Presentation is loading. Please wait.

Presentation is loading. Please wait.

AP CHEMISTRY Chapter 5 Gases Slides with gray backgrounds are not tested on the AP test.

Similar presentations

Presentation on theme: "AP CHEMISTRY Chapter 5 Gases Slides with gray backgrounds are not tested on the AP test."— Presentation transcript:

1 AP CHEMISTRY Chapter 5 Gases Slides with gray backgrounds are not tested on the AP test.

2 This is a schematic diagram showing gas molecules (purple) in a container. The molecules are constantly moving in random directions. When a molecule hits the container wall (green), it exerts a tiny force on the wall. The sum of these tiny forces, divided by the interior surface area of the container, is the pressure. (Longer arrows indicate higher velocities, shorter arrows indicate lower velocities) Gas Pressure

3 Barometer - invented by Evangelista Torricelli in 1643; uses the height of a column of mercury to measure gas pressure (especially atmospheric)

4 The Manometer a device for measuring the pressure of a gas in a container The pressure of the gas is given by h [the difference in mercury levels] in units of torr (equivalent to mm Hg).

5 a) Gas pressure = atmospheric pressure – h b) Gas pressure = atmospheric pressure + h Open-tube manometer

6 1 mm of Hg = 1 torr 760.00 mm Hg = 760.00 torr = 1.00 atm = 101.325 kPa Know these! kPa is not tested on the AP exam

7 SI unit of pressure is N/m 2 or Pascal (Pa)

8 Boyle’s Law 1 st quantitative study of gases, 1600s. Pressure and volume are inversely related. P 1 = V 2 P 1 V 1 = P 2 V 2 P 2 V 1 Memory Trick: We Boyle Peas and Vegetables

9 An ideal gas is expected to have a constant value of PV, as shown by the dotted line. CO 2 shows the largest change in PV, and this change is actually quite small. Ideal gas- gas that obeys Boyle’s law


11 Charles’ Law 1700’s -volume of a gas is directly proportional to Kelvin temperature V 1 = V 2 Temp must be in Kelvin!!!!!! T 1 T 2 Memory Trick: Charlie Brown’s Xmas is on TV

12 0 The volume of a gas at absolute zero is zero.

13 Avogadro’s Law -equal volumes of gases at the same temperature and pressure contain the same # of particles -for a gas at constant temp. and pressure, the volume is directly proportional to the # of moles of gas


15 Gay-Lussac’s Law Pressure of a gas is directly proportional to Kelvin temperature P 1 = P 2 Temp must be in Kelvin!!!!!! T 1 T 2 “GayLe drives a PT cruiser”


17 Combined Gas Law P 1 V 1 = P 2 V 2 T 1 T 2 “Peas and Vegetables on the Table” This can be used to come up with Boyle ’ s, Charles ’, or Gay - Lussac ’ s Law. Simply cover up the factors that are constant.

18 Ideal Gas Law Combining Boyle’s, Charles’, and Avogadro’s laws we get PV= nRT. R = 0.08206 (L  atm)/(K  mol) (proportionality constant) Most gases behave ideally at pressures less than 1 atm.

19 We can use the ideal gas law for all gas law problems by putting changing variables on one side and the constant on the other. Ex. If P&V change w/ others constant: P 1 V 1 = nRT and P 2 V 2 = nRT so P 1 V 1 = P 2 V 2 If V&T change with others constant: V 1 = nR and V 2 = nR so V 1 = V 2 T 1 P T 2 P T 1 T 2

20 The gas pressure inside an aerosol can is 1.5 atm at 25 o C. Assuming that the gas is ideal, what would the pressure be if the can were heated to 452 o C? P 1 =1.5 atm T 1 = 25 o C+273 = 298K P 2 = ? T 2 = 452 o C + 273 = 725K P 2 = 3.6 atm

21 A quantity of helium gas occupies a volume of 16.5 L at 78 ° C and 45.6 atm. What is the volume at STP? P 1 = 45.6 atm V 1 = 16.5L T 1 = 78 ° C + 273 = 351K P 2 = 1 atm V 2 = ? T 2 = 0 ° C + 273 = 273K V 2 = 585L

22 Many gases are shipped in high-pressure containers. If a steel tank whose volume is 50.0L contains O 2 gas at a total pressure of 1550 kPa at 23 o C, what mass of oxygen does it contain? P = 1550 kPa/101.3 = 15.3atm V = 50.0L n = ? R = 0.08206(Latm)/(Kmol) T = 23 o C + 273 = 296K PV = nRT (15.3)(50.0) = n (0.08206)(296) n = 31.5mol O 2 1010 g O 2

23 Molar Volume = 22.42 L of an ideal gas at STP (Some gases behave more ideally than others.) STP = 0 o C and 1 atm

24 CaH 2 reacts with H 2 O to produce H 2 gas. CaH 2 (s) + 2H 2 O(l)  2H 2 (g) + Ca 2+ (aq) + 2OH - (aq) Assuming complete rxn with water, how many grams of CaH 2 are required to fill a balloon to a total pressure of 1.12 atm at 15 o C if its volume is 5.50 L? P = 1.12 atm V= 5.50 L T= 15 o C + 273 = 288K n= ? R = 0.08206 Latm/Kmol PV=nRT (1.12)(5.50)=n(0.08206)288 n = 0.2606 mol H 2 0.2606 mol H 2 1 mol CaH 2 42.10g CaH 2 2 mol H 2 1 mol CaH 2 = 5.49g CaH 2

25 How many liters of N 2 are required to produce 115 g of NH 3 at STP? N 2 + 3H 2  2NH 3 115g NH 3 1 mol NH 3 1 mol N 2 22.42L N 2 17.04g NH 3 2 mol NH 3 1 mol N 2 = 75.7 L N 2

26 Molecular Weight and Density of a Gas n = mass so P = mRT MW V(MW) Since m/V = density (g/L), P = dRT MW MW = dRT P “Molecular Weight Kitty Cat” Meow = dirt/pee

27 Calculate the molar mass of a gas if 0.608g occupies 750 mL at 385 mm Hg and 35 o C. V = 750 mL = 0.75L P = 385/760 = 0.507 atm T = 35 o C + 273 = 308 K d = 0.608g/0.75L = 0.811g/L MW = dRT/P MW = (0.811)(0.08206)(308) = 40.4 g/mol 0.507


29 Dalton’s Law of Partial Pressures For a mixture of gases in a container, the total pressure exerted is the sum of the pressures that each gas would exert if it were alone. P tot = P 1 + P 2 + P 3 + … P tot = n 1 RT + n 2 RT + n 3 RT +… V V V P tot = n total (RT) (It doesn’t matter what the gas is.) V

30 Mole Fraction -the ratio of the number of moles of a given component in a mixture to the total number of moles in the mixture.  is used to symbolize mole fraction.  1 = n 1 n 1 + n 2 + n 3 +...

31 The partial pressure of a particular component of a gaseous mixture is the mole fraction of that component times the total pressure. P 1 =  1 (P total )

32 When gases are collected over water, we must adjust for the pressure of the water vapor. P H2O + P gas = P total

33 If a 0.20 L sample of O 2 at 0 o C and 1.0 atm pressure and a 0.10 L sample of N 2 at 0 o C and 2.0 atm pressure are both placed in a 0.40 L container at 0 o C, what is the total pressure in the container? P 1 V 1 =P 2 V 2 1.0(0.20) = P 2 (0.40) P 2 = 0.50 atm = P O2 (2.0)(0.10) = P 2 (0.40) P 2 = 0.50 atm = P N2 P O2 + P N2 = P total 0.50 + 0.50 = 1.00 atm

34 Kinetic Molecular Theory of Gases -a simple model that attempts to explain properties of an ideal gas

35 Gases consist of particles which have the following properties: 1. The particles are so small compared to the distances between them that the volume of the individual particles can be assumed to be negligible (zero). 2. The particles are in constant motion. The collisions of the particles with the walls of the container are the cause of the pressure exerted by the gas.

36 3. The particles are assumed to exert no forces on each other; they are assumed neither to attract nor to repel each other. 4. The average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas. -real gases don’t conform to these assumptions!!!

37 Kelvin temp. is an index of the random motions of the particles of a gas, with higher temp. meaning greater motion. KE (avg) = 3/2 RT R = 8.314 J/K mol Units of KE are J/mol KE = 1/2 mv 2 Remember that mass has to be in kg and velocity in m/s! Since we are working with energy, we need the energy R, not the gas R. Temp must be in Kelvin. Formula is not on AP exam.

38 Real gases have many collisions between particles. The average distance a particle travels between collisions in a particular gas sample is called the mean free path. These collisions produce a huge variation in velocities. As temperature increases, the range of velocities is greater.

39 Path of One Particle in a Gas

40 A Plot of the Relative Number of O 2 Molecules That Have a Given Velocity at STP

41 A Plot of the Relative Number of N 2 Molecules That Have a Given Velocity at Three Temperatures This is called a Boltzmann Distribution graph

42 Effusion and Diffusion Diffusion- mixing of gases Effusion- the passage of a gas through a tiny orifice into an evacuated chamber

43 The Effusion of a Gas into an Evacuated Chamber

44 Graham’s Law of Effusion -The rate of effusion of a gas is inversely proportional to the square root of the mass of its particles. ____ Rate of effusion for gas 1 =  MW 2 Rate of effusion for gas 2  MW 1 MW 1 and MW 2 represent the molar masses of the gases. These calculations are not tested on the AP test.

45 -lighter gases effuse & diffuse faster than heavier gases

46 HCl(g) and NH 3 (g) Meet in a Tube Because NH 3 has a lower molar mass than HCl, it moves faster and farther in the tube. Solid NH 4 Cl is formed closer to the HCl.

47 Real gases *No gas exactly follows the ideal gas law. *A real gas exhibits behavior closest to ideal behavior at low pressures and high temperatures. Students also behave most ideally under these conditions. (Summer vacation!)

48 At high temperatures, there is less interaction between particles because they are moving too fast. At high concentrations, gases have much greater attractive forces between particles. This causes particles to hit the walls of the container with less force (producing less pressure than expected).

49 At high pressure (small volume), the volume of the particles becomes significant, so that the volume available to the gas is significantly less than the container volume. Attractive forces are greatest for large, complex molecules and polar molecules.

50 Volume Taken up by Gas Particles

51 We can use the Van der Waals equation to adjust for departures from ideal conditions. PV = nRT becomes: [P obs + a(n/V) 2 ]V-nb = nRT corrected corrected pressure volume You don’t need to memorize this!

Download ppt "AP CHEMISTRY Chapter 5 Gases Slides with gray backgrounds are not tested on the AP test."

Similar presentations

Ads by Google