1}}{N_A}\right)\left(1.38 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}\right)(298 \mathrm{~K})}\right)^{1 / 2} \\
& =2.25 \times 10^{-11} \mathrm{~m}
\end{aligned}
$$
Using this value for the thermal wavelength, the entropy becomes
$$
\begin{aligned}
S & =R \ln \left(\frac{0.0244 \mathrm{~m}^3}{\left(2.25 \times 10^{-11} \mathrm{~m}\right)^3}\right)-52.25 R \\
& =69.84 R-52.25 R=17.59 R=146 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}
\end{aligned}
$$
| thermo |
|
$10^{11} \mathrm{~s}$ | 18.4 | 4.86 | 4.86 | Carbon-14 is a radioactive nucleus with a half-life of 5760 years. Living matter exchanges carbon with its surroundings (for example, through $\mathrm{CO}_2$ ) so that a constant level of ${ }^{14} \mathrm{C}$ is maintained, corresponding to 15.3 decay events per minute. Once living matter has died, carbon contained in the matter is not exchanged with the surroundings, and the amount of ${ }^{14} \mathrm{C}$ that remains in the dead material decreases with time due to radioactive decay. Consider a piece of fossilized wood that demonstrates $2.4{ }^{14} \mathrm{C}$ decay events per minute. How old is the wood? | The ratio of decay events yields the amount of ${ }^{14} \mathrm{C}$ present currently versus the amount that was present when the tree died:
$$
\frac{\left[{ }^{14} \mathrm{C}\right]}{\left[{ }^{14} \mathrm{C}\right]_0}=\frac{2.40 \mathrm{~min}^{-1}}{15.3 \mathrm{~min}^{-1}}=0.157
$$
The rate constant for isotope decay is related to the half-life as follows:
$$
k=\frac{\ln 2}{t_{1 / 2}}=\frac{\ln 2}{5760 \text { years }}=\frac{\ln 2}{1.82 \times 10^{11} \mathrm{~s}}=3.81 \times 10^{-12} \mathrm{~s}^{-1}
$$
With the rate constant and ratio of isotope concentrations, the age of the fossilized wood is readily determined:
$$
\begin{aligned}
\frac{\left[{ }^{14} \mathrm{C}\right]}{\left[{ }^{14} \mathrm{C}\right]_0} & =e^{-k t} \\
\ln \left(\frac{\left[{ }^{14} \mathrm{C}\right]}{\left[{ }^{14} \mathrm{C}\right]_0}\right) & =-k t \\
-\frac{1}{k} \ln \left(\frac{\left[{ }^{14} \mathrm{C}\right]}{\left[{ }^{14} \mathrm{C}\right]_0}\right) & =-\frac{1}{3.81 \times 10^{-12} \mathrm{~s}} \ln (0.157)=t \\
4.86 \times 10^{11} \mathrm{~s} & =t
\end{aligned}
$$
This time corresponds to an age of roughly 15,400 years. | thermo |
|
$10^{11} \mathrm{~s}$ | 18.4 | 4.86 | 4.86 | Carbon-14 is a radioactive nucleus with a half-life of 5760 years. Living matter exchanges carbon with its surroundings (for example, through $\mathrm{CO}_2$ ) so that a constant level of ${ }^{14} \mathrm{C}$ is maintained, corresponding to 15.3 decay events per minute. Once living matter has died, carbon contained in the matter is not exchanged with the surroundings, and the amount of ${ }^{14} \mathrm{C}$ that remains in the dead material decreases with time due to radioactive decay. Consider a piece of fossilized wood that demonstrates $2.4{ }^{14} \mathrm{C}$ decay events per minute. How old is the wood? | The ratio of decay events yields the amount of ${ }^{14} \mathrm{C}$ present currently versus the amount that was present when the tree died:
$$
\frac{\left[{ }^{14} \mathrm{C}\right]}{\left[{ }^{14} \mathrm{C}\right]_0}=\frac{2.40 \mathrm{~min}^{-1}}{15.3 \mathrm{~min}^{-1}}=0.157
$$
The rate constant for isotope decay is related to the half-life as follows:
$$
k=\frac{\ln 2}{t_{1 / 2}}=\frac{\ln 2}{5760 \text { years }}=\frac{\ln 2}{1.82 \times 10^{11} \mathrm{~s}}=3.81 \times 10^{-12} \mathrm{~s}^{-1}
$$
With the rate constant and ratio of isotope concentrations, the age of the fossilized wood is readily determined:
$$
\begin{aligned}
\frac{\left[{ }^{14} \mathrm{C}\right]}{\left[{ }^{14} \mathrm{C}\right]_0} & =e^{-k t} \\
\ln \left(\frac{\left[{ }^{14} \mathrm{C}\right]}{\left[{ }^{14} \mathrm{C}\right]_0}\right) & =-k t \\
-\frac{1}{k} \ln \left(\frac{\left[{ }^{14} \mathrm{C}\right]}{\left[{ }^{14} \mathrm{C}\right]_0}\right) & =-\frac{1}{3.81 \times 10^{-12} \mathrm{~s}} \ln (0.157)=t \\
4.86 \times 10^{11} \mathrm{~s} & =t
\end{aligned}
$$
This time corresponds to an age of roughly 15,400 years. | thermo |
|
$10^{-5}~\mathrm{m^2~s^{-1}}$ | 17.1 | 1.1 | 1.1 | Determine the diffusion coefficient for Ar at $298 \mathrm{~K}$ and a pressure of $1.00 \mathrm{~atm}$. |
$$
\begin{aligned}
D_{Ar} &= \frac{1}{3} \nu_{ave, Ar} \lambda_{Ar} \\
&= \frac{1}{3} \left(\frac{8RT}{\pi M_{Ar}}\right)^{\frac{1}{2}} \left(\frac{RT}{PN_A\sqrt{2}\sigma_{Ar}}\right) \\
&= \frac{1}{3} \left(\frac{8(8.314~\mathrm{J~mol^{-1}~K^{-1}}) \times 298~\mathrm{K}}{\pi(0.040~\mathrm{kg~mol^{-1}})}\right)^{\frac{1}{2}} \\
&\quad \times \left(\frac{(8.314~\mathrm{J~mol^{-1}~K^{-1}}) \times 298~\mathrm{K}}{(101,325~\mathrm{Pa}) \times (6.022 \times 10^{23}~\mathrm{mol^{-1}})} \times \frac{1}{\sqrt{2}(3.6 \times 10^{-19}~\mathrm{m^2})}\right) \\
&= \frac{1}{3} \times (397~\mathrm{m~s^{-1}}) \times (7.98 \times 10^{-8}~\mathrm{m}) \\
&= 1.1 \times 10^{-5}~\mathrm{m^2~s^{-1}}
\end{aligned}
$$
| thermo |
|
$10^6 \mathrm{~s}$ | 17.7 | 4.49 | 4.49 | Gas cylinders of $\mathrm{CO}_2$ are sold in terms of weight of $\mathrm{CO}_2$. A cylinder contains $50 \mathrm{lb}$ (22.7 $\mathrm{kg}$ ) of $\mathrm{CO}_2$. How long can this cylinder be used in an experiment that requires flowing $\mathrm{CO}_2$ at $293 \mathrm{~K}(\eta=146 \mu \mathrm{P})$ through a 1.00-m-long tube (diameter $\left.=0.75 \mathrm{~mm}\right)$ with an input pressure of $1.05 \mathrm{~atm}$ and output pressure of $1.00 \mathrm{~atm}$ ? The flow is measured at the tube output. |
The gas flow rate $\Delta V / \Delta t$ is
$$
\begin{aligned}
\frac{\Delta V}{\Delta t}= & \frac{\pi r^4}{16 \eta L P_0}\left(P_2^2-P_1^2\right) \\
= & \frac{\pi\left(0.375 \times 10^{-3} \mathrm{~m}\right)^4}{16\left(1.46 \times 10^{-5} \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-1}\right)(1.00 \mathrm{~m})(101,325 \mathrm{~Pa})} \\
& \times\left((106,391 \mathrm{~Pa})^2-(101,325 \mathrm{~Pa})^2\right) \\
= & 2.76 \times 10^{-6} \mathrm{~m}^3 \mathrm{~s}^{-1}
\end{aligned}
$$
Converting the $\mathrm{CO}_2$ contained in the cylinder to the volume occupied at $298 \mathrm{~K}$ and 1 atm pressure, we get
$$
\begin{aligned}
n_{\mathrm{CO}_2} & =22.7 \mathrm{~kg}\left(\frac{1}{0.044 \mathrm{~kg} \mathrm{~mol}^{-1}}\right)=516 \mathrm{~mol} \\
V & =\frac{n R T}{P}=1.24 \times 10^4 \mathrm{~L}\left(\frac{10^{-3} \mathrm{~m}^3}{\mathrm{~L}}\right)=12.4 \mathrm{~m}^3
\end{aligned}
$$
Given the effective volume of $\mathrm{CO}_2$ contained in the cylinder, the duration over which the cylinder can be used is
$$
\frac{12.4 \mathrm{~m}^3}{2.76 \times 10^{-6} \mathrm{~m}^3 \mathrm{~s}^{-1}}=4.49 \times 10^6 \mathrm{~s}
$$
This time corresponds to roughly 52 days.
| thermo |
|
13.5 | 0.086 | 0.086 | The vibrational frequency of $I_2$ is $208 \mathrm{~cm}^{-1}$. What is the probability of $I_2$ populating the $n=2$ vibrational level if the molecular temperature is $298 \mathrm{~K}$ ? | Molecular vibrational energy levels can be modeled as harmonic oscillators; therefore, this problem can be solved by employing a strategy identical to the one just presented. To evaluate the partition function $q$, the "trick" used earlier was to write the partition function as a series and use the equivalent series expression:
$$
\begin{aligned}
q & =\sum_n e^{-\beta \varepsilon_n}=1+e^{-\beta h c \widetilde{\nu}}+e^{-2 \beta h c \tilde{\nu}}+e^{-3 \beta h c \widetilde{\nu}}+\ldots \\
& =\frac{1}{1-e^{-\beta h c \widetilde{\nu}}}
\end{aligned}
$$
Since $\tilde{\nu}=208 \mathrm{~cm}^{-1}$ and $T=298 \mathrm{~K}$, the partition function is
$$
\begin{aligned}
q & =\frac{1}{1-e^{-\beta h c \widetilde{\nu}}} \\
& =\frac{1}{1-e^{-h c \widetilde{\nu} / k T}} \\
& =\frac{1}{1-\exp \left[-\left(\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3.00 \times 10^{10} \mathrm{~cm} \mathrm{~s}^{-1}\right)\left(208 \mathrm{~cm}^{-1}\right)}{\left(1.38 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}\right)(298 \mathrm{~K})}\right)\right]} \\
& =\frac{1}{1-e^{-1}}=1.58
\end{aligned}
$$
This result is then used to evaluate the probability of occupying the second vibrational state $(n=2)$ as follows:
$$
\begin{aligned}
p_2 & =\frac{e^{-2 \beta h c \tilde{\nu}}}{q} \\
& =\frac{\exp \left[-2\left(\frac{\left(6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}^{-1}\right)\left(3.00 \times 10^{10} \mathrm{~cm} \mathrm{~s}^{-1}\right)\left(208 \mathrm{~cm}^{-1}\right)}{\left(1.38 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}\right)(298 \mathrm{~K})}\right)\right]}{1.58} \\
& =0.086
\end{aligned}
$$ | thermo |
||
$\mathrm{~kJ} \mathrm{~mol}^{-1}$ | 6.4 | 355.1 | 355.1 | The value of $\Delta G_f^{\circ}$ for $\mathrm{Fe}(g)$ is $370.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $298.15 \mathrm{~K}$, and $\Delta H_f^{\circ}$ for $\mathrm{Fe}(g)$ is $416.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at the same temperature. Assuming that $\Delta H_f^{\circ}$ is constant in the interval $250-400 \mathrm{~K}$, calculate $\Delta G_f^{\circ}$ for $\mathrm{Fe}(g)$ at 400. K. | $$
\begin{aligned}
\Delta G_f^{\circ}\left(T_2\right) & =T_2\left[\frac{\Delta G_f^{\circ}\left(T_1\right)}{T_1}+\Delta H_f^{\circ}\left(T_1\right) \times\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\right] \\
& =400 . \mathrm{K} \times\left[\begin{array}{l}
\left.\frac{370.7 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1}}{298.15 \mathrm{~K}}+416.3 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1}\right] \\
\times\left(\frac{1}{400 . \mathrm{K}}-\frac{1}{298.15 \mathrm{~K}}\right)
\end{array}\right] \\
\Delta G_f^{\circ}(400 . \mathrm{K}) & =355.1 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$$ | thermo |
|
19.6 | 0.396 | 0.396 | The reactant 1,3-cyclohexadiene can be photochemically converted to cis-hexatriene. In an experiment, $2.5 \mathrm{mmol}$ of cyclohexadiene are converted to cis-hexatriene when irradiated with 100. W of 280. nm light for $27.0 \mathrm{~s}$. All of the light is absorbed by the sample. What is the overall quantum yield for this photochemical process? | First, the total photon energy absorbed by the sample, $E_{a b s}$, is
$$
E_{a b s}=(\text { power }) \Delta t=\left(100 . \mathrm{Js}^{-1}\right)(27.0 \mathrm{~s})=2.70 \times 10^3 \mathrm{~J}
$$
Next, the photon energy at $280 . \mathrm{nm}$ is
$$
\mathrm{E}_{p h}=\frac{h c}{\lambda}=\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(2.998 \times 10^8 \mathrm{~ms}^{-1}\right)}{2.80 \times 10^{-7} \mathrm{~m}}=7.10 \times 10^{-19} \mathrm{~J}
$$
The total number of photons absorbed by the sample is therefore
$$
\frac{E_{a b s}}{E_{p h}}=\frac{2.70 \times 10^3 \mathrm{~J}}{7.10 \times 10^{-19} \mathrm{~J} \text { photon }^{-1}}=3.80 \times 10^{21} \text { photons }
$$
Dividing this result by Avogadro's number results in $6.31 \times 10^{-3}$ Einsteins or moles of photons. Therefore, the overall quantum yield is
$$
\phi=\frac{\text { moles }_{\text {react }}}{\text { moles }_{\text {photon }}}=\frac{2.50 \times 10^{-3} \mathrm{~mol}}{6.31 \times 10^{-3} \mathrm{~mol}}=0.396 \approx 0.40
$$ | thermo |
||
$\mathrm{~J} \mathrm{~K}^{-1}$ | 5.5 | 48.6 | 48.6 | In this problem, $2.50 \mathrm{~mol}$ of $\mathrm{CO}_2$ gas is transformed from an initial state characterized by $T_i=450 . \mathrm{K}$ and $P_i=1.35$ bar to a final state characterized by $T_f=800 . \mathrm{K}$ and $P_f=$ 3.45 bar. Using Equation (5.23), calculate $\Delta S$ for this process. Assume ideal gas behavior and use the ideal gas value for $\beta$. For $\mathrm{CO}_2$,
$$
\frac{C_{P, m}}{\mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}}=18.86+7.937 \times 10^{-2} \frac{T}{\mathrm{~K}}-6.7834 \times 10^{-5} \frac{T^2}{\mathrm{~K}^2}+2.4426 \times 10^{-8} \frac{T^3}{\mathrm{~K}^3}
$$ |
Consider the following reversible process in order to calculate $\Delta S$. The gas is first heated reversibly from 450 . to 800 . $\mathrm{K}$ at a constant pressure of 1.35 bar. Subsequently, the gas is reversibly compressed at a constant temperature of 800 . $\mathrm{K}$ from a pressure of 1.35 bar to a pressure of 3.45 bar.
$$
\begin{aligned}
& \Delta S=\int_{T_i}^{T_f} \frac{C_P}{T} d T-\int_{P_i}^{P_f} V \beta d P=\int_{T_i}^{T_f} \frac{C_P}{T} d T-n R \int_{P_i}^{P_f} \frac{d P}{P}=\int_{T_i}^{T_f} \frac{C_P}{T} d T-n R \ln \frac{P_f}{P_i} \\
& =2.50 \times \int_{450 .}^{800 .} \frac{\left(18.86+7.937 \times 10^{-2} \frac{T}{\mathrm{~K}}-6.7834 \times 10^{-5} \frac{T^2}{\mathrm{~K}^2}+2.4426 \times 10^{-8} \frac{T^3}{\mathrm{~K}^3}\right)}{\frac{T}{\mathrm{~K}}} d \frac{T}{\mathrm{~K}} \\
& -2.50 \mathrm{~mol} \times 8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \times \ln \frac{3.45 \mathrm{bar}}{1.35 \mathrm{bar}} \\
& =27.13 \mathrm{~J} \mathrm{~K}^{-1}+69.45 \mathrm{~J} \mathrm{~K}^{-1}-37.10 \mathrm{~J} \mathrm{~K}^{-1}+8.57 \mathrm{~J} \mathrm{~K}^{-1} \\
& -19.50 \mathrm{~J} \mathrm{~K}^{-1} \\
& =48.6 \mathrm{~J} \mathrm{~K}^{-1} \\
&
\end{aligned}
$$
| thermo |
|
$\mathrm{~J} \mathrm{~K}^{-1}$ | 5.4 | 24.4 | 24.4 | One mole of $\mathrm{CO}$ gas is transformed from an initial state characterized by $T_i=320 . \mathrm{K}$ and $V_i=80.0 \mathrm{~L}$ to a final state characterized by $T_f=650 . \mathrm{K}$ and $V_f=120.0 \mathrm{~L}$. Using Equation (5.22), calculate $\Delta S$ for this process. Use the ideal gas values for $\beta$ and $\kappa$. For CO,
$$
\frac{C_{V, m}}{\mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}}=31.08-0.01452 \frac{T}{\mathrm{~K}}+3.1415 \times 10^{-5} \frac{T^2}{\mathrm{~K}^2}-1.4973 \times 10^{-8} \frac{T^3}{\mathrm{~K}^3}
$$ |
For an ideal gas,
$$
\begin{gathered}
\beta=\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_P=\frac{1}{V}\left(\frac{\partial[n R T / P]}{\partial T}\right)_P=\frac{1}{T} \text { and } \\
\kappa=-\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T=-\frac{1}{V}\left(\frac{\partial[n R T / P]}{\partial P}\right)_T=\frac{1}{P}
\end{gathered}
$$
Consider the following reversible process in order to calculate $\Delta S$. The gas is first heated reversibly from 320 . to $650 . \mathrm{K}$ at a constant volume of $80.0 \mathrm{~L}$. Subsequently, the gas is reversibly expanded at a constant temperature of 650 . $\mathrm{K}$ from a volume of $80.0 \mathrm{~L}$ to a volume of $120.0 \mathrm{~L}$. The entropy change for this process is obtained using the integral form with the values of $\beta$ and $\kappa$ cited earlier. The result is
$$
\Delta S=\int_{T_i}^{T_f} \frac{C_V}{T} d T+n R \ln \frac{V_f}{V_i}
$$
$\Delta S=$
$$
\begin{aligned}
1 \mathrm{~mol} \times & \int_{320 .}^{650 .} \frac{\left(31.08-0.01452 \frac{T}{\mathrm{~K}}+3.1415 \times 10^{-5} \frac{T^2}{\mathrm{~K}^2}-1.4973 \times 10^{-8} \frac{T^3}{\mathrm{~K}^3}\right)}{\mathrm{K}} d \frac{T}{\mathrm{~K}} \\
& +1 \mathrm{~mol} \times 8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \times \ln \frac{120.0 \mathrm{~L}}{80.0 \mathrm{~L}} \\
= & 22.025 \mathrm{~J} \mathrm{~K}^{-1}-4.792 \mathrm{~J} \mathrm{~K}^{-1}+5.028 \mathrm{~J} \mathrm{~K}^{-1} \\
& -1.207 \mathrm{~J} \mathrm{~K}^{-1}+3.371 \mathrm{~J} \mathrm{~K}^{-1} \\
= & 24.4 \mathrm{~J} \mathrm{~K}^{-1}
\end{aligned}
$$
| thermo |
|
$\mathrm{~V}$ | 11.3 | -0.041 | -0.041 | You are given the following reduction reactions and $E^{\circ}$ values:
$$
\begin{array}{ll}
\mathrm{Fe}^{3+}(a q)+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+}(a q) & E^{\circ}=+0.771 \mathrm{~V} \\
\mathrm{Fe}^{2+}(a q)+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe}(s) & E^{\circ}=-0.447 \mathrm{~V}
\end{array}
$$
Calculate $E^{\circ}$ for the half-cell reaction $\mathrm{Fe}^{3+}(a q)+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe}(s)$. | We calculate the desired value of $E^{\circ}$ by converting the given $E^{\circ}$ values to $\Delta G^{\circ}$ values, and combining these reduction reactions to obtain the desired equation.
$$
\begin{gathered}
\mathrm{Fe}^{3+}(a q)+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+}(a q) \\
\Delta G^{\circ}=-n F E^{\circ}=-1 \times 96485 \mathrm{C} \mathrm{mol}^{-1} \times 0.771 \mathrm{~V}=-74.39 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
\mathrm{Fe}^{2+}(a q)+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe}(s) \\
\Delta G^{\circ}=-n F E^{\circ}=-2 \times 96485 \mathrm{C} \mathrm{mol}^{-1} \times(-0.447 \mathrm{~V})=86.26 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{gathered}
$$
We next add the two equations as well as their $\Delta G^{\circ}$ to obtain
$$
\begin{gathered}
\mathrm{Fe}^{3+}(a q)+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe}(s) \\
\Delta G^{\circ}=-74.39 \mathrm{~kJ} \mathrm{~mol}^{-1}+86.26 \mathrm{~kJ} \mathrm{~mol}^{-1}=11.87 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
E_{F e^{3+} / F e}^{\circ}=-\frac{\Delta G^{\circ}}{n F}=\frac{-11.87 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1}}{3 \times 96485 \mathrm{C} \mathrm{mol}^{-1}}=-0.041 \mathrm{~V}
\end{gathered}
$$
The $E^{\circ}$ values cannot be combined directly, because they are intensive rather than extensive quantities.
The preceding calculation can be generalized as follows. Assume that $n_1$ electrons are transferred in the reaction with the potential $E_{A / B}^{\circ}$, and $n_2$ electrons are transferred in the reaction with the potential $E_{B / C}^{\circ}$. If $n_3$ electrons are transferred in the reaction with the potential $E_{A / C}^{\circ}$, then $n_3 E_{A / C}^{\circ}=n_1 E_{A / B}^{\circ}+n_2 E_{B / C}^{\circ}$. | thermo |
|
$10^4 \mathrm{bar}$ | 6.13 | 1.51 | 1.51 | At $298.15 \mathrm{~K}, \Delta G_f^{\circ}(\mathrm{C}$, graphite $)=0$, and $\Delta G_f^{\circ}(\mathrm{C}$, diamond $)=2.90 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Therefore, graphite is the more stable solid phase at this temperature at $P=P^{\circ}=1$ bar. Given that the densities of graphite and diamond are 2.25 and $3.52 \mathrm{~kg} / \mathrm{L}$, respectively, at what pressure will graphite and diamond be in equilibrium at $298.15 \mathrm{~K}$ ? | At equilibrium $\Delta G=G(\mathrm{C}$, graphite $)-G(\mathrm{C}$, diamond $)=0$. Using the pressure dependence of $G,\left(\partial G_m / \partial P\right)_T=V_m$, we establish the condition for equilibrium:
$$
\begin{gathered}
\Delta G=\Delta G_f^{\circ}(\mathrm{C}, \text { graphite })-\Delta G_f^{\circ}(\mathrm{C}, \text { diamond }) \\
+\left(V_m^{\text {graphite }}-V_m^{\text {diamond }}\right)(\Delta P)=0 \\
0=0-2.90 \times 10^3+\left(V_m^{\text {graphite }}-V_m^{\text {diamond }}\right)(P-1 \mathrm{bar}) \\
P=1 \mathrm{bar}+\frac{2.90 \times 10^3}{M_C\left(\frac{1}{\rho_{\text {graphite }}}-\frac{1}{\rho_{\text {diamond }}}\right)} \\
=1 \mathrm{bar}+\frac{2.90 \times 10^3}{12.00 \times 10^{-3} \mathrm{~kg} \mathrm{~mol}^{-1} \times\left(\frac{1}{2.25 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}}-\frac{1}{3.52 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}}\right)}\\
=10^5 \mathrm{~Pa}+1.51 \times 10^9 \mathrm{~Pa}=1.51 \times 10^4 \mathrm{bar}
\end{gathered}
$$
Fortunately for all those with diamond rings, although the conversion of diamond to graphite at $1 \mathrm{bar}$ and $298 \mathrm{~K}$ is spontaneous, the rate of conversion is vanishingly small.
| thermo |
|
12.8 | 0.11 | 0.11 | Imagine tossing a coin 50 times. What are the probabilities of observing heads 25 times (i.e., 25 successful experiments)? | The trial of interest consists of 50 separate experiments; therefore, $n=50$. Considering the case of 25 successful experiments where $j=25$. The probability $\left(P_{25}\right)$ is
$$
\begin{aligned}
P_{25} & =C(n, j)\left(P_E\right)^j\left(1-P_E\right)^{n-j} \\
& =C(50,25)\left(P_E\right)^{25}\left(1-P_E\right)^{25} \\
& =\left(\frac{50 !}{(25 !)(25 !)}\right)\left(\frac{1}{2}\right)^{25}\left(\frac{1}{2}\right)^{25}=\left(1.26 \times 10^{14}\right)\left(8.88 \times 10^{-16}\right)=0.11
\end{aligned}
$$ | thermo |
||
$\mathrm{~K}$ | 14.4 | 4943 | 4943 | In a rotational spectrum of $\operatorname{HBr}\left(B=8.46 \mathrm{~cm}^{-1}\right)$, the maximum intensity transition in the R-branch corresponds to the $J=4$ to 5 transition. At what temperature was the spectrum obtained? | The information provided for this problem dictates that the $J=4$ rotational energy level was the most populated at the temperature at which the spectrum was taken. To determine the temperature, we first determine the change in occupation number for the rotational energy level, $a_J$, versus $J$ as follows:
$$
\begin{aligned}
a_J & =\frac{N(2 J+1) e^{-\beta h c B J(J+1)}}{q_R}=\frac{N(2 J+1) e^{-\beta h c B J(J+1)}}{\left(\frac{1}{\beta h c B}\right)} \\
& =N \beta h c B(2 J+1) e^{-\beta h c B J(J+1)}
\end{aligned}
$$
Next, we take the derivative of $a_J$ with respect to $J$ and set the derivative equal to zero to find the maximum of the function:
$$
\begin{aligned}
\frac{d a_J}{d J} & =0=\frac{d}{d J} N \beta h c B(2 J+1) e^{-\beta h c B J(J+1)} \\
0 & =\frac{d}{d J}(2 J+1) e^{-\beta h c B J(J+1)} \\
0 & =2 e^{-\beta h c B J(J+1)}-\beta h c B(2 J+1)^2 e^{-\beta h c B J(J+1)} \\
0 & =2-\beta h c B(2 J+1)^2 \\
2 & =\beta h c B(2 J+1)^2=\frac{h c B}{k T}(2 J+1)^2 \\
T & =\frac{(2 J+1)^2 h c B}{2 k}
\end{aligned}
$$
Substitution of $J=4$ into the preceding expression results in the following temperature at which the spectrum was obtained:
$$
\begin{aligned}
T & =\frac{(2 J+1)^2 h c B}{2 k} \\
& =\frac{(2(4)+1)^2\left(6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)\left(3.00 \times 10^{10} \mathrm{~cm} \mathrm{~s}^{-1}\right)\left(8.46 \mathrm{~cm}^{-1}\right)}{2\left(1.38 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}\right)} \\
& =4943 \mathrm{~K}
\end{aligned}
$$ | thermo |