5.2 Gas Mixtures: Dalton's Law of Partial Pressures & Graham's Law

5.2 Gas Mixtures: Dalton's Law of Partial Pressures & Graham's Law

Gaseous systems frequently exist as multi-component mixtures. In ideal mixtures, gas particles are widely separated and exert no intermolecular forces, meaning each gas behaves independently within the shared volume.

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Dalton's Law of Partial Pressures

In 1801, John Dalton formulated the law governing gas mixtures:

The total pressure of a mixture of non-reacting gases equals the sum of the partial pressures exerted by each individual gas occupying the same container volume at the same temperature.

For a mixture of $k$ distinct gases: $P_{\text{total}} = P_1 + P_2 + P_3 + \dots + P_k = \sum_{i=1}^k P_i$

Microscopic Origin

Because ideal gas particles do not interact, each species strikes container walls independently. The observed total pressure is the sum of wall forces imparted per unit area by all individual components: $P_i = \frac{n_i RT}{V} \quad \implies \quad P_{\text{total}} = \sum \frac{n_i RT}{V} = \left(\sum n_i\right)\frac{RT}{V} = n_{\text{total}}\frac{RT}{V}$

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Mole Fraction and Partial Pressure

The relative abundance of each component is expressed by its mole fraction ($X_i$): $X_i = \frac{n_i}{n_{\text{total}}}$

Mole fraction is dimensionless, ranging from $0$ to $1$, and the sum of all mole fractions in a mixture equals unity ($\sum X_i = 1.000$).

Taking the ratio of component partial pressure to total pressure: $\frac{P_i}{P_{\text{total}}} = \frac{n_i RT / V}{n_{\text{total}} RT / V} = \frac{n_i}{n_{\text{total}}} = X_i$

Rearranging gives the core relationship: $P_i = X_i P_{\text{total}}$

Atmospheric Composition Table

At sea level ($P_{\text{total}} = 1.000\text{ atm} = 760.0\text{ mmHg}$), dry air illustrates component partial pressures:

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Collecting Gases Over Water

In laboratory experiments, insoluble or sparingly soluble gases are collected by pneumatic displacement of water in an inverted eudiometer.

Vapor Pressure Correction

Liquid water in the collection trough evaporates until liquid-vapor dynamic equilibrium is established. The collected gas is saturated with water vapor ($\text{H}_2\text{O}(g)$). By Dalton's Law: $P_{\text{total}} = P_{\text{gas}} + P_{\text{H}_2\text{O}}$

To isolate the pressure exerted solely by the dry gas, subtract the water vapor pressure: $P_{\text{gas}} = P_{\text{total}} - P_{\text{H}_2\text{O}}$

When water levels inside and outside the eudiometer are equalized, $P_{\text{total}} = P_{\text{barometric}}$. Water vapor pressure depends strictly on temperature:

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Diffusion vs Effusion

Gas transport involves two distinct phenomena driven by thermal motion:

1. Diffusion: The spontaneous intermingling of one gas through another via random motion. Due to frequent molecular collisions (short mean free path, $\sim 60\text{ nm}$ at STP), net diffusion is slow despite high molecular speeds.
2. Effusion: The escape of gas molecules through a microscopic orifice into an evacuated chamber. Orifice diameter must be smaller than the mean free path, ensuring particles pass through independently without intermolecular collisions during transit.

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Graham's Law of Effusion

In 1846, Thomas Graham showed that effusion rates at constant temperature and pressure are inversely proportional to the square root of molar mass ($\mathcal{M}$) or density ($d$): $\frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{\mathcal{M}_2}{\mathcal{M}_1}} = \sqrt{\frac{d_2}{d_1}}$

Relationship to Effusion Time

Because rate is inversely proportional to elapsed time ($t$) for a fixed volume ($\text{Rate} \propto 1/t$): $\frac{t_1}{t_2} = \frac{\text{Rate}_2}{\text{Rate}_1} = \sqrt{\frac{\mathcal{M}_1}{\mathcal{M}_2}}$ Lighter gases effuse faster, requiring less time to discharge a specified volume.

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Practical Applications

Uranium Isotope Enrichment

Graham's Law enabled the separation of fissile $^{235}\text{U}$ ($0.71\%$ natural abundance) from non-fissile $^{238}\text{U}$ ($99.28\%$) by converting uranium into gaseous uranium hexafluoride ($\text{UF}_6$):

• $^{235}\text{UF}_6$: $\mathcal{M} = 235.04 + 6(19.00) = 349.04\text{ g/mol}$
• $^{238}\text{UF}_6$: $\mathcal{M} = 238.05 + 6(19.00) = 352.05\text{ g/mol}$

The single-stage separation factor is: $\alpha = \frac{\text{Rate}(^{235}\text{UF}_6)}{\text{Rate}(^{238}\text{UF}_6)} = \sqrt{\frac{352.05}{349.04}} \approx 1.0043$ Because $^{235}\text{UF}_6$ effuses only $0.43\%$ faster, thousands of barrier diffusion stages in series were required to produce enriched uranium.

Gas Leak Detection

Methane ($\text{CH}_4, \mathcal{M} = 16.0\text{ g/mol}$) effuses $\sqrt{29.0/16.0} = 1.35$ times faster than average air molecules, allowing rapid detection of micro-punctures in natural gas lines.

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Worked Numerical Examples

Example 1: Gas Collection Over Water

Oxygen gas generated from $\text{KClO}_3$ decomposition is collected over water at $24.0^\circ\text{C}$ in a $385.0\text{ mL}$ eudiometer under $752.4\text{ mmHg}$ barometric pressure. Calculate the dry $\text{O}_2$ mass produced.

1. Dalton correction ($P_{\text{H}_2\text{O}} = 22.4\text{ mmHg}$ at $24.0^\circ\text{C}$): $P_{\text{O}_2} = 752.4 - 22.4 = 730.0\text{ mmHg} = 0.9605\text{ atm}$
2. Convert parameters: $V = 0.3850\text{ L}$, $T = 297.15\text{ K}$.
3. Calculate moles and mass: $n_{\text{O}_2} = \frac{(0.9605)(0.3850)}{(0.08206)(297.15)} = 0.01516\text{ mol}$ $m = 0.01516\text{ mol} \times 32.00\text{ g/mol} = 0.485\text{ g}$

Example 2: Effusion Time and Molar Mass

An unknown gas effuses through an aperture in $108.0\text{ s}$, while an equal volume of methane ($\text{CH}_4, \mathcal{M} = 16.04\text{ g/mol}$) effuses in $48.0\text{ s}$ under identical conditions. Find the unknown molar mass.

$\frac{t_{\text{unknown}}}{t_{\text{CH}_4}} = \frac{108.0}{48.0} = 2.25 = \sqrt{\frac{\mathcal{M}_{\text{unknown}}}{16.04}}$ $\mathcal{M}_{\text{unknown}} = (2.25)^2 \times 16.04 = 5.0625 \times 16.04 = 81.2\text{ g/mol}$