8.1 Ideal and Real Gases and the Kinetic Molecular Theory

8.1 Ideal and Real Gases and the Kinetic Molecular Theory

Quick Answer: Physical Chemistry gives 10 TOS items to ideal and real gases. Kinetic molecular theory is half of another 10-item area, shared with electrolyte solutions. Expect ideal-gas and partial-pressure calculations, gas density and molar mass, molecular speeds and Graham's law, collision frequency and mean free path, and real-gas corrections with the van der Waals equation and critical constants. Section 8.5 extends real gases to fugacity.

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The Ideal Gas Law

$pV = nRT, \qquad R = 8.314\text{ J mol}^{-1}\text{ K}^{-1} = 0.08206\text{ L atm mol}^{-1}\text{ K}^{-1} = 0.08314\text{ L bar mol}^{-1}\text{ K}^{-1}$

The ideal gas law combines Boyle's law ($pV$ constant at fixed $T$), Charles's law ($V \propto T$ at fixed $p$), and Avogadro's principle ($V \propto n$). Always use kelvin.

Gas density and molar mass: rearranging gives $\rho = \frac{pM}{RT}$ and $M = \frac{\rho RT}{p}$. A gas with a density of 1.25 g/L at 0 C and 1 atm has $M = (1.25)(0.08206)(273.15)/1 = 28.0$ g/mol, consistent with $N_2$ or $CO$.

Dalton's Law of Partial Pressures

$p_{total} = \sum_i p_i, \qquad p_i = x_i\,p_{total}$

Gas collected over water contains water vapour, so $p_{gas} = p_{atm} - p_{H_2O}$ (water's vapour pressure is 23.8 mmHg at 25 C).

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Kinetic Molecular Theory

Postulates: molecules are point masses in random straight-line motion, collide elastically, and exert no forces on each other between collisions. The derivation gives $pV = \frac{1}{3}nMu_{rms}^2$, which matches the ideal gas law when

$\bar{E}_{k} = \frac{3}{2}RT\ \text{per mole} \qquad (\tfrac{3}{2}kT\ \text{per molecule})$

Average kinetic energy depends only on temperature: at the same $T$, $H_2$ and $O_2$ molecules have the same average kinetic energy, but $H_2$ moves faster.

The Maxwell-Boltzmann Speed Distribution

Use $M$ in kg/mol so the units come out in m/s. Raising the temperature or lowering the molar mass broadens the distribution and shifts it to higher speeds.

Graham's Law of Effusion

$\frac{\text{rate}_1}{\text{rate}_2} = \sqrt{\frac{M_2}{M_1}}$

Hydrogen effuses $\sqrt{32.00/2.016} = 3.98$ times faster than oxygen. The enrichment factor per stage for $^{235}UF_6$ over $^{238}UF_6$ is only $\sqrt{352.04/349.03} = 1.0043$, so gaseous diffusion needed over a thousand stages.

Collisions and Mean Free Path

For molecules of diameter $d$ (collision cross-section $\sigma = \pi d^2$) at number density $\mathcal{N} = p/kT$:

$z = \sqrt{2}\,\sigma\,\bar{u}\,\mathcal{N}, \qquad \lambda = \frac{\bar{u}}{z} = \frac{kT}{\sqrt{2}\,\sigma\,p}$

Worked example: For $N_2$ ($d = 0.37$ nm, $\sigma = 0.43\text{ nm}^2$) at 298 K and 1 atm, $\lambda \approx 67$ nm and $z \approx 7 \times 10^9\text{ s}^{-1}$. The mean free path is inversely proportional to pressure, so at $10^{-6}$ atm it grows to about 7 cm. That is why mass spectrometers need high vacuum.

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Real Gases

The compression factor $Z = pV_m/RT$ equals 1 for an ideal gas. Attractions make $Z < 1$ at moderate pressures, while repulsions (finite molecular size) make $Z > 1$ at high pressures.

The van der Waals Equation

$\left(p + \frac{an^2}{V^2}\right)(V - nb) = nRT$

The constant $a$ corrects for attractions and $b$ for the excluded volume of the molecules. For $CO_2$, $a = 3.640\text{ L}^2\text{ bar mol}^{-2}$ and $b = 0.04267\text{ L mol}^{-1}$.

Worked example: For 1.00 mol $CO_2$ in 0.500 L at 300 K:

• Ideal: $p = \frac{(0.08314)(300)}{0.500} = 49.9$ bar
• van der Waals: $p = \frac{(0.08314)(300)}{0.500 - 0.04267} - \frac{3.640}{(0.500)^2} = 54.5 - 14.6 = 40.0$ bar

The attraction term dominates, so the real pressure is about 20% below the ideal value.

Critical Constants and Corresponding States

Above the critical temperature a gas cannot be liquefied by pressure alone. The van der Waals equation predicts:

$V_c = 3b, \qquad T_c = \frac{8a}{27Rb}, \qquad p_c = \frac{a}{27b^2}, \qquad Z_c = \frac{3}{8}$

For $CO_2$, these give $T_c = 304$ K and $p_c = 74.0$ bar, close to the measured 304.1 K and 73.8 bar. Real gases have $Z_c$ of about 0.27-0.31, so the predicted 0.375 is only approximate. The principle of corresponding states says that gases at the same reduced variables ($p_r = p/p_c$, $T_r = T/T_c$) have similar compression factors.

The Virial Equation and Boyle Temperature

$Z = 1 + \frac{B(T)}{V_m} + \frac{C(T)}{V_m^2} + \cdots$

From the van der Waals equation, $B = b - a/RT$. $B$ is negative at low $T$ (attractions dominate) and positive at high $T$. At the Boyle temperature $T_B = a/Rb$, $B = 0$ and the gas behaves nearly ideally over a range of pressures. For $CO_2$ the van der Waals estimate is about 1026 K.