Physical Chemistry for Grade 10 4th qt.pptx

It is a ppt presentation good for grade 10 during their 4th grading

1. Physical Chemistry

2. GAS IT UP! Republic Act no. 9729, also known as Chemical Change Act 2009.

3. R.A 9729 It is the policy of the State to afford full protection and the advancement of the right of the people to a healthful ecology in accord with the rhythm and harmony of nature.

4. VALUES Show concern and respect for others and the environment Develop an open mind Use science to improve your way of life

5. Avogadro’s Law Boyle’s Law IDEAL GAS LAW Charles’s Law Gay-Lussac’s Law

6. Comparison of Solid, Liquid, and Gas molecule Property Solid Liquid Gas Shape Has definite shape Takes the shape of its container Takes the shape of its container Volume Has definite volume Has definite volume Takes the volume of the container Arrangement of particles Closely packed Close to one another, but with no regular arrangement Very far apart from one another Compressibility Incompressible Compressible Very compressible Density High Slight dense Low Movement of particles Slow Fast Very fast Forces of attraction Strong Moderate Almost neglible

7. The Kinetic Molecular Theory The word kinetic refers to the motion of the molecules as they constantly move in all direction.

8. 4 Principal assumptions of the kinetic molecular theory Gases consist of very tiny molecule. The distance between these molecules is significantly larger than the size of the molecules themselves. Therefore, the volume taken up by the gas molecules is assumed to be negligible relative to the volume of their container.

9. There are no force of attraction between and among the gas molecules.

10. Gas molecules are in a constant, random, and linear motion. They collide frequently with one another and with the walls of their container. These collisions are perfectly elastic- that is, no energy is lost nor gained when gas molecules collide.

11. The average kinetic energy of gas molecules is the same for all gases at the same temperature, and its value is directly proportional to the Kelvin temperature.

12. Basic Assumption of the KMT of Gases Assumption Concept Particle Volume A gas consist of a large collection of individual particles. The volume of an individual particles are extremely small, as compared to the volume of its container. Particle Motion Gas particles are in constant, random, straight-line motion, except when they collide with the walls of the container or with one another. Particle Collision Collision of gas particles are elastic. The colliding gas particles or molecules exchange energy, but they do not lose energy through friction. This means that the total kinetic energy is constant. Between collisions, the gas particles or molecules do not influence each other.

13. Physical Qualities by Gas Molecules VOLUME Space occupied by a sample matter. Units: L, mL, m³, cm³

14. TEMPERATURE the measure of hotness or coldness expressed in terms of any of several scales, including Fahrenheit and Celsius.

15. AMOUNT OF SUBSTANCE This physical quality is measured in moles (mol).

16. PRESSURE The force exerted per unit area. Units: atm, Pa, kPa

17. BOYLE’S LAW

18. This  e mpirical  relation, formulated by the physicist  Robert Boyle  in 1662

19. A gas l aw that states that a gas's pressure and volume are inversely proportional

20. At a cylinder of a scuba tank has a volume of 1500 mL at a pressure of 0.75 atm. What volume of air is needed to fill a cylinder to a pressure of 100 atm at constant temperature? Given: V 1 = 1,500 mL P 1 = 0.75 atm V 2 = ? P 2 = 100 atm P 1 V 1 =P 2 V 2 0.75 atm(1,500 mL)= 100 atm(V 2 ) 0.75 atm(1,500 mL) = 100 atm(V 2 ) 100 atm 100 atm V 2 = 1,125 mL = 11.25 mL 100

21. At a cylinder of a scuba tank has a volume of 1500 mL at a pressure of 0.75 atm. What volume of air is needed to fill a cylinder to a pressure of 100 atm at constant temperature? Given: V 1 = 1,500 mL P 1 = 0.75 atm V 2 = ? P 2 = 100 atm P 1 V 1 =P 2 V 2 0.75 atm(1,500 mL)= 100 atm(V 2 ) 0.75 atm(1,500 mL) = 100 atm(V 2 ) 100 atm 100 atm V 2 = 1,125 mL = 11.25 mL 100 STEP 1 Identify the given values

22. At a cylinder of a scuba tank has a volume of 1500 mL at a pressure of 0.75 atm. What volume of air is needed to fill a cylinder to a pressure of 100 atm at constant temperature? Given: V 1 = 1,500 mL P 1 = 0.75 atm V 2 = ? P 2 = 100 atm P 1 V 1 =P 2 V 2 0.75 atm(1,500 mL)= 100 atm(V 2 ) 0.75 atm(1,500 mL) = 100 atm(V 2 ) 100 atm 100 atm V 2 = 1,125 mL = 11.25 mL 100 STEP 2 Identify the FORMULA that you’ll be needing

23. At a cylinder of a scuba tank has a volume of 1500 mL at a pressure of 0.75 atm. What volume of air is needed to fill a cylinder to a pressure of 100 atm at constant temperature? Given: V 1 = 1,500 mL P 1 = 0.75 atm V 2 = ? P 2 = 100 atm P 1 V 1 =P 2 V 2 0.75 atm(1,500 mL)= 100 atm(V 2 ) 0.75 atm(1,500 mL) = 100 atm(V 2 ) 100 atm 100 atm V 2 = 1,125 mL = 11.25 mL 100 STEP 3 Substitute the values to the formula

24. At a cylinder of a scuba tank has a volume of 1500 mL at a pressure of 0.75 atm. What volume of air is needed to fill a cylinder to a pressure of 100 atm at constant temperature? Given: V 1 = 1,500 mL P 1 = 0.75 atm V 2 = ? P 2 = 100 atm P 1 V 1 =P 2 V 2 0.75 atm(1,500 mL)= 100 atm(V 2 ) 0.75 atm(1,500 mL) = 100 atm(V 2 ) 100 atm 100 atm V 2 = 1,125 mL = 11.25 mL 100 Cross-out to get what you’re looking for Cross out the same unit

25. At a cylinder of a scuba tank has a volume of 1500 mL at a pressure of 0.75 atm. What volume of air is needed to fill a cylinder to a pressure of 100 atm at constant temperature? Given: V 1 = 1,500 mL P 1 = 0.75 atm V 2 = ? P 2 = 100 atm P 1 V 1 =P 2 V 2 0.75 atm(1,500 mL)= 100 atm(V 2 ) 0.75 atm(1,500 mL) = 100 atm(V 2 ) 100 atm 100 atm V 2 = 1,125 mL = 11.25 mL 100 Simplify

26. At a cylinder of a scuba tank has a volume of 1500 mL at a pressure of 0.75 atm. What volume of air is needed to fill a cylinder to a pressure of 100 atm at constant temperature? Given: V 1 = 1,500 mL P 1 = 0.75 atm V 2 = ? P 2 = 100 atm P 1 V 1 =P 2 V 2 0.75 atm(1,500 mL)= 100 atm(V 2 ) 0.75 atm(1,500 mL) = 100 atm(V 2 ) 100 atm 100 atm V 2 = 1,125 mL = 11.25 mL 100 You got the answer! V 2 = 11.25 mL

27. TRY THIS! A gas, which is occupying a volume of 700 mL at a pressure of 0.85 atm, is allowed to expand (at constant temperature) until the pressure becomes 0.35 atm. What will be the final volume of the gas? A sample of air occupies 3.0 L when the pressure is 1.3 atm. If the temperature is kept constant, what pressure is needed to reach the final volume of 1.25 L?

28. Charles’s Law

29. This  empirical  relation was first suggested by the French physicist  J acque Charles  about 1787 and was later placed on a sound empirical footing by the chemist  Joseph-Louis Gay-Lussac .

30. A statement that the volume occupied by a fixed amount of gas is directly proportional to its absolute temperature, if the pressure remains constant.

31. The gas in a balloon has a volume of 15 L at 50° C. what will be the temperature if the volume is increased to 30 L? Given: V 1 = 15 L T 1 = 50°C +273.15= 323.15 K V 2 = 30 L V 1 = V 2 15 L(T 2 ) = 30 L (323.15K) T 2 T 2 15 L 15 L 15 L = 30 L T 2 = 9,694.5 323.15K T 2 15 15 L = 30 L T 2 = 646.3 323.15K T 2 15 L(T 2 ) =30 L (323.15K)

32. The gas in a balloon has a volume of 15 L at 50° C. what will be the temperature if the volume is increased to 30 L? Given: V 1 = 15 L T 1 = 50°C V 2 = 30 L V 1 = T 1 15 L(T 2 ) = 30 L (50 °C) V 2 T 2 15 L 15 L 15 L = 30 L T 2 = 1,500 50°C T 2 15 15 L = 30 L T 2 = 100°C 50°C T 2 15 L(T 2 ) =30 L (50 °C) Identify the given values

33. The gas in a balloon has a volume of 15 L at 50° C. what will be the temperature if the volume is increased to 30 L? Given: V 1 = 15 L T 1 = 50°C V 2 = 30 L V 1 = T 1 15 L(T 2 ) = 30 L (50 °C) V 2 T 2 15 L 15 L 15 L = 30 L T 2 = 1,500 50°C T 2 15 15 L = 30 L T 2 = 100°C 50°C T 2 15 L(T 2 ) =30 L (50 °C) Identify the formula to be used

34. The gas in a balloon has a volume of 15 L at 50° C. what will be the temperature if the volume is increased to 30 L? Given: V 1 = 15 L T 1 = 50°C V 2 = 30 L V 1 = T 1 15 L(T 2 ) = 30 L (50 °C) V 2 T 2 15 L 15 L 15 L = 30 L T 2 = 1,500 50°C T 2 15 15 L = 30 L T 2 = 100°C 50°C T 2 15 L(T 2 ) =30 L (50 °C) Substitute the value to the formula

35. The gas in a balloon has a volume of 15 L at 50° C. what will be the temperature if the volume is increased to 30 L? Given: V 1 = 15 L T 1 = 50°C V 2 = 30 L V 1 = T 1 15 L(T 2 ) = 30 L (50 °C) V 2 T 2 15 L 15 L 15 L = 30 L T 2 = 1,500 50°C T 2 15 15 L = 30 L T 2 = 100°C 50°C T 2 15 L(T 2 ) =30 L (50 °C) Cross multiply

36. The gas in a balloon has a volume of 15 L at 50° C. what will be the temperature if the volume is increased to 30 L? Given: V 1 = 15 L T 1 = 50°C V 2 = 30 L V 1 = T 1 15 L(T 2 ) = 30 L (50 °C) V 2 T 2 15 L 15 L 15 L = 30 L T 2 = 1,500 50°C T 2 15 15 L = 30 L T 2 = 100°C 50°C T 2 15 L(T 2 ) =30 L (50 °C) Divide and cancel out the same unit.

37. The gas in a balloon has a volume of 15 L at 50° C. what will be the temperature if the volume is increased to 30 L? Given: V 1 = 15 L T 1 = 50°C V 2 = 30 L V 1 = T 1 15 L(T 2 ) = 30 L (50 °C) V 2 T 2 15 L 15 L 15 L = 30 L T 2 = 1,500 50°C T 2 15 15 L = 30 L T 2 = 100°C 50°C T 2 15 L(T 2 ) =30 L (50 °C) Simplify

38. The gas in a balloon has a volume of 15 L at 50° C. what will be the temperature if the volume is increased to 30 L? Given: V 1 = 15 L T 1 = 50°C V 2 = 30 L V 1 = T 1 15 L(T 2 ) = 30 L (50 °C) V 2 T 2 15 L 15 L 15 L = 30 L T 2 = 1,500 50°C T 2 15 15 L = 30 L T 2 = 100°C 50°C T 2 15 L(T 2 ) =30 L (50 °C) Answer: T 2 is 100 °C

39. TRY THIS! A 35 L volume of carbon dioxide gas is heated from 35°C to 75°C, at constant pressure. What is the final volume of the gas? Under constant pressure, a sample of nitrogen gas, initially at 77°C at 7.7 L, is cooled until its final volume is 2.5 L. What is the final temperature of the gas?

40. Gay-Lussac’s Law

41. Joseph-Louis Gay-Lussac  (born December 6, 1778, Saint-Léonard-de- Noblat , France—died May 9, 1850, Paris) was a French chemist and physicist who pioneered investigations into the behavior of gases, established new techniques for analysis, and made notable advances in applied  chemistry .

42. A gas law which states that the pressure exerted by a gas (of a given mass and kept at a constant volume) varies directly with the absolute temperature of the gas.

43. Determine the pressure change when a constant volume of gas at 1.00 atm is heated from 20.0 °C to 30.0 °C. Given: P 1 = 1.00 atm T 1 = 20 °C+273.15=293.15k P 2 = T 2 = 30 °C+273.15=303.15k P 1 = P 2 P 2 (293.15k) = 1.00 atm(303.15k) T 1 T 2 293.15k 293.15 k 1.00 atm = P 2 P 2 = 303.15 293.15k 303.15k 293.15 1.00 atm = P 2 P 2 = 1.03 atm 293.15k 303.15k P 2 (293.15k)= 1.00 atm(303.15k)

44. Determine the pressure change when a constant volume of gas at 1.00 atm is heated from 20.0 °C to 30.0 °C. Given: P 1 = 1.00 atm T 1 = 20 °C P 2 = T 2 = 30 °C P 1 = P 2 P 2 (20.0°C) = 1.00 atm(30.0°C) T 1 T 2 20.0 °C 20.0 °C 1.00 atm = 20.0°C P 2 = 30 P 2 30.0°C 20 1.00 atm = 20.0°C P 2 = 1.5 atm P 2 30.0°C P 2 (20.0°C)= 1.00 atm(30.0°C) Identify the given values

45. Determine the pressure change when a constant volume of gas at 1.00 atm is heated from 20.0 °C to 30.0 °C. Given: P 1 = 1.00 atm T 1 = 20 °C P 2 = T 2 = 30 °C P 1 = P 2 P 2 (20.0°C) = 1.00 atm(30.0°C) T 1 T 2 20.0 °C 20.0 °C 1.00 atm = 20.0°C P 2 = 30 P 2 30.0°C 20 1.00 atm = 20.0°C P 2 = 1.5 atm P 2 30.0°C P 2 (20.0°C)= 1.00 atm(30.0°C) Identify the formula to be used

46. Determine the pressure change when a constant volume of gas at 1.00 atm is heated from 20.0 °C to 30.0 °C. Given: P 1 = 1.00 atm T 1 = 20 °C P 2 = T 2 = 30 °C P 1 = P 2 P 2 (20.0°C) = 1.00 atm(30.0°C) T 1 T 2 20.0 °C 20.0 °C 1.00 atm = 20.0°C P 2 = 30 P 2 30.0°C 20 1.00 atm = 20.0°C P 2 = 1.5 atm P 2 30.0°C P 2 (20.0°C)= 1.00 atm(30.0°C) Substitute the value to the formula

47. Determine the pressure change when a constant volume of gas at 1.00 atm is heated from 20.0 °C to 30.0 °C. Given: P 1 = 1.00 atm T 1 = 20 °C P 2 = T 2 = 30 °C P 1 = P 2 P 2 (20.0°C) = 1.00 atm(30.0°C) T 1 T 2 20.0 °C 20.0 °C 1.00 atm = 20.0°C P 2 = 30 P 2 30.0°C 20 1.00 atm = 20.0°C P 2 = 1.5 atm P 2 30.0°C P 2 (20.0°C)= 1.00 atm(30.0°C) Cross multiply

48. Determine the pressure change when a constant volume of gas at 1.00 atm is heated from 20.0 °C to 30.0 °C. Given: P 1 = 1.00 atm T 1 = 20 °C P 2 = T 2 = 30 °C P 1 = P 2 P 2 (20.0°C) = 1.00 atm(30.0°C) T 1 T 2 20.0 °C 20.0 °C 1.00 atm = 20.0°C P 2 = 30 P 2 30.0°C 20 1.00 atm = 20.0°C P 2 = 1.5 atm P 2 30.0°C P 2 (20.0°C)= 1.00 atm(30.0°C) Divide and cancel out the same unit.

49. Determine the pressure change when a constant volume of gas at 1.00 atm is heated from 20.0 °C to 30.0 °C. Given: P 1 = 1.00 atm T 1 = 20 °C P 2 = T 2 = 30 °C P 1 = P 2 P 2 (20.0°C) = 1.00 atm(30.0°C) T 1 T 2 20.0 °C 20.0 °C 1.00 atm = 20.0°C P 2 = 30 P 2 30.0°C 20 1.00 atm = 20.0°C P 2 = 1.5 atm P 2 30.0°C P 2 (20.0°C)= 1.00 atm(30.0°C) Simplify

50. Determine the pressure change when a constant volume of gas at 1.00 atm is heated from 20.0 °C to 30.0 °C. Given: P 1 = 1.00 atm T 1 = 20 °C P 2 = T 2 = 30 °C P 1 = P 2 P 2 (20.0°C) = 1.00 atm(30.0°C) T 1 T 2 20.0 °C 20.0 °C 1.00 atm = 20.0°C P 2 = 30 P 2 30.0°C 20 1.00 atm = 20.0°C P 2 = 1.5 atm P 2 30.0°C P 2 (20.0°C)= 1.00 atm(30.0°C) Answer: P 2 = 1.5 atm

51. TRY THIS! If a gas in a closed container is pressurized from 15.0 atmospheres to 16.0 atmospheres and its original temperature was 25.0 °C, what would the final temperature of the gas be in degrees Celsius? Determine the pressure change when a constant volume of gas at 3.00 atm is heated from 10.0 °C to 60.0 °C.