1. BEHAVIOR OF GASES Chapter 12

2. THREE STATES OF MATTER Vocabulary Review Boiling point- Temperature at which the vapor pressure of a liquid is just equal to the external pressure on the liquid. Vaporization – The conversion of a liquid to a gas or a vapor. Condensation – The conversion of a gas to a liquid

3. KINETIC MOLECULAR THEORY (KMT) Theory used to explain gas laws. KMT assumptions are Gases consist of molecules in constant, random motion.Gases consist of molecules in constant, random motion. Pressure arises from collisions with container walls.Pressure arises from collisions with container walls. No attractive or repulsive forces between molecules. Collisions elastic.No attractive or repulsive forces between molecules. Collisions elastic. Volume of molecules is negligible.Volume of molecules is negligible.

4. Properties of Gases Gas properties can be modeled using math. Model depends on— V = volume of the gas (L)V = volume of the gas (L) T = temperature (Kelvin)T = temperature (Kelvin) P = pressure (atmospheres)P = pressure (atmospheres)

5. Kinetic Molecular Theory Because we assume molecules are in motion, they have a kinetic energy. At the same Temperature, all gases have the same average KE. As Temperature goes up, KE also increases — and so does speed of the molecules. Temperature = measure of the average kinetic energy of the particles of a substance

6. Ch. 12 Notes-- Behavior of Gases Here is the _____________ relationship between the # of gas particles in a container and the volume and pressure of the container: As the # of gas particles _____________, the volume of a flexible container will ____________ if the temperature and pressure of the container remain constant. # particles ___, V ___ *Example: Blowing ______ air into a balloon makes it larger. As the # of gas particles ____________, the pressure of a rigid container will ____________ if the temperature and volume of the container remain constant. # particles ___, P ___ *Examples: Pushing the button on an aerosol can releases the gas and ___________ the pressure in the container. Adding too much gas into a rigid container could make it ___________ from too much pressure! qualitative increase ↑↑ more increase ↑↑ decreases explode

7. # of Gas Particles vs. Pressure

8. Pressure and the number of molecules are directly related More molecules means more collisions. Fewer molecules means fewer collisions. Gases naturally move from areas of high pressure to low pressure because there is empty space to move in.

9. Volume of Gas In a smaller container, molecules have less room to move. Hit the sides of the container more often. As volume decreases, pressure increases. (think of a syringe)

10. Here is the qualitative relationship between the pressure, temperature, and volume of a constant # of gas particles in a container: (1) ___________ Law: At a constant temperature, as the volume of a container __________ the pressure of the container will ___________. V___, P ___ *Example: Compressing the gas in a flexible container will _________ its volume. Gas Laws Pressure Volume Boyle’s ↑↓ decreasesincrease decrease

11. Boyle’s Law If Temperature is constant, then P 1 V 1 = P 2 V 2 This means that as Pressure goes up as Volume goes down. Pressure and Volume are indirectly related Robert Boyle (1627-1691). Son of Early of Cork, Ireland. ’? 3> How you doin’? 3>

12. Boyle’s Law A bicycle pump is a good example of Boyle’s law. As the volume of the air trapped in the pump is reduced, its pressure goes up, and air is forced into the tire.

13. P 1 = initial pressure V 1 = initial volume P 2 = final pressure V 2 = final volume STP = standard temperature (273 Kelvin) and pressure (1 atmosphere) Pressure can be measured in atmospheres (atm), torrs (torr), millimeters mercury (mmHg) or kilopascals (kpa) Conversions: 1 atm = 760 torr = 760 mmHg = 101.325 kpa Boyle’s Law P 1 V 1 = P 2 V 2

14. 1.If we have 4 L of methane gas at a pressure of 1.50 x 10 3 mmHg. What will the pressure of the gas be if the volume is decreased to 2.5 L? P 1 = V 1 = P 2 = V 2 = Boyle’s Law Practice Problems 4 liters ? 1.50 x 10 3 mmHg 2.5 liters Plug the #’s into the equation and solve for P 2. (1.05 x 10 3 )(4)(2.5)(P 2 ) = (1.50 x 10 3 ) (4) = P 2 2.5 2400 mmHg = P 2

15. 2. You have a car with an internal volume of 12,000 L. If you drive your car into the river and it implodes, what will be the volume of the gas when the pressure goes from 1.0 atm to 1.4 atm? P 1 = V 1 = P 2 = V 2 = Boyle’s Law Practice Problems 12,000 L ? 1 atm 1.4 atm Plug the #’s into the equation and solve for V 2. (1)(12,000)(V2)(V2)(1.4) = (1) (12,000) = V 2 1.4 8571.43 L = V 2

16.  If P1 and P2 are given, they must both be in the same unit. If they are not, you must make a conversion. Usually the conversion is to atm’s since it is the measure of standard pressure. Make the following conversions using: 1 atm = 760 torr = 760 mmHg = 101.325 kpa Practice Problems: (1) Convert 6.5 atm to torr (2) Convert 98 torr to mmHg (3) Convert 177 mmHg to atm 760 torr 1 atm 6.5 atm x= 4940 torr 760 mmHg 760 torr 98 torr x= 98 mmHg 1 atm 177 mmHg x =.23 atm 760 mmHg Boyle’s Law Practice Problems

17. (2) ____________ Law: At a constant pressure, as the temperature of a container __________ the volume of the container will ___________. T___, V ___ *Examples: Heating a balloon will cause it to ___________. Taking a balloon outside on a cold winter day will cause it to _____________. If you could keep a gas from condensing, you could cool it off to absolute zero and the volume of the gas would be _________! Gas Laws (continued) Volume Temperature (K) Charles’s increasesincrease ↑ inflate shrink zero ↑

18. Charles’s Law If Pressure is constant, then V 1 T 2 = V 2 T 1 This means as Volume goes up so does Temperature. This means Volume and Temperature are directly related. Jacques Charles (1746- 1823). Isolated boron and studied gases. Balloonist. Hey baby! You need a date?

19. Charles’s original balloon Modern long-distance balloon

20. Charles’s Law

21. V 1 = initial volume T 2 = final temperature V 2 = final volume T 1 = initial temperature STP = standard temperature (273 Kelvin) and pressure (1 atmosphere) Conversions: Kelvin = degrees Celsius + 273 Charles’s Law V 1 T 2 = V 2 T 1

22. 1.If we have 2 L of methane gas at a temperature of 40 degrees Celsius, what will the volume of the gas be if we heat the gas to 80 degrees Celsius? V 1 = T 2 = V 2 = T 1 = Charles’s Law Practice Problems 80 ̊ C + 273 = 353 K ? 2 liters 40 ̊ C + 273 = 313 K Plug the #’s into the equation and solve for V 2. (2)(353)(313)(V 2 )= (2) (353) = V 2 313 2.26 L = V 2

23. 1.If you have 45 L of helium in a balloon at 25 ̊ C and you increase the temperature of the balloon to 55 ̊ C, what will the new volume of the balloon be? V 1 = T 2 = V 2 = T 1 = Charles’s Law Practice Problems 55 ̊ C + 273 = 328 K ? 45 liters 25 ̊ C + 273 = 298 K Plug the #’s into the equation and solve for V 2. (45)(328)(298)(V 2 )= (45) (328) = V 2 298 49.53 L = V 2

24. (3) ____________ Law: At a constant volume, as the temperature of a container __________ the pressure of the container will ___________. P 1 T 2 = P 2 T 1 T___, P ___ *Example: Heating a rigid container causes the gas inside to move __________ which causes _________ pressure. Be careful! Too much heat will make it explode! Gas Laws (continued) Pressure Temperature (K) Guy-Lussac’s increasesincrease ↑ fastermore ↑

25. The Combined Gas Law

26. Combining the gas laws So far we have seen three gas laws: Jacques CharlesRobert Boyle P1V1P1V1 =P2V2P2V2 V1V1 T1T1 =V2V2 T2T2 These are all subsets of a more encompassing law: the combined gas law P1P1 T1T1 =P2P2 T2T2 P 1 V 1 T 2 P 2 V 2 T 1 = Joseph Louis Gay-Lussac Pick me baby! Bachelor #1Bachelor #2 Bachelor #3 I’m the hottest pick me! Check me out!

27. The Combined Gas Law This equation combines all of the previous three laws into one convenient form. Boyles Law: Guy-Lussac’s Law: Pressure and temperature Charles’s Law: Volume and temperature Pressure and volume Combined Gas Law P 1 x V 1 x T 2 P 2 x V 2 x T 1 = (initial conditions) = (final conditions) Using the Combined Gas Law requires you to have the temperature in _____________ units. The pressure and volume units can be anything as long as the initial and final units are ______ __________. Kelvin the same

28. Often the volume of a gas is needed at “standard conditions.” For scientists, this means “STP”. Standard temperature is ______K, and standard pressure is ____________ atmosphere (atm) 1 atmosphere (atm) = 760 mm Hg = 760 mmHg = 101.3 kPa = 14.7 lbs/in 2 (psi) Practice Problems: 1) 80.0 mL of helium is in a balloon at 25˚C. What will the new volume of the balloon be if the temp. is raised to 100˚C? Standard Temperature and Pressure: (STP) P 1 = ______ V 1 = ______ T K 1 = ______ P 2 = ______ V 2 = ______ T K 2 = ______ 80.0 mL 298 K373 K ??? Plug the #’s into the equation and solve for V 2. (80.0) (298) = (V 2 ) (373) V 2 = 100 mL 273 1 (Since pressure is not mentioned, it can be assumed that it was constant. You can thrown it out of our equation.)

29. Practice Problems (continued): 2) A rigid steel container is filled with neon under a pressure of 760 mm Hg and a temperature of 325 K. If the temperature is reduced to standard temperature, what will the new pressure be? P 1 = ______ V 1 = ______ T K 1 = ______ P 2 = ______ V 2 = ______ T K 2 = ______ 760 mm 325 K273 K ??? Plug the #’s into the equation and solve for P 2. (760) (325) =(P 2 ) (273) P 2 = 638 mm Hg 3) A balloon at a pressure of 4.5 atmospheres, 300 K, and a volume of 35.0 liters is changed to STP conditions. What will the new volume of the balloon become? P 1 = ______ V 1 = ______ T K 1 = ______ P 2 = ______ V 2 = ______ T K 2 = ______ 4.5 atm 300 K273 K 1 atm Plug the #’s into the equation and solve for V 2. (4.5)(35.0)(300)= (1)(V 2 ) (273) V 2 = 143 L 35.0 L??? Volume is not mentioned, so assume it is constant.

31. Dalton’s Law of Partial Pressure What happens to the pressure of a gas as we mix different gases in the container? The ______ of each individual gas pressure equals the _______ gas pressure of the container. P (total) = P 1 +P 2 +P 3 … sum total

33. 33 1) A container has oxygen, nitrogen, and helium in it. The total pressure of the container is 2.4 atmospheres. If all the partial pressures are equal to one another, what ate the partial pressures of each gas? Total number of gases = Dalton’s Law of Partial Pressures total  P 2.4 atm 3 P gas = 2.4 atm ÷ 3 = 0.8 atm

34. 34 2) Two flasks are connected with a rubber hose. The first flask contains N 2 at a pressure of 0.75 atm., and the second flask contains O 2 at a pressure of 1.25 atm. When the two flasks are opened and mix, what will the pressure be in the resulting mixture? Dalton’s Law of Partial Pressures P 2 = P 1 = 0.75 atm1.25 atm P total = 0.75 atm + 1.25 atm = 2.0 atm

35. 35 3) A mixture of neon and argon gases exerts a total pressure of 2.39 atm. The partial pressure of the neon alone is 1.84 atm. What is the partial pressure of the argon? Dalton’s Law of Partial Pressures P 1 = P total = 2.39 atm1.84 atm P 2 = 2.39 atm – 1.84 atm = 0.55 atm

36. Avogadro’s hypothesis states that ________ volumes of gases (under the same temp. and pressure conditions) contain _______ number of particles. If containers have the same ____, ____, and ___, then they will have the same ____ of particles regardless of the _________ of the gas particle. You might think that a small gas molecule would take up ______ space than a large gas molecule, but it ___________ at the same _________________ and ______________!! Avogadro’s Hypothesis equal T P V # size less doesn’t temperaturepressure

37. A mole is a term for a certain ______________ of objects. 1 mole = 6.02 x 10 23 objects *Other Examples: 1 pair = __ objects; 1 dozen = __ objects 1 gross = ____ objects; 1 _______ = 24 objects Since this value is so huge, it is used to measure very small objects like ___________ and _______________. Gas Conversions Factors At STP conditions, 1 mole of any gas occupies 22.4 Liters of space. Here are the conversion factors: 1 mole= __________ particles= _____L (at STP)= gram-formula mass The Mole Concept number 212 144case atomsmolecules 6.02 x 10 23 22.4

38. Gram-Formula Mass The # of grams that 6.02 x 10 23 particles, (or ___ mole), weighs is called the gram formula mass. The mass is found from the weights of the elements on the ____________ __________. *Examples: He = ____ g/mole H 2 = ____ g/mole H 2 O= ____ g/mole CO 2 = _____ g/mole Practice Problems: (1) Convert 3 moles of Helium to Liters (at STP). (2) Convert 50 grams of ammonia gas (NH 3 ) to # of molecules. (3) Convert 3.01x10 23 atoms of Neon to grams of Neon. 1 periodic table 4.02.0 18.044.0 22.4 L 1 mole 3 moles x = 67.2 L 6.02 x 10 23 molecules 17.0 g 50 g x = 17.7 x 10 23 molecules 20.2 grams 3.01x 10 23 atoms x = 10.1 grams 6.02 x 10 23 atoms

39. The Ideal Gas Law An equation used to calculate the __________ of gas in a container (in units of _________.) PV=nRT The units for T= __________, V = _________, n = # of moles R = Ideal Gas Constant The value of R changes depending on the unit of ____________ used in the equation: R = 62.4 (mm Hg)(L)/(mole)(K) R = 8.31 (kPa)(L)/(mole)(K) R = 0.0821 (atm.)(L)/(mole)(K) R = 2.45 (in. Hg)(L)/(mole)(K) amount moles KelvinLiters pressure

40. The Ideal Gas Law Practice Problems: 1) 6.5 moles of a gas has a pressure of 1.30 atmospheres and it has a temperature of 20˚Celsius. What is the volume of the gas? 2) How many moles of gas are there in a 7.3 liter balloon with a pressure of 847 mm Hg and temperature of 395 K? ( ) ( ) = ( ) ( ) ( )1.30V6.50.0821293 K V = 120 L ( ) ( ) = ( ) ( ) ( )8477.3n62.4395 K n = 0.25 moles

41. Dalton’s Law of Partial Pressure The ______ of each individual gas pressure equals the _______ gas pressure of the container. P (total) = P 1 +P 2 +P 3 … Practice Problem: A container has oxygen, nitrogen, and helium in it. The total pressure of the container is 2.4 atmospheres. If all of the partial pressures are the equal to one another, what are the partial pressures of the gases? sumtotal P gas = 2.4 atm ÷ 3 = 0.8 atm

43. Diffusion vs. Effusion The spreading out of a gas from _______ to _____ concentrations is called diffusion. *Example: ___________ in a room spreads out A gas escaping through a ______ _______ in a container is called effusion. As the size of a molecule _____________, the effusion speed and diffusion rate ______________...(inverse relationship.) Effusion highlow Perfume tiny hole increases decrease

44. “Ideal” Gases Real gases, (like nitrogen), will eventually ___________ into a liquid when the temperature gets too ____ or the pressure gets too _____. If you want a gas to act more ideally, keep the temperature _____ and the pressure ______. That way, they will act more like an ideal gas and never have a chance of _______________. The best real gas that acts like an ideal gas is __________. It doesn’t condense until the temperature gets to ______K. Real Gas condense low high low condensing helium 4

45. Ideal Gases vs. Real Gases