Chapter 12: Kinetic Theory (Kinetic Theory)

Radius $r=1.5\text{A˚}=1.5\times 10^{-10}$ m. Volume of one molecule $=\frac{4}{3}\pi r^3=1.41\times 10^{-29}{\text{m}}^3$.

For one mole: $1.41\times 10^{-29}\times 6.02\times 10^{23}=8.5\times 10^{-6}{\text{m}}^3$. Actual (molar) volume at STP $=22.4\times 10^{-3}{\text{m}}^3$.

Fraction $=\frac{8.5\times 10^{-6}}{22.4\times 10^{-3}}=3.8\times 10^{-4}$.

About 4 × 10⁻⁴

$V=\frac{nRT}{P}=\frac{1\times 8.31\times 273}{1.013\times 10^5}=0.0224{\text{m}}^3=22.4$ litres.

22.4 L. Shown.

(a) For an ideal gas $\frac{PV}{T}=nR$ is constant at all pressures: the dotted line shows ideal gas behaviour.

(b) A real gas behaves more ideally at higher temperature. The T₁ curve is closer to the dotted line, so T₁ > T₂.

(c) At P → 0 every gas is ideal: $\frac{PV}{T}=nR=\frac{1.00\times 10^{-3}}{32.0\times 10^{-3}}\times 8.31=0.26{\text{J K}}^{-1}$.

(d) No: 1.00 g of hydrogen contains more moles. The same value needs the same n $=0.03125$ mol: mass $=0.03125\times 2.02\times 10^{-3}=6.3\times 10^{-5}$ kg.

(a) Ideal gas behaviour (b) T₁ > T₂ (c) 0.26 J K⁻¹ (d) No; 6.3 × 10⁻⁵ kg of H₂

Absolute pressures: 16 atm and 12 atm; $V=0.030{\text{m}}^3$.

$n_1=\frac{16\times 1.013\times 10^5\times 0.030}{8.31\times 300}=19.5$ mol; $n_2=\frac{12\times 1.013\times 10^5\times 0.030}{8.31\times 290}=15.1$ mol.

Mass removed $=(19.5-15.1)\times 32=140$ g.

About 0.14 kg

At the bottom: $P_1=1.013\times 10^5+40\times 10^3\times 9.8=4.93\times 10^5$ Pa, $T_1=285$ K. At the top: $P_2=1.013\times 10^5$ Pa, $T_2=308$ K.

$V_2=V_1\frac{P_1{T}_2}{P_2{T}_1}=1.0\times \frac{4.93\times 10^5\times 308}{1.013\times 10^5\times 285}=5.3{\text{cm}}^3$.

5.3 cm³ (5.3 × 10⁻⁶ m³)

$N=\frac{PV}{k_{B}T}=\frac{1.013\times 10^5\times 25.0}{1.38\times 10^{-23}\times 300}$

About 6.1 × 10²⁶ molecules

Helium is monatomic: average energy $=\frac{3}{2}{k}_{B}T$.

(i) $1.5\times 1.38\times 10^{-23}\times 300=6.2\times 10^{-21}$ J (ii) $\times 6000$ K: $1.24\times 10^{-19}$ J (iii) $\times 10^7$ K: $2.1\times 10^{-16}$ J.

(i) 6.2 × 10⁻²¹ J (ii) 1.24 × 10⁻¹⁹ J (iii) 2.1 × 10⁻¹⁶ J

Yes: by Avogadro's law, equal volumes at the same P and T contain equal numbers of molecules ($N=\frac{PV}{k_{B}T}$).

No: $v_{rms}=\sqrt{\frac{3RT}{M}}$ depends on molar mass. Neon (20) is lightest compared with Cl₂ (71) and UF₆ (352), so its $v_{rms}$ is the largest.

Equal numbers; v_rms differs and is largest for neon.

Equal $v_{rms}$ ⇒ equal $\frac{T}{M}$: $T_{Ar}=T_{He}\frac{M_{Ar}}{M_{He}}=253\times \frac{39.9}{4.0}$.

About 2.52 × 10³ K

Number density $n=\frac{P}{k_{B}T}=\frac{2.0\times 1.013\times 10^5}{1.38\times 10^{-23}\times 290}=5.1\times 10^{25}{\text{m}}^{-3}$; diameter $d=2.0\times 10^{-10}$ m.

$\lambda =\frac{1}{\sqrt{2}n\pi d^2}=\frac{1}{1.414\times 5.1\times 10^{25}\times 3.14\times 4\times 10^{-20}}\approx 1.1\times 10^{-7}$ m.

$v_{rms}=\sqrt{\frac{3\times 8.31\times 290}{0.028}}=5.1\times 10^2{\text{m s}}^{-1}$.

Collision frequency $=\frac{v_{rms}}{\lambda }\approx 5\times 10^9{\text{s}}^{-1}$.

Time for a collision $\approx \frac{d}{v_{rms}}=\frac{2\times 10^{-10}}{510}=4\times 10^{-13}$ s; time between collisions $=\frac{\lambda }{v_{rms}}\approx 2\times 10^{-10}$ s, about 500 times longer. So a molecule moves freely most of the time.

λ ≈ 1 × 10⁻⁷ m; ≈ 5 × 10⁹ collisions per second; free time ≈ 500 × collision time