Charles's Law.pptx

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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first published by French natural philosopher Joseph Louis Gay-Lussac in 1802, although he credited the discovery to unpublished work from the 1780s by Jacques Charles, hence the name.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: V 1 / T 1 = V 2 /T 2

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 V 1 = 3.5L T 1 = 250°K V 2 = ? T 2 = 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 : V 1 /T 1 = V 2 /T 2

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 V 1 /T 1 = V 2 /T 2 will become 3.5/250 = V 2 /400` (3.5)(400) = (V 2 )(250) (3.5)(400) = 1,400 (V 2 )(250) = 250 V 2

14. PROBLEM - SOLVING Solution We now have 1400 = 250 V 2 For us to get the V 2 , we will divide both sides to its T 1, which is 250. 1400/250 = 5.6 and 250 V 2 /250 = V 2 We now have 5.6 = V 2 .

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 : V 2 = 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 V 1 = 275 mL T 1 = 25°C V 2 = ? T 2 = 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 V 1 = 275 mL T 1 = 25°C + 273 = 298°K V 2 = ? T 2 = 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 : V 1 /T 1 = V 2 /T 2

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 V 1 /T 1 = V 2 /T 2 will become 275/298 = V 2 /323 (275)(323) = (298)(V 2 ) (275)(323) = 88,825 (298)(V 2 ) = 298 V 2

22. PROBLEM - SOLVING Solution We now have 88,825 = 298 V 2 For us to get the V 2 , we will divide both sides to its T 1, which is 298. 88,835/298 = 298.07 and 298 V 2 /298= V 2 We now have 298.07 = V 2 .

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 : V 2 = 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 V 1 = 300 mL T 1 = 17°C V 2 = ? T 2 = 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 V 1 = 300 mL T 1 = 17°C + 273 = 290°K V 2 = ? T 2 = 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 : V 1 /T 1 = V 2 /T 2

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 V 1 /T 1 = V 2 /T 2 will become 300/290 = V 2 /283 (300)(283) = (290)(V 2 ) (300)(283) = 84,900 (290)(V 2 ) = 290 V 2

29. PROBLEM - SOLVING Solution We now have 84,900 = 290 V 2 For us to get the V 2 , we will divide both sides to its T 1, which is 290. 84,900/290 = 292.76 and 290 V 2 /290 = V 2 We now have 292.76 = V 2 .

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 : V 2 = 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 V 1 = 500 mL T 1 = 80°C V 2 = 0.24L T 2 = ?

33. REMINDER: In Charles’s Law, we can use any unit for our volume as long as V 1 and V 2 has the same unit. If not, the answer will be inaccurate. In this case, we will have to convert and match the units of V 1 and V 2.

34. PROBLEM - SOLVING The unit of the V 1 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 V 1 = 500 mL = 0.5L T 1 = 80°C V 2 = 0.24L T 2 = ?

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 V 1 = 0.5 L T 1 = 80°C = 353°K V 2 = 0.24L T 2 = ?

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 : V 1 /T 1 = V 2 /T 2

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 V 1 /T 1 = V 2 /T 2 will become 0.5/353 = 0.24/T 2 (0.5)(T 2 ) = (0.24)(353) (0.5)(T 2 ) = 0.5 T 2 (0.24)(353) = 84.72

38. PROBLEM - SOLVING Solution We now have 0.5 T 2 = 84.72 For us to get the T 2 , we will divide both sides to its V 1, which is 0.5. 0.5 T 2 /0.5 = T 2 and 84.72/0.5 = 169.44 We now have T 2 = 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 : T 2 = -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

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