Term: 2^{nd} Term
Week: 8
Class: Senior Secondary School 1
Age: 15 years
Duration: 40 minutes of 5 periods each
Date:
Subject: Chemistry
Topic: Gas laws III
SPECIFIC OBJECTIVES: At the end of the lesson, pupils should be able to
INSTRUCTIONAL TECHNIQUES: Identification, explanation, questions and answers, demonstration, videos from source
INSTRUCTIONAL MATERIALS: Videos, loud speaker, textbook, pictures
INSTRUCTIONAL PROCEDURES
PERIOD 12
PRESENTATION 
TEACHER’S ACTIVITY 
STUDENT’S ACTIVITY 
STEP 1 INTRODUCTION 
The teacher reviews the previous lesson on GayLussac and Avogadro’s law and the ideal gas laws 
Students pay attention 
STEP 2 EXPLANATION 
He states, Graham’s law, Avogadro’s numbers, and explains the mole concept. .

Students pay attention and participates 
STEP 3 DEMONSTRATION 
He carries out calculations on the molar volume of gases 
Students pay attention and participate 
STEP 4 NOTE TAKING 
The teacher writes a summarized note on the board 
The students copy the note in their books 
NOTE
GAS LAWS
Graham’s Law
Graham’s Law which is popularly known as Graham’s Law of Effusion was formulated by Thomas Graham in the year 1848. Thomas Graham experimented with the effusion process and discovered an important feature: gas molecules that are lighter will travel faster than the heavier gas molecules.
According to Graham’s Law, at constant pressure and temperature, molecules or atoms with lower molecular mass will effuse faster than the higher molecular mass molecules or atoms. Thomas even found out the rate at which they escape through diffusion. In other words, it states that the rate of effusion of a gas is inversely proportional to the square root of its molecular mass. This formula is generally used while comparing the rates of two different gases at equal pressures and temperatures. The formula can be written as
Rate_{1} =√ M_{2}
Rate _{2} M_{1}
M_{1} is the molar mass of gas 1
M_{2} is the molar mass of gas 2
Rate_{1} is the rate of effusion of the first gas
Rate_{2} is the rate of effusion for the second gas
It states that the rate of diffusion or effusion is inversely proportional to its molecular mass.
Avogadro’s constant (N_{o}) = 6.02 x 10^{23} that represent the number of particles of any gas in 22.4dm^{3} at a standard temperature of 273K and pressure of 760mmHg.
Avogadro’s constant allows for calculations of pure substances in moles involving stoichiometric principles. With this the molecular
masses of gases can be calculated by comparing the masses of equal volume of gases
CALCULATIONS BASED ON GAS LAWS
Molar Volume of Gas
It is the volume occupied by one mole of gas. All gases at room temperature and pressure (r.t.p) has a volume of 22.4dm^{3} or 22400cm^{3} were 1dm^{3}= 1000cm^{3}
Formula: Volume of a gas = Number of moles (n) x Molar volume (M_{v})
Example
What is the number of moles of 240cm^{3} of Cl_{2} at r.t.p?
Solution
n = volume /M_{v}
= 240 / 22400
= 0107 mol.
Molar Volume and Molar Mass
Moles to Grams
The following formula shows how to change from moles to g, which is another common unit of measure:
Moles = grams
Molar mass
To figure out what a substance’s molar mass is, you have to use the useful periodic table. It can be worked out by adding up the masses of all the atoms in the substance. For example, if you need to figure out the molar mass of NaCl, you would:
Na has a mass number of 22.99 g/mol.
Cl’s mass number is 35.45 g/mol.
So, the molar mass of NaCl is 22.99 plus 35.45, which equals 58.44 g/mol.
Examples
Find the volume of 7g of N_{2} gas at r.t.p. (N = 14)
Solution
Step 1: Find the number of moles from the mass of nitrogen n= mass/molar mass
= 7/28
= 25 mol
Step 2: Find the volume of nitrogen, now with formula of gas 0.25 mol =
Volume of gas = 0.25 mol x 22.4 = 5.6 dm^{3} (or 5600cm^{3})
Concentration of solutions tells the number of solute in a given volume of solution. Concentration (C) Calculating the Amount of Solute
Moles of solute (n) = Concentration (C) x Volume of solution in (dm^{3})
Examples
What is the mass of solute in 600cm^{3} of 1.5 Molar NaOH solution?
Solution
Volume of solution in dm^{3} = 0.60 dm^{3} n = 1.5 x 0.60= 0.9 mol Mass = 0.9 x 40= 36g
EVALUATION: 1. State the Graham’s Law
a.0.100 L of CO_{2} at 307 torr and 26 °C
b. 8.75 L of C_{2}H_{4}, at 378.3 kPa and 483 K
c. 221 mL of Ar at 0.23 torr and –54 °C
4. A high altitude balloon is filled with 1.41 × 10^{4}L of hydrogen at a temperature of 21 °C and a pressure of 745 torr. What is the volume of the balloon at a height of 20km, where the temperature is –48 °C and the pressure is 63.1 torr?
5. While resting, the average 70kg human male consumes 14 L of pure O_{2} per hour at 25 °C and 100 kPa. How many moles of O_{2} are consumed by a 70 kg man while resting for 1.0 h?
6. A puncture takes away half of the volume of a tire with 10 moles of air and a 40litre volume. How much air is left in a tire that has been deflated?
7. At 25°C and 2.00 atm, a sample of 6.0 L holds 0.5 moles of a gas. What is the final volume of the gas if 0.25 moles more are added at the same pressure and temperature?
CLASSWORK: As in evaluation
CONCLUSION: The teacher commends the students positively