Properties of Gases. Occupy the entire volume of their container Compressible Flow readily and mix easily Have low densities, low molecular weight

Transcription

1 Chapter 5 Gases

2 Properties of Gases Occupy the entire volume of their container Compressible Flow readily and mix easily Have low densities, low molecular weight

3 Atmospheric Pressure Atmospheric pressure Pressure from earth s atmosphere at sea level Equal to atmosphere Unit is atm 3 Barometer Used to measure atmospheric pressure Height in mm Hg At sea level height is exactly 760mm Other units: atm is equal to mmhg 760 torr.035 kpa 9.9 in. Hg

4 Measuring Experimental Gas Pressure A manometer measures pressure of gases Below atm Above atm Closed tube manometer Closed tube manometer P gas = Height = mm Hg P gas = 760mm Hg + height 4

5 Gas Laws 5

6 Boyle s Law: Pressure/Volume Relationship For a fixed amount of gas at constant temperature, volume decreases as pressure increases 6 Rearranging equation PV = constant so P V = P V P PV V V PV P pressure: volume Cuts space between molecules volume: pressure Molecule-wall collisions increase

7 Charles Law: Volume/Temp. Relationship For a fixed amount of gas at constant pressure, volume increases as temperature increases 7 Rearranging equation V/T= constant so V /T = V /T VT V T TV T V Temp. in kelvins (T), not Celsius (t) T (K) = t (0ºC) As temperature increases: Molecules move faster They hit the wall harder Volume increases to hold pressure

8 Avogadro s law: Moles/Volume Relationship At fixed temperature and pressure, the volume of a gas depends on the # of moles of gas present 8 Rearranging equation V/n= constant so V /n = V /n V Vn n n nv V More molecules need more space to maintain the same pressure and temperature

9 A 4.50-L cylinder contains He(g) at an unknown pressure. It is now connected to a 9.5-L evacuated cylinder. When the connecting valve between the two cylinders is opened, the pressure falls to.40 atm. What was the pressure in the 4.50-L cylinder? Initial Conditions (P V ) Final Conditions (P V ) 9 V =4.50L P =? atm V =4.50L+9.5L=97.0L P =.40 atm PV.40atm 97.0L P x x 30. atm V 4.50L

10 Decreasing the temperature of 0.00 L of H (g) from 5ºC to -77ºC decreases its volume to what value? Find your data: T = 5ºC = 98K (convert to K) T = -77ºC = 96K (convert to K) V is given in problem = 0.00L V =?L 0 From Charles s Law: V T V T VT 0.00Lx96K V 6. 58L T 98K

11 The Combined Gas Law and the Ideal Gas Constant Combines all three laws by means of varying the volume Boyles Law Charles s Law Avogadro s Law constant = PV constant = V / T constant = V / n STP: Standard Temperature and Pressure = atm 73K P= PVatm V=.4L n T atmx.4l molx73k 0.08Latm molk Ideal Gas Constant: R=. Acts as side of combined gas law R P V n T PV n T

12 Using the Ideal Gas Law Ideal Gas Law PV = nrt Must convert units to match R R PV nt 0.08Latm molk Identify the noble gas if 3.6 g exerts a pressure of 93 torr at 37ºC in a 7.5-L container? Convert T and P: T = 37 ºC = 30.5 K P= =.70atm Solve for moles Find molar mass MM=.. = 0.g/mol Identify noble gas.70atm 7.5L molk n x x x. 7mol 0.08Latm 30.5K Neon: MM0.7g/mol x

13 The Law of Combining Volumes At the same T and P, the volume of a gas is directly related to the moles of gas and the number of molecules of a gas. 3 The ratio is the same as that in the chemical equation

14 Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many liters of NO are obtained from.0 liter of ammonia at the same temperature and pressure? 4 Write the balanced chemical equation 4NH 3 (g) + 5O (g) 4NO (g)+ 6H O (g) Write the balanced chemical equation Mole ratio: 4 NH 3 4NO Volume ratio: 4L of NH 3 4Lof NO Solve for liter of NH 3 Mole ratio: NH 3 NO Volume ratio: L of NH 3 Lof NO produced

15 What is the Mass of 6. L of SF 6 at STP? Gather your Data V = 6.L V =.4L n =? n = mole Equation needed: nv.00mol 6.L n x x 0. 73moles V.4L 5 Convert between moles and mass using molar mass Molar Mass of SF 6 = g/mol 0.73moles 46.05g? g x 06g mol

16 Determine the volume in liters of NO(g) that will be produced from 5.5 L of 0 (g) and excess NH 3 (g) at constant pressure and temperature. 4 NH 3 (g) + 5 O (g) 4 NO (g) + 6 H O (l) 6 Using Avogadro s Law n O O NO n 0.3mol O x V.4LO 5molO V xv n mol.4l NO NO V 4. n molno Using the Law of Combining Volumes? x5.5l x0.87mol 5.5L 4L O NO LNO x LO L L NO NO 4mol mol NO

17 A 6.4-g sample of a gas at 5.0 lb/in and 5.0ºC is confined in a 7.5 L container. This gas is moved to a 8.5 L container at 00.0ºC and 4.0 lb/in. How much gas was removed or added? Get equation: P V PV n T PV n nt nt TPV 7 Use gas masses, not moles n = 6.4g n =? Convert temp. to Kelvin T = 98K T = 373K Pressure units cancel: P = 5.0 lb/in P = 4.0 lb/in Volume as written: V = 7.5L V = 8.5L 6.4g 98K 4.0lb / in 8.5L n x x x. 9g 373K 5.0lb / in 7.5L 6.4g of gas at start -.9g after moving it = 4.5g removed

18 Stoichiometry and Gas Laws What volume in liters of CO (g) at 50ºC and atm is produced from 453.6g Fe O 3 (s) in the following reaction: Fe O 3 (s) + 3C (s) Fe (s) +3 CO (g) Determine # moles of CO (g) produced to get n? mol CO 453.6g Fe O molfe O 3mol 3 3 CO x x 8. 5mol 59.6g mol Fe O 3 Fe O 3 CO n 8 Use the ideal gas law to determine the volume V = nrt n = 8.5 mol R = 0.08 L atm / mol K P T = 53.5 K P = atm 8.5mol 0.08Latm 53.5K V x x x 405L molk.904atm

19 9 Dalton s Law of Partial Pressures

20 Mixtures of Gases: Dalton's Law of Partial Pressures The Total Pressure is the sum of the partial pressures. P total = P + P + P The Partial Pressure is the pressure exerted by each individual gas in the container. P = (n RT)/ V; P = (n RT)/ V; and so on The Mole Fraction is the fraction of all the molecules in a mixture that are of a given type. P /P total = n /n total = x P = x. P total

21 Mixtures of Gases Dalton's Law of Partial Pressures

22 A sample of natural gas contains 8.4 moles of CH 4, 0.4 moles of C H 6, and 0.6 moles of C 3 H 8. If the total pressure of the gases is.37 atm, what is the partial pressure of propane (C 3 H 8 )? P i = X i P T Solving for X i, the mole fraction: X C 8.4mol 4 0.6mol 3 8 3H 8 CH 0.4mol C C H H 6 0.6mol C 3 H Solving for P i : P i = X i P T = 0.03 x.37atm = 0.08atm

23 Collection of Gases Over Water An gas is collected into a container of water The water saturated gas rises and displaces liquid water The vapor pressure is the partial pressure of the water vapor It must be deducted from the measured pressure of the gas 3 P total P atm = P gas + P water(g) P gas = P atm -P water(g)

24 The Kinetic Molecular Theory 4

25 Kinetic Molecular Theory The molecules are in constant random motion 5 Completely fill the container Vast spaces between molecules Molecules considered volumeless Move easily when force is applied Elastic Collisions No gain or loss of energy Molecules move faster as the temperature is raised Temperature increases the average kinetic energy Pressure is created by molecules hitting the walls Assume no interaction between molecules

26 Molar Mass & Temperature Effects 6 u urms Higher temperatures Molecules move faster 3RT MolarMass Small molecules move faster than big ones

27 Diffusion and Effusion 7 Diffusion is the process by which one gas mixes with another as a result of molecular movement r r t t MolarMass MolarMass NH 3 7g/mol NH 4 Cl HCl 36g/mol NH 4 Cl collects closer to HCl HCl molecules larger than NH 3 molecules Effusion is the process in which a gas under pressure escapes from its container through a small hole.

28 Deviation from Ideal Behavior 8

29 The Van de Waal s Equation 9 Assumptions of the kinetic-molecular theory not exact At high pressure/density gases act more like liquids/solids. Attractive forces between molecules increase. Molecular volume no longer negligible Vanderwaal s Equation: (P + an /V )(V-nb) = nrt Molecular attraction correction coefficient (a) Actual P is higher than the measured P P actual = P + an /V Volume correction coefficient (b) Actual V is lower than the container V V actual = V-nb