GAS LAWS

Day 1: Physical Properties - Unit 7 - Gases

I. Kinetic Molecular Theory

• Ideal Gas Particles:

  • Have no volume.

  • Participate in elastic collisions.

    • Participating in elastic collisions means that when gas particles collide with each other or with the walls of their container, they do so without losing any kinetic energy. In an elastic collision, the total kinetic energy before and after the collision remains the same, which is a key assumption of the Kinetic Molecular Theory for ideal gases. This implies that the particles bounce off each other without any deformation or generation of heat, maintaining their energy and speed.

  • Are in constant, random, straight-line motion.

  • Do not attract or repel each other.

  • Average kinetic energy (KE) is directly related to Kelvin temperature.

II. Real Gases

• Real Gas Particles:

  • Have volume.

  • Attract each other.

• Behavior is most ideal under:

  • Low pressures

  • High temperatures

  • Nonpolar atoms/molecules

III. Characteristics of Gases

A. Expansion

• Gases expand to fill any container (random motion, no attraction).

B. Fluidity

• Gases are fluids, similar to liquids (no attraction).

C. Density

• Gases have very low densities (no volume, lots of empty space).

IV. Compressibility and Movement

A. Compressibility

• Gases are compressible (lots of empty space).

B. Diffusion & Effusion

• Gases undergo diffusion (random motion distributing evenly) and effusion (passing through tiny openings).

V. Temperature Conversions

• Temperature conversions:

  • Fahrenheit (°F)

  • Celsius (°C)

  • Kelvin (K)

• Important temperatures:

  • Absolute zero: -273°C = 0 K

  • Freezing point of water: 0°C = 273 K

  • Boiling point of water: 100°C = 373 K

• Always use absolute temperature (Kelvin) when working with gases.

VI. Pressure

A. Measurement

• Barometers: measure atmospheric pressure.

  • Types:

    • Mercury Barometer

    • Aneroid Barometer

B. Key Units at Sea Level

• 101.325 kPa (kilopascal)

• 1 atm

• 760 mm Hg

• 760 torr

• 14.7 psi

VII. Standard Temperature and Pressure (STP)

• STP Conditions:

  • Temperature: 0°C = 273 K

  • Pressure: 1 atm = 101.325 kPa

Day 2: The Gas Laws

I. Boyle’s Law

• Relationship: Pressure (P) and Volume (V) are inversely related at constant mass and temperature.

• Equation: PV = k

II. Charles’ Law

• Relationship: Volume (V) and Absolute Temperature (T) are directly related at constant mass and pressure.

III. Gay-Lussac’s Law

• Relationship: Pressure (P) and Absolute Temperature (T) are directly related at constant mass and volume.

IV. Combined Gas Law

• Equation: P1V1/T1 = P2V2/T2

V. Gas Law Problems

A. Charles’ Law Example

• Given: V1 = 473 cm3, T1 = 36°C = 309K, T2 = 94°C = 367K

• Calculate: V2 = 562 cm3

B. Boyle’s Law Example

• Given: V1 = 100. mL, P1 = 150. kPa, P2 = 200. kPa

• Calculate: V2 = 75.0 mL

C. Combined Gas Law Example

• Given: V1 = 7.84 cm3, P1 = 71.8 kPa, T1 = 25°C = 298 K, P2 = 101.325 kPa, T2 = 273 K

• Calculate: V2 = 5.09 cm3 (STP)

D. Gay-Lussac’s Law Example

• Given: P1 = 765 torr, T1 = 23°C = 296K, P2 = 560. torr

• Calculate: T2 = 217 K or -56.3°C

Day 3: Gas Laws Continued

I. Avogadro’s Principle

• States that equal volumes of gases contain equal numbers of moles at constant temperature and pressure.

II. Dalton’s Law

• Total pressure of a mixture of gases equals the sum of partial pressures.

  • Equation: Ptotal = P1 + P2 + ...

A. Example of Dalton’s Law

• Given: Ptotal = 94.4 kPa, PH2O = 2.72 kPa

• Calculate: PH2 = 91.7 kPa (pressure of dry hydrogen gas)

III. Graham’s Law

A. Diffusion and Effusion

• Diffusion: spreading of gas molecules throughout a container.

• Effusion: passing of gas through a tiny opening.

B. Rate of Diffusion/Effusion

• Speed of diffusion is inversely related to the square root of its molar mass. Ratio: vA/vB = √(MB/MA)

C. Examples of Graham’s Law

• If Kr diffuses 1.381 times faster than Br2, calculate molar masses based on speeds.

• Determine average speed of hydrogen molecules compared to oxygen under the same conditions.

IV. Ideal Gas Law

• Relationship: PV = nRT

• R (Gas Constant):

  • R = 0.0821 L·atm/mol·K

  • R = 8.315 dm³·kPa/mol·K

A. Example Calculations

• Calculate pressure of 0.412 mol of He at 16°C occupying 3.25 L: P = 3.01 atm.

• Find volume of 85 g of O2 at 25°C and 104.5 kPa.

V. Gas Stoichiometry

A. Volume-Moles Relationship

• For STP, use 22.4 L/mol; for non-STP, use ideal gas law.

B. Example Problem

• From 5.25 g of CaCO3 at 103 kPa & 25ºC, find volume of CO2.

  • Use ideal gas law to find liters after calculating moles.

VI. Final Calculation Examples

• How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C.

  • Use stoichiometry to convert moles to grams.

Participating in elastic collisions means that when gas particles collide with each other or with the walls of their container, they do so without losing any kinetic energy. In an elastic collision, the total kinetic energy before and after the collision remains the same, which is a key assumption of the Kinetic Molecular Theory for ideal gases. This implies that the particles bounce off each other without any deformation or generation of heat, maintaining their energy and speed.