Gas Laws and Their Applications

Boyle’s Law

Boyle’s Law describes the inverse relationship between the pressure and volume of a gas at constant temperature.

• Definition: For a fixed amount of gas at constant temperature, the pressure and volume are inversely proportional.

• Equation:

• Key Point: If pressure increases, volume decreases, and vice versa.

• Example: A gas at 2.0 atm has a volume of 8.0 L. At 1.0 atm, the volume is L.

Charles’s Law

Charles’s Law relates the volume and temperature of a gas at constant pressure.

• Definition: For a fixed amount of gas at constant pressure, the volume is directly proportional to its absolute temperature (in Kelvin).

• Equation:

• Key Point: As temperature increases, volume increases.

• Example: 600 mL at 30°C (303 K) to 90°C (363 K): mL.

Avogadro’s Law

Avogadro’s Law states that the volume of a gas is directly proportional to the number of moles at constant temperature and pressure.

• Equation: or

• Key Point: More moles of gas mean greater volume.

• Example: 2.0 mol in 10 L, so 6.0 mol occupies L.

Gay-Lussac’s Law

This law relates the pressure and temperature of a gas at constant volume.

• Equation:

• Key Point: As temperature increases, pressure increases.

• Example: atm.

Ideal Gas Law

The Ideal Gas Law combines the simple gas laws into one equation, relating pressure, volume, temperature, and moles.

• Equation:

• R (Gas Constant): L·atm/(mol·K)

• Applications: Used to calculate moles, mass, molar mass, and number of molecules.

• Example: mol.

STP Conditions

Standard Temperature and Pressure (STP) are reference conditions for gases.

• STP: 1 atm, 273 K

• Molar Volume: 1 mol gas = 22.4 L at STP

• Example: Mass of O2 in 44.8 L at STP: $2\times$ 32 g/mol = 64 g.

Gas Properties & Pressure Units

Pressure Units

• Common Units: atm, mmHg (torr), kPa, psi

• Conversions: 1 atm = 760 mmHg = 101.3 kPa = 14.7 psi

• Example: $755\times \frac{1\ \text{atm}}{760\ \text{torr}} = 0.993$ atm

Density of a Gas at STP

• Equation:

• Example: CO2 (44 g/mol): g/L

Solutions & Electrolytes

Molarity (M)

Molarity is the concentration of a solution, defined as moles of solute per liter of solution.

• Equation:

• Example: 12.0 g NaCl in 0.500 L: mol; M

Dilutions

• Equation:

• Example: mL

Electrolytes

• Strong Electrolytes: Strong acids (e.g., HCl), strong bases (e.g., NaOH), soluble salts

• Weak Electrolytes: Weak acids (e.g., HNO2), weak bases

• Nonelectrolytes: Molecular compounds (e.g., sugar)

• Example: HNO2 is a weak acid.

Concentration of Ions

• Key Point: Multiply the solution molarity by the subscript of the ion in the formula.

• Example: 0.400 M CaCl2 gives 0.800 M Cl– ions.

Chemical Reactions & Equations

Types of Reactions

• Precipitation: Formation of an insoluble solid

• Acid–Base: Transfer of H+ ions

• Redox: Transfer of electrons

Net Ionic Equations

• Steps:

  1. Write the complete ionic equation

  2. Remove spectator ions

  3. Write the net ionic equation with only reacting species

• Example: Sodium phosphate + calcium chloride:

  • Molecular: 2 Na3PO4(aq) + 3 CaCl2(aq) → Ca3(PO4)2(s) + 6 NaCl(aq)

  • Net ionic: 2 PO43–(aq) + 3 Ca2+(aq) → Ca3(PO4)2(s)

Spectator Ions

• Definition: Ions that do not participate in the chemical reaction.

• Example: In K2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2 KCl(aq), K+ and Cl– are spectator ions.

Oxidation States & Redox Reactions

Assigning Oxidation States

• Rules:

  • Oxygen: –2

  • Hydrogen: +1

  • Group 1 metals: +1

  • Sum of oxidation states equals overall charge

• Example: Cr in CrO42–:

Oxidizing and Reducing Agents

• Oxidizing Agent: Gets reduced (gains electrons)

• Reducing Agent: Gets oxidized (loses electrons)

• Example: Zn + Cu2+ → Zn2+ + Cu; Cu2+ is the oxidizing agent, Zn is the reducing agent.

Stoichiometry & Percent Yield

Percent Yield

• Equation:

• Example:

Gas Stoichiometry

• Key Point: At same T & P, volume ratios equal mole ratios.

• Example: N2 + 3 H2 → 2 NH3; 3 L N2 produces 6 L NH3.

Gas Law Graphs

• Boyle’s Law: P vs. V is a hyperbola (inverse)

• Charles’s Law: V vs. T is a straight line (direct)

• Avogadro’s Law: V vs. n is a straight line (direct)

• Gay-Lussac’s Law: P vs. T is a straight line (direct)

Writing Balanced Molecular Equations

• Key Steps: Write correct formulas, balance atoms, include physical states.

• Examples:

  • Calcium carbonate + nitric acid → CaCO3(s) + 2 HNO3(aq) → Ca(NO3)2(aq) + CO2(g) + H2O(l)

  • Lithium bromide + silver nitrate → LiBr(aq) + AgNO3(aq) → AgBr(s) + LiNO3(aq)

  • Sodium phosphate + aluminum chloride → 2 Na3PO4(aq) + 3 AlCl3(aq) → AlPO4(s) + 6 NaCl(aq)

Net Ionic Equations

• Steps:

  1. Split aqueous strong electrolytes into ions

  2. Identify and remove spectator ions

  3. Write net ionic equation with only reacting particles

• Example: FeCl2(aq) + 2 NaOH(aq) → Fe(OH)2(s) + 2 NaCl(aq)

  • Net ionic: Fe2+(aq) + 2 OH–(aq) → Fe(OH)2(s)

Derivations of the Ideal Gas Law

• Solving for Variables:

  • For P:

  • For V:

  • For n:

  • For T:

• Molar Mass:

• Density:

• Example: Number of molecules: , then multiply by Avogadro’s number ( molecules/mol).

• Example: Molar mass from mass, volume, P, T:

Gas Stoichiometry (Not at STP)

• Key Point: Use volume ratios only if T & P are constant; otherwise, use the ideal gas law.

• Example: C7H16 + 11 O2 → 7 CO2 + 8 H2O; 5.00 L C7H16 produces 35.0 L CO2 at same T & P.

Percent Yield, Limiting Reactant, and Theoretical Yield

• Limiting Reactant: The reactant that is completely consumed first, limiting the amount of product formed.

• Theoretical Yield: Maximum amount of product possible from limiting reactant.

• Percent Yield:

• Example: C6H6 + 3 Cl2 → C6H3Cl3 + 3 HCl; 20.0 g benzene, 40.0 g Cl2:

  • a) Identify limiting reactant by comparing mole ratios

  • b) Calculate theoretical yield from limiting reactant

  • c) Percent yield if actual yield is 18.4 g:

Acid–Base Stoichiometry

• Neutralization: Acid reacts with base to form water and a salt.

• Steps: Convert given quantities to moles, use mole ratios, convert to desired units.

• Example: How many mL of 0.150 M H2SO4 to neutralize 35.0 mL of 0.200 M KOH?

  • Balanced: H2SO4 + 2 KOH → K2SO4 + 2 H2O

  • Calculate moles KOH, use stoichiometry to find moles H2SO4, then volume.

Dilutions

• Equation:

• Example: Volume of 3.0 M KCl to make 250 mL of 0.75 M KCl: mL

Summary Table: Gas Laws and Equations

Additional info: This guide expands on the original points with academic context, definitions, and worked examples to ensure clarity and completeness for exam preparation.