Chapter 13

$\text{102}\text{º}\text{F}$

$\text{20}\text{.}0\text{º}\text{C}$ and $\text{25}\text{.}6\text{º}\text{C}$

$\text{9890}\text{º}\text{F}$

(a) $\text{22}\text{.}2\text{º}\text{C}$

(b) $\begin{array}{lll}\text{Δ}T\left(\text{º}\text{F}\right)& =& T_2\left(\text{º}\text{F}\right)-T_1\left(\text{º}\text{F}\right)\\ & =& \frac{9}{5}{T}_2\left(\text{º}\text{C}\right)+\text{32}\text{.}0\text{º}-\left(\frac{9}{5}{T}_1\left(\text{º}\text{C}\right)+\text{32}\text{.}0\text{º}\right)\\ & =& \frac{9}{5}\left(T_2\left(\text{º}\text{C}\right)-T_1\left(\text{º}\text{C}\right)\right)=\frac{9}{5}\text{Δ}T\left(\text{º}\text{C}\right)\end{array}$

169.98 m

$5\text{.}4\times {\text{10}}^{-6}\text{m}$

Because the area gets smaller, the price of the land DECREASES by $\text{~}\$\text{17},\text{000}\text{.}$

$\begin{array}{lll}V& =& V_0+\text{Δ}V=V_0(1+\beta \text{Δ}T)\\ & =& (\text{60}\text{.}\text{00 L})\left[1+\left(\text{950}\times {\text{10}}^{-6}/\text{º}\text{C}\right)\left(\text{35}\text{.}0\text{º}\text{C}-\text{15}\text{.}0\text{º}\text{C}\right)\right]\\ & =& \text{61}\text{.}1\text{L}\end{array}$

(a) 9.35 mL

(b) 7.56 mL

0.832 mm

We know how the length changes with temperature: $\text{Δ}L={\mathrm{\alpha L}}_0\text{Δ}T$. Also we know that the volume of a cube is related to its length by $V=L^3$, so the final volume is then $V=V_0+\text{Δ}V={\left(L_0+\text{Δ}L\right)}^3$. Substituting for $\text{Δ}L$ gives

$V={\left(L_0+{\mathrm{\alpha L}}_0\text{Δ}T\right)}^3=L_0^3{\left(1+\alpha \text{Δ}T\right)}^3\text{.}$

Now, because $\alpha \text{Δ}T$ is small, we can use the binomial expansion:

$V\approx L_0^3\left(1+\mathrm{3\alpha \Delta T}\right)=L_0^3+\mathrm{3\alpha }{L}_0^3\text{Δ}T.$

So writing the length terms in terms of volumes gives $V=V_0+\text{Δ}V\approx V_0+{\mathrm{3\alpha }V}_0\text{Δ}T,$ and so

$\text{Δ}V={\mathrm{\beta V}}_0\text{Δ}T\approx {\mathrm{3\alpha }V}_0\text{Δ}T,\text{or}\beta \approx \mathrm{3\alpha }.$

1.62 atm

(a) 0.136 atm

(b) 0.135 atm. The difference between this value and the value from part (a) is negligible.

(a) $\text{nRT}=(\text{mol})(\text{J/mol}\cdot \text{K})(\text{K})=\text{J}$

(b) $\text{nRT}=(\text{mol})(\text{cal/mol}\cdot \text{K})(\text{K})=\text{cal}$

(c) $\begin{array}{lll}\text{nRT}& =& (\text{mol})(\text{L}\cdot \text{atm/mol}\cdot \text{K})(\text{K})\\ & =& \text{L}\cdot \text{atm}=({\text{m}}^3)({\text{N/m}}^2)\\ & =& \text{N}\cdot \text{m}=\text{J}\end{array}$

$7\text{.}\text{86}\times {\text{10}}^{-2}\text{mol}$

(a) $6\text{.}\text{02}\times {\text{10}}^5{\text{km}}^3$

(b) $6\text{.}\text{02}\times {\text{10}}^8\text{km}$

$-\text{73}\text{.}9\text{º}\text{C}$

(a) $9\text{.}\text{14}\times {\text{10}}^6{\text{N/m}}^2$

(b) $8\text{.}\text{23}\times {\text{10}}^6{\text{N/m}}^2$

(c) 2.16 K

(d) No. The final temperature needed is much too low to be easily achieved for a large object.

41 km

(a) $3\text{.}7\times {\text{10}}^{-\text{17}}\text{Pa}$

(b) $6\text{.}0\times {\text{10}}^{\text{17}}{\text{m}}^3$

(c) $8\text{.}4\times {\text{10}}^2\text{km}$

$1\text{.}\text{25}\times {\text{10}}^3\text{m/s}$

(a) $1\text{.}\text{20}\times {\text{10}}^{-\text{19}}\text{J}$

(b) $1\text{.}\text{24}\times {\text{10}}^{-\text{17}}\text{J}$

$\text{458}\text{K}$

$1\text{.}\text{95}\times {\text{10}}^7\text{K}$

$6\text{.}\text{09}\times {\text{10}}^5\text{m/s}$

$7\text{.}\text{89}\times {\text{10}}^4\text{Pa}$

(a) $1\text{.}\text{99}\times {\text{10}}^5\text{Pa}$

(b) 0.97 atm

$3\text{.}\text{12}\times {\text{10}}^4\text{Pa}$

78.3%

(a) $2\text{.}\text{12}\times {\text{10}}^4\text{Pa}$

(b) $1\text{.}\text{06}\text{\%}$

(a) $8\text{.}\text{80}\times {\text{10}}^{-2}\text{g}$

(b) $6\text{.}\text{30}\times {\text{10}}^3\text{Pa}$; the two values are nearly identical.

82.3%

$4\text{.}\text{77}\text{º}\text{C}$

$\text{38}\text{.}3\text{m}$

$\frac{\left(F_{\text{B}}/w_{\text{Cu}}\right)}{{\left(F_{\text{B}}/w_{\text{Cu}}\right)}^{\prime }}=1\text{.}\text{02}$. The buoyant force supports nearly the exact same amount of force on the copper block in both circumstances.

(a) $4\text{.}\text{41}\times {\text{10}}^{\text{10}}{\text{mol/m}}^3$

(b) It’s unreasonably large.

(c) At high pressures such as these, the ideal gas law can no longer be applied. As a result, unreasonable answers come up when it is used.

(a) $7\text{.}\text{03}\times {\text{10}}^8\text{m/s}$

(b) The velocity is too high—it’s greater than the speed of light.

(c) The assumption that hydrogen inside a supernova behaves as an idea gas is responsible, because of the great temperature and density in the core of a star. Furthermore, when a velocity greater than the speed of light is obtained, classical physics must be replaced by relativity, a subject not yet covered.

(a), (c)

(d)

(b)

(a) 7.29 × 10^-21 J; (b) 352K or 79ºC