Gases: Pressure, Laws, and Kinetic Molecular Theory

Gases: Pressure, Laws, and Kinetic Molecular Theory

Pressure and Pressure Unit Conversions

Pressure is a fundamental property of gases, defined as the force exerted per unit area. In chemistry, several units are commonly used to measure pressure, including Pascals (Pa), torr (mm Hg), and atmospheres (atm).

• Definition of Pressure: Pressure ($P$) is given by the equation: $P = \frac{F}{A}$ where $F$ is force and $A$ is area.

• Common Pressure Units:

  • Pascals (Pa): SI unit; $1 \text{ Pa} = 1 \text{ N/m}^2$

  • Atmospheres (atm): Standard atmospheric pressure; $1 \text{ atm} = 101,325 \text{ Pa}$

  • Torr (mm Hg): $1 \text{ atm} = 760 \text{ torr} = 760 \text{ mm Hg}$

• Conversion Examples:

  • To convert $2.00 \text{ atm}$ to torr: $2.00 \text{ atm} \times 760 \text{ torr/atm} = 1520 \text{ torr}$

  • To convert $500 \text{ mm Hg}$ to atm: $500 \text{ mm Hg} \times \frac{1 \text{ atm}}{760 \text{ mm Hg}} = 0.658 \text{ atm}$

The Ideal Gas Law and Combined Gas Laws

The behavior of gases under various conditions is described by several gas laws. The ideal gas law relates pressure, volume, temperature, and amount of gas.

• Ideal Gas Law: $PV = nRT$ where $P$ = pressure, $V$ = volume, $n$ = moles, $R$ = gas constant ($0.0821 \text{ L atm mol}^{-1} \text{K}^{-1}$), $T$ = temperature in Kelvin.

• Combined Gas Law: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$ Used when the amount of gas is constant but conditions change.

• Example: Calculate the volume occupied by $2.00$ mol of an ideal gas at $1.00$ atm and $273$K:$V = \frac{nRT}{P} = \frac{2.00 \times 0.0821 \times 273}{1.00} = 44.8 \text{ L}$

Density and Molar Mass of Gases

The density of a gas can be related to its molar mass using the ideal gas law.

• Density Formula: $d = \frac{PM}{RT}$ where $d$ = density, $P$ = pressure, $M$ = molar mass, $R$ = gas constant, $T$ = temperature.

• Calculating Molar Mass: Rearranging the formula: $M = \frac{dRT}{P}$

• Example: Find the density of $O_2$ gas ($M = 32.00 \text{ g/mol}$) at $1.00$ atm and $273$K:$d = \frac{1.00 \times 32.00}{0.0821 \times 273} = 1.43 \text{ g/L}$

Stoichiometry of Gas Reactions

Stoichiometric calculations involving gases often use the ideal gas law to relate moles of gas to volume under specific conditions.

• Key Steps:

  1. Write the balanced chemical equation.

  2. Use molar ratios to relate reactants and products.

  3. Apply the ideal gas law to convert between moles and volume.

• Example: How many liters of $CO_2$ are produced at STP from $10.0$ g of $CaCO_3$? $CaCO_3 (s) \rightarrow CaO (s) + CO_2 (g)$ Moles $CaCO_3 = \frac{10.0}{100.09} = 0.100$ mol Moles $CO_2 = 0.100$ mol Volume $CO_2 = 0.100 \times 22.4 = 2.24$ L

Partial Pressures and Dalton's Law

In a mixture of gases, each gas exerts a pressure independently of the others. The total pressure is the sum of the partial pressures.

• Dalton's Law of Partial Pressures: $P_{\text{total}} = P_1 + P_2 + P_3 + \ldots$

• Calculating Partial Pressure: $P_i = X_i P_{\text{total}}$ where $X_i$ is the mole fraction of gas $i$.

• Example: If a mixture contains $2$mol$N_2$and $1$ mol $O_2$ at a total pressure of $3$atm:$X_{N_2} = \frac{2}{3}$,$P_{N_2} = \frac{2}{3} \times 3 = 2$ atm

Kinetic Molecular Theory of Gases

The kinetic molecular theory explains the macroscopic properties of gases by considering their molecular motion.

• Main Postulates:

  • Gases consist of tiny particles in constant, random motion.

  • Collisions between particles and container walls are elastic.

  • The volume of gas particles is negligible compared to the container.

  • No intermolecular forces act between particles.

  • The average kinetic energy is proportional to temperature: $KE_{avg} = \frac{3}{2} k_B T$ where $k_B$ is Boltzmann's constant.

• Explaining Gas Laws:

  • Pressure results from collisions with container walls.

  • Temperature reflects average kinetic energy.

• Effusion and Diffusion:

  • Effusion: Movement of gas through a small hole.

  • Diffusion: Mixing of gases due to random motion.

  • Graham's Law of Effusion: $\frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{M_2}{M_1}}$ where $M_1$ and $M_2$ are molar masses.

• Example: Hydrogen ($M = 2$ g/mol) effuses faster than oxygen ($M = 32$ g/mol): $\frac{\text{Rate}_{H_2}}{\text{Rate}_{O_2}} = \sqrt{\frac{32}{2}} = 4$

Additional info: These notes expand on the learning objectives by providing definitions, formulas, and examples for each concept.