Chapter 10: The Gas Laws - Dalton's Law, Ideal Gas Law, and Graham's Law

Dalton’s Law of Partial Pressures

Definition & Fundamental Principles: Dalton’s Law states that the total pressure exerted inside a container is equal to the sum of the partial pressures exerted by each individual gas within that container.
Partial Pressure: This is defined as the specific contribution each individual gas makes toward the total pressure of the mixture.
Mathematical Representation: The law is expressed by the formula: $P_{Total} = P_1 + P_2 + P_3 + …$
Environmental Constants: In any gas mixture, all components share the exact same temperature ( $T$ ) and volume ( $V$ ).
Conceptual Examples: * Atmospheric Air: Air is a mixture of different gases (primarily nitrogen and oxygen) rather than a pure substance (element or compound). Each gas contributes to the atmospheric pressure we experience. * Water Bottle Scenario: In a water bottle, the pressure inside is affected by the mixture of gases present above the liquid surface.
Pressure and Moles Summation: * Adding Pressures: If three containers have pressures of $2\,atm$ , $1\,atm$ , and $3\,atm$ respectively, combining them into a fourth container results in a total pressure of $6\,atm$ ( $2 + 1 + 3 = 6$ ). * Adding Moles: Similarly, the total number of moles ( $n_{total}$ ) is the sum of the moles of each individual component: $n_1 + n_2 + n_3 = n_{total}$ . For example: $12\,moles + 8\,moles + 12\,moles = 32\,moles$ .
Example Problem A: Calculation of total pressure in a balloon. * Given: $\text{Oxygen Pressure } (P_1) = 170\,mmHg$ ; $\text{Nitrogen Pressure } (P_2) = 620\,mmHg$ . * Identification: $P_1 = 170\,mmHg$ , $P_2 = 620\,mmHg$ , $P_{tot} = X$ . * Formula: $P_1 + P_2 = P_{tot}$ . * Substitution: $170\,mmHg + 620\,mmHg = X$ . * Answer: $X = 790\,mmHg$ .
Example Problem B: Solving for an individual gas pressure. * Given: $\text{Total Pressure } (P_{tot}) = 1.3\,atm$ ; $\text{Nitrogen Pressure } (P_2) = 720\,mmHg$ . Find Oxygen pressure in $mmHg$ . * Identification: $P_1 = X\,mmHg$ , $P_2 = 720\,mmHg$ , $P_{tot} = 1.3\,atm$ . * Formula: Convert $atm$ to $mmHg$ for $P_{tot}$ , then use $P_1 + P_2 = P_{tot}$ . * Substitution: $1.3\,atm \times 760\,mmHg / 1\,atm = 988\,mmHg$ $X\,mmHg + 720\,mmHg = 988\,mmHg$ * Answer: $X = 268\,mmHg$ .
Practice Problem: A container with three gases (A, B, C) has a total pressure of $125\,kPa$ . $P_A = 83\,kPa$ and $P_B = 28\,kPa$ . What is the pressure of gas C? * Answer: $14\,kPa$ .

Mole Fractions in Gas Mixtures

Mole Fraction ( $x_i$ ): The mole fraction of an individual gas component in an ideal gas mixture relates its partial pressure to the total pressure and its moles to the total moles.
Formulas for Mole Fraction: $x_i = \frac{P_i}{P_{total}} = \frac{n_i}{n_{total}}$ * $x_i$ = Mole fraction. * $P_i$ = Partial pressure of a specific gas. * $P_{total}$ = Total gas pressure of the container. * $n_i$ = Moles of a specific gas. * $n_{total}$ = Total number of moles of gas in the container.
Example Problem C: Finding partial pressures from masses. * Given: $V = 2.2\,L$ , $0.800\,g\text{ of } Cl_2$ , $0.211\,g\text{ of } SO_3$ , and $P_{tot} = 1025\,mmHg$ . Find partial pressures of each. * Step 1: Convert Mass to Moles: $0.800\,g\,Cl_2 \times \frac{1\,mol\,Cl_2}{70.90\,g\,Cl_2} = 0.0113\,mol\,Cl_2$ $0.211\,g\,SO_3 \times \frac{1\,mol\,SO_3}{80.07\,g\,SO_3} = 0.00264\,mol\,SO_3$ * Step 2: Calculate Total Moles: $0.0113 + 0.00264 = 0.0139\,mol_{total}$ * Step 3: Solve for $P_{Cl_2}$ (X): $\frac{X}{1025\,mmHg} = \frac{0.0113\,mol}{0.0139\,mol}$ $X = 833\,mmHg$ * Step 4: Solve for $P_{SO_3}$ (Y): $833\,mmHg + Y = 1025\,mmHg$ $Y = 192\,mmHg$ * Final Answer: $P_{Cl_2} = 833\,mmHg$ ; $P_{SO_3} = 192\,mmHg$ .

Avogadro’s Law and The Ideal Gas Law

Avogadro’s Law/Hypothesis: Volume ( $V$ ) is directly proportional to the number of molecules (moles, $n$ ) at a constant temperature ( $T$ ) and pressure ( $P$ ). * $V = K \times n$
The Ideal Gas Law Equation: $P \times V = n \times R \times T$ * Variables: Pressure times Volume equals the number of moles times the Ideal Gas Constant ( $R$ ) times the temperature in Kelvin. * Temperature: Must always be in Kelvin ( $K$ ) for this law. * The Ideal Gas Constant ( $R$ ): Unlike other variables, $R$ is a true constant. Its value depends on the units used for pressure: * $R = 0.0821\,\frac{L \times atm}{mol \times K}$ * $R = 62.4\,\frac{L \times mmHg}{K \times mol}$ * $R = 8.315\,\frac{L \times kPa}{K \times mol}$
Application: This law allows for the counting of moles by measuring pressure, volume, and temperature, liberating calculations from the constraints of STP (Standard Temperature and Pressure).
Example Problem A (Finding Moles): Moles of air in a $2.0\,L$ bottle at $19^\circ C$ and $747\,mmHg$ . * ID: $V = 2.0\,L$ , $P = 747\,mmHg$ , $T = 19^\circ C$ , $n = X$ , $R = 62.4\,\frac{L \times mmHg}{K \times mol}$ . * Formula: Convert $T$ to $K$ ( $19 + 273 = 292\,K$ ), then use $PV = nRT$ . * Substitution: $(747\,mmHg)(2.0\,L) = X(62.4)(292\,K)$ . * Answer: $X = 0.082\,moles\text{ air}$ .
Example Problem B (Finding Pressure): Pressure of $1.8\,g\text{ of } H_2$ gas in a $4.3\,L$ balloon at $27^\circ C$ . * ID: $V = 4.3\,L$ , $P = X$ , $T = 27^\circ C$ ( $300\,K$ ), $m = 1.8\,g\,H_2$ , $R = 62.4$ . * Conversion: $1.8\,g\,H_2 \times \frac{1\,mol\,H_2}{2.02\,g\,H_2} = 0.89\,mol\,H_2$ . * Substitution: $(X\,mmHg)(4.3\,L) = (0.89\,mol)(62.4)(300\,K)$ . * Answer: $X = 3900\,mmHg$ .

Gases and Stoichiometry

Background: Gases are difficult to mass; therefore, we find their quantity using moles through measured volume, temperature, and pressure.
Review of STP (Standard Temperature and Pressure): * Conditions: $0^\circ C$ and $1\,atm$ pressure. * Molar Volume: At STP, $1\,mole$ of any gas occupies $22.4\,L$ . * Avogadro’s Hypothesis: Equal volumes of gas at the same temperature and pressure contain the same number of particles.
Stoichiometry Not At STP: * Since chemical reactions occur in mole ratios, if a gas is not at STP, use the Ideal Gas Law ( $PV = nRT$ ) to convert between volume and moles.
Example Problem C: Volume of $CO_2$ formed from $5.25\,g\text{ of } CaCO_3$ at $103\,kPa$ and $25^\circ C$ . * Reaction Ratio: $CaCO_3 \rightarrow CaO + CO_2$ ( $1$ to $1$ ratio). * Step 1 (Stoichiometry): $5.25\,g\,CaCO_3 \times \frac{1\,mol\,CaCO_3}{100.09\,g\,CaCO_3} \times \frac{1\,mol\,CO_2}{1\,mol\,CaCO_3} = 0.0525\,mol\,CO_2$ . * Step 2 (Ideal Gas Law): $P=103\,kPa$ , $n=0.0525\,mol$ , $T=298\,K$ , $R=8.315$ . * Substitution: $(103\,kPa)V = (0.0525\,mol)(8.315)(298\,K)$ . * Answer: $V = 1.26\,L\,CO_2$ .
Example Problem D: Grams of $Al_2O_3$ formed from $15.0\,L\text{ of } O_2$ at $97.3\,kPa$ and $21^\circ C$ . * Reaction: $4Al + 3O_2 \rightarrow 2Al_2O_3$ . * Step 1 (Ideal Gas Law): Find moles of $O_2$ . $(97.3\,kPa)(15.0\,L) = n(8.315)(294\,K)$ $n = 0.597\,mol\,O_2$ * Step 2 (Stoichiometry): $0.597\,mol\,O_2 \times \frac{2\,mol\,Al_2O_3}{3\,mol\,O_2} \times \frac{101.96\,g\,Al_2O_3}{1\,mol\,Al_2O_3} = 40.6\,g\,Al_2O_3$ . * Answer: $40.6\,g\,Al_2O_3$ .

Diffusion and Effusion (Graham’s Law)

Diffusion: The movement of molecules from areas of high concentration to areas of low concentration (e.g., perfume molecules spreading across a room).
Effusion: The diffusion of gas through a tiny hole in a container.
Graham’s Law: The rate of effusion and diffusion is inversely proportional to the square root of the molar mass ( $M$ ) of the molecules.
Kinetic Energy Connection: $KE = \frac{1}{2}mv^2$ , where $m = \text{mass}$ and $v = \text{velocity}$ .
Summary of Molecular Behavior: * At the same temperature, large molecules move slower than small molecules. * Large molecules effuse and diffuse slower than small molecules. * Case Study: Helium vs. Air: Helium escapes from a balloon faster than air because Helium has a smaller mass. The lecture uses the analogy that race cars (small mass) move faster than semi-trucks (large mass).

Questions & Discussion

Video Prompt: There is a reference to a video covering "partial pressure and vapor pressure."
Visual Review: Students are encouraged to identify gas laws via images/diagrams.
Brain Breaks (Jokes): * Atom Joke: Two atoms talk on a street; one lost an electron and is "positive" about it. * Sick Chemist: If you can't "helium" or "curium," you must "barium." * Problem Solving: Chemists are great at solving problems because they have all the "solutions."