Term: 2^nd Term

Week: 6

Class: Senior Secondary School 1

Age: 15 years

Duration: 40 minutes of 5 periods each

Date:

Subject: Chemistry

Topic:- Gas laws I

SPECIFIC OBJECTIVES: At the end of the lesson, pupils should be able to

Define gas laws
State Boyle’s law
State Charles’ law
State the general gas law equation
Solve problems on all the laws learnt

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 the laws of constant composition, multiple proportions and balancing chemical equations	Students pay attention
STEP 2 EXPLANATION	He defines gas laws. He further states and explains Boyle and Charles’ law, stating their formulae repetitively.	Students pay attention and participates
STEP 3 DEMONSTRATION	He outlines the general gas equations and solves problems on all the gas laws learnt	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
Gas laws deals with the relationship between the Temperature (T), Pressure (P) and Volume (V) of gases.

Boyle’s law

It states that the volume of a fixed mass of gas is inversely proportional to its pressure at constant temperature. That is when temperature is not changing, the volume of a given gas increases as the pressure decrease and vice versa

For a gas, the relationship between volume and pressure (at constant mass and temperature) can be expressed mathematically as follows.

P ∝ (1/V)

Where P is the pressure exerted by the gas and V is the volume occupied by it. This proportionality can be converted into an equation by adding a constant, k.

P = k*(1/V) ⇒ PV = k

The pressure v/s volume curve for a fixed amount of gas kept at constant temperature is illustrated below.

Graphically,

Formula and Derivation

As per Boyle’s law, any change in the volume occupied by a gas (at constant quantity and temperature) will result in a change in the pressure exerted by it. In other words, the product of the initial pressure and the initial volume of a gas is equal to the product of its final pressure and final volume (at constant temperature and number of moles). This law can be expressed mathematically as follows:

P₁V₁ = P₂V₂

Where,

P₁ is the initial pressure exerted by the gas
V₁ is the initial volume occupied by the gas
P₂ is the final pressure exerted by the gas
V₂ is the final volume occupied by the gas

This expression can be obtained from the pressure-volume relationship suggested by Boyle’s law. For a fixed amount of gas kept at a constant temperature, PV = k. Therefore,

P₁V₁ = k (initial pressure * initial volume)

P₂V₂ = k (final pressure * final volume)

∴ P₁V₁ = P₂V₂

This equation can be used to predict the increase in the pressure exerted by a gas on the walls of its container when the volume of its container is decreased (and its quantity and absolute temperature remain unchanged)

SOLVED EXERCISES ON BOYLE’S LAW

Exercise 1

A fixed amount of a gas occupies a volume of 1L and exerts a pressure of 400 kPa on the walls of its container. What would be the pressure exerted by the gas if it is completely transferred into a new container having a volume of 3 liters (assuming the temperature and quantity of gas remains constant)?

Solution

Given,

Initial volume (V₁) = 1L

Initial pressure (P₁) = 400 kPa

Final volume (V₂) = 3L

As per Boyle’s law, P₁V₁ = P₂V₂ ⇒ P₂ = (P₁V₁)/V₂

P₂ = (1L * 400 kPa)/3L = 133.33 kPa

Therefore, the gas exerts a pressure of 133.33 kPa on the walls of the 3-liter container.

Exercise 2

A gas exerts a pressure of 3 kPa on the walls of container 1. When container 1 is emptied into a 10-liter container, the pressure exerted by the gas increases to 6 kPa. Find the volume of container 1. Assume that the temperature and quantity of the gas remain constant.

Solution

Given,

Initial pressure, P₁ = 3kPa

Final pressure, P₂ = 6kPa

Final volume, V₂ = 10L

According to Boyle’s law, V₁ = (P₂V₂)/P₁

V₁ = (6 kPa * 10 L)/3 kPa = 20 L

Therefore, the volume of container 1 is 20 L.

Charles’ Law

It states that the volume of a given mass of gas is directly proportional to its absolute temperature at constant pressure

V ∝ T i.e V / T = constant

Or,

V₁ = V₂

T₁ T₂

Where,

V1 and V2 are the Initial Volumes and Final Volume respectively. T1 refers to the Initial Temperature and T2 refers to the Final Temperature. Both the temperatures are in the units of Kelvin.

Graphically,

Jacques Charles, a French scientist, in 1787, discovered that keeping the pressure constant, the volume of a gas varies on changing its temperature. Later, Joseph Gay-Lussac, in 1802, modified and generalized the concept as Charles’s law. At very high temperatures and low pressures, gases obey Charles’ law.

Solved Examples

Question 1:

A gas occupies a volume of 400cm³ at 0°C and 780 mm Hg. What volume (in litres) will it occupy at 80°C and 780 mm Hg?

Solution:

Given,

V1= 400 cm³

V2 =?

T1= 0°C= 0+273 = 273 K

T2= 80°C= 80+273 = 353 K

Here the pressure is constant and only the temperature is changed.

Using Charles Law,

V₁ = V₂

T₁ T₂

400 = V₂

273 353

V₂=400×353

273

V₂=517.21cm³

1 cubic centimeter = 0.001 litre =1 x 10-3 litre

∴ 517.21 cubic centimeter = 517.21 x 10-3 = 0.517 litres

Question 2

Find the initial volume of a gas at 150 K, if the final volume is 6 L at 100 K

Solution:

Given,

V1=?
V2 =6 L
T1= 150 K

T2= 100 K

Using Charles Law,

V₁ = V₂

T₁ T₂

V₁ = 6

150 100

V₁ =6×150

100

V₁ =9L

The initial volume of a gas at 150 K is 9 litres.

General Gas Law

It has no definition statement rather it deals with change in pressure and volume of a gas at a constant temperature from initial to final. And the change in volume and temperature at constant pressure from initial to final. It is the combination of Boyles and Charles law

P₁ to P₂ initial pressure to final pressure

V₁ to V₂ initial volume to final volume

T₁ to T₂ initial temperature to final temperature

Thus at constant temperature

P₁V₁ = P₂V_x

V_x = P₁V₁ / P₂ …………..eqn 1 from Boyle’s law

At constant pressure

V_x / T₁ = V₂ / T₂

V_x = V₂T₁ / T_{2 …………………..}eqn. 2 from Charle’s law

If P_x in eqn 1 = P_x in eqn 2 then

P₁V₁/P₂ = V₂T₁ / T₂

P₁V₁ / T₁ = P₂V₂ / T₂

Problems Related to Gas Law

(1) A sealed jar whose volume is exactly 1 L, which contains 1 mole of air at a temperature of 20 degrees Celcius, assuming that the air behaves as an ideal gas. So, what is the pressure inside the jar in Pa?

Solution:

By solving with the help of the ideal gas equation,

PV = nRT

(1) By rearranging the equation, we can get,

P = nRT/V

(2) Write down all the values which are known in the SI unit.

n= 1

R= 8.314J/K/mol

T= 20degree celcius=(20+273.15)K=293.15K

V=1L=0.001m³

(3) Put all the values in the equation

P= nRT/V

P=(1*8.314*293.15)/0.001

P= 2,437,249

P=2.437*10^6 Pa

The pressure is almost 24atm.

EVALUATION: 1. State Boyle’s law

State Charle’s law
3. A gas occupies 221cm³at a temperature of 0 C and pressure of 760 mm Hg. What will be the volume at 100°C?
A sample of oxygen gas has a volume of 225 mL when its pressure is 1.12 atm. What will the volume of the gas be at a pressure of 0.98 atm if the temperature remains constant?
An ideal gas occupying a 2.0 L flask at 760 torrs is allowed to expand to a volume of 6,000 mL. Calculate the final pressure
A gas is initially in a 5 L piston with a pressure of 1 atm. What is the new volume if the pressure changes to 3.5 atm by moving the piston down?
A balloon of volume 0.666 L at 1.03atm is placed in a pressure chamber where the pressure becomes 5.68atm. Determine the new volume.
A gas in a 30.0 mL container is at a pressure of 1.05 atm and is compressed to a volume of 15.0 mL. What is the new pressure of the container?
9. Differentiate between Boyle’s law and Charle’s law.
10. A gas occupies a volume of 500.0 mL at a temperature of 10.0 °C. What will be its volume at 50.0°C?
11. A gas occupies a volume of 100.0 mL at a temperature of 27.0 °C. What is the volume at 10.0 °C?

CLASSWORK: As in evaluation

CONCLUSION: The teacher commends the students positively

Lesson Notes By Weeks and Term - Senior Secondary 1