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

$P<em>1V</em>1 = P<em>2V</em>2$

• 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: $P<em>1 = \frac{P</em>2 V<em>2}{V</em>1}$
   • Calculation: $P_1 = \frac{(1195 ext{ torr}) (1.93 ext{ L})}{10.8 ext{ L}} = 214 ext{ torr}$

2. Solve for Missing Volume

   • Given: Radius = 19.6 cm, Pressure₁ = 918 mmHg, Pressure₂ = 21.4 psi
   • Volume of sphere: $V = \frac{4}{3}πr^3 = \frac{4}{3} π (0.196 m)^3 = 31.5 L$
   • Formula: $P<em>1V</em>1 = P<em>2V</em>2$
   • 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

$\frac{V<em>1}{T</em>1} = \frac{V<em>2}{T</em>2}$

• 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: $T_2 = -167 + 273 = 106 K$
   • $V<em>2 = V</em>1 \frac{T<em>2}{T</em>1}$ 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

$\frac{P<em>1}{T</em>1} = \frac{P<em>2}{T</em>2}$

• 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 $T<em>2$ leads to find: 428 K then $T</em>2 - 273 = -230 °C$.

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

$\frac{V<em>1}{n</em>1} = \frac{V<em>2}{n</em>2}$

• 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.