The Workbook

1. At constant temperature the pressure of a fixed mass of gas is inversely proportional to its volume: \(PV = \) constant.

2. \(A\): \(4 \times 0.5 = 2\). \(B\): \(2 \times 1 = 2\). \(C\): \(1 \times 2 = 2\) atm L. The product is the same at all three points.

3. \(P = \dfrac{2}{0.25} = 8\) atm.

At constant pressure the volume of a fixed mass of gas is proportional to its absolute temperature: \(\dfrac{V}{T} = \) constant.

\(V_2 = V_1\dfrac{T_2}{T_1} = 600 \times \dfrac{400}{300} = 800\) mL.

Volume is proportional to the absolute temperature, not to the Celsius reading. \(\dfrac{127}{27}\) would give \(2822\) mL, which is wrong. Only the Kelvin scale starts from the point where \(V\) would be zero.

At constant volume the pressure of a fixed mass of gas is proportional to its absolute temperature: \(\dfrac{P}{T} = \) constant.

\(P_2 = 2 \times \dfrac{360}{300} = 2.4\) atm.

1. At constant pressure \(V \propto T\), so \(V\) is a linear function of \(t\): \(V = V_0(1 + t/273)\). Every such line reaches \(V = 0\) at \(t = -273^\circ\)C, which is absolute zero.

2. \(P_2\). At any temperature the red line gives the smaller volume. By Boyle's law a smaller volume at the same temperature means a larger pressure.

\(\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}\), so \(V_2 = V_1\dfrac{P_1}{P_2}\dfrac{T_2}{T_1} = 6 \times \dfrac{1}{2} \times \dfrac{400}{300} = 4\) L.

Equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules.

\(5.6\) L is \(\dfrac{5.6}{22.4} = 0.25\) mol. Molecules \(= 0.25 \times 6 \times 10^{23} = 1.5 \times 10^{23}\). Mass \(= 0.25 \times 32 = 8\) g.

\(P\) is the pressure (Pa), \(V\) the volume (m\(^3\)) and \(n\) the number of moles. \(T\) is the absolute temperature (K) and \(R = 8.3\) J/mol K is the gas constant.

\(V = \dfrac{nRT}{P} = \dfrac{1 \times 8.3 \times 300}{10^5} = 0.0249\) m\(^3 = 24.9\) L.

\(n = \dfrac{PV}{RT} = \dfrac{4 \times 10^5 \times 12.45 \times 10^{-3}}{8.3 \times 300} = \dfrac{4980}{2490} = 2\) mol.

\(n = \dfrac{m}{M}\), so \(PV = \dfrac{m}{M}RT\) and \(\rho = \dfrac{m}{V} = \dfrac{PM}{RT}\).

\(\rho = \dfrac{10^5 \times 0.029}{8.3 \times 300} = \dfrac{2900}{2490} = 1.16\) kg/m\(^3\).

\(n = \dfrac{N}{N_A}\), so \(PV = \dfrac{N}{N_A}RT = NkT\) with \(k = \dfrac{R}{N_A}\), the Boltzmann constant.

\(k = \dfrac{8.3}{6 \times 10^{23}} = 1.38 \times 10^{-23}\) J/K.

1. \(A\): \(200 \times 12.45 = 2490\) kPa L. \(B\): \(100 \times 24.9 = 2490\) kPa L. Both points lie on one isotherm, so \(PV = nRT\) has the same value.

2. \(PV = 2490\) kPa L \(= 2490\) J. \(T_1 = \dfrac{PV}{nR} = \dfrac{2490}{1 \times 8.3} = 300\) K.

3. \(T_2 = \dfrac{4980}{8.3} = 600\) K. Twice the \(PV\) product means twice the temperature.

In a mixture of gases that do not react, each gas exerts the pressure it would exert alone in the same vessel. The total pressure is the sum of these partial pressures.

\(p = \dfrac{nRT}{V}\) with \(\dfrac{RT}{V} = \dfrac{8.3 \times 300}{0.0249} = 10^5\) Pa per mole. Hydrogen: \(2 \times 10^5\) Pa. Oxygen: \(10^5\) Pa. Total: \(3 \times 10^5\) Pa.

At constant volume \(P \propto T\). Doubling \(P\) doubles \(T\): \(T = 2 \times 300 = 600\) K \(= 327^\circ\)C.

At constant pressure \(V \propto T\). \(V = 2 \times \dfrac{330}{300} = 2.2\) L.

Boyle's law: \(P_1V_1 = P_2V_2\), so \(V_2 = 1 \times \dfrac{4}{1} = 4\) cm\(^3\).

1. \(P = \dfrac{nR}{V}T\). With \(n\) and \(V\) fixed, \(P\) is proportional to \(T\), so the graph is a straight line through the origin with slope \(\dfrac{nR}{V}\).

2. The slope is \(\dfrac{nR}{V}\), so the less steep line has the larger volume: \(B\). At \(300\) K, \(P_A = 200\) kPa and \(P_B = 100\) kPa, so \(\dfrac{V_A}{V_B} = \dfrac{P_B}{P_A} = \dfrac{1}{2}\).

3. \(V_A = \dfrac{nRT}{P} = \dfrac{1 \times 8.3 \times 300}{2 \times 10^5} = 0.01245\) m\(^3 = 12.45\) L.

At fixed \(V\) and \(T\), \(P \propto n\). The moles left are \(\dfrac{15}{20} = \dfrac{3}{4}\) of the original, so \(\dfrac{1}{4}\) of the gas has escaped.

The total number of moles is fixed: \(\dfrac{P_AV_A + P_BV_B}{RT} = \dfrac{P(V_A + V_B)}{RT}\).

\(P = \dfrac{3 \times 2 + 1.5 \times 4}{2 + 4} = \dfrac{12}{6} = 2\) atm.

One mole, \(32\) g, fills \(22.4\) L: \(\rho = \dfrac{32}{22.4} = 1.43\) g/L. Since \(1\) g/L \(= 1\) kg/m\(^3\), this is \(1.43\) kg/m\(^3\).

1. The piston is free to move. It settles where the gas pressure balances the load and the air above: \(P = P_0 + \dfrac{mg}{A}\). The load does not change, so \(P\) does not change.

2. Charles's law, at constant pressure. \(V = 2 \times \dfrac{450}{300} = 3\) L.

The molecules are points: their own volume is negligible compared with the volume of the container. They exert no forces on each other except during collisions. Collisions with each other and with the walls are perfectly elastic and take negligible time.

\(PV = NkT\), so \(\dfrac{N}{V} = \dfrac{P}{kT} = \dfrac{10^5}{1.38 \times 10^{-23} \times 300} = 2.4 \times 10^{25}\) per m\(^3\).

\(P\) is the pressure, \(\rho\) the density of the gas and \(v_{rms}\) the root mean square speed of its molecules, \(v_{rms} = \sqrt{\overline{v^2}}\).

\(\rho = \dfrac{Nm}{V}\), so \(P = \dfrac{1}{3}\dfrac{Nm}{V}\overline{v^2} = \dfrac{Nm\overline{v^2}}{3V}\).

1. The collision is elastic, so \(v_x\) reverses and the other components do not change. Change in momentum \(= mv_x - (-mv_x) = 2mv_x\).

2. After a collision the molecule crosses the box and comes back, a distance \(2L\) at speed \(v_x\): time \(= \dfrac{2L}{v_x}\). Average force \(= \dfrac{\text{momentum per collision}}{\text{time between collisions}} = \dfrac{2mv_x}{2L/v_x} = \dfrac{mv_x^2}{L}\).

3. Total force on the wall \(= \sum \dfrac{mv_x^2}{L} = \dfrac{Nm\overline{v_x^2}}{L} = \dfrac{Nm\overline{v^2}}{3L}\). Pressure \(= \dfrac{\text{force}}{L^2} = \dfrac{Nm\overline{v^2}}{3L^3}\).

\(v_{rms} = \sqrt{\dfrac{3P}{\rho}} = \sqrt{\dfrac{3 \times 10^5}{1.2}} = \sqrt{2.5 \times 10^5} = 500\) m/s.

\(P = \dfrac{1}{3}\dfrac{m_{gas}}{V}v_{rms}^2\), so \(PV = \dfrac{1}{3}m_{gas}v_{rms}^2\). With \(PV = nRT\) and \(m_{gas} = nM\): \(\dfrac{1}{3}nMv_{rms}^2 = nRT\), so \(v_{rms} = \sqrt{\dfrac{3RT}{M}}\).

Dividing \(R\) and \(M\) by \(N_A\): \(v_{rms} = \sqrt{\dfrac{3kT}{m}}\).

Oxygen: \(v_{rms} = \sqrt{\dfrac{3 \times 8.3 \times 300}{0.032}} = \sqrt{2.33 \times 10^5} = 483\) m/s.

At the same \(T\), \(v_{rms} \propto \dfrac{1}{\sqrt{M}}\). \(\dfrac{v_H}{v_O} = \sqrt{\dfrac{32}{2}} = 4\), so the ratio is \(4 : 1\).

\(v_H = 4 \times 483 = 1932\) m/s, about \(1.93\) km/s.

\(\sqrt{\dfrac{3RT_O}{32}} = \sqrt{\dfrac{3R \times 300}{2}}\), so \(\dfrac{T_O}{32} = \dfrac{300}{2}\) and \(T_O = 16 \times 300 = 4800\) K.

\(v_{rms} \propto \sqrt{T}\). Doubling \(v_{rms}\) needs four times the absolute temperature: \(T = 4 \times 300 = 1200\) K \(= 927^\circ\)C.

\(\overline{KE} = \dfrac{3}{2}kT = 1.5 \times 1.38 \times 10^{-23} \times 300 = 6.2 \times 10^{-21}\) J.

One mole: \(N_A \times \dfrac{3}{2}kT = \dfrac{3}{2}RT = 1.5 \times 8.3 \times 300 = 3735\) J.

The average kinetic energy is \(\dfrac{3}{2}kT\) and depends only on the temperature, so it is the same for both gases.

The rms speed is \(\sqrt{3kT/m}\). A hydrogen molecule is \(16\) times lighter, so for the same kinetic energy it moves \(\sqrt{16} = 4\) times faster.

1. \(P\) is at the peak: the most probable speed \(v_p\). \(Q\) is the average speed \(v_{avg}\) and \(R\) is the rms speed \(v_{rms}\). They come in that order: \(v_p < v_{avg} < v_{rms}\).

2. \(v_p = \sqrt{\dfrac{2RT}{M}}\), \(v_{avg} = \sqrt{\dfrac{8RT}{\pi M}}\), \(v_{rms} = \sqrt{\dfrac{3RT}{M}}\).

3. The height of the curve at \(v\) gives the number of molecules with speeds near \(v\). The area under the whole curve is the total number of molecules.

All three have the common factor \(\sqrt{RT/M}\), so the ratio is \(\sqrt{2} : \sqrt{8/\pi} : \sqrt{3} = 1.414 : 1.596 : 1.732\).

Dividing by \(1.414\): \(1 : 1.13 : 1.22\). The rms speed is always the largest, because squaring gives more weight to the fast molecules.

1. \(B\). Its peak is at a higher speed, and \(v_p = \sqrt{2RT/M}\) grows with \(T\).

2. The peak falls. The molecules spread over a wider range of speeds, and the total number is fixed, so fewer molecules share any one speed.

3. The area under each curve: it is the total number of molecules, which does not change on heating.

1. \(Y\). At the same temperature all molecules have the same average kinetic energy. So the lighter molecules move faster and their peak lies further to the right.

2. \(v_p \propto \dfrac{1}{\sqrt{M}}\). \(\dfrac{v_Y}{v_X} = 2 = \sqrt{\dfrac{M_X}{M_Y}}\), so \(M_X : M_Y = 4 : 1\).

3. \(M_X = 4 \times 4 = 16\) g/mol (methane, for example).

\(v_{avg} = \dfrac{1 + 2 + 3 + 4 + 5}{5} = 3\) km/s.

\(v_{rms} = \sqrt{\dfrac{1 + 4 + 9 + 16 + 25}{5}} = \sqrt{11} = 3.32\) km/s.

The rms speed is larger. It is always at least as large as the average, because squaring weights the larger speeds more. The two are equal only if all the speeds are equal.

\(v_{avg} = v_{rms}\sqrt{\dfrac{8}{3\pi}} = 500 \times 0.921 = 461\) m/s.

\(v_p = v_{rms}\sqrt{\dfrac{2}{3}} = 500 \times 0.816 = 408\) m/s.

\(P = \dfrac{1}{3}\rho v_{rms}^2 = \dfrac{1}{3} \times 0.09 \times 1840^2 = 0.03 \times 3.39 \times 10^6 = 1.02 \times 10^5\) Pa.

This is \(1\) atm within \(1\%\). The kinetic theory gives the right pressure from the measured density and speed.

At fixed \(V\), \(P \propto T\), so the pressure doubles. \(v_{rms} \propto \sqrt{T}\), so the rms speed becomes \(\sqrt{2}\) times, about \(1.41\) times.

Faster molecules hit the walls harder, with \(\sqrt{2}\) times the momentum, and more often, \(\sqrt{2}\) times as frequently. The two factors together double the force on the walls.

1. \(v_{rms}^2 = \dfrac{3RT}{M}\) is proportional to \(T\), with slope \(\dfrac{3R}{M}\) and zero intercept.

2. Slope \(= \dfrac{2.5 \times 10^5}{300} = 833\) m\(^2\)/s\(^2\) K. \(M = \dfrac{3R}{\text{slope}} = \dfrac{3 \times 8.3}{833} = 0.030\) kg/mol \(= 30\) g/mol.

3. At \(600\) K, \(v_{rms}^2 = 5 \times 10^5\) m\(^2\)/s\(^2\) and \(v_{rms} = 707\) m/s, which is \(\sqrt{2} \times 500\).

Left side: \([P] = \dfrac{\text{N}}{\text{m}^2} = \dfrac{\text{kg m s}^{-2}}{\text{m}^2} = \) kg m\(^{-1}\) s\(^{-2}\).

Right side: \([\rho v^2] = \) kg m\(^{-3} \times\) m\(^2\) s\(^{-2} = \) kg m\(^{-1}\) s\(^{-2}\). The two sides agree.

\(1\) g of helium is \(\dfrac{1}{4} = 0.25\) mol. \(KE = \dfrac{3}{2}nRT = 1.5 \times 0.25 \times 8.3 \times 300 = 934\) J.

\(P = \dfrac{1}{3}\dfrac{N}{V}m\overline{v^2} = \dfrac{2}{3}\dfrac{N}{V}\left(\dfrac{1}{2}m\overline{v^2}\right)\).

Put \(\dfrac{1}{2}m\overline{v^2} = \dfrac{3}{2}kT\): \(P = \dfrac{2}{3}\dfrac{N}{V} \times \dfrac{3}{2}kT = \dfrac{NkT}{V}\), so \(PV = NkT\).

Heating raises the average kinetic energy, so the molecules move faster. Each collision with a wall now transfers more momentum, and the collisions come more often because the molecules cross the vessel more quickly. Both effects increase the force on the walls, so the pressure rises.

The degrees of freedom are the number of independent ways a molecule can store energy. This is the number of independent coordinates needed to describe its motion.

Monatomic: \(3\) (translation along \(x\), \(y\), \(z\)). Rigid diatomic: \(3 + 2 = 5\) (two rotations about axes perpendicular to the bond). Non-linear triatomic: \(3 + 3 = 6\).

The atoms are almost points on the bond axis, so the moment of inertia about that axis is negligible. Rotation about it stores no energy and is not counted.

In thermal equilibrium the energy is shared equally among all degrees of freedom: each has an average energy \(\dfrac{1}{2}kT\) per molecule.

Diatomic, \(f = 5\): energy \(= \dfrac{5}{2}kT = 2.5 \times 1.38 \times 10^{-23} \times 300 = 1.04 \times 10^{-20}\) J.

\(U = N\dfrac{f}{2}kT = \dfrac{f}{2}nRT\).

Helium (\(f = 3\)): \(U = 1.5 \times 2 \times 8.3 \times 300 = 7470\) J. Nitrogen (\(f = 5\)): \(U = 2.5 \times 2 \times 8.3 \times 300 = 12450\) J.

At constant volume no work is done, so all the heat goes into \(U\): \(nC_V\Delta T = \Delta U = \dfrac{f}{2}nR\Delta T\), giving \(C_V = \dfrac{f}{2}R\).

At constant pressure the gas also expands: \(Q = \Delta U + P\Delta V\). From \(PV = nRT\) at constant \(P\), \(P\Delta V = nR\Delta T\). So \(nC_P\Delta T = nC_V\Delta T + nR\Delta T\), giving \(C_P - C_V = R\).

\(\gamma = \dfrac{C_P}{C_V} = \dfrac{C_V + R}{C_V} = 1 + \dfrac{R}{fR/2} = 1 + \dfrac{2}{f}\).

\(f = 3\): \(\gamma = \dfrac{5}{3} = 1.67\). \(f = 5\): \(\gamma = \dfrac{7}{5} = 1.4\). \(f = 6\): \(\gamma = \dfrac{4}{3} = 1.33\).

Monatomic: \(C_V = \dfrac{3}{2}R = 12.45\) J/mol K, \(C_P = \dfrac{5}{2}R = 20.75\) J/mol K.

Diatomic: \(C_V = \dfrac{5}{2}R = 20.75\) J/mol K, \(C_P = \dfrac{7}{2}R = 29.05\) J/mol K.

1. Translation only: \(C_V = \dfrac{3}{2}R\), so \(f = 3\). At such low temperatures the molecules do not have enough energy to rotate.

2. Near \(100\) K rotation begins and \(C_V\) rises to \(\dfrac{5}{2}R\). Near \(2000\) K the atoms begin to vibrate along the bond and \(C_V\) rises to \(\dfrac{7}{2}R\).

3. \(f = 3\), \(5\) and \(7\). Vibration adds two degrees of freedom, one for kinetic and one for potential energy.

\(\gamma = 1 + \dfrac{2}{f} = 1.4\), so \(\dfrac{2}{f} = 0.4\) and \(f = 5\). \(C_V = \dfrac{5}{2}R = 20.75\) J/mol K. The gas is diatomic, such as nitrogen or oxygen.

Oxygen is diatomic: \(C_V = 20.75\) and \(C_P = 29.05\) J/mol K.

Constant volume: \(Q = nC_V\Delta T = 2 \times 20.75 \times 10 = 415\) J. Constant pressure: \(Q = nC_P\Delta T = 2 \times 29.05 \times 10 = 581\) J.

The extra \(166\) J \(= nR\Delta T\) is the work done by the gas as it pushes back the surroundings while it expands.

\(A\), a single atom: \(3\) translational, \(0\) rotational, \(f = 3\), \(\gamma = \dfrac{5}{3}\).

\(B\), a rigid dumbbell: \(3\) translational, \(2\) rotational (the two dashed axes), \(f = 5\), \(\gamma = \dfrac{7}{5}\).

\(C\), a bent molecule: \(3\) translational, \(3\) rotational, \(f = 6\), \(\gamma = \dfrac{4}{3}\).

\(C_V\) of a mixture is the total heat capacity divided by the total moles: \(C_V = \dfrac{n_1C_{V1} + n_2C_{V2}}{n_1 + n_2} = \dfrac{1 \times \frac{3}{2}R + 1 \times \frac{5}{2}R}{2} = 2R\).

\(C_P = C_V + R = 3R\). \(\gamma = \dfrac{3R}{2R} = 1.5\).

\(C_V = \dfrac{3 \times \frac{3}{2}R + 1 \times \frac{5}{2}R}{4} = \dfrac{7R}{4} = 1.75R = 14.5\) J/mol K.

\(C_P = 2.75R\), so \(\gamma = \dfrac{2.75}{1.75} = \dfrac{11}{7} = 1.57\).

The equivalent molar mass is the total mass divided by the total number of moles: \(M = \dfrac{n_1M_1 + n_2M_2}{n_1 + n_2}\).

\(M = \dfrac{2 \times 4 + 1 \times 32}{3} = \dfrac{40}{3} = 13.3\) g/mol.

Helium: \(c_V = \dfrac{12.45}{0.004} = 3112\) J/kg K. Nitrogen: \(c_V = \dfrac{20.75}{0.028} = 741\) J/kg K.

One kilogram of helium contains seven times as many molecules as one kilogram of nitrogen. Each molecule needs its share of energy, so the light gas needs far more heat per kilogram.

1. \(P\): slope \(= \dfrac{3700}{300} = 12.3\) J/K \(\approx \dfrac{3}{2}R\). \(Q\): slope \(= \dfrac{6200}{300} = 20.7\) J/K \(\approx \dfrac{5}{2}R\). The slope of \(U\) against \(T\) for \(1\) mol is \(C_V\).

2. \(Q\). Its \(C_V = \dfrac{5}{2}R\) means \(f = 5\), the value for a rigid diatomic molecule. \(P\) with \(f = 3\) is monatomic.

3. \(U \propto f\) at the same \(T\) and \(n\), so \(U_Q : U_P = 5 : 3\).

\(C_P - C_V = 29.1 - 20.8 = 8.3\) J/mol K \(= R\), as Mayer's relation requires.

\(\gamma = \dfrac{29.1}{20.8} = 1.4 = \dfrac{7}{5}\), so \(f = 5\) and the gas is diatomic.

1. The slopes are the molar heat capacities. \(C_V = \dfrac{2075}{100} = 20.75\) J/mol K and \(C_P = \dfrac{2905}{100} = 29.05\) J/mol K.

2. \(C_V = \dfrac{f}{2}R\), so \(f = \dfrac{2 \times 20.75}{8.3} = 5\).

3. \(W = Q_P - Q_V = 2905 - 2075 = 830\) J, which is \(nR\Delta T = 1 \times 8.3 \times 100\).

\(f = 5 + 2 = 7\). \(C_V = \dfrac{7}{2}R = 29.05\) J/mol K. \(\gamma = 1 + \dfrac{2}{7} = \dfrac{9}{7} = 1.29\).

A monatomic molecule has only translational energy, so its total energy is \(\dfrac{3}{2}kT\).

\(n\) moles contain \(N = nN_A\) molecules: \(U = nN_A \times \dfrac{3}{2}kT = \dfrac{3}{2}n(N_Ak)T = \dfrac{3}{2}nRT\).

1. The gas in the rigid vessel. It does no work, so all of \(Q\) raises its internal energy. Under the piston part of \(Q\) is spent pushing the piston up.

2. \(Q = nC_V\Delta T_V = nC_P\Delta T_P\), so \(\dfrac{\Delta T_V}{\Delta T_P} = \dfrac{C_P}{C_V} = \gamma = 1.4\) for a diatomic gas.

Helium is monatomic. Constant pressure: \(Q = nC_P\Delta T = 1 \times 20.75 \times 100 = 2075\) J. Constant volume: \(Q = nC_V\Delta T = 1 \times 12.45 \times 100 = 1245\) J.

\(Q = nC_P\Delta T\) and \(\Delta U = nC_V\Delta T\), so \(\dfrac{\Delta U}{Q} = \dfrac{C_V}{C_P} = \dfrac{5}{7}\). The rest, \(\dfrac{R}{C_P} = \dfrac{2}{7}\), is work done by the gas.

The mean free path is the average distance a molecule travels between two successive collisions. \(d\) is the diameter of a molecule and \(n\) is the number of molecules per unit volume.

\(\lambda \propto \dfrac{1}{n}\), so doubling \(n\) halves \(\lambda\). \(\lambda \propto \dfrac{1}{d^2}\), so doubling \(d\) reduces \(\lambda\) to a quarter.

1. The molecule hits any molecule whose centre lies within \(d\) of its own path. It sweeps a cylinder of radius \(d\) and length \(vt\): volume \(= \pi d^2 vt\).

2. Collisions \(= n \times \pi d^2 vt\). The distance travelled is \(vt\), so \(\lambda = \dfrac{vt}{n\pi d^2 vt} = \dfrac{1}{\pi d^2 n}\).

3. The other molecules are not at rest. Their relative speed is on average \(\sqrt{2}\) times \(v\), so collisions happen \(\sqrt{2}\) times as often and \(\lambda\) is \(\sqrt{2}\) times shorter.

\(\pi d^2 = \pi \times 9 \times 10^{-20} = 2.83 \times 10^{-19}\) m\(^2\). \(\sqrt{2}\pi d^2 n = 1.414 \times 2.83 \times 10^{-19} \times 2.5 \times 10^{25} = 10^7\) per m.

\(\lambda = 10^{-7}\) m \(= 100\) nm. This is \(\dfrac{10^{-7}}{3 \times 10^{-10}} \approx 330\) molecular diameters: a gas is mostly empty space.

\(PV = NkT\) gives \(n = \dfrac{N}{V} = \dfrac{P}{kT}\). So \(\lambda = \dfrac{kT}{\sqrt{2}\pi d^2 P}\).

\(\lambda \propto \dfrac{1}{P}\): doubling \(P\) halves \(\lambda\). \(\lambda \propto T\): doubling \(T\) at fixed \(P\) doubles \(\lambda\), because the gas expands and the molecules are further apart.

Mean of the five free paths \(= \dfrac{2 + 4 + 3 + 5 + 6}{5} = 4\) units \(= 4 \times 25 = 100\) nm.

1. \(200\) nm at \(0.5\) atm and \(400\) nm at \(0.25\) atm. \(\lambda \propto \dfrac{1}{P}\), so the graph is a hyperbola.

2. \(\lambda = 100 \text{ nm} \times \dfrac{1}{10^{-3}} = 10^5\) nm \(= 0.1\) mm. In a good vacuum the free path can exceed the size of the chamber.

Collision frequency \(= \dfrac{v}{\lambda} = \dfrac{500}{10^{-7}} = 5 \times 10^9\) per second.

Time between collisions \(= \dfrac{\lambda}{v} = \dfrac{10^{-7}}{500} = 2 \times 10^{-10}\) s.

A molecule travels only about \(100\) nm before it collides and changes direction. It makes billions of collisions per second, so its path is a random zigzag. Its net progress across the room is a tiny fraction of the distance it actually travels, and diffusion is slow.

At high pressure the molecules are close together. The assumption that they occupy no volume fails: van der Waals replaced \(V\) by \(V - nb\), the space actually free for motion.

The assumption that they exert no forces on each other also fails. The attraction between molecules reduces the pressure on the walls, so he added \(\dfrac{an^2}{V^2}\) to the measured pressure.

\(\left(P + \dfrac{an^2}{V^2}\right)(V - nb) = nRT\).

\(a\) measures the strength of the attraction between molecules; \(b\) measures the volume excluded by the molecules themselves per mole.

A molecule about to hit the wall is pulled back by the others, so the measured \(P\) is less than the ideal value. The term is added to correct it. The molecules cannot enter the volume taken up by other molecules, so \(nb\) is subtracted from \(V\).

\(\dfrac{an^2}{V^2}\) is a pressure, so \(a = \dfrac{\text{Pa} \times \text{m}^6}{\text{mol}^2}\): Pa m\(^6\)/mol\(^2\).

\(nb\) is a volume, so \(b\) is in m\(^3\)/mol.

1. \(PV = nRT\), so \(Z = 1\) at every pressure: the green line.

2. Attraction between molecules. It lowers the pressure below the ideal value, so \(PV\) is less than \(nRT\). This is the \(a\) correction.

3. The finite size of the molecules. The free volume \(V - nb\) is less than \(V\), so the measured \(V\) is more than the ideal value and \(Z > 1\). This is the \(b\) correction. At \(B\) the two corrections cancel and the gas happens to obey \(PV = nRT\).

Volume of one molecule \(= \dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi(1.5 \times 10^{-10})^3 = 1.41 \times 10^{-29}\) m\(^3\).

\(b \approx 4 \times 6 \times 10^{23} \times 1.41 \times 10^{-29} = 3.4 \times 10^{-5}\) m\(^3\)/mol \(= 34\) cm\(^3\)/mol. Real values are of this size: \(27\) cm\(^3\)/mol for hydrogen, \(43\) for carbon dioxide.

Ideal: \(P = \dfrac{nRT}{V} = \dfrac{8.3 \times 300}{0.0249} = 10^5\) Pa.

Van der Waals: \(P = \dfrac{nRT}{V - nb} - \dfrac{an^2}{V^2} = \dfrac{2490}{0.0249 - 0.00003} - \dfrac{0.1}{(0.0249)^2} = 100120 - 161 = 99959\) Pa.

The two differ by \(0.04\%\). The size correction raises \(P\) by \(120\) Pa and the attraction lowers it by \(161\) Pa. Both are tiny at ordinary pressure, so the gas is almost ideal.

Low pressure: the molecules are far apart, so their own volume is negligible and the attraction between them is weak.

High temperature: the molecules move fast, so their kinetic energy is much larger than the potential energy of attraction. The attraction hardly affects their motion.

A layer of air of thickness \(dh\) has weight \(\rho g\,dh\) per unit area, so the pressure falls by that much going up: \(dP = -\rho g\,dh\).

Put \(\rho = \dfrac{PM}{RT}\): \(\dfrac{dP}{P} = -\dfrac{Mg}{RT}dh\).

With \(T\) constant, integrate from \(0\) to \(h\): \(\ln\dfrac{P}{P_0} = -\dfrac{Mgh}{RT}\), so \(P = P_0\,e^{-Mgh/RT}\).

\(H = \dfrac{RT}{Mg} = \dfrac{8.3 \times 290}{0.029 \times 10} = 8300\) m \(= 8.3\) km.

\(H\) is the height over which the pressure falls by a factor \(e\). It is also the height the whole atmosphere would have if its density stayed at the sea-level value.

\(P = P_0\,e^{-h/H}\). At \(h = H\): \(P = P_0 e^{-1} = 0.37P_0\). At \(h = 2H\): \(P = P_0 e^{-2} = 0.135P_0\).

\(P = \dfrac{P_0}{2}\) when \(e^{-h/H} = \dfrac{1}{2}\), so \(h = H\ln 2 = 8.3 \times 0.693 = 5.75\) km.

1. At \(8.3\) km, \(P/P_0 = 0.37\); at \(16.6\) km, \(0.135\).

2. \(24.9\) km is \(3H\): \(P/P_0 = e^{-3} = 0.05\).

3. Each \(8.3\) km divides the pressure by \(e = 2.72\). An exponential falls by the same factor for every equal step.

\(\rho = \dfrac{PM}{RT}\) with \(T\) constant, so \(\rho \propto P\) and \(\rho = \rho_0\,e^{-h/H}\).

At \(8.3\) km: \(\rho = 1.2 \times 0.37 = 0.44\) kg/m\(^3\).

The atmosphere is not isothermal: the temperature falls by about \(6^\circ\)C per kilometre in the lower atmosphere, so \(H\) is smaller at height. Also \(g\) and the composition of the air change slowly with height. The formula is a fair guide for the first \(10\) km only.

\(\dfrac{P}{P_0} = e^{-8.8/8.3} = e^{-1.06} = 0.35\). The measured value is about \(0.33\).

The air is a third as dense, so each breath brings in only a third of the oxygen. The body cannot get enough oxygen from the thin air.