Kinetic Theory

Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms:
\begin{array}{|l|c|c|} \hline \text { Substance } & \text { Atomic Mass (u) } & \begin{array}{l} \text { Density }\left(10^{3}\right. \\ \left.\mathrm{Kg} \mathrm{m}^{-3}\right) \end{array} \\ \hline \text { Carbon (diamond) } & 12.01 & 2.22 \\ \hline \text { Gold } & 197.00 & 19.32 \\ \hline \text { Nitrogen (liquid) } & 14.01 & 1.00 \\ \hline \text { Lithium } & 6.94 & 0.53 \\ \hline \text { Fluorine (liquid) } & 19.00 & 1.14 \\ \hline \end{array}
[Hint: Assume the atoms to be ‘tightly packed’ in a solid or liquid phase, and use the known value of Avogadro’s number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few \AA].

If $r$ is the radius of the atom then the volume of each atom will be $(4 / 3) \pi r^{3}$ Volume of all the substance will be $=(4 / 3) \pi r^{3} \times N=M / \rho$ $M=$ atomic mass of the substance...

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A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have a uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres
\mathbf{n}_{2}=\mathbf{n}_{1} \exp \left[-m g\left(h_{2}-h_{1}\right) / k_{B} T\right]
where n_{2}, n_{1} refer to number density at heights h_{2} and h_{1} respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column: n_{2}=n_{1} \exp \left[-m g N_{A}\left(\rho-\rho^{\prime}\right)\left(h_{2}-h_{1}\right) /(\rho R T)\right]
where \rho is the density of the suspended particle, and \rho ‘, that of surrounding medium. [ \mathrm{N}_{\mathrm{A}} is Avogadro’s number, and \mathbf{R} the universal gas constant.] [Hint: Use Archimedes principle to find the apparent weight of the suspended particle.]

Law of atmosphere states that, $\mathrm{n}_{2}=\mathrm{n}_{1} \exp \left[-\mathrm{mg}\left(\mathrm{h}_{2}-\mathrm{h}_{1}\right) / \mathrm{k}_{\mathrm{B}} T\right]$ According to Archimedes principle,...

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From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 \mathrm{~cm}^{3} \mathrm{~s} 1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 \mathrm{~cm}^{3} \mathrm{~s}^{-1}. Identify the gas.
[Hint: Use Graham’s law of diffusion: R_{1} / R_{2}=\left(M_{2} / M_{1}\right)^{1 / 2}, where R_{1}, R_{2} are diffusion rates of gases 1 and 2 , and \mathbf{M}_{1} and \mathbf{M}_{2} their respective molecular masses. The law is a simple consequence of kinetic theory.]

Rate of diffusion of hydrogen is given as $R_{1}=28.7 \mathrm{~cm}^{3} \mathrm{~s}^{-1}$ Rate of diffusion of another gas is given as $R_{2}=7.2 \mathrm{~cm}^{3} \mathrm{~s}^{-1}$ According to...

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Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17^{\circ} \mathrm{C}. Take the radius of a nitrogen molecule to be roughly 1.0 A. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of \mathrm{N}_{2}=28.0 \mathrm{u} ).

Mean free path is given as $1.11\times10^{-7}$ Collision frequency is given as $4.58\times10^{9}s^{-1}$ Successive collision time ≅ 500 x (Collision time) Pressure inside the cylinder containing...

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Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is \mathbf{V}_{\text {rms }} the largest?

All three vessels are the same size and have the same capacity. As a result, the pressure, volume, and temperature of each gas are the same. The three vessels will each contain an equal quantity of...

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An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27^{\circ} \mathrm{C} . After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17^{\circ} \mathrm{C}. Estimate the mass of oxygen taken out of the cylinder \left(\mathbf{R}=8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right., molecular mass of \left.\mathrm{O}_{2}=32 \mathrm{u}\right).

Volume of gas is given as $V_{1}=30$ litres $=30 \times 10^{-3} \mathrm{~m}^{3}$ Gauge pressure is given as $\mathrm{P}_{1}=15 \mathrm{~atm}=15 \times 1.013 \times 10^{5} \mathrm{P}$ a Temperature...

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