Chapter 13: Temperature, Kinetic Theory, and the Gas LawsContent check — College Physics (AP), Chapter 13: Temperature, Kinetic Theory, and the Gas Laws

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

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

(a) 9.35 mL

(b) 7.56 mL

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 $V=$ $V_0+$ $\text{Δ}V=$ ${L_0+\text{Δ}L}^3$. Substituting for $\text{Δ}L$ gives

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

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

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

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

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

(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.

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

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

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

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

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

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

(b) 0.97 atm

(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.

(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.