2. Objectives  The Kinetic Molecular Theory of Gases  Quantities That Describe a Gas  Factors that Affect Gas Pressure  The Gas Laws

4. Kinetic Molecular Theory of Gases 1. A gas consists of individual particles in constant and random motion. 2. The volume of the individual gas molecules is small when compared to the distance between gas molecules. 3. Collisions between molecules are perfectly elastic. 4. The individual particles do not attract or repel one another in any way. 5. The pressure of the gas is due entirely to the force of the collisions of the gas particles with the walls of the container.

5. Quantities That Affect a Gas  There are four quantities needed to describe gases. 1. Pressure (P) 2. Amount of Gas (in moles) 3. Volume (V) 4. Temperature (T)

6. Pressure  Is defined as the force exerted on an object divided by the area over which it is exerted.  The units of pressure are: mm Hg (millimeters of mercury) atm (standard atmospheres) kPa (kilopascals) 1 atm = 760 mm Hg = 101.3kPa

8. Factors That Affect Gas Pressure 1. Amount of Gas The amount of gas and gas pressure are proportional. If you increase the amount of gas, you increase the gas pressure When you add a gas you increase the amount of gas particles which increases the collision rate between gas particles. Is measured in moles (n).

10. Factors That Affect Gas Pressure 2. Volume Volume and gas pressure are inversely proportional. By decreasing the volume, gas pressure is increased. By decreasing the volume, you decrease the amount of space gas particles have to move. In turn, you will increase the amount of the collision of gas particles. The more collisions you have, the higher the gas pressure. Is measured in L, mL

12. Factors That Affect Gas Pressure 3. Temperature Temperature and gas pressure are proportional. An increase in temperature, will result in an increase in gas pressure. As gas particles absorb the thermal energy from the temperature increase, they become faster moving and impact the walls of their containers and other gas particles. The result of this is an increase in gas pressure. is measured in Kelvin (K). K = °C + 273

14. Gas Law Units  In calculating gas law problems it is important to know which units are used to describe what quantities. Pressure = atm, kPa, mmHg, torr, Temperature = Kelvin Volume = L n= moles

15. STP (Standard Temperature & Pressure)  When comparing volumes of gases, scientist use a common reference point called standard temperature and pressure or STP.  The values for STP are Temp.= 273 K or 0  C Pressure= 101.3kPa 1 atm 760 mm Hg 760 torr

16. Combined Gas Law  Formula: P 1 x V 1 = P 2 x V 2 T 1 T 2  A single expression that shows the relationship between temperature (T), pressure (P), and volume (V).  It is a useful way to recall Boyle’s, Charles’ and Gay-Lussac’s gas laws.

17. To Solve Gas Law Problems 1. Determine the formula and the gas law necessary to solve the problems. 2. Determine all the given values and which variable they describe. 3. Plug known values into the formula and solve for the missing variable. 4. Be sure you use the correct significant figures, unit, and label.

18. Boyle’s Law Robert Boyle (1627-1691) did a series of experiments to determine the relationship between volume and pressure.  States that for a given mass held at constant temperature, the volume of the gas varies inversely with pressure.  In other words if the pressure is increased two times, the volume decreases two times.  Use the combined gas law formula to determine the formula for Boyle’s Law Formula: P 1 x V 1 = P 2 x V 2

19. http://www.asc-csa.gc.ca/images/neemo_graph_boyles_law.jpg

20. http://users.humboldt.edu/rpaselk/ChemSu pp/Images/Boyles_Law_animated.gif

21. Boyle’s Law  For Example, if a volume of 1.0 L (V 1 ) is at a pressure of 100 kPa (P 1 ). If you increase the volume to 2.0 L (V 2 ), the pressure decrease to 50kPa (P 2 ). 100kPa x 1.0 L = 50kPa x 2.0L Formula: P 1 x V 1 = P 2 x V 2

22. Sample Problem  A high-altitude balloon contains 30.0 L of helium gas at 103kPa. What is the volume when the balloon rises to an altitude where the pressure is only 25.0kPa (assume the temperature remains constant)? V 1 = 30.0L V 2 = ? P 1 = 103kPa P 2 = 25.0 kPa V 2 = (103kPa)(30.0L) (25.0kPa) Formula: P 1 x V 1 = P 2 x V 2 V 2 = 124 L

23. Charles Law  J.A.C. Charles observed the effect of temperature on the volume of gas in 1787.  Charles’ Law states that the volume of a fixed mass of gas is directly proportional to its Kelvin temperature if the pressure is kept constant.  So if you increase the volume of a gas the temperature increases as well.  Determine the formula using the combined gas law. Formula: V 1 = V 2 T 1 T 2

24. Charles’ Law Graph

25. Charles’ Law Simulation

26. Charles Law  For example, a 1.0L (V 1 ) sample of gas is at a temperature of 300K(T 1 ). When the temperature is increased to 600K(T 2 ), the volume increases to 2.0L(V 2 ). 1.0L = 2.0L 300K 600K Formula: V 1 = V 2 T 1 T 2

27. Sample Problem  A balloon inflated in a room at 24°C has a volume of 4.00L. The balloon is then heated to a temperature of 58°C. What is the new volume if the pressure remains constant? Formula: V 1 = V 2 T 1 T 2 V 1 = 4.00LV 2 = ? T 1 =24°C+273T 2 = 58°C + 273 = 297K = 331K

28. Charles Law  4.0 L = V 2 297K 331K  (4.0L)(331K) = (297K)(V 2 )  (4.0L)(331K) = (V 2 ) (297K)  4.5L= V 2 Formula: V 1 = V 2 T 1 T 2

29. Sample Problem AA gas has a pressure of 6.58kPa at 538K. What will be the pressure at 211 K if the volume does not change? PP 1 = 6.58 kPaP 2 = ? T 1 = 538KT 2 = 211 K Formula: P 1 = P 2 T 1 T 2

30. Sample Problem  6.58kPa = P 2 538K 211K  (6.58kPa)(211K)=(538K)(P 2 )  2.58kPa = P 2

31. Gay-Lussac’s Law  Is a form of Charles’ Law that describes the relationship between pressure and temperature.  States the pressure of a gas is directly proportional to the Kelvin temperature if the volume remains constant.  In other words, if you increase the temperature, you increase the pressure.  Determine the formula using the combined gas law. Formula: P 1 = P 2 T 1 T 2

32. Sample Problem AA gas has a pressure of 6.58kPa at 538K. What will be the pressure at 211 K if the volume does not change? PP 1 = 6.58 kPaP 2 = ? T 1 = 538KT 2 = 211 K

33. Sample Problem  6.58kPa = P 2 538K 211K  (6.58kPa)(211K)=(538K)(P 2 )  2.58kPa = P 2

34. Combined Gas Law Continued  There are some gas law problems that will use the combined gas law formula with all of the variables.

35. Sample Problem  The volume of a gas filled balloon is 30.0L at 40°C and 153kPa pressure. What volume will the balloon have at standard temperature and pressure (STP)?  NOTE: STP = 273K and 101.3kPa  V 1 = 30LV 2 = ? T 1 = 313KT 2 = 273K P 1 = 153kPaP 2 = 101.3kPa Formula: P 1 x V 1 = P 2 x V 2 T 1 T 2

36. Sample Problem  (153kPa)(30.0L) = (101.3kPa)(V 2 ) 313K 273K  (153kPa)(30.0L)(273K)= (101.3kPa)(V 2 )(313K)  (153kPa)(30.0L)(273K) = V 2 (101.3kPa)(313K)  40 L = V 2 Formula: P 1 x V 1 = P 2 x V 2 T 1 T 2

39. Kinetic Molecular Theory of Gases Revisited 1. In the Kinetic Molecular Theory of Gases it is stated that an ideal gas has no volume or IMFs (intermolecular forces) 2. Boyle’s, Charles’, and Gay-Lussac’s gas laws are based upon the behavior of an ideal gas and not a real gas.

40. Ideal Gases versus Real Gases  Based upon the reading selection provided by your instructor, answer the following question: How does the behavior of an ideal gas differ from that of a real gas? PLEASE NOTE THIS IS NOT A ONE PART ANSWER. BE PREPARED TO SHARE YOUR FINDINGS.

41. Ideal Gas Law The ideal gas equation explains how gases should behave under ideal conditions according to the Kinetic Molecular Theory of Gases. PV= nRT  P = pressure  V = volume in liters  n = number of mol  R = ideal gas constant (depends on the unit of pressure) (8.314 L x kPa /mol-K or.0821 L x atm /mol-K)  T = temperature in Kelvin

42. Sample Problem  You fill a rigid steel cylinder that has a volume of 20.0L with nitrogen gas (N 2 )(g) to a final pressure of 2.00 x 10 4 kPa at 28°C. How many moles of (N 2 )(g) does the cylinder contain? P= 2.00 x 10 4 kPa V= 20.0L R= 8.31 (L x kPa) T= 28°C + 273= 301K (K x mol) Formula P x V = nRT

43. Sample Problem 2.00 x 10 4 kPa x 20.0L = (n) (8.31)(301K) 2.00 x 10 4 kPa x 20.0L = (n) (8.31)(301K) 1.6 x 10 2 moles = n

44. Ideal Gas Law Video

45. Dalton’s Law  If gases behave according the KMT there should be no difference in the P-V-T relationships whether the gas molecules are the same or different.  In a mixture of gases, each gas should exert a pressure that is independent of the other gases. These pressures are referred to as partial pressures.  Dalton’s Law is the sum of the partial pressures exerted by each gas in a mixture of gases. Formula: P total = P 1 + P 2 + P 3 + …..

46. Dalton’s Law

47.  Air contains oxygen, nitrogen, carbon dioxide, and trace amounts of other gases. What is the partial pressure of oxygen at 101.3 kPa of total pressure if the partial pressure of nitrogen, carbon dioxide, and other gases are 79.10kPa, 0.040kPa, and 0.94kPa, respectively. Formula: P total = P 1 + P 2 + P 3 + …..

48. Dalton’s Law  P N 2 = 79.10 kPaP CO 2 = 0.040 kPa P others = 0.94 kPaP total = 101.3 kPa P O 2 = ?  101.3kPa = (79.10 + 0.040 + 0.94)kPa + P O 2  101.3 -(79.10 + 0.040 + 0.94) = P O 2  21.2 kPa = P O 2 Formula: P total = P 1 + P 2 + P 3 + …..