Gases and Gas Laws Notes

Unit 8: Gases and Gas Laws

Learning Intention

  • Study of gas laws:
    • Boyle's Law
    • Charles' Law
    • Gay-Lussac's Law
    • Avogadro's Law

Success Criteria

  • Able to use Boyle's, Charles', Gay-Lussac's, and Avogadro's gas laws to solve for missing variables.
  • Units must match when plugging into gas law formulas.

Properties of a Gas

  • Pressure (P): The force a gas exerts on its container.
  • Volume (V): The three-dimensional space occupied by the gas.
  • Temperature (T): Measurement of the thermal state of the gas.

Units

  • Pressure: mmHg, atm, Pa, kPa, torr, psi
  • Volume: L
  • Temperature: K

Boyle's Law

  • Relates: Pressure (P) and Volume (V) with Temperature and Moles constant.
  • Type of Relationship: Inverse
    • As volume decreases, pressure increases.

Boyle's Law Equation

P1V1 = P2V2

  • 1 = Initial conditions, 2 = Final conditions
  • Required Units:
    • Pressure: atm, torr, mmHg, Pa, kPa (based on problem)
    • Volume: L

Boyle's Law Graph

  • Shape: Curve representing inverse relationship.

Example Problems

  1. Solve for Missing Pressure

    • Given: Volume = 10.8 L, V₂ = 1.93 L, P₂ = 1195 torr
    • Formula: P1 = rac{P2 V2}{V1}
    • Calculation: P_1 = rac{(1195 ext{ torr}) (1.93 ext{ L})}{10.8 ext{ L}} = 214 ext{ torr}

  2. Solve for Missing Volume

    • Given: Radius = 19.6 cm, Pressure₁ = 918 mmHg, Pressure₂ = 21.4 psi
    • Volume of sphere: V = rac{4}{3}πr^3 = rac{4}{3} π (0.196 m)^3 = 31.5 L
    • Formula: P1V1 = P2V2
    • Calculation yields new volume: 26.1 L.

Charles' Law

  • Relates: Temperature (T) and Volume (V) with Pressure and Moles constant.
  • Type of Relationship: Direct
    • As temperature increases, volume increases.

Charles' Law Equation

\frac{V1}{T1} = \frac{V2}{T2}

  • Required Units:
    • Volume: L
    • Temperature: K

Example Problems

  1. Solve for Missing Temperature

    • Given: Initial Volume = 2.26 L, Temperature₁ = 291°C
    • Convert Temperature: T = 291 + 273 = 564 K
    • Calculation: For new volume V₂ = 0.349, \frac{(564 K) (0.349 L)}{V_1} = ?? yields 87.1 °C.

  2. Solve for Missing Volume

    • Given: Moles of gas = 8.60 mol, Initial Temperature = 0.00 °C, Final Temperature = -167°C
    • Convert: T_2 = -167 + 273 = 106 K
    • V2 = V1 \frac{T2}{T1} leads to calculated final volume: 749 L.

Gay-Lussac's Law

  • Relates: Pressure (P) and Temperature (T) with Volume and Moles constant.
  • Type of Relationship: Direct.

Gay-Lussac's Law Equation

\frac{P1}{T1} = \frac{P2}{T2}

  • Required Units:
    • Pressure: atm, torr, mmHg, Pa, kPa
    • Temperature: K

Example Problems

  1. Solve for Missing Temperature

    • Given: Temperature₁ = 61.0 °C, Pressure₁ = 29.4 psi, Pressure₂ = 0.256 atm
    • Convert: T_1 = 61.0 + 273 = 334 K
    • Calculation for T2 leads to find: 428 K then T2 - 273 = -230 °C.

  2. Solve for Missing Pressure

    • Given: Pressure₁ = 3.04 atm, Temperature₁ = -100 °C, Temperature₂ = 276 K
    • Convert: T_1 = -100 + 273 = 173 K
    • Solve for final pressure using the law equation.

Avogadro's Law

  • Relates: Volume (V) and Moles (n) while keeping Temperature and Pressure constant.
  • Type of Relationship: Direct.
    • Increase in moles leads to an increase in volume.

Avogadro's Law Equation

\frac{V1}{n1} = \frac{V2}{n2}

  • Required Units:
    • Volume: L
    • Moles: mol

Example Problems

  1. Solve for Missing Volume
    • Given: 0.950 mol of O2 at 20.0 L
    • Calculate for 1.35 mol of O2 at the same conditions:
    • Calculation yields: V_2 = \frac{(20.0 L)(1.35 mol)}{0.950 mol}.

Practice Problems

  1. Calculate pressure in a balloon compressed from 650 mL to 400 mL.
  2. Determine new temperature when a gas sample cools from 275 K to a new volume of 195 mL.
  3. Find resulting volume when oxygen is heated from 310 K to 405 K with an initial volume of 70.0 L.