<< Chapter < Page | Chapter >> Page > |
Find the total number of collisions between molecules in 1.00 s in 1.00 L of nitrogen gas at standard temperature and pressure ( $0\phantom{\rule{0.2em}{0ex}}\text{\xb0C}$ , 1.00 atm). Use $1.88\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-10}}\phantom{\rule{0.2em}{0ex}}\text{m}$ as the effective radius of a nitrogen molecule. (The number of collisions per second is the reciprocal of the collision time.) Keep in mind that each collision involves two molecules, so if one molecule collides once in a certain period of time, the collision of the molecule it hit cannot be counted.
(a) Estimate the specific heat capacity of sodium from the Law of Dulong and Petit. The molar mass of sodium is 23.0 g/mol. (b) What is the percent error of your estimate from the known value, $1230\phantom{\rule{0.2em}{0ex}}\text{J/kg}\phantom{\rule{0.2em}{0ex}}\xb7\text{\xb0}\text{C}$ ?
a. $1080\phantom{\rule{0.2em}{0ex}}\text{J/kg}\phantom{\rule{0.2em}{0ex}}\text{\xb0C}$ ; b. $12\%$
A sealed, perfectly insulated container contains 0.630 mol of air at $20.0\phantom{\rule{0.2em}{0ex}}\text{\xb0C}$ and an iron stirring bar of mass 40.0 g. The stirring bar is magnetically driven to a kinetic energy of 50.0 J and allowed to slow down by air resistance. What is the equilibrium temperature?
Find the ratio $f({v}_{\text{p}})\text{/}f({v}_{\text{rms}})$ for hydrogen gas $(M=2.02\phantom{\rule{0.2em}{0ex}}\text{g/mol})$ at a temperature of 77.0 K.
$2\sqrt{e}\text{/}3$ or about 1.10
Unreasonable results . (a) Find the temperature of 0.360 kg of water, modeled as an ideal gas, at a pressure of $1.01\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{Pa}$ if it has a volume of $0.615\phantom{\rule{0.2em}{0ex}}{\text{m}}^{3}$ . (b) What is unreasonable about this answer? How could you get a better answer?
Unreasonable results . (a) Find the average speed of hydrogen sulfide, ${\text{H}}_{2}\text{S}$ , molecules at a temperature of 250 K. Its molar mass is 31.4 g/mol (b) The result isn’t very unreasonable, but why is it less reliable than those for, say, neon or nitrogen?
a. 411 m/s; b. According to [link] , the ${C}_{V}$ of ${\text{H}}_{2}\text{S}$ is significantly different from the theoretical value, so the ideal gas model does not describe it very well at room temperature and pressure, and the Maxwell-Boltzmann speed distribution for ideal gases may not hold very well, even less well at a lower temperature.
An airtight dispenser for drinking water is $25\phantom{\rule{0.2em}{0ex}}\text{cm}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}10\phantom{\rule{0.2em}{0ex}}\text{cm}$ in horizontal dimensions and 20 cm tall. It has a tap of negligible volume that opens at the level of the bottom of the dispenser. Initially, it contains water to a level 3.0 cm from the top and air at the ambient pressure, 1.00 atm, from there to the top. When the tap is opened, water will flow out until the gauge pressure at the bottom of the dispenser, and thus at the opening of the tap, is 0. What volume of water flows out? Assume the temperature is constant, the dispenser is perfectly rigid, and the water has a constant density of $1000\phantom{\rule{0.2em}{0ex}}{\text{kg/m}}^{3}$ .
Eight bumper cars, each with a mass of 322 kg, are running in a room 21.0 m long and 13.0 m wide. They have no drivers, so they just bounce around on their own. The rms speed of the cars is 2.50 m/s. Repeating the arguments of Pressure, Temperature, and RMS Speed , find the average force per unit length (analogous to pressure) that the cars exert on the walls.
29.5 N/m
Verify that ${v}_{p}=\sqrt{\frac{2{k}_{\text{B}}T}{m}}$ .
Verify the normalization equation ${\int}_{0}^{\infty}f(v)dv=1}.$ In doing the integral, first make the substitution $u=\sqrt{\frac{m}{2{k}_{\text{B}}T}}v=\frac{v}{{v}_{p}}.$ This “scaling” transformation gives you all features of the answer except for the integral, which is a dimensionless numerical factor. You’ll need the formula
${\int}_{0}^{\infty}{x}^{2}{e}^{\text{\u2212}{x}^{2}}}dx=\frac{\sqrt{\pi}}{4$
to find the numerical factor and verify the normalization.
Substituting
$v=\sqrt{\frac{2{k}_{\text{B}}T}{m}}u$ and
$dv=\sqrt{\frac{2{k}_{\text{B}}T}{m}}du$ gives
$\begin{array}{cc}{\displaystyle {\int}_{0}^{\infty}\frac{4}{\sqrt{\pi}}{\left(\frac{m}{2{k}_{\text{B}}T}\right)}^{3\text{/}2}{v}^{2}{e}^{\text{\u2212}m{v}^{2}\text{/}2{k}_{\text{B}}T}}dv\hfill & ={\displaystyle {\int}_{0}^{\infty}\frac{4}{\sqrt{\pi}}{\left(\frac{m}{2{k}_{\text{B}}T}\right)}^{3\text{/}2}\left(\frac{2{k}_{\text{B}}T}{m}\right){u}^{2}{e}^{\text{\u2212}{u}^{2}}\sqrt{\frac{2{k}_{\text{B}}T}{m}}}du\hfill \\ & ={\displaystyle {\int}_{0}^{\infty}\frac{4}{\sqrt{\pi}}{u}^{2}{e}^{\text{\u2212}{u}^{2}}du}=\frac{4}{\sqrt{\pi}}\phantom{\rule{0.2em}{0ex}}\frac{\sqrt{\pi}}{4}=1\hfill \end{array}$
Verify that $\overline{v}=\sqrt{\frac{8}{\pi}\phantom{\rule{0.2em}{0ex}}\frac{{k}_{\text{B}}T}{m}.}$ Make the same scaling transformation as in the preceding problem.
Verify that ${v}_{\text{rms}}=\sqrt{\stackrel{\text{\u2013}}{{v}^{2}}}=\sqrt{\frac{3{k}_{\text{B}}T}{m}}$ .
Making the scaling transformation as in the previous problems, we find that
$\stackrel{\text{\u2013}}{{v}^{2}}={\displaystyle {\int}_{0}^{\infty}\frac{4}{\sqrt{\pi}}{\left(\frac{m}{2{k}_{\text{B}}T}\right)}^{3\text{/}2}{v}^{2}{v}^{2}{e}^{\text{\u2212}m{v}^{2}\text{/}2{k}_{\text{B}}T}}dv={\displaystyle {\int}_{0}^{\infty}\frac{4}{\sqrt{\pi}}\phantom{\rule{0.2em}{0ex}}\frac{2{k}_{\text{B}}T}{m}{u}^{4}{e}^{\text{\u2212}{u}^{2}}du}.$
As in the previous problem, we integrate by parts:
${\int}_{0}^{\infty}{u}^{4}{e}^{\text{\u2212}{u}^{2}}du}={\left[\text{\u2212}\frac{1}{2}{u}^{3}{e}^{\text{\u2212}{u}^{2}}\right]}_{0}^{\infty}+\frac{3}{2}{\displaystyle {\int}_{0}^{\infty}{u}^{2}{e}^{\text{\u2212}{u}^{2}}du}.$
Again, the first term is 0, and we were given in an earlier problem that the integral in the second term equals
$\frac{\sqrt{\pi}}{4}$ . We now have
$\stackrel{\text{\u2013}}{{v}^{2}}=\frac{4}{\sqrt{\pi}}\phantom{\rule{0.2em}{0ex}}\frac{2{k}_{\text{B}}T}{m}\phantom{\rule{0.2em}{0ex}}\frac{3}{2}\phantom{\rule{0.2em}{0ex}}\frac{\sqrt{\pi}}{4}=\frac{3{k}_{\text{B}}T}{m}.$
Taking the square root of both sides gives the desired result:
${v}_{\text{rms}}=\sqrt{\frac{3{k}_{\text{B}}T}{m}}$ .
Notification Switch
Would you like to follow the 'University physics volume 2' conversation and receive update notifications?