Chapter 12: Kinetic Theory

Idea: The volume actually occupied by the molecules themselves (the molecular volume) is essentially the volume the same molecules occupy when packed together as a liquid, because in the liquid phase the molecules are nearly touching.

Take a fixed mass $$M$$ of water. In the liquid state its volume is $$V_{liquid} = \dfrac{M}{\rho_{water}} = \dfrac{M}{1000}$$.

As vapour, the same mass occupies $$V_{vapour} = \dfrac{M}{\rho_{vapour}} = \dfrac{M}{0.6}$$.

The ratio of the vapour volume to the liquid volume is $$\dfrac{V_{vapour}}{V_{liquid}} = \dfrac{M/0.6}{M/1000} = \dfrac{1000}{0.6} \approx 1.7 \times 10^{3}$$.

In the liquid the molecules are closely packed, so the molecular volume is essentially equal to the liquid volume; the fraction of molecular volume to total volume in the liquid is taken as $$1$$.

When the same molecules spread out into the vapour, the total volume increases by the factor $$\dfrac{1000}{0.6}$$, while the molecular volume itself does not change. Hence the fraction becomes smaller by the same factor:

$$\text{Fraction} = \dfrac{V_{molecular}}{V_{vapour}} = \dfrac{V_{liquid}}{V_{vapour}} = \dfrac{0.6}{1000} = 6 \times 10^{-4}$$

So only about $$0.06\%$$ of the volume of water vapour is actually occupied by the molecules — the rest is empty space.

In the liquid (or solid) phase the water molecules are closely packed, so the density of a single molecule is roughly equal to the density of bulk water, $$\rho = 1000 \, \mathrm{kg \, m^{-3}}$$.

To find the volume of one molecule we first need the mass of one molecule. One mole of water has mass $$M_0 = (2 + 16) \, \mathrm{g} = 18 \, \mathrm{g} = 0.018 \, \mathrm{kg}$$ and contains $$N_A = 6.02 \times 10^{23}$$ molecules.

Mass of one water molecule:

$$m = \dfrac{M_0}{N_A} = \dfrac{0.018}{6.02 \times 10^{23}} \approx 3 \times 10^{-26} \, \mathrm{kg}$$

Volume of one water molecule:

$$V = \dfrac{m}{\rho} = \dfrac{3 \times 10^{-26}}{1000} = 3 \times 10^{-29} \, \mathrm{m^3}$$

Treating the molecule as a tiny sphere of radius $$r$$, we have $$V = \dfrac{4}{3}\pi r^3$$, so

$$r = \left(\dfrac{3V}{4\pi}\right)^{1/3} = \left(\dfrac{3 \times 3 \times 10^{-29}}{4 \times 3.14}\right)^{1/3} \approx 2 \times 10^{-10} \, \mathrm{m} = 2 \, \mathrm{\AA}$$

So a water molecule has a volume of about $$3 \times 10^{-29} \, \mathrm{m^3}$$, corresponding to a radius of about $$2 \, \mathrm{\AA}$$.

This example estimates the average spacing between molecules in water vapour, using the data of the previous examples.

From Example 12.1, a given mass of water in the vapour state occupies $$\dfrac{1000}{0.6} \approx 1.7 \times 10^{3}$$ times the volume it occupies as a liquid. So the volume available to each molecule in the vapour is about $$10^{3}$$ times larger than in the liquid.

If the volume per molecule increases by a factor of $$10^{3}$$, the linear dimension (which scales as the cube root of volume) increases by

$$\left(10^{3}\right)^{1/3} = 10 \text{ times}$$

In the liquid the molecules are essentially touching, and the radius of a molecule is about $$2 \, \mathrm{\AA}$$ (Example 12.2). In the vapour the corresponding linear size available per molecule becomes

$$10 \times 2 \, \mathrm{\AA} = 20 \, \mathrm{\AA}$$

The average distance between two neighbouring molecules is twice this value:

$$\text{Interatomic distance} \approx 2 \times 20 \, \mathrm{\AA} = 40 \, \mathrm{\AA} = 4 \times 10^{-9} \, \mathrm{m}$$

Thus in water vapour the molecules are, on average, about $$40 \, \mathrm{\AA}$$ apart — roughly ten times the molecular size itself.

At a fixed temperature the average kinetic energy of a molecule, $$\dfrac{1}{2} m \langle v^2 \rangle = \dfrac{3}{2}k_B T$$, is the same for all molecules. So the lighter molecule moves faster.

The two species are uranium hexafluoride molecules, $$\mathrm{UF_6}$$, each containing one of the two uranium isotopes. With the atomic mass of fluorine $$= 19$$, the molecular masses are:

For $${}^{235}\mathrm{U}$$: $$M_1 = 235 + 6 \times 19 = 235 + 114 = 349 \, \mathrm{u}$$

For $${}^{238}\mathrm{U}$$: $$M_2 = 238 + 6 \times 19 = 238 + 114 = 352 \, \mathrm{u}$$

Since $$v \propto \dfrac{1}{\sqrt{M}}$$, the molecule containing the lighter isotope $${}^{235}\mathrm{U}$$ (molecular mass $$349 \, \mathrm{u}$$) has the larger average speed.

The ratio of the speeds is

$$\dfrac{v_{349}}{v_{352}} = \sqrt{\dfrac{M_2}{M_1}} = \sqrt{\dfrac{352}{349}} = \sqrt{1.0086} = 1.0044$$

The percentage difference in speeds is

$$\dfrac{\Delta v}{v} \times 100 = (1.0044 - 1) \times 100 \approx 0.44\%$$

So the molecules carrying $${}^{235}\mathrm{U}$$ are the faster ones, by about $$0.44\%$$. (This tiny difference is exactly what is exploited to separate the uranium isotopes by gaseous diffusion.)

Step 1 — Find the number of moles. At standard temperature and pressure, 1 mole of any ideal gas occupies $$22.4 \, \mathrm{litres}$$. The cylinder holds $$44.8 \, \mathrm{litres}$$ of helium at STP, so

$$\mu = \dfrac{44.8}{22.4} = 2 \, \mathrm{mol}$$

Step 2 — Choose the right specific heat. The cylinder has a fixed capacity, so the volume of the gas is constant. The heat needed is therefore governed by the molar specific heat at constant volume, $$C_v$$.

Helium is monatomic, so

$$C_v = \dfrac{3}{2}R$$

Step 3 — Compute the heat. The heat required to raise the temperature by $$\Delta T = 15.0^\circ\mathrm{C} = 15.0 \, \mathrm{K}$$ is

$$Q = \mu \, C_v \, \Delta T = \mu \times \dfrac{3}{2}R \times \Delta T$$

$$Q = 2 \times \dfrac{3}{2} \times R \times 15.0 = 45 \, R$$

$$Q = 45 \times 8.31 \, \mathrm{J} = 374 \, \mathrm{J}$$

About $$374 \, \mathrm{J}$$ of heat is needed.

The mean free path is given by

where $$n$$ is the number density of molecules and $$d$$ their diameter.

The size $$d$$ of a water-vapour molecule is taken to be the same as that of an air molecule, $$d \approx 2 \times 10^{-10} \, \mathrm{m}$$. For air at STP it was found (Eq. 12.41) that the number density is $$n_0 = 2.7 \times 10^{25} \, \mathrm{m^{-3}}$$ and the mean free path is $$l_0 \approx 2.9 \times 10^{-7} \, \mathrm{m}$$.

Adjust the number density to 373 K. At fixed pressure, $$n = \dfrac{P}{k_B T}$$, so the number density is inversely proportional to the absolute temperature. Going from $$273 \, \mathrm{K}$$ to $$373 \, \mathrm{K}$$,

$$n = 2.7 \times 10^{25} \times \dfrac{273}{373} \approx 2 \times 10^{25} \, \mathrm{m^{-3}}$$

Find the mean free path. Since $$l \propto \dfrac{1}{n}$$,

$$l = l_0 \times \dfrac{n_0}{n} = 2.9 \times 10^{-7} \times \dfrac{2.7 \times 10^{25}}{2 \times 10^{25}} \approx 4 \times 10^{-7} \, \mathrm{m}$$

The mean free path of a water molecule in vapour at 373 K is about $$4 \times 10^{-7} \, \mathrm{m}$$ — roughly 100 times the interatomic distance ($$\sim 40 \, \mathrm{\AA}$$) found in Example 12.3. It is this large value that allows a gas to behave as a gas, unable to hold together without a container.

At STP, 1 mole of oxygen gas occupies the molar volume $$V_{gas} = 22.4 \, \mathrm{litres} = 22.4 \times 10^{-3} \, \mathrm{m^3}$$ and contains $$N_A = 6.02 \times 10^{23}$$ molecules.

Volume of one molecule. Taking the oxygen molecule as a sphere of diameter $$d = 3 \, \mathrm{\AA} = 3 \times 10^{-10} \, \mathrm{m}$$, its radius is $$r = 1.5 \times 10^{-10} \, \mathrm{m}$$, so

$$v = \dfrac{4}{3}\pi r^3 = \dfrac{4}{3} \times 3.14 \times (1.5 \times 10^{-10})^3$$

$$v = \dfrac{4}{3} \times 3.14 \times 3.375 \times 10^{-30} = 1.41 \times 10^{-29} \, \mathrm{m^3}$$

Total molecular volume of 1 mole.

$$V_{molecular} = N_A \, v = 6.02 \times 10^{23} \times 1.41 \times 10^{-29} = 8.51 \times 10^{-6} \, \mathrm{m^3}$$

Required fraction.

$$\dfrac{V_{molecular}}{V_{gas}} = \dfrac{8.51 \times 10^{-6}}{22.4 \times 10^{-3}} = 3.8 \times 10^{-4}$$

So only about $$3.8 \times 10^{-4}$$ (less than $$0.04\%$$) of the volume occupied by oxygen gas at STP is actually taken up by the molecules themselves; the rest is empty space.

The molar volume is the volume occupied by 1 mole of an ideal gas. We obtain it from the ideal gas equation

$$PV = \mu R T$$

For 1 mole, $$\mu = 1$$, so

$$V = \dfrac{RT}{P}$$

At STP the standard conditions are:

$$T = 0^\circ\mathrm{C} = 273 \, \mathrm{K}$$
$$P = 1 \, \mathrm{atm} = 1.013 \times 10^{5} \, \mathrm{Pa}$$
$$R = 8.31 \, \mathrm{J \, mol^{-1} \, K^{-1}}$$

Substituting these values:

$$V = \dfrac{R T}{P} = \dfrac{8.31 \times 273}{1.013 \times 10^{5}}$$

$$V = \dfrac{2268.6}{1.013 \times 10^{5}} = 2.24 \times 10^{-2} \, \mathrm{m^3}$$

Converting to litres ($$1 \, \mathrm{m^3} = 10^{3} \, \mathrm{litre}$$):

$$V = 2.24 \times 10^{-2} \times 10^{3} \, \mathrm{litre} = 22.4 \, \mathrm{litre}$$

Hence the molar volume of any ideal gas at STP is $$22.4 \, \mathrm{litres}$$, as required.

Use the ideal gas equation in the form $$PV = \mu RT = \dfrac{m}{M}RT$$, so that the mass of gas is $$m = \dfrac{PVM}{RT}$$. We find the number of moles before and after, and take the difference.

Data. The gauge pressure must be converted to absolute pressure by adding 1 atm.

Initial state: $$P_1 = (15 + 1) \, \mathrm{atm} = 16 \, \mathrm{atm} = 16 \times 1.013 \times 10^{5} = 1.621 \times 10^{6} \, \mathrm{Pa}$$, and $$T_1 = 27^\circ\mathrm{C} = 300 \, \mathrm{K}$$.

Final state: $$P_2 = (11 + 1) \, \mathrm{atm} = 12 \, \mathrm{atm} = 12 \times 1.013 \times 10^{5} = 1.216 \times 10^{6} \, \mathrm{Pa}$$, and $$T_2 = 17^\circ\mathrm{C} = 290 \, \mathrm{K}$$.

Volume $$V = 30 \, \mathrm{litre} = 30 \times 10^{-3} \, \mathrm{m^3}$$, molar mass $$M = 32 \, \mathrm{g} = 32 \times 10^{-3} \, \mathrm{kg \, mol^{-1}}$$.

Initial number of moles.

$$\mu_1 = \dfrac{P_1 V}{R T_1} = \dfrac{1.621 \times 10^{6} \times 30 \times 10^{-3}}{8.31 \times 300} = \dfrac{4.863 \times 10^{4}}{2493} = 19.5 \, \mathrm{mol}$$

Final number of moles.

$$\mu_2 = \dfrac{P_2 V}{R T_2} = \dfrac{1.216 \times 10^{6} \times 30 \times 10^{-3}}{8.31 \times 290} = \dfrac{3.647 \times 10^{4}}{2409.9} = 15.1 \, \mathrm{mol}$$

Oxygen removed. The number of moles withdrawn is

$$\Delta\mu = \mu_1 - \mu_2 = 19.5 - 15.1 = 4.4 \, \mathrm{mol}$$

The corresponding mass is

$$\Delta m = \Delta\mu \times M = 4.4 \times 32 \, \mathrm{g} \approx 140 \, \mathrm{g} = 0.14 \, \mathrm{kg}$$

About $$0.14 \, \mathrm{kg}$$ of oxygen is taken out of the cylinder.

The amount of gas in the bubble stays fixed as it rises, so we apply the combined gas law

$$\dfrac{P_1 V_1}{T_1} = \dfrac{P_2 V_2}{T_2}$$

where subscript 1 denotes the bottom of the lake and 2 the surface.

Pressure at the bottom (state 1). This is atmospheric pressure plus the pressure of the 40 m column of water above the bubble:

$$P_1 = P_{atm} + \rho g h = 1.013 \times 10^{5} + (1000)(9.8)(40)$$

$$P_1 = 1.013 \times 10^{5} + 3.92 \times 10^{5} = 4.933 \times 10^{5} \, \mathrm{Pa}$$

Pressure at the surface (state 2).

$$P_2 = P_{atm} = 1.013 \times 10^{5} \, \mathrm{Pa}$$

Temperatures. $$T_1 = 12^\circ\mathrm{C} = 285 \, \mathrm{K}$$ and $$T_2 = 35^\circ\mathrm{C} = 308 \, \mathrm{K}$$.

Solve for the final volume.

$$V_2 = V_1 \times \dfrac{P_1}{P_2} \times \dfrac{T_2}{T_1}$$

$$V_2 = 1.0 \times \dfrac{4.933 \times 10^{5}}{1.013 \times 10^{5}} \times \dfrac{308}{285}$$

$$V_2 = 1.0 \times 4.87 \times 1.081 = 5.3 \, \mathrm{cm^3}$$

The bubble grows to about $$5.3 \, \mathrm{cm^3}$$ — roughly five times its original size — by the time it reaches the surface.

Treat the air in the room as an ideal gas. The most convenient form of the ideal gas law for finding the number of molecules $$N$$ is

$$PV = N k_B T \quad\Rightarrow\quad N = \dfrac{PV}{k_B T}$$

Data.

$$P = 1 \, \mathrm{atm} = 1.013 \times 10^{5} \, \mathrm{Pa}$$
$$V = 25.0 \, \mathrm{m^3}$$
$$T = 27^\circ\mathrm{C} = 300 \, \mathrm{K}$$
$$k_B = 1.38 \times 10^{-23} \, \mathrm{J \, K^{-1}}$$

Compute.

$$N = \dfrac{(1.013 \times 10^{5})(25.0)}{(1.38 \times 10^{-23})(300)}$$

$$N = \dfrac{2.533 \times 10^{6}}{4.14 \times 10^{-21}} = 6.1 \times 10^{26}$$

The room contains about $$6.1 \times 10^{26}$$ air molecules.

Number of molecules. By Avogadro's law (which follows from the ideal gas law $$PV = N k_B T$$), equal volumes of gas at the same temperature and pressure contain equal numbers of molecules.

The three vessels have equal capacity (the same volume $$V$$) and the gases are at the same $$T$$ and $$P$$, so

$$N = \dfrac{PV}{k_B T}$$

is the same for all three. Yes — all three vessels contain equal numbers of molecules (neon, chlorine and uranium hexafluoride).

RMS speed. The root mean square speed is

$$v_{rms} = \sqrt{\dfrac{3 k_B T}{m}}$$

At the same temperature, $$v_{rms} \propto \dfrac{1}{\sqrt{m}}$$, so it depends on the molecular mass. The three gases have very different molecular masses:

Neon $$\approx 20.2 \, \mathrm{u}$$, chlorine $$(\mathrm{Cl_2}) \approx 70.9 \, \mathrm{u}$$, uranium hexafluoride $$(\mathrm{UF_6}) \approx 352 \, \mathrm{u}$$.

So the rms speeds are not the same. The lightest molecule has the largest rms speed; since neon has the smallest molecular mass, neon has the largest $$v_{rms}$$.

The root mean square speed of a gas atom is

$$v_{rms} = \sqrt{\dfrac{3RT}{M}}$$

where $$M$$ is the molar mass. We require the rms speed of argon at some temperature $$T_{\mathrm{Ar}}$$ to equal that of helium at $$T_{\mathrm{He}} = -20^\circ\mathrm{C} = 253 \, \mathrm{K}$$:

$$\sqrt{\dfrac{3R\,T_{\mathrm{Ar}}}{M_{\mathrm{Ar}}}} = \sqrt{\dfrac{3R\,T_{\mathrm{He}}}{M_{\mathrm{He}}}}$$

Squaring both sides and cancelling the common factor $$3R$$:

$$\dfrac{T_{\mathrm{Ar}}}{M_{\mathrm{Ar}}} = \dfrac{T_{\mathrm{He}}}{M_{\mathrm{He}}}$$

Solving for $$T_{\mathrm{Ar}}$$:

$$T_{\mathrm{Ar}} = T_{\mathrm{He}} \times \dfrac{M_{\mathrm{Ar}}}{M_{\mathrm{He}}} = 253 \times \dfrac{39.9}{4.0}$$

$$T_{\mathrm{Ar}} = 253 \times 9.975 = 2523.7 \, \mathrm{K}$$

So at about $$T_{\mathrm{Ar}} \approx 2.52 \times 10^{3} \, \mathrm{K}$$ the argon atoms have the same rms speed as helium atoms at $$-20^\circ\mathrm{C}$$. (Argon must be much hotter because its atoms are about 10 times heavier than helium atoms.)

Step 1 — Number density. From $$PV = N k_B T$$, the number of molecules per unit volume is

$$n = \dfrac{P}{k_B T}$$

With $$P = 2.0 \, \mathrm{atm} = 2.0 \times 1.013 \times 10^{5} = 2.026 \times 10^{5} \, \mathrm{Pa}$$ and $$T = 17^\circ\mathrm{C} = 290 \, \mathrm{K}$$:

$$n = \dfrac{2.026 \times 10^{5}}{(1.38 \times 10^{-23})(290)} = \dfrac{2.026 \times 10^{5}}{4.00 \times 10^{-21}} = 5.06 \times 10^{25} \, \mathrm{m^{-3}}$$

Step 2 — Mean free path. The molecular diameter is $$d = 2r = 2 \times 1.0 \, \mathrm{\AA} = 2.0 \times 10^{-10} \, \mathrm{m}$$. The mean free path is

$$l = \dfrac{1}{\sqrt{2}\,\pi d^2 n}$$

$$l = \dfrac{1}{1.414 \times 3.14 \times (2.0 \times 10^{-10})^2 \times 5.06 \times 10^{25}}$$

$$l = \dfrac{1}{1.414 \times 3.14 \times 4.0 \times 10^{-20} \times 5.06 \times 10^{25}} = \dfrac{1}{8.99 \times 10^{6}}$$

$$l = 1.11 \times 10^{-7} \, \mathrm{m}$$

Step 3 — RMS speed. With molar mass $$M = 28.0 \, \mathrm{g} = 28.0 \times 10^{-3} \, \mathrm{kg \, mol^{-1}}$$:

$$v_{rms} = \sqrt{\dfrac{3RT}{M}} = \sqrt{\dfrac{3 \times 8.31 \times 290}{28.0 \times 10^{-3}}}$$

$$v_{rms} = \sqrt{\dfrac{7229.7}{0.028}} = \sqrt{2.58 \times 10^{5}} = 508 \, \mathrm{m \, s^{-1}}$$

Step 4 — Collision frequency. This is the number of collisions per second, $$\nu = \dfrac{v_{rms}}{l}$$:

$$\nu = \dfrac{508}{1.11 \times 10^{-7}} = 4.58 \times 10^{9} \, \mathrm{s^{-1}}$$

Step 5 — Collision time versus free-travel time. The time the molecule moves freely between two successive collisions is

$$\tau_{free} = \dfrac{l}{v_{rms}} = \dfrac{1.11 \times 10^{-7}}{508} = 2.18 \times 10^{-10} \, \mathrm{s}$$

The collision time — roughly the time the molecule takes to cross its own diameter during a collision — is

$$\tau_{coll} \approx \dfrac{d}{v_{rms}} = \dfrac{2.0 \times 10^{-10}}{508} = 3.9 \times 10^{-13} \, \mathrm{s}$$

The ratio of the two times is

$$\dfrac{\tau_{free}}{\tau_{coll}} = \dfrac{l/v_{rms}}{d/v_{rms}} = \dfrac{l}{d} = \dfrac{1.11 \times 10^{-7}}{2.0 \times 10^{-10}} \approx 500$$

So a nitrogen molecule spends about $$500$$ times as long travelling freely between collisions as it does in a collision. The molecule is in free flight for the overwhelming majority of the time.