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.