### Challenge Problems

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.

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.

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 ${v}_{\text{rms}}=\sqrt{\stackrel{\text{\u2013}}{{v}^{2}}}=\sqrt{\frac{3{k}_{\text{B}}T}{m}}$.