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
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 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:
P_{1}V_{1} = P_{2}V_{2}
Where,
This expression can be obtained from the pressurevolume relationship suggested by Boyle’s law. For a fixed amount of gas kept at a constant temperature, PV = k. Therefore,
P_{1}V_{1} = k (initial pressure * initial volume)
P_{2}V_{2} = k (final pressure * final volume)
∴ P_{1}V_{1} = P_{2}V_{2}
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 (V_{1}) = 1L
Initial pressure (P_{1}) = 400 kPa
Final volume (V_{2}) = 3L
As per Boyle’s law, P_{1}V_{1} = P_{2}V_{2} ⇒ P_{2} = (P_{1}V_{1})/V_{2}
P_{2} = (1L * 400 kPa)/3L = 133.33 kPa
Therefore, the gas exerts a pressure of 133.33 kPa on the walls of the 3liter container.
Exercise 2
A gas exerts a pressure of 3 kPa on the walls of container 1. When container 1 is emptied into a 10liter 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, P_{1} = 3kPa
Final pressure, P_{2} = 6kPa
Final volume, V_{2} = 10L
According to Boyle’s law, V_{1} = (P_{2}V_{2})/P_{1}
V_{1} = (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,
V_{1} = V_{2}
T_{1} T_{2}
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 GayLussac, 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 400cm^{3} 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,
V_{1} = V_{2}
T_{1} T_{2}
400 = V_{2}
273 353
V_{2}=400×353
273
V_{2}=517.21cm^{3}
1 cubic centimeter = 0.001 litre =1 x 103 litre
∴ 517.21 cubic centimeter = 517.21 x 103 = 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,
V_{1} = V_{2}
T_{1} T_{2}
V_{1} = 6
150 100
V_{1} =6×150
100
V_{1} =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
P_{1} to P_{2} initial pressure to final pressure
V_{1} to V_{2} initial volume to final volume
T_{1} to T_{2} initial temperature to final temperature
Thus at constant temperature
P_{1}V_{1} = P_{2}V_{x}
V_{x} = P_{1}V_{1} / P_{2} …………..eqn 1 from Boyle’s law
At constant pressure
V_{x} / T_{1} = V_{2} / T_{2}
V_{x} = V_{2}T_{1} / T_{2 …………………..}eqn. 2 from Charle’s law
If P_{x} in eqn 1 = P_{x} in eqn 2 then
P_{1}V_{1}/P_{2} = V_{2}T_{1} / T_{2}
P_{1}V_{1} / T_{1} = P_{2}V_{2} / T_{2}
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=0.001m^{3}
(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
CLASSWORK: As in evaluation
CONCLUSION: The teacher commends the students positively