Unit 5 - The Gas State and Gas Laws

To-Do List

• Check if I have extra time
• Study for lab exam on Tuesday

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Factors Determining the State of a Substance

• Forces holding particles together
• Kinetic energy (which tends to pull them apart)

Forces Between Particles

Force Type	Typical State
Ionic	Usually Solid
Polar (Intermolecular)	Can be Liquid or Gas
Dispersion	Can be Liquid or Gas

Kinetic Molecular Theory - Brownian Motion

• States: All substances contain entities that are in constant random motion.
  • Robert Brown observed pollen grains moving under a microscope.
  • Explanation: This motion explains how scents travel.

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Temperature

• Temperature is a measure of the average kinetic energy of particles.
• Higher Temperature: More energy, more likely to overcome attractive forces.
• Lower Temperature: More likely to be in a liquid or solid state.

How Entities Move

• Vibrational Motion:
  • Strongest type of motion.
  • Occurs in solids.
• Rotational Motion:
  • Occurs in liquids, gases, and soft solids (e.g., candle wax).
• Translational Motion:
  • Occurs in liquids and gases.

Kinetic Molecular Theory (KMT)

• Volume of Gas Molecules:
  • The volume of individual gas molecules is negligible compared to the volume of the container.
• Forces Between Molecules:
  • No attractive or repulsive forces between gas molecules.
• Collisions:
  • Collisions between gas molecules are perfectly elastic (no loss of kinetic energy).
• Temperature and Kinetic Energy:
  • The average kinetic energy of gas molecules is directly related to temperature. The higher the temperature, the greater the average kinetic energy of the molecules.

KMT and the Ideal Gas

• Ideal Gas:
  • A hypothetical gas where particles occupy no space and do not attract each other.
• Real Gases:
  • No gas is truly ideal, but the ideal gas model allows for accurate mathematical predictions under various conditions.

Ideal gas law

$$\begin{array}{rl}& R=8.31\\ & T=(kelvin)\\ & PV=nRT\end{array}$$

Dalton’s Law of partial pressure

The total pressure of a mixture of non reacting gases is the sum of partial presses of individual gasses

$$P_{tot}=P_1+P_2\dots P_n$$

Law of contstant pressure

Random work

Gas collection over water

Partial pressures of atomospheric gases

n2: 79.09504
o2: 21.22235
Ar: 0.94209 kpa
co2: 0.03039 kpa

Catalytic converter

2 CO + 1 O2 → 2 CO2
V = 65.0L

$$\begin{array}{rl}& V_o=65\times \frac{1}{2}=32.5L\end{array}$$

6.04g

$C{h}_4+2O_2\to C{O}_2+2H_{2}O$
m = 6.04g v=?
T = 35c
P=100kPa

(100)(V) = (8.31)(35)
n = PV/RT

0.39875(8.31)(35) ) / 100

(0.39875(8.314)(308.15 ) / 100
= 10.2L
… volume of co2 is

Airbag

$$\begin{array}{rl}& 2Na{N}_3\to 2Na+3N_2\\ & \\ & V=67LP=105kPaT=305.15k\\ & \\ & n=\frac{PV}{RT}\\ & =\frac{(105kPa)(67L)}{8.314(305.15)}\\ & =2.772941\\ & \\ & 2.772941\ast 2/3\\ & 65.0099\times 1.848627=120g\end{array}$$

Density and molar mass

$$\begin{array}{rl}& \frac{PV}{RT}=\frac{m}{M}\\ & M=\frac{mRt}{PV}\end{array}$$

Notes: Gas Laws and Calculations

Problem 1: Molar Mass of a Gas

Given:

• Mass of gas: 7.723 g
• Volume: 2.0 L
• Temperature: 27°C = 300.15 K
• Pressure: 1.8 atm

Formula:
Calculation:
Result: The molar mass of the gas is approximately 44.03 g/mol.
Gas Identity: Based on the molar mass, the gas is likely carbon dioxide (CO**)**.

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Problem 2: Molar Mass of a Gas Using Density

Given:

• Density: 3.023 g/L
• Temperature: 25°C = 298.15 K
• Pressure: 2.31 atm

Formula:
Calculation:
Result: The molar mass of the gas is approximately 32.00 g/mol.
Gas Identity: Based on the molar mass, the gas is likely oxygen (O**)**.

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Problem 3: Volume of Carbon Dioxide Produced

Given:

• Mass of methane (CH): 6.40 g
• Excess oxygen (O)
• Temperature: 35.0°C = 308.15 K
• Pressure: 100.0 kPa

Reaction:
Step 1: Moles of CH
Step 2: Moles of CO (1:1 ratio)
Step 3: Volume of CO using Ideal Gas Law
Result: The volume of CO produced is 10.2 L.

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Key Formulas and Notes

1. Ideal Gas Law:
2. Density Relation:
3. Molar Mass Using Density:
4. Combined Gas Law:
5. Graham’s Law of Effusion:

What volume of carbon dioxide is produced when 6.40 g of methane gas, CH4, reacts with excess oxygen? All gases are at 35.0 ̊C and 100.0 kPawh

$$\begin{array}{rl}& \frac{6.40}{16.04}=0.399002493molC{H}_4\\ & 273.15c+33c=306.15k\\ & C{H}_4+2O_2\to C{O}_2+2H_{2}O\\ & \frac{1C{H}_4}{1C{O}_2}\times 0.399002493=0.399002493molC{O}_2\\ & V=\frac{0.399002493(8.314)(306.15)}{100kPa}\\ & =10.1559345441L\\ & =10.2L\\ & \dots \text{the volume of}C{O}_2\text{Produced is 10.2L}\end{array}$$