unit
stringlengths 0
63
| problemid
stringlengths 3
29
| comment
stringclasses 32
values | answer_latex
stringlengths 1
49
| answer_number
stringlengths 1
18
| problem_text
stringlengths 39
1.2k
| solution
stringclasses 112
values | source
stringclasses 10
values |
|---|---|---|---|---|---|---|---|
$\text{K}$ | e2.20(a)(b) | 205 | 205 | Assume all gases are perfect unless stated otherwise. Unless otherwise stated, thermodynamic data are for 298.15 K. When $120 \mathrm{mg}$ of naphthalene, $\mathrm{C}_{10} \mathrm{H}_8(\mathrm{~s})$, was burned in a bomb calorimeter the temperature rose by $3.05 \mathrm{~K}$. By how much will the temperature rise when $10 \mathrm{mg}$ of phenol, $\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}(\mathrm{s})$, is burned in the calorimeter under the same conditions? | atkins |
||
$\text{kPA}$ | e2.11(a) | 22 | 22 | Assume all gases are perfect unless stated otherwise. Unless otherwise stated, thermodynamic data are for 298.15 K. Calculate the final pressure of a sample of carbon dioxide that expands reversibly and adiabatically from $57.4 \mathrm{kPa}$ and $1.0 \mathrm{dm}^3$ to a final volume of $2.0 \mathrm{dm}^3$. Take $\gamma=1.4$. | atkins |
||
$\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}$ | e2.12(a)(a) | 30 | 30 | Assume all gases are perfect unless stated otherwise. Unless otherwise stated, thermodynamic data are for 298.15 K. When $229 \mathrm{~J}$ of energy is supplied as heat to $3.0 \mathrm{~mol} \mathrm{Ar}(\mathrm{g})$ at constant pressure, the temperature of the sample increases by $2.55 \mathrm{~K}$. Calculate the molar heat capacities at constant volume. | atkins |
||
$\mathrm{kJ} \mathrm{mol}^{-1}$ | e3.20(a) | -0.55 | -0.55 | Assume that all gases are perfect and that data refer to 298.15 K unless otherwise stated. The fugacity coefficient of a certain gas at $200 \mathrm{~K}$ and 50 bar is 0.72. Calculate the difference of its molar Gibbs energy from that of a perfect gas in the same state. | atkins |
||
$\mathrm{kJ} \mathrm{mol}^{-1}$ | e2.23(a)(b) | -111.92 | -111.92 | Assume all gases are perfect unless stated otherwise. Unless otherwise stated, thermodynamic data are for 298.15 K. Given the reactions (1) and (2) below, determine $\Delta_{\mathrm{r}} U^{\ominus}$ for reaction (3).
(1) $\mathrm{H}_2(\mathrm{g})+\mathrm{Cl}_2(\mathrm{g}) \rightarrow 2 \mathrm{HCl}(\mathrm{g})$, $\Delta_{\mathrm{r}} H^{\ominus}=-184.62 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(2) $2 \mathrm{H}_2(\mathrm{g})+\mathrm{O}_2(\mathrm{g}) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(\mathrm{g})$, $\Delta_{\mathrm{r}} H^\ominus=-483.64 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(3) $4 \mathrm{HCl}(\mathrm{g})+\mathrm{O}_2(\mathrm{g}) \rightarrow 2 \mathrm{Cl}_2(\mathrm{g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g})$ | atkins |
||
$\mathrm{~Pa} \mathrm{~m}^3 \mathrm{~mol}^{-1}$ | 2.2 | -0.28 | -0.28 | The change in molar internal energy when $\mathrm{CaCO}_3(\mathrm{~s})$ as calcite converts to another form, aragonite, is $+0.21 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Calculate the difference between the molar enthalpy and internal energy changes when the pressure is 1.0 bar given that the densities of the polymorphs are $2.71 \mathrm{~g} \mathrm{~cm}^{-3}$ and $2.93 \mathrm{~g} \mathrm{~cm}^{-3}$, respectively. | The change in enthalpy when the transition occurs is
$$
\begin{aligned}
\Delta H_{\mathrm{m}} & =H_{\mathrm{m}}(\text { aragonite })-H_{\mathrm{m}}(\text { calcite }) \\
& =\left\{U_{\mathrm{m}}(\mathrm{a})+p V_{\mathrm{m}}(\mathrm{a})\right\}-\left\{U_{\mathrm{m}}(\mathrm{c})+p V_{\mathrm{m}}(\mathrm{c})\right\} \\
& =\Delta U_{\mathrm{m}}+p\left\{V_{\mathrm{m}}(\mathrm{a})-V_{\mathrm{m}}(\mathrm{c})\right\}
\end{aligned}
$$
where a denotes aragonite and c calcite. It follows by substituting $V_{\mathrm{m}}=M / \rho$ that
$$
\Delta H_{\mathrm{m}}-\Delta U_{\mathrm{m}}=p M\left(\frac{1}{\rho(\mathrm{a})}-\frac{1}{\rho(\mathrm{c})}\right)
$$
Substitution of the data, using $M=100 \mathrm{~g} \mathrm{~mol}^{-1}$, gives
$$
\begin{aligned}
\Delta H_{\mathrm{m}}-\Delta U_{\mathrm{m}} & =\left(1.0 \times 10^5 \mathrm{~Pa}\right) \times\left(100 \mathrm{~g} \mathrm{~mol}^{-1}\right) \times\left(\frac{1}{2.93 \mathrm{~g} \mathrm{~cm}^{-3}}-\frac{1}{2.71 \mathrm{~g} \mathrm{~cm}^{-3}}\right) \\
& =-2.8 \times 10^5 \mathrm{~Pa} \mathrm{~cm}{ }^3 \mathrm{~mol}^{-1}=-0.28 \mathrm{~Pa} \mathrm{~m}^3 \mathrm{~mol}^{-1}
\end{aligned}
$$ | atkins |
|
$\mathrm{dm}^3\mathrm{~mol}^{-1}$ | 1.4 | 0.366 | 0.366 | Estimate the molar volume of $\mathrm{CO}_2$ at $500 \mathrm{~K}$ and 100 atm by treating it as a van der Waals gas. |
According to Table $$
\begin{array}{lll}
\hline & \boldsymbol{a} /\left(\mathbf{a t m} \mathbf{d m}^6 \mathrm{~mol}^{-2}\right) & \boldsymbol{b} /\left(\mathbf{1 0}^{-2} \mathrm{dm}^3 \mathrm{~mol}^{-1}\right) \\
\hline \mathrm{Ar} & 1.337 & 3.20 \\
\mathrm{CO}_2 & 3.610 & 4.29 \\
\mathrm{He} & 0.0341 & 2.38 \\
\mathrm{Xe} & 4.137 & 5.16 \\
\hline
\end{array}
$$, $a=3.610 \mathrm{dm}^6 \mathrm{~atm} \mathrm{~mol}^{-2}$ and $b=4.29 \times 10^{-2} \mathrm{dm}^3$ $\mathrm{mol}^{-1}$. Under the stated conditions, $R T / p=0.410 \mathrm{dm}^3 \mathrm{~mol}^{-1}$. The coefficients in the equation for $V_{\mathrm{m}}$ are therefore
$$
\begin{aligned}
& b+R T / p=0.453 \mathrm{dm}^3 \mathrm{~mol}^{-1} \\
& a / p=3.61 \times 10^{-2}\left(\mathrm{dm}^3 \mathrm{~mol}^{-1}\right)^2 \\
& a b / p=1.55 \times 10^{-3}\left(\mathrm{dm}^3 \mathrm{~mol}^{-1}\right)^3
\end{aligned}
$$
Therefore, on writing $x=V_{\mathrm{m}} /\left(\mathrm{dm}^3 \mathrm{~mol}^{-1}\right)$, the equation to solve is
$$
x^3-0.453 x^2+\left(3.61 \times 10^{-2}\right) x-\left(1.55 \times 10^{-3}\right)=0
$$
The acceptable root is $x=0.366$, which implies that $V_{\mathrm{m}}=0.366 \mathrm{dm}^3 \mathrm{~mol}^{-1}$. | atkins |
|
$\mathrm{kN} \mathrm{mol}^{-1}$ | 20.3 | 17 | 17 | Suppose the concentration of a solute decays exponentially along the length of a container. Calculate the thermodynamic force on the solute at $25^{\circ} \mathrm{C}$ given that the concentration falls to half its value in $10 \mathrm{~cm}$. |
The concentration varies with position as
$$
c=c_0 \mathrm{e}^{-x / \lambda}
$$
where $\lambda$ is the decay constant. Therefore,
$$
\frac{\mathrm{d} c}{\mathrm{~d} x}=-\frac{c}{\lambda}
$$
From Equation $$
\mathcal{F}=-\frac{R T}{c}\left(\frac{\partial c}{\partial x}\right)_{p, T}
$$ then implies that
$$
\mathcal{F}=\frac{R T}{\lambda}
$$
We know that the concentration falls to $\frac{1}{2} c_0$ at $x=10 \mathrm{~cm}$, so we can find $\lambda$ from $\frac{1}{2}=\mathrm{e}^{-(10 \mathrm{~cm}) / \lambda}$. That is $\lambda=(10 \mathrm{~cm} / \ln 2)$. It follows that
$$
\mathcal{F}=\left(8.3145 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right) \times(298 \mathrm{~K}) \times \ln 2 /\left(1.0 \times 10^{-1} \mathrm{~m}\right)=17 \mathrm{kN} \mathrm{mol}^{-1}
$$
where we have used $1 \mathrm{~J}=1 \mathrm{~N} \mathrm{~m}$. | atkins |
|
$\mathrm{~kJ}$ | 5.2 | -6.9 | -6.9 | A container is divided into two equal compartments. One contains $3.0 \mathrm{~mol} \mathrm{H}_2(\mathrm{~g})$ at $25^{\circ} \mathrm{C}$; the other contains $1.0 \mathrm{~mol} \mathrm{~N}_2(\mathrm{~g})$ at $25^{\circ} \mathrm{C}$. Calculate the Gibbs energy of mixing when the partition is removed. Assume perfect behaviour. | Given that the pressure of nitrogen is $p$, the pressure of hydrogen is $3 p$; therefore, the initial Gibbs energy is
$$
G_{\mathrm{i}}=(3.0 \mathrm{~mol})\left\{\mu^{\ominus}\left(\mathrm{H}_2\right)+R T \ln 3 p\right\}+(1.0 \mathrm{~mol})\left\{\mu^{\ominus}\left(\mathrm{N}_2\right)+R T \ln p\right\}
$$
When the partition is removed and each gas occupies twice the original volume, the partial pressure of nitrogen falls to $\frac{1}{2} p$ and that of hydrogen falls to $\frac{3}{2} p$. Therefore, the Gibbs energy changes to
$$
G_{\mathrm{f}}=(3.0 \mathrm{~mol})\left\{\mu^{\ominus}\left(\mathrm{H}_2\right)+R T \ln \frac{3}{2} p\right\}+(1.0 \mathrm{~mol})\left\{\mu^{\ominus}\left(\mathrm{N}_2\right)+R T \ln \frac{1}{2} p\right\}
$$
The Gibbs energy of mixing is the difference of these two quantities:
$$
\begin{aligned}
\Delta_{\text {mix }} G & =(3.0 \mathrm{~mol}) R T \ln \left(\frac{\frac{3}{2} p}{3 p}\right)+(1.0 \mathrm{~mol}) R T \ln \left(\frac{\frac{1}{2} p}{p}\right) \\
& =-(3.0 \mathrm{~mol}) R T \ln 2-(1.0 \mathrm{~mol}) R T \ln 2 \\
& =-(4.0 \mathrm{~mol}) R T \ln 2=-6.9 \mathrm{~kJ}
\end{aligned}
$$
| atkins |
|
$\mathrm{~m} \mathrm{~s}^{-1}$ | 20.1 | 475 | 475 | What is the mean speed, $\bar{c}$, of $\mathrm{N}_2$ molecules in air at $25^{\circ} \mathrm{C}$ ? | The integral required is
$$
\begin{aligned}
\bar{c} & =4 \pi\left(\frac{M}{2 \pi R T}\right)^{3 / 2} \int_0^{\infty} v^3 \mathrm{e}^{-M v^2 / 2 R T} \mathrm{~d} v \\
& =4 \pi\left(\frac{M}{2 \pi R T}\right)^{3 / 2} \times \frac{1}{2}\left(\frac{2 R T}{M}\right)^2=\left(\frac{8 R T}{\pi M}\right)^{1 / 2}
\end{aligned}
$$
where we have used the standard result from tables of integrals (or software) that
$$
\int_0^{\infty} x^3 \mathrm{e}^{-a x^2} \mathrm{~d} x=\frac{1}{2 a^2}
$$
Substitution of the data then gives
$$
\bar{c}=\left(\frac{8 \times\left(8.3141 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right) \times(298 \mathrm{~K})}{\pi \times\left(28.02 \times 10^{-3} \mathrm{~kg} \mathrm{~mol}^{-1}\right)}\right)^{1 / 2}=475 \mathrm{~m} \mathrm{~s}^{-1}
$$
where we have used $1 \mathrm{~J}=1 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2}$.
| atkins |
|
$\mathrm{kPa}$ | 20.2 | 8.7 | 8.7 | Caesium (m.p. $29^{\circ} \mathrm{C}$, b.p. $686^{\circ} \mathrm{C}$ ) was introduced into a container and heated to $500^{\circ} \mathrm{C}$. When a hole of diameter $0.50 \mathrm{~mm}$ was opened in the container for $100 \mathrm{~s}$, a mass loss of $385 \mathrm{mg}$ was measured. Calculate the vapour pressure of liquid caesium at $500 \mathrm{~K}$. |
The mass loss $\Delta m$ in an interval $\Delta t$ is related to the collision flux by
$$
\Delta m=Z_{\mathrm{W}} A_0 m \Delta t
$$
where $A_0$ is the area of the hole and $m$ is the mass of one atom. It follows that
$$
Z_{\mathrm{W}}=\frac{\Delta m}{A_0 m \Delta t}
$$
Because $Z_{\mathrm{W}}$ is related to the pressure by eqn $$
Z_{\mathrm{W}}=\frac{p}{(2 \pi m k T)^{1 / 2}}
$$, we can write
$$
p=\left(\frac{2 \pi R T}{M}\right)^{1 / 2} \frac{\Delta m}{A_0 \Delta t}
$$
Because $M=132.9 \mathrm{~g} \mathrm{~mol}^{-1}$, substitution of the data gives $p=8.7 \mathrm{kPa}$ (using $1 \mathrm{~Pa}=$ $\left.1 \mathrm{~N} \mathrm{~m}^{-2}=1 \mathrm{~J} \mathrm{~m}^{-1}\right)$.
| atkins |
|
8.2 | 0.23 | 0.23 | To get an idea of the distance dependence of the tunnelling current in STM, suppose that the wavefunction of the electron in the gap between sample and needle is given by $\psi=B \mathrm{e}^{-\kappa x}$, where $\kappa=\left\{2 m_{\mathrm{e}}(V-E) / \hbar^2\right\}^{1 / 2}$; take $V-E=2.0 \mathrm{eV}$. By what factor would the current drop if the needle is moved from $L_1=0.50 \mathrm{~nm}$ to $L_2=0.60 \mathrm{~nm}$ from the surface? |
When $L=L_1=0.50 \mathrm{~nm}$ and $V-E=2.0 \mathrm{eV}=3.20 \times 10^{-19} \mathrm{~J}$ the value of $\kappa L$ is
$$
\begin{aligned}
\kappa L_1 & =\left\{\frac{2 m_{\mathrm{e}}(V-E)}{\hbar^2}\right\}^{1 / 2} L_1 \\
& =\left\{\frac{2 \times\left(9.109 \times 10^{-31} \mathrm{~kg}\right) \times\left(3.20 \times 10^{-19} \mathrm{~J}\right)}{\left(1.054 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)^2}\right\}^{1 / 2} \times\left(5.0 \times 10^{-10} \mathrm{~m}\right) \\
& =\left(7.25 \times 10^9 \mathrm{~m}^{-1}\right) \times\left(5.0 \times 10^{-10} \mathrm{~m}\right)=3.6
\end{aligned}
$$
Because $\kappa L_1>1$, we use eqn $T\approx 16 \varepsilon(1-\varepsilon) \mathrm{e}^{-2 \kappa L}$ to calculate the transmission probabilities at the two distances. It follows that
$$
\text { current at } \begin{aligned}
\frac{L_2}{\text { current at } L_1} & =\frac{T\left(L_2\right)}{T\left(L_1\right)}=\frac{16 \varepsilon(1-\varepsilon) \mathrm{e}^{-2 \kappa L_2}}{16 \varepsilon(1-\varepsilon) \mathrm{e}^{-2 \kappa L_1}}=\mathrm{e}^{-2 \kappa\left(L_2-L_1\right)} \\
& =\mathrm{e}^{-2 \times\left(7.25 \times 10^{-9} \mathrm{~m}^{-1}\right) \times\left(1.0 \times 10^{-10} \mathrm{~m}\right)}=0.23
\end{aligned}
$$ | atkins |
||
$\mathrm{pm}$ | 19.4 | 226 | 226 | Calculate the typical wavelength of neutrons that have reached thermal equilibrium with their surroundings at $373 \mathrm{~K}$.
| From the equipartition principle, we know that the mean translational kinetic energy of a neutron at a temperature $T$ travelling in the $x$-direction is $E_{\mathrm{k}}=\frac{1}{2} k T$. The kinetic energy is also equal to $p^2 / 2 m$, where $p$ is the momentum of the neutron and $m$ is its mass. Hence, $p=(m k T)^{1 / 2}$. It follows from the de Broglie relation $\lambda=h / p$ that the neutron's wavelength is
$$
\lambda=\frac{h}{(m k T)^{1 / 2}}
$$
Therefore, at $373 \mathrm{~K}$,
$$
\begin{aligned}
\lambda & =\frac{6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}}{\left\{\left(1.675 \times 10^{-27} \mathrm{~kg}\right) \times\left(1.381 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}\right) \times(373 \mathrm{~K})\right\}^{1 / 2}} \\
& =\frac{6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}}{\left(1.675 \times 1.381 \times 373 \times 10^{-50}\right)^{1 / 2}\left(\mathrm{~kg}^2 \mathrm{~m}^2 \mathrm{~s}^{-2}\right)^{1 / 2}} \\
& =2.26 \times 10^{-10} \mathrm{~m}=226 \mathrm{pm}
\end{aligned} | atkins |
|
$\mathrm{~nm}$ | 19.1 | 0.21 | 0.21 | Calculate the separation of the $\{123\}$ planes of an orthorhombic unit cell with $a=0.82 \mathrm{~nm}, b=0.94 \mathrm{~nm}$, and $c=0.75 \mathrm{~nm}$. |
For the first part, simply substitute the information into eqn $$
\frac{1}{d_{h k l}^2}=\frac{h^2}{a^2}+\frac{k^2}{b^2}+\frac{l^2}{c^2}
$$. For the second part, instead of repeating the calculation, note that, if all three Miller indices are multiplied by $n$, then their separation is reduced by that factor:
$$
\frac{1}{d_{n h, n k, n l}^2}=\frac{(n h)^2}{a^2}+\frac{(n k)^2}{b^2}+\frac{(n l)^2}{c^2}=n^2\left(\frac{h^2}{a^2}+\frac{k^2}{b^2}+\frac{l^2}{c^2}\right)=\frac{n^2}{d_{h k l}^2}
$$
which implies that
$$
d_{n h, n k, n l}=\frac{d_{h k l}}{n}
$$
Answer Substituting the indices into eqn $$
\frac{1}{d_{h k l}^2}=\frac{h^2}{a^2}+\frac{k^2}{b^2}+\frac{l^2}{c^2}
$$ gives
$$
\frac{1}{d_{123}^2}=\frac{1^2}{(0.82 \mathrm{~nm})^2}+\frac{2^2}{(0.94 \mathrm{~nm})^2}+\frac{3^2}{(0.75 \mathrm{~nm})^2}=0.22 \mathrm{~nm}^{-2}
$$
Hence, $d_{123}=0.21 \mathrm{~nm}$.
| atkins |
|
$10^{-47} \mathrm{~kg} \mathrm{~m}^2$ | 12.1 | 1.91 | 1.91 | Calculate the moment of inertia of an $\mathrm{H}_2 \mathrm{O}$ molecule around the axis defined by the bisector of the $\mathrm{HOH}$ angle (3). The $\mathrm{HOH}$ bond angle is $104.5^{\circ}$ and the bond length is $95.7 \mathrm{pm}$. |
$$
I=\sum_i m_i x_i^2=m_{\mathrm{H}} x_{\mathrm{H}}^2+0+m_{\mathrm{H}} x_{\mathrm{H}}^2=2 m_{\mathrm{H}} x_{\mathrm{H}}^2
$$
If the bond angle of the molecule is denoted $2 \phi$ and the bond length is $R$, trigonometry gives $x_{\mathrm{H}}=R \sin \phi$. It follows that
$$
I=2 m_{\mathrm{H}} R^2 \sin ^2 \phi
$$
Substitution of the data gives
$$
\begin{aligned}
I & =2 \times\left(1.67 \times 10^{-27} \mathrm{~kg}\right) \times\left(9.57 \times 10^{-11} \mathrm{~m}\right)^2 \times \sin ^2\left(\frac{1}{2} \times 104.5^{\circ}\right) \\
& =1.91 \times 10^{-47} \mathrm{~kg} \mathrm{~m}^2
\end{aligned}
$$
Note that the mass of the $\mathrm{O}$ atom makes no contribution to the moment of inertia for this mode of rotation as the atom is immobile while the $\mathrm{H}$ atoms circulate around it.
| atkins |
|
$\mathrm{~kJ} \mathrm{~mol}^{-1}$ | 6.3 | +80 | +80 | The data below show the temperature variation of the equilibrium constant of the reaction $\mathrm{Ag}_2 \mathrm{CO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{Ag}_2 \mathrm{O}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})$. Calculate the standard reaction enthalpy of the decomposition.
$\begin{array}{lllll}T / \mathrm{K} & 350 & 400 & 450 & 500 \\ K & 3.98 \times 10^{-4} & 1.41 \times 10^{-2} & 1.86 \times 10^{-1} & 1.48\end{array}$ |
We draw up the following table:
$\begin{array}{lllll}T / \mathrm{K} & 350 & 400 & 450 & 500 \\ \left(10^3 \mathrm{~K}\right) / T & 2.86 & 2.50 & 2.22 & 2.00 \\ -\ln K & 7.83 & 4.26 & 1.68 & -0.39\end{array}$
By plotting these points, the slope of the graph is $+9.6 \times 10^3$, so
$$
\Delta_{\mathrm{r}} H^{\ominus}=\left(+9.6 \times 10^3 \mathrm{~K}\right) \times R=+80 \mathrm{~kJ} \mathrm{~mol}^{-1}
$$
| atkins |
|
$10^2 \mathrm{~kg} \mathrm{~mol}^{-1}$ | 5.4 | 1.2 | 1.2 | The osmotic pressures of solutions of poly(vinyl chloride), PVC, in cyclohexanone at $298 \mathrm{~K}$ are given below. The pressures are expressed in terms of the heights of solution (of mass density $\rho=0.980 \mathrm{~g} \mathrm{~cm}^{-3}$ ) in balance with the osmotic pressure. Determine the molar mass of the polymer.
$\begin{array}{llllll}c /\left(\mathrm{g} \mathrm{dm}^{-3}\right) & 1.00 & 2.00 & 4.00 & 7.00 & 9.00 \\ h / \mathrm{cm} & 0.28 & 0.71 & 2.01 & 5.10 & 8.00\end{array}$
|
The data give the following values for the quantities to plot:
$$
\begin{array}{llllll}
c /\left(\mathrm{g} \mathrm{dm}^{-3}\right) & 1.00 & 2.00 & 4.00 & 7.00 & 9.00 \\
(h / c) /\left(\mathrm{cm} \mathrm{g}^{-1} \mathrm{dm}^3\right) & 0.28 & 0.36 & 0.503 & 0.729 & 0.889
\end{array}
$$
By plotting these points,the intercept is at 0.21 . Therefore,
$$
\begin{aligned}
M & =\frac{R T}{\rho g} \times \frac{1}{0.21 \mathrm{~cm} \mathrm{~g}^{-1} \mathrm{dm}^3} \\
& =\frac{\left(8.3145 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right) \times(298 \mathrm{~K})}{\left(980 \mathrm{~kg} \mathrm{~m}^{-1}\right) \times\left(9.81 \mathrm{~m} \mathrm{~s}^{-2}\right)} \times \frac{1}{2.1 \times 10^{-3} \mathrm{~m}^4 \mathrm{~kg}^{-1}} \\
& =1.2 \times 10^2 \mathrm{~kg} \mathrm{~mol}^{-1}
\end{aligned}
$$ | atkins |
|
$10^{-12} \mathrm{~m}$ | 7.2 | 6.1 | 6.1 | Estimate the wavelength of electrons that have been accelerated from rest through a potential difference of $40 \mathrm{kV}$. | The expression $p^2 / 2 m_{\mathrm{e}}=e \Delta \phi$ solves to $p=\left(2 m_{\mathrm{e}} e \Delta \phi\right)^{1 / 2}$; then, from the de Broglie relation $\lambda=h / p$,
$$
\lambda=\frac{h}{\left(2 m_{\mathrm{e}} e \Delta \phi\right)^{1 / 2}}
$$
Substitution of the data and the fundamental constants (from inside the front cover) gives
$$
\begin{aligned}
\lambda & =\frac{6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}}{\left\{2 \times\left(9.109 \times 10^{-31} \mathrm{~kg}\right) \times\left(1.602 \times 10^{-19} \mathrm{C}\right) \times\left(4.0 \times 10^4 \mathrm{~V}\right)\right\}^{1 / 2}} \\
& =6.1 \times 10^{-12} \mathrm{~m}
\end{aligned}
$$
| atkins |
|
$\mathrm{~kJ} \mathrm{~mol}^{-1}$ | 2.6 | -242.6 | -242.6 | The standard enthalpy of formation of $\mathrm{H}_2 \mathrm{O}(\mathrm{g})$ at $298 \mathrm{~K}$ is $-241.82 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Estimate its value at $100^{\circ} \mathrm{C}$ given the following values of the molar heat capacities at constant pressure: $\mathrm{H}_2 \mathrm{O}(\mathrm{g}): 33.58 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} ; \mathrm{H}_2(\mathrm{~g}): 28.82 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} ; \mathrm{O}_2(\mathrm{~g})$ : $29.36 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$. Assume that the heat capacities are independent of temperature. | The reaction is $\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g})$, so
$$
\Delta_{\mathrm{r}} C_p^{\ominus}=C_{p, \mathrm{~m}}^{\ominus}\left(\mathrm{H}_2 \mathrm{O}, \mathrm{g}\right)-\left\{C_{p, \mathrm{~m}}^{\ominus}\left(\mathrm{H}_2, \mathrm{~g}\right)+\frac{1}{2} C_{p, \mathrm{~m}}^{\ominus}\left(\mathrm{O}_2, \mathrm{~g}\right)\right\}=-9.92 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}
$$
It then follows that
$$
\Delta_{\mathrm{r}} H^{\ominus}(373 \mathrm{~K})=-241.82 \mathrm{~kJ} \mathrm{~mol}^{-1}+(75 \mathrm{~K}) \times\left(-9.92 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)=-242.6 \mathrm{~kJ} \mathrm{~mol}^{-1}
$$
| atkins |
|
$\mathrm{~kJ} \mathrm{~mol}^{-1}$ | 6.5 | -21.2 | -21.2 | The standard potential of the cell $\mathrm{Pt}(\mathrm{s})\left|\mathrm{H}_2(\mathrm{~g})\right| \mathrm{HBr}(\mathrm{aq})|\operatorname{AgBr}(\mathrm{s})| \mathrm{Ag}(\mathrm{s})$ was measured over a range of temperatures, and the data were found to fit the following polynomial:
$$
E_{\text {cell }}^{\bullet} / \mathrm{V}=0.07131-4.99 \times 10^{-4}(T / \mathrm{K}-298)-3.45 \times 10^{-6}(\mathrm{~T} / \mathrm{K}-298)^2
$$
The cell reaction is $\operatorname{AgBr}(\mathrm{s})+\frac{1}{2} \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{Ag}(\mathrm{s})+\mathrm{HBr}(\mathrm{aq})$. Evaluate the standard reaction enthalpy at $298 \mathrm{~K}$. |
At $T=298 \mathrm{~K}, E_{\text {cell }}^{\ominus}=+0.07131 \mathrm{~V}$, so
$$
\begin{aligned}
\Delta_{\mathrm{r}} G^{\ominus} & =-v F E_{\text {cell }}^{\ominus}=-(1) \times\left(9.6485 \times 10^4 \mathrm{Cmol}^{-1}\right) \times(+0.07131 \mathrm{~V}) \\
& =-6.880 \times 10^3 \mathrm{~V} \mathrm{Cmol}^{-1}=-6.880 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$$
The temperature coefficient of the cell potential is
$$
\frac{\mathrm{d} E_{\text {cell }}^{\ominus}}{\mathrm{d} T}=-4.99 \times 10^{-4} \mathrm{~V} \mathrm{~K}^{-1}-2\left(3.45 \times 10^{-6}\right)(T / \mathrm{K}-298) \mathrm{V} \mathrm{K}^{-1}
$$
At $T=298 \mathrm{~K}$ this expression evaluates to
$$
\frac{\mathrm{d} E_{\text {cell }}^{\ominus}}{\mathrm{d} T}=-4.99 \times 10^{-4} \mathrm{~V} \mathrm{~K}^{-1}
$$
So, from eqn $$
\frac{\mathrm{d} E_{\text {cell }}^{\Theta}}{\mathrm{d} T}=\frac{\Delta_{\mathrm{r}} S^{\Theta}}{v F}
$$ , the reaction entropy is
$$
\begin{aligned}
\Delta_{\mathrm{r}} S^{\ominus} & =1 \times\left(9.6485 \times 10^4 \mathrm{Cmol}^{-1}\right) \times\left(-4.99 \times 10^{-4} \mathrm{~V} \mathrm{~K}^{-1}\right) \\
& =-48.1 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}
\end{aligned}
$$
The negative value stems in part from the elimination of gas in the cell reaction. It then follows that
$$
\begin{aligned}
\Delta_{\mathrm{r}} H^{\ominus} & =\Delta_{\mathrm{r}} G^{\ominus}+T \Delta_{\mathrm{r}} S^{\ominus}=-6.880 \mathrm{~kJ} \mathrm{~mol}^{-1}+(298 \mathrm{~K}) \times\left(-0.0482 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right) \\
& =-21.2 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$$
| atkins |
|
$\mathrm{atm}$ | 1.2 | 167 | 167 | In an industrial process, nitrogen is heated to $500 \mathrm{~K}$ in a vessel of constant volume. If it enters the vessel at $100 \mathrm{~atm}$ and $300 \mathrm{~K}$, what pressure would it exert at the working temperature if it behaved as a perfect gas? | Cancellation of the volumes (because $V_1=V_2$ ) and amounts (because $\left.n_1=n_2\right)$ on each side of the combined gas law results in
$$
\frac{p_1}{T_1}=\frac{p_2}{T_2}
$$
which can be rearranged into
$$
p_2=\frac{T_2}{T_1} \times p_1
$$
Substitution of the data then gives
$$
p_2=\frac{500 \mathrm{~K}}{300 \mathrm{~K}} \times(100 \mathrm{~atm})=167 \mathrm{~atm}
$$
| atkins |
|
$\mathrm{kg}/\mathrm{s}$ | 16.7.43 | 0 | 0 | A fluid has density $870 \mathrm{~kg} / \mathrm{m}^3$ and flows with velocity $\mathbf{v}=z \mathbf{i}+y^2 \mathbf{j}+x^2 \mathbf{k}$, where $x, y$, and $z$ are measured in meters and the components of $\mathbf{v}$ in meters per second. Find the rate of flow outward through the cylinder $x^2+y^2=4$, $0 \leqslant z \leqslant 1$.
| calculus |
||
$\mathrm{cm}$ | 6.4.9(b) | 10.8 | 10.8 | Suppose that $2 \mathrm{~J}$ of work is needed to stretch a spring from its natural length of $30 \mathrm{~cm}$ to a length of $42 \mathrm{~cm}$. How far beyond its natural length will a force of $30 \mathrm{~N}$ keep the spring stretched? | calculus |
||
$\mathrm{J}$ | 12.3.49 | 144 | 144 | Find the work done by a force $\mathbf{F}=8 \mathbf{i}-6 \mathbf{j}+9 \mathbf{k}$ that moves an object from the point $(0,10,8)$ to the point $(6,12,20)$ along a straight line. The distance is measured in meters and the force in newtons. | calculus |
||
$\mathrm{m}/\mathrm{s}$ | 13.4.25 | 30 | 30 | A ball is thrown at an angle of $45^{\circ}$ to the ground. If the ball lands $90 \mathrm{~m}$ away, what was the initial speed of the ball?
| calculus |
||
$\mathrm{J}$ | 6.4.17 | 3857 | 3857 | Show how to approximate the required work by a Riemann sum. Then express the work as an integral and evaluate it. A leaky 10-kg bucket is lifted from the ground to a height of $12 \mathrm{~m}$ at a constant speed with a rope that weighs $0.8 \mathrm{~kg} / \mathrm{m}$. Initially the bucket contains $36 \mathrm{~kg}$ of water, but the water leaks at a constant rate and finishes draining just as the bucket reaches the 12-m level. How much work is done?
| calculus |
||
6.2.55 | 24 | 24 | Find the volume of the described solid $S$. The base of $S$ is an elliptical region with boundary curve $9 x^2+4 y^2=36$. Cross-sections perpendicular to the $x$-axis are isosceles right triangles with hypotenuse in the base. | calculus |
|||
$\mathrm{ft}^3$ | 15.4.35 | $1800\pi$ | 5654.86677646 | A swimming pool is circular with a $40-\mathrm{ft}$ diameter. The depth is constant along east-west lines and increases linearly from $2 \mathrm{ft}$ at the south end to $7 \mathrm{ft}$ at the north end. Find the volume of water in the pool. | calculus |
||
$\mathrm{AU}$ | 10.6.27 | 35.64 | 35.64 | The orbit of Halley's comet, last seen in 1986 and due to return in 2062, is an ellipse with eccentricity 0.97 and one focus at the sun. The length of its major axis is $36.18 \mathrm{AU}$. [An astronomical unit (AU) is the mean distance between the earth and the sun, about 93 million miles.] By finding a polar equation for the orbit of Halley's comet, what is the maximum distance from the comet to the sun? | calculus |
||
$\mathrm{ft} / \mathrm{s}$ | 2.1.5(a) | -32 | -32 | If a ball is thrown into the air with a velocity of $40 \mathrm{ft} / \mathrm{s}$, its height in feet $t$ seconds later is given by $y=40 t-16 t^2$. Find the average velocity for the time period beginning when $t=2$ and lasting 0.5 second. | calculus |
||
$\mathrm{cm}^3$ | 6.2.43 | 1110 | 1110 | A CAT scan produces equally spaced cross-sectional views of a human organ that provide information about the organ otherwise obtained only by surgery. Suppose that a CAT scan of a human liver shows cross-sections spaced $1.5 \mathrm{~cm}$ apart. The liver is $15 \mathrm{~cm}$ long and the cross-sectional areas, in square centimeters, are $0,18,58,79,94,106,117,128,63, 39, 0$. Use the Midpoint Rule to estimate the volume of the liver.
| calculus |
||
$\mathrm{in}$ | 8.1.39 | 29.36 | 29.36 | A manufacturer of corrugated metal roofing wants to produce panels that are $28 \mathrm{in}$. wide and $2 \mathrm{in}$. thick by processing flat sheets of metal as shown in the figure. The profile of the roofing takes the shape of a sine wave. Verify that the sine curve has equation $y=\sin (\pi x / 7)$ and find the width $w$ of a flat metal sheet that is needed to make a 28-inch panel. (Use your calculator to evaluate the integral correct to four significant digits.)
| calculus |
||
$\mathrm{L}/\mathrm{min}$ | 8.4.17 | 6.6 | 6.6 | The dye dilution method is used to measure cardiac output with $6 \mathrm{mg}$ of dye. The dye concentrations, in $\mathrm{mg} / \mathrm{L}$, are modeled by $c(t)=20 t e^{-0.6 t}, 0 \leqslant t \leqslant 10$, where $t$ is measured in seconds. Find the cardiac output. | calculus |
||
$\mathrm{h}$ | 9.RP.11(a) | Review Plus Problem | 9.8 | 9.8 | A planning engineer for a new alum plant must present some estimates to his company regarding the capacity of a silo designed to contain bauxite ore until it is processed into alum. The ore resembles pink talcum powder and is poured from a conveyor at the top of the silo. The silo is a cylinder $100 \mathrm{ft}$ high with a radius of $200 \mathrm{ft}$. The conveyor carries ore at a rate of $60,000 \pi \mathrm{~ft}^3 / \mathrm{h}$ and the ore maintains a conical shape whose radius is 1.5 times its height. If, at a certain time $t$, the pile is $60 \mathrm{ft}$ high, how long will it take for the pile to reach the top of the silo? | calculus |
|
$\mathrm{min}$ | 12.2.39 | 20.2 | 20.2 | A boatman wants to cross a canal that is $3 \mathrm{~km}$ wide and wants to land at a point $2 \mathrm{~km}$ upstream from his starting point. The current in the canal flows at $3.5 \mathrm{~km} / \mathrm{h}$ and the speed of his boat is $13 \mathrm{~km} / \mathrm{h}$. How long will the trip take?
| calculus |
||
7.R.73 | 2 | 2 | Find the area bounded by the curves $y=\cos x$ and $y=\cos ^2 x$ between $x=0$ and $x=\pi$. | calculus |
|||
$\mathrm{ft-lb}$ | 12.3.51 | $2400\cos({40}^{\circ})$ | 1838.50666349 | A sled is pulled along a level path through snow by a rope. A 30-lb force acting at an angle of $40^{\circ}$ above the horizontal moves the sled $80 \mathrm{ft}$. Find the work done by the force. | calculus |
||
$\Omega$ | 14.4.39 | $\frac{1}{17}$ | 0.05882352941 | If $R$ is the total resistance of three resistors, connected in parallel, with resistances $R_1, R_2, R_3$, then
$$
\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}
$$
If the resistances are measured in ohms as $R_1=25 \Omega$, $R_2=40 \Omega$, and $R_3=50 \Omega$, with a possible error of $0.5 \%$ in each case, estimate the maximum error in the calculated value of $R$. | calculus |
||
$\mathrm{cm^2}$ | 14.4.33 | 5.4 | 5.4 | The length and width of a rectangle are measured as $30 \mathrm{~cm}$ and $24 \mathrm{~cm}$, respectively, with an error in measurement of at most $0.1 \mathrm{~cm}$ in each. Use differentials to estimate the maximum error in the calculated area of the rectangle.
| calculus |
||
$\mathrm{10^7} \mathrm{~km}$ | 10.6.29 | 7 | 7 | The planet Mercury travels in an elliptical orbit with eccentricity 0.206 . Its minimum distance from the sun is $4.6 \times 10^7 \mathrm{~km}$. Find its maximum distance from the sun. | calculus |
||
$\mathrm{cm^3}$ | 14.4.35 | 16 | 16 | Use differentials to estimate the amount of tin in a closed tin can with diameter $8 \mathrm{~cm}$ and height $12 \mathrm{~cm}$ if the tin is $0.04 \mathrm{~cm}$ thick. | calculus |
||
$\mathrm{ft-lb}$ | 6.4.15 | 650000 | 650000 | Show how to approximate the required work by a Riemann sum. Then express the work as an integral and evaluate it. A cable that weighs $2 \mathrm{~lb} / \mathrm{ft}$ is used to lift $800 \mathrm{~lb}$ of coal up a mine shaft $500 \mathrm{~ft}$ deep. Find the work done.
| calculus |
||
$\mathrm{mg}$ | 11.2.69(a) | 157.875 | 157.875 | A patient takes $150 \mathrm{mg}$ of a drug at the same time every day. Just before each tablet is taken, 5$\%$ of the drug remains in the body. What quantity of the drug is in the body after the third tablet? | calculus |
||
$\mathrm{ft-lb}$ | 6.4.1(b) | 7200 | 7200 | A $360-\mathrm{lb}$ gorilla climbs a tree to a height of $20 \mathrm{~ft}$. Find the work done if the gorilla reaches that height in 5 seconds. | calculus |
||
$\mathrm{cm^2}$ | D.89 | 14.34457 | 14.34457 | Find the area of triangle $A B C$, correct to five decimal places, if
$$
|A B|=10 \mathrm{~cm} \quad|B C|=3 \mathrm{~cm} \quad \angle A B C=107^{\circ}
$$ | calculus |
||
16.R.33 | $-\frac{1}{2}$ | -0.5 | Use Stokes' Theorem to evaluate $\int_C \mathbf{F} \cdot d \mathbf{r}$, where $\mathbf{F}(x, y, z)=x y \mathbf{i}+y z \mathbf{j}+z x \mathbf{k}$, and $C$ is the triangle with vertices $(1,0,0),(0,1,0)$, and $(0,0,1)$, oriented counterclockwise as viewed from above.
| calculus |
|||
$\mathrm{m}$ | 8.1.37 | 209.1 | 209.1 | A hawk flying at $15 \mathrm{~m} / \mathrm{s}$ at an altitude of $180 \mathrm{~m}$ accidentally drops its prey. The parabolic trajectory of the falling prey is described by the equation
$$
y=180-\frac{x^2}{45}
$$
until it hits the ground, where $y$ is its height above the ground and $x$ is the horizontal distance traveled in meters. Calculate the distance traveled by the prey from the time it is dropped until the time it hits the ground. Express your answer correct to the nearest tenth of a meter. | calculus |
||
7.7.43 | 59.4 | 59.4 | The intensity of light with wavelength $\lambda$ traveling through a diffraction grating with $N$ slits at an angle $\theta$ is given by $I(\theta)=N^2 \sin ^2 k / k^2$, where $k=(\pi N d \sin \theta) / \lambda$ and $d$ is the distance between adjacent slits. A helium-neon laser with wavelength $\lambda=632.8 \times 10^{-9} \mathrm{~m}$ is emitting a narrow band of light, given by $-10^{-6}<\theta<10^{-6}$, through a grating with 10,000 slits spaced $10^{-4} \mathrm{~m}$ apart. Use the Midpoint Rule with $n=10$ to estimate the total light intensity $\int_{-10^{-6}}^{10^{-6}} I(\theta) d \theta$ emerging from the grating. | calculus |
|||
$\%$ | 14.4.41 | 2.3 | 2.3 | A model for the surface area of a human body is given by $S=0.1091 w^{0.425} h^{0.725}$, where $w$ is the weight (in pounds), $h$ is the height (in inches), and $S$ is measured in square feet. If the errors in measurement of $w$ and $h$ are at most $2 \%$, use differentials to estimate the maximum percentage error in the calculated surface area. | calculus |
||
16.7.47 | $1248\pi$ | 3920.70763168 | The temperature at the point $(x, y, z)$ in a substance with conductivity $K=6.5$ is $u(x, y, z)=2 y^2+2 z^2$. Find the rate of heat flow inward across the cylindrical surface $y^2+z^2=6$, $0 \leqslant x \leqslant 4$ | calculus |
|||
$\mathrm{ft} / \mathrm{s}$ | 2.7.13 | -24 | -24 | If a ball is thrown into the air with a velocity of $40 \mathrm{ft} / \mathrm{s}$, its height (in feet) after $t$ seconds is given by $y=40 t-16 t^2$. Find the velocity when $t=2$. | calculus |
||
$\mathrm{mi}/\mathrm{h}$ | 12.2.35 | $\sqrt{493}$ | 22.2036033112 | A woman walks due west on the deck of a ship at $3 \mathrm{mi} / \mathrm{h}$. The ship is moving north at a speed of $22 \mathrm{mi} / \mathrm{h}$. Find the speed of the woman relative to the surface of the water. | calculus |
||
$\mathrm{ft-lb}$ | 6.4.1(a) | 7200 | 7200 | A $360-\mathrm{lb}$ gorilla climbs a tree to a height of $20 \mathrm{~ft}$. Find the work done if the gorilla reaches that height in 10 seconds. | calculus |
||
$\mathrm{ft}/\mathrm{s}$ | 13.4.31 | $10\sqrt{93}$ | 96.4365076099 | A ball is thrown eastward into the air from the origin (in the direction of the positive $x$-axis). The initial velocity is $50 \mathrm{i}+80 \mathrm{k}$, with speed measured in feet per second. The spin of the ball results in a southward acceleration of $4 \mathrm{ft} / \mathrm{s}^2$, so the acceleration vector is $\mathbf{a}=-4 \mathbf{j}-32 \mathbf{k}$. What speed does the ball land?
| calculus |
||
$\$$ | 8.R.17 | 7166.67 | 7166.67 | The demand function for a commodity is given by
$$
p=2000-0.1 x-0.01 x^2
$$
Find the consumer surplus when the sales level is 100 . | calculus |
||
$\mathrm{~kg} / \mathrm{m}$ | 6.5.19 | 6 | 6 | The linear density in a rod $8 \mathrm{~m}$ long is $12 / \sqrt{x+1} \mathrm{~kg} / \mathrm{m}$, where $x$ is measured in meters from one end of the rod. Find the average density of the rod. | calculus |
||
$\mathrm{ft-lb}$ | 6.4.3 | 4.5 | 4.5 | A variable force of $5 x^{-2}$ pounds moves an object along a straight line when it is $x$ feet from the origin. Calculate the work done in moving the object from $x=1 \mathrm{~ft}$ to $x=10 \mathrm{~ft}$. | calculus |
||
$\mathrm{days}$ | 9.R.19 | 15 | 15 | One model for the spread of an epidemic is that the rate of spread is jointly proportional to the number of infected people and the number of uninfected people. In an isolated town of 5000 inhabitants, 160 people have a disease at the beginning of the week and 1200 have it at the end of the week. How long does it take for $80 \%$ of the population to become infected? | calculus |
||
$\mathrm{cm}^3$ | 6.2.53 | 10 | 10 | Find the volume of the described solid S. A tetrahedron with three mutually perpendicular faces and three mutually perpendicular edges with lengths $3 \mathrm{~cm}$, $4 \mathrm{~cm}$, and $5 \mathrm{~cm}$ | calculus |
||
6.R.23 | review problem | 36 | 36 | The base of a solid is a circular disk with radius 3 . Find the volume of the solid if parallel cross-sections perpendicular to the base are isosceles right triangles with hypotenuse lying along the base. | calculus |
||
$\mathrm{m}/\mathrm{s}$ | 13.4.23(c) | 200 | 200 | A projectile is fired with an initial speed of $200 \mathrm{~m} / \mathrm{s}$ and angle of elevation $60^{\circ}$. Find the speed at impact. | calculus |
||
$\mathrm{J}$ | 6.R.27 | review problem | 3.2 | 3.2 | A force of $30 \mathrm{~N}$ is required to maintain a spring stretched from its natural length of $12 \mathrm{~cm}$ to a length of $15 \mathrm{~cm}$. How much work is done in stretching the spring from $12 \mathrm{~cm}$ to $20 \mathrm{~cm}$ ?
| calculus |
|
$\mathrm{10^{-4}} \mathrm{~cm}^3/\mathrm{s}$ | 8.4.15 | 1.19 | 1.19 | Use Poiseuille's Law to calculate the rate of flow in a small human artery where we can take $\eta=0.027, R=0.008 \mathrm{~cm}$, $I=2 \mathrm{~cm}$, and $P=4000$ dynes $/ \mathrm{cm}^2$. | calculus |
||
7.7.2 | with solution | 41 | 41 | How large should we take $n$ in order to guarantee that the Trapezoidal and Midpoint Rule approximations for $\int_1^2(1 / x) d x$ are accurate to within 0.0001 ? | We saw in the preceding calculation that $\left|f^{\prime \prime}(x)\right| \leqslant 2$ for $1 \leqslant x \leqslant 2$, so we can take $K=2$, $a=1$, and $b=2$ in 3 . Accuracy to within 0.0001 means that the size of the error should be less than 0.0001 . Therefore we choose $n$ so that
$$
\frac{2(1)^3}{12 n^2}<0.0001
$$
Solving the inequality for $n$, we get
or
$$
\begin{aligned}
& n^2>\frac{2}{12(0.0001)} \\
& n>\frac{1}{\sqrt{0.0006}} \approx 40.8
\end{aligned}
$$
Thus $n=41$ will ensure the desired accuracy. | calculus |
|
10.4.4 | with solution | 8 | 8 | Find the length of the cardioid $r=1+\sin \theta$. |
The cardioid's full length is given by the parameter interval $0 \leqslant \theta \leqslant 2 \pi$, $$
\begin{aligned}
L & =\int_0^{2 \pi} \sqrt{r^2+\left(\frac{d r}{d \theta}\right)^2} d \theta=\int_0^{2 \pi} \sqrt{(1+\sin \theta)^2+\cos ^2 \theta} d \theta \\
& =\int_0^{2 \pi} \sqrt{2+2 \sin \theta} d \theta
\end{aligned}
$$
We could evaluate this integral by multiplying and dividing the integrand by $\sqrt{2-2 \sin \theta}$, or we could use a computer algebra system. In any event, we find that the length of the cardioid is $L=8$.
| calculus |
|
15.1.1 | with solution | 34 | 34 | Estimate the volume of the solid that lies above the square $R=[0,2] \times[0,2]$ and below the elliptic paraboloid $z=16-x^2-2 y^2$. Divide $R$ into four equal squares and choose the sample point to be the upper right corner of each square $R_{i j}$. |
The paraboloid is the graph of $f(x, y)=16-x^2-2 y^2$ and the area of each square is $\Delta A=1$. Approximating the volume by the Riemann sum with $m=n=2$, we have
$$
\begin{aligned}
V & \approx \sum_{i=1}^2 \sum_{j=1}^2 f\left(x_i, y_j\right) \Delta A \\
& =f(1,1) \Delta A+f(1,2) \Delta A+f(2,1) \Delta A+f(2,2) \Delta A \\
& =13(1)+7(1)+10(1)+4(1)=34
\end{aligned}
$$ | calculus |
|
6.5.1 | with equation | 2 | 2 | Find the average value of the function $f(x)=1+x^2$ on the interval $[-1,2]$. | With $a=-1$ and $b=2$ we have
$$
\begin{aligned}
f_{\text {ave }} & =\frac{1}{b-a} \int_a^b f(x) d x=\frac{1}{2-(-1)} \int_{-1}^2\left(1+x^2\right) d x \\
& =\frac{1}{3}\left[x+\frac{x^3}{3}\right]_{-1}^2=2
\end{aligned}
$$ | calculus |
|
6.1.2 | with solution | $\frac{1}{3}$ | 0.333333333333333 | Find the area of the region enclosed by the parabolas $y=x^2$ and $y=2 x-x^2$ |
We first find the points of intersection of the parabolas by solving their equations simultaneously. This gives $x^2=2 x-x^2$, or $2 x^2-2 x=0$. Thus $2 x(x-1)=0$, so $x=0$ or 1 . The points of intersection are $(0,0)$ and $(1,1)$. By analyzing these two equations, we can see that the top and bottom boundaries are
$$
y_T=2 x-x^2 \quad \text { and } \quad y_B=x^2
$$
The area of a typical rectangle is
$$
\left(y_T-y_B\right) \Delta x=\left(2 x-x^2-x^2\right) \Delta x
$$
and the region lies between $x=0$ and $x=1$. So the total area is
$$
\begin{aligned}
A & =\int_0^1\left(2 x-2 x^2\right) d x=2 \int_0^1\left(x-x^2\right) d x \\
& =2\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1=2\left(\frac{1}{2}-\frac{1}{3}\right)=\frac{1}{3}
\end{aligned}
$$
| calculus |
|
6.2.4 | with solution | $\frac{2\pi}{15}$ | 0.41887902047 | The region $\mathscr{R}$ enclosed by the curves $y=x$ and $y=x^2$ is rotated about the $x$-axis. Find the volume of the resulting solid. |
The curves $y=x$ and $y=x^2$ intersect at the points $(0,0)$ and $(1,1)$. A cross-section in the plane $P_x$ has the shape of a washer (an annular ring) with inner radius $x^2$ and outer radius $x$, so we find the cross-sectional area by subtracting the area of the inner circle from the area of the outer circle:
$$
A(x)=\pi x^2-\pi\left(x^2\right)^2=\pi\left(x^2-x^4\right)
$$
Therefore we have
$$
\begin{aligned}
V & =\int_0^1 A(x) d x=\int_0^1 \pi\left(x^2-x^4\right) d x \\
& =\pi\left[\frac{x^3}{3}-\frac{x^5}{5}\right]_0^1=\frac{2 \pi}{15}
\end{aligned}
$$
| calculus |
|
7.7.4 | with solution | 0.693150 | 0.693150 | Use Simpson's Rule with $n=10$ to approximate $\int_1^2(1 / x) d x$. | Putting $f(x)=1 / x, n=10$, and $\Delta x=0.1$ in Simpson's Rule, we obtain
$$
\begin{aligned}
\int_1^2 \frac{1}{x} d x & \approx S_{10} \\
& =\frac{\Delta x}{3}[f(1)+4 f(1.1)+2 f(1.2)+4 f(1.3)+\cdots+2 f(1.8)+4 f(1.9)+f(2)] \\
& =\frac{0.1}{3}\left(\frac{1}{1}+\frac{4}{1.1}+\frac{2}{1.2}+\frac{4}{1.3}+\frac{2}{1.4}+\frac{4}{1.5}+\frac{2}{1.6}+\frac{4}{1.7}+\frac{2}{1.8}+\frac{4}{1.9}+\frac{1}{2}\right) \\
& \approx 0.693150
\end{aligned}
$$
| calculus |
|
15.1.3 | -11.875 | -11.875 | Use the Midpoint Rule with $m=n=2$ to estimate the value of the integral $\iint_R\left(x-3 y^2\right) d A$, where $R=\{(x, y) \mid 0 \leqslant x \leqslant 2,1 \leqslant y \leqslant 2\}$. |
In using the Midpoint Rule with $m=n=2$, we evaluate $f(x, y)=x-3 y^2$ at the centers of the four subrectangles. So $\bar{X}_1=\frac{1}{2}, \bar{X}_2=\frac{3}{2}, \bar{y}_1=\frac{5}{4}$, and $\bar{y}_2=\frac{7}{4}$. The area of each subrectangle is $\Delta A=\frac{1}{2}$. Thus
$$
\begin{aligned}
\iint_R\left(x-3 y^2\right) d A & \approx \sum_{i=1}^2 \sum_{j=1}^2 f\left(\bar{x}_i, \bar{y}_j\right) \Delta A \\
& =f\left(\bar{x}_1, \bar{y}_1\right) \Delta A+f\left(\bar{x}_1, \bar{y}_2\right) \Delta A+f\left(\bar{x}_2, \bar{y}_1\right) \Delta A+f\left(\bar{x}_2, \bar{y}_2\right) \Delta A \\
& =f\left(\frac{1}{2}, \frac{5}{4}\right) \Delta A+f\left(\frac{1}{2}, \frac{7}{4}\right) \Delta A+f\left(\frac{3}{2}, \frac{5}{4}\right) \Delta A+f\left(\frac{3}{2}, \frac{7}{4}\right) \Delta A \\
& =\left(-\frac{67}{16}\right) \frac{1}{2}+\left(-\frac{139}{16}\right) \frac{1}{2}+\left(-\frac{51}{16}\right) \frac{1}{2}+\left(-\frac{123}{16}\right) \frac{1}{2} \\
& =-\frac{95}{8}=-11.875
\end{aligned}
$$
Thus we have
$$
\iint_R\left(x-3 y^2\right) d A \approx-11.875
$$
| calculus |
||
$\mathrm{~cm}^3$ | 14.4.5 | with solution | $20\pi$ | 62.8318530718 | The base radius and height of a right circular cone are measured as $10 \mathrm{~cm}$ and $25 \mathrm{~cm}$, respectively, with a possible error in measurement of as much as $0.1 \mathrm{~cm}$ in each. Use differentials to estimate the maximum error in the calculated volume of the cone. | The volume $V$ of a cone with base radius $r$ and height $h$ is $V=\pi r^2 h / 3$. So the differential of $V$ is
$$
d V=\frac{\partial V}{\partial r} d r+\frac{\partial V}{\partial h} d h=\frac{2 \pi r h}{3} d r+\frac{\pi r^2}{3} d h
$$
Since each error is at most $0.1 \mathrm{~cm}$, we have $|\Delta r| \leqslant 0.1,|\Delta h| \leqslant 0.1$. To estimate the largest error in the volume we take the largest error in the measurement of $r$ and of $h$. Therefore we take $d r=0.1$ and $d h=0.1$ along with $r=10, h=25$. This gives
$$
d V=\frac{500 \pi}{3}(0.1)+\frac{100 \pi}{3}(0.1)=20 \pi
$$
Thus the maximum error in the calculated volume is about $20 \pi \mathrm{cm}^3 \approx 63 \mathrm{~cm}^3$.
| calculus |
$\mathrm{~J}$ | 6.4.3 | with equation | 1.56 | 1.56 | A force of $40 \mathrm{~N}$ is required to hold a spring that has been stretched from its natural length of $10 \mathrm{~cm}$ to a length of $15 \mathrm{~cm}$. How much work is done in stretching the spring from $15 \mathrm{~cm}$ to $18 \mathrm{~cm}$ ? | According to Hooke's Law, the force required to hold the spring stretched $x$ meters beyond its natural length is $f(x)=k x$. When the spring is stretched from $10 \mathrm{~cm}$ to $15 \mathrm{~cm}$, the amount stretched is $5 \mathrm{~cm}=0.05 \mathrm{~m}$. This means that $f(0.05)=40$, so
$$
0.05 k=40 \quad k=\frac{40}{0.05}=800
$$
Thus $f(x)=800 x$ and the work done in stretching the spring from $15 \mathrm{~cm}$ to $18 \mathrm{~cm}$ is
$$
\begin{aligned}
W & \left.=\int_{0.05}^{0.08} 800 x d x=800 \frac{x^2}{2}\right]_{0.05}^{0.08} \\
& =400\left[(0.08)^2-(0.05)^2\right]=1.56 \mathrm{~J}
\end{aligned}
$$ | calculus |
nm | 1-38 | 0.123 | 0.123 | Calculate the de Broglie wavelength for an electron with a kinetic energy of $100 \mathrm{eV}$ | chemmc |
||
$10^{-19} \mathrm{~J}$ | 1-18 | Only the first part, the work function is taken | 3.52 | 3.52 | The threshold wavelength for potassium metal is $564 \mathrm{~nm}$. What is its work function?
| chemmc |
|
D-7 | Math Part D (after chapter 4) | 3 / 2 | 1.5 | Evaluate the series
$$
S=\sum_{n=0}^{\infty} \frac{1}{3^n}
$$ | chemmc |
||
$10^{-23} \mathrm{~s}$ | 1-49 | 2.9 | 2.9 | The relationship introduced in Problem $1-48$ has been interpreted to mean that a particle of mass $m\left(E=m c^2\right)$ can materialize from nothing provided that it returns to nothing within a time $\Delta t \leq h / m c^2$. Particles that last for time $\Delta t$ or more are called real particles; particles that last less than time $\Delta t$ are called virtual particles. The mass of the charged pion, a subatomic particle, is $2.5 \times 10^{-28} \mathrm{~kg}$. What is the minimum lifetime if the pion is to be considered a real particle? | chemmc |
||
$\mathrm{~K}$
| 1-17 | 5300 | 5300 | A household lightbulb is a blackbody radiator. Many lightbulbs use tungsten filaments that are heated by an electric current. What temperature is needed so that $\lambda_{\max }=550 \mathrm{~nm}$ ? | chemmc |
||
D-6 | Math Part D (after chapter 4) | 1 | 1 | Evaluate the series
$$
S=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots
$$
| chemmc |
||
V | 1-40 | 0.082 | 0.082 | Through what potential must a proton initially at rest fall so that its de Broglie wavelength is $1.0 \times 10^{-10} \mathrm{~m}$ ? | chemmc |
||
$\mathrm{~N} \cdot \mathrm{m}^{-1}$ | 5-9 | 479 | 479 | Example 5-3 shows that a Maclaurin expansion of a Morse potential leads to
$$
V(x)=D \beta^2 x^2+\cdots
$$
Given that $D=7.31 \times 10^{-19} \mathrm{~J} \cdot$ molecule ${ }^{-1}$ and $\beta=1.81 \times 10^{10} \mathrm{~m}^{-1}$ for $\mathrm{HCl}$, calculate the force constant of $\mathrm{HCl}$. | chemmc |
||
1-25 | no units | 3 | 3 | A line in the Lyman series of hydrogen has a wavelength of $1.03 \times 10^{-7} \mathrm{~m}$. Find the original energy level of the electron. | chemmc |
||
$10^{14} \mathrm{~Hz}$ | 1-15 | just the first part is taken: frequency of light | 4.738 | 4.738 | A helium-neon laser (used in supermarket scanners) emits light at $632.8 \mathrm{~nm}$. Calculate the frequency of this light. | chemmc |
|
$10^{-23} \mathrm{~kg} \cdot \mathrm{m} \cdot \mathrm{s}^{-1}$ | 1-47 | discard the second part of the answer | 6.6 | 6.6 | What is the uncertainty of the momentum of an electron if we know its position is somewhere in a $10 \mathrm{pm}$ interval? | chemmc |
|
$\mathrm{eV}$ | 1-34 | 54.394 | 54.394 | Using the Bohr theory, calculate the ionization energy (in electron volts and in $\mathrm{kJ} \cdot \mathrm{mol}^{-1}$ ) of singly ionized helium. | chemmc |
||
$10^{-22} \mathrm{~J}$ | 1-51 | 7 | 7 | When an excited nucleus decays, it emits a $\gamma$ ray. The lifetime of an excited state of a nucleus is of the order of $10^{-12} \mathrm{~s}$. What is the uncertainty in the energy of the $\gamma$ ray produced? | chemmc |
||
nm | 1-28 | only the first part of the question, the wavelength | 91.17 | 91.17 | Calculate the wavelength and the energy of a photon associated with the series limit of the Lyman series. | chemmc |
|
$10^{-25} \mathrm{~J}$ | 1-50 | 7 | 7 | Given a context information that there is also an uncertainty principle for energy and time:
$$
\Delta E \Delta t \geq h
$$, another application of the relationship has to do with the excitedstate energies and lifetimes of atoms and molecules. If we know that the lifetime of an excited state is $10^{-9} \mathrm{~s}$, then what is the uncertainty in the energy of this state? | chemmc |
||
$\mathrm{K}$ | 1-42 | 2500 | 2500 | One of the most powerful modern techniques for studying structure is neutron diffraction. This technique involves generating a collimated beam of neutrons at a particular temperature from a high-energy neutron source and is accomplished at several accelerator facilities around the world. If the speed of a neutron is given by $v_{\mathrm{n}}=\left(3 k_{\mathrm{B}} T / m\right)^{1 / 2}$, where $m$ is the mass of a neutron, then what temperature is needed so that the neutrons have a de Broglie wavelength of $50 \mathrm{pm}$ ? | chemmc |
||
$10^{-10} \mathrm{~m}$
| 1-8 | 3 | 3 | The temperature of the fireball in a thermonuclear explosion can reach temperatures of approximately $10^7 \mathrm{~K}$. What value of $\lambda_{\max }$ does this correspond to? | chemmc |
||
D-21 | Math Part D (after chapter 4) | -1/2 | -0.5 | Show that l'Hôpital's rule amounts to forming a Taylor expansion of both the numerator and the denominator. Evaluate the limit
$$
\lim _{x \rightarrow 0} \frac{\ln (1+x)-x}{x^2}
$$
both ways and report the final result. | chemmc |
||
D-8 | Math Part D (after chapter 4) | 1/3 | 0.3333333 | Evaluate the series
$$
S=\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{2^n}
$$ | chemmc |
||
% | D-4 | Math Part D (after chapter 4) | 0.249 | 0.249 | Calculate the percentage difference between $\ln (1+x)$ and $x$ for $x=0.0050$ | chemmc |
|
1-30 | no units | 7.00 | 7.00 | Calculate the reduced mass of a nitrogen molecule in which both nitrogen atoms have an atomic mass of 14.00. | chemmc |
||
mm | 1-45 | 0.139 | 0.139 | Two narrow slits are illuminated with red light of wavelength $694.3 \mathrm{~nm}$ from a laser, producing a set of evenly placed bright bands on a screen located $3.00 \mathrm{~m}$ beyond the slits. If the distance between the bands is $1.50 \mathrm{~cm}$, then what is the distance between the slits?
| chemmc |
||
$10^{-18} \mathrm{~J} / \alpha \text {-particle}$ | 1-41 | 1.3 | 1.3 | Calculate the energy associated with an $\alpha$ particle that has fallen through a potential difference of $4.0 \mathrm{~V}$. Take the mass of an $\alpha$ particle to be $6.64 \times 10^{-27} \mathrm{~kg}$. | chemmc |
||
$10^{16}$ photons | 1-13 | part (a) only | 1.07 | 1.07 | Calculate the number of photons in a $2.00 \mathrm{~mJ}$ light pulse at (a) $1.06 \mu \mathrm{m}$
| chemmc |
|
$\mathrm{~cm}^{-1}$ | 5-14 | 556 | 556 | The force constant of ${ }^{35} \mathrm{Cl}^{35} \mathrm{Cl}$ is $319 \mathrm{~N} \cdot \mathrm{m}^{-1}$. Calculate the fundamental vibrational frequency | chemmc |
||
$10^{-15} \mathrm{~J}$ | 1-11 | 2 | 2 | $$
\text {Calculate the energy of a photon for a wavelength of } 100 \mathrm{pm} \text { (about one atomic diameter). }
$$
| chemmc |
||
$10^{17} \mathrm{~Hz}$ | 1-37 | only the ground state energy is there | 1.69 | 4.59 | A proton and a negatively charged $\mu$ meson (called a muon) can form a short-lived species called a mesonic atom. The charge of a muon is the same as that on an electron and the mass of a muon is $207 m_{\mathrm{e}}$. Assume that the Bohr theory can be applied to such a mesonic atom and calculate the frequency associated with the $n=1$ to $n=2$ transition in a mesonic atom. | chemmc |