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1. CHARLES’S : LAW
SCIENCE GRADE 10 : 4TH QUARTER

2. CHARLES’S LAW
• It was first published by
French natural philosopher
Joseph Louis Gay-Lussac in
1802, although he credited
the discovery to unpublished
work from the 1780s by
Jacques Charles, hence the
name.

3. CHARLES’S LAW
• It is an experimental gas law which
describes how gases tend to expand
when heated.
• It is also known as the “law of volumes.”.

4. MODERN STATEMENT
of Charles’s Law
“When the pressure on a sample of a
dry gas is held at constant, the Kelvin
temperature and the volume will be
directly related.”

5. BASIC IDEA
of Charles’s Law
When the temperature of the
gas rises, the volume in an
object will increase.

6. Basic Idea of Charles’s Law
Example:
I have here a balloon, which is:
V = 2L and T = 300°K
But what do you think will
happen to the volume if the
temperature rose at 600°K?

7. Basic Idea of Charles’s Law
Example:
If the temperature
of the gas rose, the
volume of the
balloon will
increase.

8. PROBLEM - SOLVING
in Charles’s Law

9. PROBLEM - SOLVING
in Charles’s Law
FORMULA:
V1 / T1 = V2/T2

10. PROBLEM - SOLVING
1. A 3.5 liter flexible container holds a gas at 250°K. What
will the new volume be, if the temperature is increased to
400°K?
we will use this format to solve the word problem:
• Given
• Equation
• Solution
• Final Answer

11. PROBLEM - SOLVING
1. A 3.5 liter flexible container holds a gas at 250°K. What
will the new volume be, if the temperature is increased to
400°K?
• Given
V1 = 3.5L
T1 = 250°K
V2 = ?
T2 = 400°K

12. PROBLEM - SOLVING
1. A 3.5 liter flexible container holds a gas at 250°K. What
will the new volume be, if the temperature is increased to
400°K?
• Equation :
V1 /T1 = V2 /T2

13. PROBLEM - SOLVING
1. A 3.5 liter flexible container holds a gas at 250°K. What
will the new volume be, if the temperature is increased to
400°K?
• Solution
• CROSS MULTIPLY V1 /T1 = V2 /T2
will become 3.5/250 = V2/400`
• (3.5)(400) = (V2)(250)
• (3.5)(400) = 1,400
• (V2)(250) = 250 V2

14. PROBLEM - SOLVING
• Solution
• We now have
1400 = 250 V2
• For us to get the V2, we will divide both sides to its T1,
which is 250.
• 1400/250 = 5.6 and 250 V2/250 = V2
• We now have 5.6 = V2.

15. PROBLEM - SOLVING
1. A 3.5 liter flexible container holds a gas at 250°K. What
will the new volume be, if the temperature is increased to
400°K?
Final Answer :
V2= 5.6 L

16. DID YOU KNOW?
That in Charles’s Law, you can only use
Kelvin for temperature. Celsius and
Fahrenheit will be inacurrate when solving
problems. So when the given temperature is
at °C or °F, you will need to convert it to °K.

17. PROBLEM - SOLVING
2. A 275 mL balloon is filled with air at 250°C. If the
temperature is increased to 50°C, what is the new volume
of the balloon?
we will use this format to solve the word problem:
• Given
• Equation
• Solution
• Final Answer

18. PROBLEM - SOLVING
2. A 275 mL balloon is filled with air at 25°C. If the
temperature is increased to 50°C, what is the new volume
of the balloon?
• Given
V1 = 275 mL
T1 = 25°C
V2 = ?
T2 = 50°C

19. PROBLEM - SOLVING
But, the V and T will only be proportional if the
temperature is in Kelvin, so we will convert it using the
formula: TK = TC +273
• Given
V1 = 275 mL
T1 = 25°C + 273 = 298°K
V2 = ?
T2 = 50°C + 273 = 323°K

20. PROBLEM - SOLVING
2. A 275 mL balloon is filled with air at 25°C. If the
temperature is increased to 50°C, what is the new volume
of the balloon?
• Equation :
V1 /T1 = V2 /T2

21. PROBLEM - SOLVING
2. A 275 mL balloon is filled with air at 250°C. If the
temperature is increased to 50°C, what is the new volume
of the balloon?
• Solution
• CROSS MULTIPLY V1 /T1 = V2 /T2
will become 275/298 = V2/323
• (275)(323) = (298)(V2)
• (275)(323) = 88,825
• (298)(V2) = 298 V2

22. PROBLEM - SOLVING
• Solution
• We now have 88,825 = 298 V2
• For us to get the V2, we will divide both sides to its T1,
which is 298.
• 88,835/298 = 298.07 and 298 V2/298= V2
• We now have 298.07 = V2.

23. PROBLEM - SOLVING
2. A 275 mL balloon is filled with air at 250°C. If the
temperature is increased to 50°C, what is the new volume
of the balloon?
Final Answer :
V2= 298.07 mL

24. PROBLEM - SOLVING
3. Given 300.0 mL of a gas at 17.0 °C. What is its volume
at 10.0 °C?
we will use this format to solve the word problem:
• Given
• Equation
• Solution
• Final Answer

25. PROBLEM - SOLVING
3. Given 300.0 mL of a gas at 17.0 °C. What is its volume
at 10.0 °C?
• Given
V1 = 300 mL
T1 = 17°C
V2 = ?
T2 = 10°C

26. PROBLEM - SOLVING
But, the V and T will only be proportional if the
temperature is in Kelvin, so we will convert it using the
formula: TK = TC +273
• Given
V1 = 300 mL
T1 = 17°C + 273 = 290°K
V2 = ?
T2 = 10°C + 273 = 283°K

27. PROBLEM - SOLVING
3. Given 300.0 mL of a gas at 17.0 °C. What is its volume
at 10.0 °C?
• Equation :
V1 /T1 = V2 /T2

28. PROBLEM - SOLVING
3. Given 300.0 mL of a gas at 17.0 °C. What is its volume
at 10.0 °C?
• Solution
• CROSS MULTIPLY V1 /T1 = V2 /T2
will become 300/290 = V2/283
• (300)(283) = (290)(V2)
• (300)(283) = 84,900
• (290)(V2) = 290 V2

29. PROBLEM - SOLVING
• Solution
• We now have 84,900 = 290 V2
• For us to get the V2, we will divide both sides to its T1,
which is 290.
• 84,900/290 = 292.76 and 290 V2/290 = V2
• We now have 292.76 = V2.

30. PROBLEM - SOLVING
3. Given 300.0 mL of a gas at 17.0 °C. What is its volume
at 10.0 °C?
Final Answer :
V2= 292.76 mL

31. PROBLEM - SOLVING
4. The volume of a 500 mL container is decreased to 0.24
L. What is the new temperature in celsius if the original
temperature is at 80°C?
we will use this format to solve the word problem:
• Given
• Equation
• Solution
• Final Answer

32. PROBLEM - SOLVING
4. The volume of a 500 mL container is decreased to 0.24
L. What is the new temperature in celsius if the original
temperature is at 80°C?
• Given
V1 = 500 mL
T1 = 80°C
V2 = 0.24L
T2 = ?

33. REMINDER:
In Charles’s Law, we can use any unit for
our volume as long as V1 and V2 has the
same unit. If not, the answer will be
inaccurate. In this case, we will have to
convert and match the units of V1 and V2.

34. PROBLEM - SOLVING
The unit of the V1 and V2 were different. We will convert
V1 from mL to L.
DIVIDE the given mL by 1,000 to convert it to L.
• Given
V1 = 500 mL = 0.5L
T1 = 80°C
V2 = 0.24L
T2 = ?

35. PROBLEM - SOLVING
But, the V and T will only be proportional if the
temperature is in Kelvin, so we will convert it using the
formula: TK = TC +273
• Given
V1 = 0.5 L
T1 = 80°C = 353°K
V2 = 0.24L
T2 = ?

36. PROBLEM - SOLVING
4. The volume of a 500 mL container is decreased to 0.24
L. What is the new temperature in celsius if the original
temperature is at 80°C?
• Equation :
V1 /T1 = V2 /T2

37. PROBLEM - SOLVING
4. The volume of a 500 mL container is decreased to 0.24
L. What is the new temperature in celsius if the original
temperature is at 80°C?
• Solution
• CROSS MULTIPLY V1 /T1 = V2 /T2
will become 0.5/353 = 0.24/T2
• (0.5)(T2) = (0.24)(353)
• (0.5)(T2) = 0.5 T2
• (0.24)(353) = 84.72

38. PROBLEM - SOLVING
• Solution
• We now have 0.5 T2 = 84.72
• For us to get the T2, we will divide both sides to its V1,
which is 0.5.
• 0.5 T2/0.5 = T2 and 84.72/0.5 = 169.44
• We now have T2 = 169.44
BUT THIS IS NOT THE FINAL ANSWER YET.

39. PROBLEM - SOLVING
4. The volume of a 500 mL container is decreased to 0.24
L. What is the new temperature in celsius if the original
temperature is at 80°C?
• We are looking for the CELSIUS, not Kelvin. So our
final answer should be in CELSIUS.
°K - 273 = °C
• 169.44 - 273 = -103.56

40. PROBLEM - SOLVING
4. The volume of a 500 mL container is decreased to 0.24
L. What is the new temperature in celsius if the original
temperature is at 80°C?
Final Answer :
T2= -103°C

41. REAL LIFE : APPLICATION
of Charles’s Law

42. 1. Hot Air Balloon
- when the hot air
balloon is heated,
it will float in the air.
Rea-life Application
of Charles’s Law

43. Rea-life Application
of Charles’s Law
2. A Helium Balloon in a Cold Night /
Day
- when you bring your helium
balloon outside in a cold night
or day, it will crumble even a
bit. But when you bring it to a
warm room, it will go back to its
original shape.

44. • 3. Bakery
• - When a pizza/bread
dough, which has yeast
on it, is placed in a
warm place for few
hours, it will double in
size.
Rea-life Application
of Charles’s Law

45. 4. Deodorant Bottles
- When it is exposed to
heat, the gas molecules
inside it expands, which
may cause the deodorant
bottle to explode.
Rea-life Application
of Charles’s Law

46. 5. Pingpong Ball
- When the pingpong
ball was dented, you will
just have to float it on hot
water for it to regain its
shape.
Rea-life Application
of Charles’s Law

47. 6. Tyres during winter or
rainy season.
- In a cold weather, the tyres
of your car requires a pressure
check from time to time because
the pressure inside it is decreasing
when the temperature is lower
than usual.
Rea-life Application
of Charles’s Law

48. THANK YOU FOR LISTENING!