# Lesson Notes By Weeks and Term - Senior Secondary 1

Gas laws III

Term: 2nd 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

1. State Avogadro’s number and the mole concept
2. State Graham’s law
3. Carry out calculations on the molar volume of gases

INSTRUCTIONAL TECHNIQUES: Identification, explanation, questions and answers, demonstration, videos from source

INSTRUCTIONAL MATERIALS: Videos, loud speaker, textbook, pictures

INSTRUCTIONAL PROCEDURES

PERIOD 1-2

 PRESENTATION TEACHER’S ACTIVITY STUDENT’S ACTIVITY STEP 1 INTRODUCTION The teacher reviews the previous lesson on Gay-Lussac 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

Rate1 =√ M2

Rate 2         M1

M1 is the molar mass of gas 1

M2 is the molar mass of gas 2

Rate1 is the rate of effusion of the first gas

Rate2 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 (No) = 6.02 x 1023 that represent the number of particles of any gas in 22.4dm3 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 22.4dm3 or 22400cm3 were 1dm3= 1000cm3

Formula: Volume of a gas = Number of moles (n) x Molar volume (Mv)

Example

What is the number of moles of 240cm3 of Cl2 at r.t.p?

Solution

n = volume /Mv

= 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 N2 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 dm3 (or 5600cm3)

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 (dm3)

Examples

What is the mass of solute in 600cm3 of 1.5 Molar NaOH solution?

Solution

Volume of solution in dm3 = 0.60 dm3 n = 1.5 x 0.60= 0.9 mol Mass = 0.9 x 40= 36g

EVALUATION:    1. State the Graham’s Law

2. How many grams of gas are present in each of the following cases?

a.0.100 L of CO2 at 307 torr and 26 °C

b. 8.75 L of C2H4, 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 × 104L 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 70-kg human male consumes 14 L of pure O2 per hour at 25 °C and 100 kPa. How many moles of O2 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 40-litre 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?

1. 0.450 mole of Fe contains how many atoms?
2. 0.200 mole of H2O contains how many molecules?
3. Calculate the number of molecules in 1.058 mole of H2O

CLASSWORK: As in evaluation

CONCLUSION: The teacher commends the students positively