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Gases Chapter 5 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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1 Gases Chapter 5 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2 Three States of Matter 1.5

3 Gases assume the volume and shape of their containers. Gases are the most compressible state of matter. Gases will mix evenly and completely when confined to the same container. Gases have much lower densities than liquids and solids. Gases have weak interaction between molecules. Gas molecules are constantly in motion. Physical Characteristics of Gases

4 Elements that exist as gases at 25 0 C and 1 atmosphere 5.1

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6 What type of compounds form gases at 25°C? Elements: H 2, N 2, O 2, O 3,F 2, Ne, Cl 2, He, Ar, Kr, Xe, Rn Compounds: HF, HCl, HBr, HI, CO, CO 2, CH4, NH 3, NO, NO 2, N 2 O, SO 2, H 2 S, HCN => Molecular Compounds Ionic Compounds do not form gases due to strong electrostatic forces => ions are not stable as gases at high concentrations Ex: NaCl

7 Gas molecules exert pressure on any surface, because molecule are constantly in motion. How do we measure pressure?

8 Units of Pressure 1 pascal (Pa) = 1 N/m 2 1 atm = 760 mmHg = 760 torr 1 atm = 101,325 Pa = 101.325 kPa 1 atm = 29.92 in Hg Barometer Pressure = Force Area (force = mass x acceleration) = kg x m/s 2 = newton (N) Pressure is a concentration unit !! Balance forces

9 Convert pressure units Convert 26.5 torr into mm Hg and atm. 26.5 torr x 1 mm Hg = 26.5 mm Hg 1 torr 26.5 torr x 1 atm = 0.0349 atm 760. torr

10 Sea level1 atm 4 miles0.5 atm 10 miles0.2 atm 5.2 Density of air decreases as we move away from Earth => Reduced gravitational forces

11 5.2 Figure 5.4 P = 0 P = atm Closed endOpen end

12 Why choose Hg for a manometer? Need a liquid High density => short distance Very low evaporation rate

13 5.2 h Difference between column heights decreases as pressure in bulb decreases.

14 Gas Laws  Macroscopic Behavior of Gases  Assume gases behave ideally  Focus on Pressure (P), Temperature (T), Volume (V) and number of moles (n) Robert Boyle – 17 th Century used manometers to study Pressure / Volume behavior of gases

15 5.3 As P (h) increases V decreases Temperature is constant. Fixed amount of gas used (same number of moles)

16 P  1/V P x V = constant P 1 x V 1 = P 2 x V 2 Boyle’s Law Constant temperature Constant amount of gas 0.6 atm x 2 L = 1.2 L atm 0.3 atm x 4 L = 1.2 L atm

17 A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL? P 1 x V 1 = P 2 x V 2 P 1 = 726 mmHg V 1 = 946 mL P 2 = ? V 2 = 154 mL P 2 = P 1 x V 1 V2V2 726 mmHg x 946 mL 154 mL = = 4460 mmHg 5.3 Can use any pressure or volume unit as long as you are consistent !!!

18 P 1 x V 1 = P 2 x V 2 This equation is similar to the equation for dilutions !! What happens when we change temperature? Jacques Charles – 18 th Century Volume / Temperature behavior of gases Boyle’s Law

19 As T increasesV increases 5.3

20 Variation of gas volume with temperature at constant pressure. V  TV  T V = constant x T V 1 /T 1 = V 2 /T 2 T (K) = t ( 0 C) + 273.15 Charles’ & Gay-Lussac’s Law Temperature must be in Kelvin !!! Why ?? V/T = constant -OR -

21 Variation of gas volume with temperature at constant pressure. Charles’ & Gay-Lussac’s Law All lines converge at one temperature - absolute zero

22 A sample of carbon monoxide gas occupies 3.20 L at 125 0 C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V 1 = 3.20 L T 1 = 398.15 K V 2 = 1.54 L T 2 = ? T 2 = V 2 x T 1 V1V1 1.54 L x 398.15 K 3.20 L = = 192 K V 1 /T 1 = V 2 /T 2 T 1 = 125 ( 0 C) + 273.15 (K) = 398.15 K Can use any volume unit as long as you are consistent !!!

23 What happens when we change temperature at constant volume? Jacques Charles & Joseph Gay-Lussac Pressure / Temperature behavior of gases => Similar behavior to Volume / Temperature relationship P  TP  T P = constant x T P 1 /T 1 = P 2 /T 2 P/T = constant -OR - Assume quantity of gas and volume are constant.

24 Avogadro’s Law V  number of moles (n) V = constant x n V 1 /n 1 = V 2 /n 2 5.3 Constant temperature Constant pressure

25 Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure? 4NH 3 + 5O 2 4NO + 6H 2 O 1 mole NH 3 1 mole NO At constant T and P 1 volume NH 3 1 volume NO 5.3

26 Ideal Gas Equation 5.4 Charles’ law: V  T  (at constant n and P) Avogadro’s law: V  n  (at constant P and T) Boyle’s law: V  (at constant n and T) 1 P V V  nT P V = constant x = R nT P P R is the gas constant PV = nRT Charles & Gay-Lussac’s law: P  T  (at constant n and V) Combine the 3 law that vary V to get:

27 The conditions 0 0 C and 1 atm are called standard temperature and pressure (STP). PV = nRT R = PV nT = (1 atm)(22.414L) (1 mol)(273.15 K) R = 0.082057 L atm / (mol K) 5.4 Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L. Watch your units !!! NOTE: 1 mole and 1 atm are exact numbers !!

28 PV = nRT R = P1V1P1V1 nT 1 All previous laws can be derived from the Ideal Gas Law = P2V2P2V2 nT 2 V 1 /n 1 = V 2 /n 2 Avogadro’s law: Boyle’s law: P 1 x V 1 = P 2 x V 2 Charles’ law: V 1 /T 1 = V 2 /T 2 Charles & Gay-Lussac’s law: P 1 /T 1 = P 2 /T 2 (at constant n and T) (at constant n and P) (at constant n and V) (at constant P and T)

29 5.3

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32 What is the volume (in liters) occupied by 49.8 g of HCl at STP? PV = nRT V = nRT P T = 0 0 C = 273.15 K P = 1 atm n = 49.8 g x 1 mol HCl 36.45 g HCl = 1.37 mol V = 1 atm 1.37 mol x 0.0821 x 273.15 K Latm molK V = 30.6 L 5.4

33 Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0 C is heated to 85 0 C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? PV = nRT n, V and R are constant nR V = P T = constant P1P1 T1T1 P2P2 T2T2 = P 1 = 1.20 atm T 1 = 291 K P 2 = ? T 2 = 358 K P 2 = P 1 x T2T2 T1T1 = 1.20 atm x 358 K 291 K = 1.48 atm 5.4

34 Density (d) Calculations d = m V = PMPM RT m is the mass of the gas in g M is the molar mass of the gas Molar Mass ( M ) of a Gaseous Substance dRT P M = d is the density of the gas in g/L 5.4

35 A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.0 0 C. What is the molar mass of the gas? 5.3 dRT P M = d = m V 4.65 g 2.10 L = = 2.21 g L M = 2.21 g L 1 atm x 0.0821 x 300.15 K Latm molK M = 54.6 g/mol

36 Gas Stoichiometry What is the volume of CO 2 produced at 37 0 C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O (l) g C 6 H 12 O 6 mol C 6 H 12 O 6 mol CO 2 V CO 2 5.60 g C 6 H 12 O 6 1 mol C 6 H 12 O 6 180 g C 6 H 12 O 6 x 6 mol CO 2 1 mol C 6 H 12 O 6 x = 0.187 mol CO 2 V = nRT P 0.187 mol x 0.0821 x 310.15 K Latm molK 1.00 atm = = 4.76 L 5.5

37 Dalton’s Law of Partial Pressures V and T are constant P1P1 P2P2 P total = P 1 + P 2 In an ideal gas, all gas molecules act the same.

38 Consider a case in which two gases, A and B, are in a container of volume V. P A = n A RT V P B = n B RT V n A is the number of moles of A n B is the number of moles of B P T = P A + P B X A = nAnA n A + n B X B = nBnB n A + n B P A = X A P T P B = X B P T P i = X i P T mole fraction (X i ) = nini nTnT What are the units on mole fraction?? What is the sum of all of the mole fractions?

39 A sample of natural gas contains 8.24 moles of CH 4, 0.421 moles of C 2 H 6, and 0.116 moles of C 3 H 8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C 3 H 8 )? P i = X i P T X propane = 0.116 8.24 + 0.421 + 0.116 P T = 1.37 atm = 0.0132 P propane = 0.0132 x 1.37 atm= 0.0181 atm 5.6

40 2KClO 3 (s) 2KCl (s) + 3O 2 (g) Bottle full of oxygen gas and water vapor P T = P O + P H O 22 5.6

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42 Chemistry in Action: Scuba Diving and the Gas Laws PV Depth (ft)Pressure (atm) 01 332 663 5.6

43 Kinetic Molecular Theory of Gases 1.A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume. 2.Gas molecules are in constant motion in random directions. Collisions among molecules are perfectly elastic. 3.Gas molecules exert neither attractive nor repulsive forces on one another. 4.The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy 5.7

44 Kinetic theory of gases and … Compressibility of Gases Boyle’s Law P  collision rate with wall Collision rate  number density Number density  1/V P  1/V Charles’ Law P  collision rate with wall Collision rate  average kinetic energy of gas molecules Average kinetic energy  T P  TP  T 5.7

45 Kinetic theory of gases and … Avogadro’s Law P  collision rate with wall Collision rate  number density Number density  n P  nP  n Dalton’s Law of Partial Pressures Molecules do not attract or repel one another P exerted by one type of molecule is unaffected by the presence of another gas P total =  P i 5.7

46 Apparatus for studying molecular speed distribution 5.7

47 The distribution of speeds for nitrogen gas molecules at three different temperatures The distribution of speeds of three different gases at the same temperature 5.7 u rms = 3RT M 

48 Chemistry in Action: Super Cold Atoms Gaseous Rb Atoms 1.7 x 10 -7 K Bose-Einstein Condensate

49 Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. 5.7 NH 3 17 g/mol HCl 36 g/mol NH 4 Cl

50 Deviations from Ideal Behavior 1 mole of ideal gas PV = nRT n = PV RT = 1.0 5.8 Repulsive Forces Attractive Forces

51 Effect of intermolecular forces on the pressure exerted by a gas. 5.8

52 Van der Waals equation nonideal gas P + (V – nb) = nRT an 2 V2V2 () } corrected pressure } corrected volume


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