School: Tangub City National High School
Demonstrator: Justine C. Pama
Teaching Date and Time: June 13-16, 2023
Grade Level: Grade 10
Subject: Chemistry
I – Content Standard: The learners demonstrate an understanding of:
A) how gases behave based on the motion and relative distances between
gas particles.
II – Objectives: At the end of the topic the students must have:
a. recognized the relationship between volume and pressure at constant
temperature,
b. illustrated the volume-pressure relationship in a graph,
c. valued the importance of reading aerosol cans.
Materials:
syringes, balloons
References: Science 10 Learner’s Module pp. 362-369
III – Subject Matter: “Boyle’s Law”
IV – Procedure: 7E’s (Grade 10 students)
Teacher’s Activity
A. ELICIT (Access prior knowledge.)
Our first activity this morning is entitled
“Pass the Box”. As you can see, I have here a box
containing rolled pieces of paper with questions
regarding on our last topic. All you have to do is
pass the box to your seatmate while I play a
music. Once the music stops, the last person
who’s holding the box will get to pick a rolled
piece of paper, read the question, and answer it.
Note: Avoid shouting during the activity. Am I
clear?
Yes ma’am.
Questions:
1. What are the four properties of gases?
Students’ Activity
Answers:
The four properties of gases are mass, volume,
pressure, and temperature.
2. What is the standard unit for volume?
The standard unit for volume is Liter (L).
3. What is the standard unit for pressure?
The standard unit for pressure is atmosphere (atm).
4. What is the
temperature?
standard
unit
for The standard unit for temperature is Kelvin (K).
Do you have any questions or clarifications
regarding on our last topic? Then let’s proceed.
None ma’am.
B. ENGAGE (Get the students’ minds focused on
the topic (short; question or picture.)
Before I will introduce our lesson for today,
let’s have an activity first.
Presentation:
Now class, do you know that there is a
certain law which states that there is a
relationship between volume and temperature?
If that’s the case, sit properly, and widen
your mind as we are going to discover the volumetemperature relationship. Everybody read.
Yes ma’am.
The balloon is being inflated with helium gas and
lighted with a match which grew bigger in its size until
it exploded.
Yes ma’am. The volume of the balloon increases
because volume is the size or the amount of space a
certain substance occupies. And the size of the balloon
from the first scenario showed that it expands.
The lighted match causes the balloon to increase in
volume because it produces heat.
Temperature ma’am.
The volume of the balloon grows smaller or decreases.
No ma’am.
“Charles’s Law”
C. EXPLORE (Provide students with a common
experience.)
Before we will start our discussion, let us
first have another activity. But before we will start
our activity, let me show you first the objectives
for today’s lesson. Everybody read.
At the end of the topic the students must have:
A) recognized the relationship between volume
and temperature at constant pressure,
B) illustrated
the
volume-temperature
relationship in a graph,
C) applied the Charles’s Law in solving problems.
Activity 1:
I will group you into 4 groups. As you can
see, I have here with me some materials that will
be used in this activity.
Materials:
(4) 700 mL beaker
(4) balloon
(4) container
Ice
Heater
(4) Thermometer
Procedures: (For hot water)
1. Inflate the balloon in a small size enough to be
put inside the beaker. (There should be
enough space for the balloon to
bulge/expand.)
2. Measure the circumference of the balloon
using a tape measure and record it.
3. Pour 300 mL of hot water in the beaker.
4. Get the temperature reading of the hot water.
5. Put the balloon inside the beaker together
with the hot water and observe.
6. After observing what happened, measure
again the circumference of the balloon. (Do
these procedures for the cold water.)
While doing this activity, you are going to fill
up the table that will be given to you and answer
the following questions:
1. What happens to the size of the balloon as
the temperature increases?
2. What happens to the size of the balloon as
temperature decreases?
3. How does the change in the temperature
relate to the volume of gas in the balloon?
Note: Handle the laboratory apparatuses
properly.
I will give you 5 minutes to do this activity. Yes ma’am.
Each group must select one representative who None ma’am.
will discuss or explain what happened to their
activity and what are the results. Am I clear?
Students’ Table:
Do you have any questions or clarifications? Set-up
Average
Average Circumference of
Then your time will start now!
Temperature
the Balloon
(After 5 minutes)
(℃)
Before After Difference
Time is up! Let’s start with group 1.
Warm
water
Cold
water
Students’ possible answer to the question:
A1: The size of the balloon becomes bigger.
A2: The size of the balloon becomes smaller.
A3: Directly proportional.
(The rest of the group presents their work.)
D. EXPLAIN (Teach the concept. Should include
interaction between teacher and students.)
To know more about the concept of the
activity that you did a while ago, let’s discuss the
relationship between the volume and
temperature at constant pressure in Charles’s
Law. But before that, let me introduce to you the
man who discovered this law.
Jacques-Alexandre-César Charles (1746-1823) –
was born in France. A French mathematician,
physicist, and inventor who was the first to
ascend in a hydrogen balloon. He developed the
Charles’s
law
concerning
the thermal
expansion of gases.
Charles’s Law: “At constant pressure, the volume
of a fixed amount of gas is directly proportional to
temperature.”
In his experiment, Jacques Charles trapped
a sample of gas in a cylinder with a movable piston
in water bath at different temperatures.
He found out that different gases decreased
their volume by factors 1/273 per ℃ of cooling.
With this rate of reduction, if gas will be cooled up
to -273 ℃, it will have a zero volume. Just like the
activity that you did a while ago. You measure the
circumference of the balloon before soaking it in
a hot water. While soaking it in a hot water, the
balloon starts to grow bigger than its original size.
While soaking it in the cold water, the balloon
starts to grow smaller than its original size.
Therefore, this law simply means that as the
temperature increases, the volume also
increases. And as the temperature decreases, the
volume also decreases at constant pressure.
Charles’s
Law
can
be
mathematically as:
𝑽 ∝ 𝑻 𝒂𝒕 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝑷
expressed
Where: V = Volume and
T = Temperature expressed in Kelvin
Why is there a need to convert
℃ 𝑡𝑜 𝐾? Kelvin is the basic unit for measuring
temperature in the International System (SI). “It
denotes the absolute temperature scale whereby
0K or absolute zero is defined as the temperature
when molecules will have the lowest energy.”
Removing the proportionality symbol (∝)
and using the equality sign (=) the equation will be
as follows:
𝑽
𝑽 = 𝒌𝑻 𝑜𝑟 𝒌 =
𝑻
Thus, in a direct proportion, the quotient of
the variable is constant.
If you are going to consider the initial and
final conditions, you will arrive at the following
equations:
𝑉1
𝑉2
= 𝑘 𝑎𝑛𝑑
=𝑘
𝑇1
𝑇2
Whereas, 𝑽1 is the initial volume and 𝑽𝟐 is the
final volume
𝑻𝟏 is the initial temperature and 𝑻𝟐 is
the final temperature
If the volume-temperature ratios are the
same in the initial and final conditions, then we
will arrive at this equation:
𝑽𝟏 𝑽𝟐
=
𝑻𝟏 𝑻𝟐
To illustrate the mathematical equations,
let’s apply Charles’s Law in solving problems
related to volume-temperature relationship in
gases.
Sample Problem:
An inflated balloon with a volume of 0.75 L
at 30 ℃ was placed inside the freezer where the
temperature is – 10℃. Find out what will happen
to the volume of the balloon if the pressure
remains constant.
Just like we did before, let’s start with the
given variables:
Initial Conditions
Final Conditions
𝑽𝟏 = 𝟎. 𝟕𝟓 𝑳
𝑽𝟐 = ?
𝑻𝟏 = 𝟑𝟎℃ = 𝟑𝟎𝟑𝑲 𝑻𝟐 = −𝟏𝟎℃ = 𝟐𝟔𝟑𝑲
Convert the temperature to Kelvin.
Initial Temperature:
𝐾 = ℃ + 273.15
= 30 + 273.15
𝑲 = 𝟑𝟎𝟑
Final Temperature:
𝐾 = ℃ + 273.15
= −10 + 273.15
𝑲 = 𝟐𝟔𝟑. 𝟏𝟓
Solve for the final volume.
(0.75𝐿)(263.15𝐾)
𝑉1 𝑇2
197.36𝐿
𝑉2 =
=
=
𝑇1
303𝐾
303
= 𝟎. 𝟔𝟓𝑳
You can see from the solution that the
volume decreases because the temperature
decreases too. In this case, the volume between
the gas molecules decreases because the kinetic
energy is also affected by the temperature. Gas
molecules move slowly at low temperature, thus
there is less collision and so it will occupy smaller
space.
Let’s have another example:
3.1 𝑐𝑚3 of a gas have a temperature of 15℃.
What temperature is required to increase the
volume to 3.5 𝑐𝑚3 with pressure remaining
constant?
Who wants to solve example #2? Yes
student 1?
Initial Conditions
Final Conditions
𝟑
𝑽𝟐 = 3.5 𝑐𝑚3
𝑽𝟏 = 𝟑. 𝟏 𝒄𝒎
𝑻𝟏 = 𝟏𝟓℃ = 𝟐𝟖𝟖. 𝟏𝟓𝑲
𝑻𝟐 = ?
Convert the temperature to Kelvin.
Initial Temperature:
𝐾 = ℃ + 273.15
= 15 + 273.15
𝑲 = 𝟐𝟖𝟖. 𝟏𝟓
Solve for the final temperature.
(3.1 𝑐𝑚3 )(288.15𝐾)
𝑉2 𝑇1
893.27
𝑇2 =
=
=
3
𝑉1
3.5 𝑐𝑚
3.5
= 𝟐𝟓𝟓. 𝟐𝟐𝑲
Yes ma’am.
Is his/her answer correct class?
Okay very good! Clap your hands for student None ma’am.
1. Do you have any questions or clarifications?
Then I will be the one who’s going to ask you some
questions.
Values Integration
Why do you think it is important for us to
know the relationship between volume and It is important for us to know the relationship between
temperature at constant pressure especially volume and temperature to understand how hot air
when we are going to ride a hot air balloon?
balloons float. Without knowing how this relationship
works, we wouldn’t be able to discover hot air balloons.
Okay very good! It is important for us to
know the relationship between volume and
temperature at constant pressure to understand
how hot air balloons float. Without further
knowledge with this kind of law, Jacques Charles
wouldn’t be able to discover hot air balloons. And
people wouldn’t get to experience riding it.
Understanding Charles’s law will give us
knowledge on how things with gases expand or
inflate such as balloons, balls, and the new
discovered sky lanterns people use to celebrate
special occasions such as Christmas or New Year.
It also gives us tips on how to restore a
dented/skewed Ping-Pong balls by doing the Yes ma’am.
same activity that you did to the balloon a while
ago. Am I understood?
Then let’s proceed to our last activity.
E. ELABORATE (Students apply the information
learned from the Explain.)
Activity 2: Illustrating Charles’s Law in a Graph
Directions: Group the students into 4. Let them
get a book and refer to page 373. They will be
given a graphic organizer and a graph for their
answers. Write their solutions in converting ℃ to
Kelvin in a one whole sheet of paper. (5 minutes
only)
Problem:
A gas cylinder was measured to have
different volumes at different temperature as
shown in Table 8. Complete the table with the
necessary information. Plot the data from Table 8
in a graph by placing the volume in the y axis and
temperature at Kelvin scale in the x axis.
Trial Volume Temperature Temperature
Reading
(℃)
(K)
(mL)
1
25
2
2
30
57
3
35
102
4
40
152
V. – EVALUATE (How will you know the students have learned the concept?)
(Problem Solving)
Directions: Solve the following problems and show your solution. Convert the final volume to
liters (L). Write your answer in a one whole sheet of paper.
1. A cylinder with a movable piston contains 250 𝑐𝑚3 air at 10℃. If the pressure is kept
constant, at what temperature would you expect the volume to be 150𝑐𝑚3 ?
2. A tank (not rigid) contains 2.3 L of helium gas at 25℃. What will be the volume of the tank
after heating it and its content to 40℃ temperature at constant pressure?
3. At 20℃, the volume of chlorine gas is 15𝑑𝑚3 . Compute for the resulting volume if the
temperature is adjusted to 318K provided that the pressure remains the same.
VI. – EXTEND (Deepen conceptual understanding through use in new context.)
Directions: Answer the question given in a ½ crosswise sheet of paper.
1. Who introduced the Gay-Lussac’s Law? (Give a little background)
2. State the Gay-Lussac’s Law and explain.
3. What is the mathematical expression of the law?