2 Kinetic Theory of Gases Particles of a gasMove rapidly in straight lines and are in constant motion.Have kinetic energy that increases with an increase in temperature.Are very far apart.Have essentially no attractive (or repulsive) forces.Have very small volumes compared to the volume of the container they occupy.
3 Properties of GasesGases are described in terms of four properties: pressure (P), volume (V), temperature (T), and amount (n).
4 BarometerA barometer measures the pressure exerted by the gases in the atmosphere.The atmospheric pressure is measured as the height in mm of the mercury column.
5 Learning CheckA. The downward pressure of the Hg in a barometer is _____ than/as the weight of the atmosphere.1) greater 2) less 3) the sameB. A water barometer is 13.6 times taller than a Hg barometer (DHg = 13.6 g/mL) because1) H2O is less dense2) H2O is heavier3) air is more dense than H2O
6 SolutionA.The downward pressure of the Hg in a barometer is 3) the same as the weight of the atmosphere.B. A water barometer is 13.6 times taller than a Hg barometer (DHg = 13.6 g/mL) because1) H2O is less dense
7 PressureA gas exerts pressure, which is defined as a force acting on a specific area. Pressure (P) = Force AreaOne atmosphere (1 atm) is 760 mm Hg.1 mm Hg = 1 torr 1.00 atm = 760 mm Hg = 760 torr
8 Units of PressureIn science, pressure is stated in atmospheres (atm), millimeters of mercury (mm Hg), and Pascals (Pa).
9 Learning CheckA. What is 475 mm Hg expressed in atm? 1) 475 atm 2) atm 3) x 105 atmB. The pressure in a tire is 2.00 atm. What is this pressure in mm Hg? 1) mm Hg 2) mm Hg 3) 22,300 mm Hg
10 Solution A. What is 475 mm Hg expressed in atm? 2) 0.638 atm 485 mm Hg x atm = atm760 mm HgB. The pressure of a tire is measured as 2.00 atm. What is this pressure in mm Hg?2) mm Hg2.00 atm x mm Hg = mm Hg1 atm
11 Boyle’s LawThe pressure of a gas is inversely related to its volume when T and n are constant.If volume decreases, the pressure increases.
12 PV Constant in Boyle’s Law The product P x V remains constant as long as T and n do not change. P1V1 = atm x L = atm L P2V2 = atm x L = atm L P3V3 = atm x L = atm LBoyle’s Law can be stated as P1V1 = P2V2 (T, n constant)
13 Solving for a Gas Law Factor The equation for Boyle’s Law can be rearranged to solve for any factor.To solve for V2, divide both sides by P2.P1V1 = P2V Boyle’s LawP2 P2P1V1 = V2P2Solving for P2P2 = P1 V1V2
14 PV in Breathing Mechanics When the lungs expand, the pressure in the lungs decreases.Inhalation occurs as air flows towards the lower pressure in the lungs.
15 PV in Breathing Mechanics When the lung volume decreases, pressure within the lungs increases.Exhalation occurs as air flows from the higher pressure in the lungs to the outside.
16 Calculation with Boyle’s Law Freon-12, CCl2F2, is used in refrigeration systems. What is the new volume (L) of a 8 L sample of Freon gas initially at 50 mm Hg after its pressure is changed to 200 mm Hg at constant T?1. Set up a data tableConditions 1 Conditions 2P1 = 50 mm Hg P2 = 200 mm HgV1 = 8 L V2 =?
17 Calculation with Boyle’s Law (continued) 2. When pressure increases, volume decreases.Solve Boyle’s Law for V2:P1V1 = P2V2V = V1P1P2V2 = 8 L x mm Hg = L200 mm Hgpressure ratio decreases volume
18 Learning CheckThe helium in a cylinder has a volume of 120 mL and a pressure of 840 mm Hg. A change in the volume results in a lower pressure inside the cylinder. Does cylinder A or B represent the final volume?At a new pressure of 420 mm Hg, what is the new volume?1) 60 mL 2) 120 mL 3) 240 mL
19 SolutionThe helium in a cylinder has a volume of 120 mL and a pressure of 840 mm Hg. A change in the volume results in a lower pressure inside the cylinder. Does cylinder A or B represent the final volume?B) If P decreases, V increases.At a new pressure of 420 mm Hg, what is the new volume of the cylinder?3) 240 mL
20 Learning CheckA sample of helium gas has a volume of 6.4 L at a pressure of atm. What is the new volume when the pressure is increased to 1.40 atm (T constant)?A) 3.2 L B) 6.4 L C) 12.8 L
21 Solution A) 3.2 L Solve for V2: P1V1 = P2V2 V2 = V1P1 P2 V2 = L x atm = L1.40 atmVolume decreases when there is an increase in the pressure (Temperature is constant).
22 Learning CheckA sample of oxygen gas has a volume of 12.0 L at 600. mm Hg. What is the new pressure when the volume changes to 36.0 L? (T and n constant.)1) mm Hg2) mm Hg3) 1200 mm Hg
23 Solution1) mm Hg Data table Conditions 1 Conditions 2 P1 = mm Hg P2 = ??? V1 = L V2 = LP2 = P1 V1V2600. mm Hg x L = mm Hg L
24 Charles’ LawThe Kelvin temperature of a gas is directly related to the volume (P and n are constant).When the temperature of a gas increases, its volume increases.
25 Charles’ Law V and T For two conditions, Charles’ Law is written V1 = V (P and n constant)T T2Rearranging Charles’ Law to solve for V2V2 = V1T2T1
26 Learning CheckSolve Charles’ Law expression for T2.V1 = V2T T2
27 Solution V1 = V2 T1 T2 Cross multiply to give V1T2 = V2T1 Isolate T2 by dividing through by V1V1T = V2T1V1 V1T2 = V2T1V1
28 Calculations Using Charles’ Law A balloon has a volume of 785 mL at 21°C. IfThe temperature drops to 0°C, what is the newvolume of the balloon (P constant)?1. Set up data table:Conditions 1 Conditions 2V1 = 785 mL V2 = ?T1 = 21°C = 294 K T2 = 0°C = 273 KBe sure that you always use the Kelvin (K)temperature in gas calculations.
29 Calculations Using Charles’ Law (continued) 2. Solve Charles’ law for V2V1 = V2T T2V2 = V1 T2T1V2 = 785 mL x K = 729 mL294 K
30 Learning CheckA sample of oxygen gas has a volume of 420 mL at a temperature of 18°C. At what temperature (in °C) will the volume of the oxygen be 640 mL (P and n constant)?1) 443°C2) 170°C3) – 82°C
31 Solution 170°C T2 = T1V2 V1 T2 = 291 K x 640 mL = 443 K 420 mL = 443 K – 273 K = 170°C
32 Gay-Lussac’s Law: P and T The pressure exerted by a gas is directly related to the Kelvin temperature of the gas at constant V and n.P1 = P2T1 T2
33 Calculation with Gay-Lussac’s Law A gas has a pressure at 2.0 atm at 18°C. Whatis the new pressure when the temperature is62°C? (V and n constant)Set up a data table.Conditions 1 Conditions 2P1 = 2.0 atm P2 =T1 = 18°C T2 = 62°C + 273= 291 K = K?
34 Calculation with Gay-Lussac’s Law (continued) 2. Solve Gay-Lussac’s Law for P2P1 = P2T T2P2 = P1 T2T1P2 = 2.0 atm x 335 K = atm291 K
35 Learning Check Use the gas laws to complete with Increases 2) DecreasesA. Pressure _______ when V decreases.B. When T decreases, V _______.C. Pressure _______ when V changesfrom 12.0 L to 24.0 L.D. Volume _______when T changes from15.0 °C to 45.0°C.
36 Solution Use the gas laws to complete with 1) Increases 2) Decreases A. Pressure 1) Increases, when V decreases.B. When T decreases, V 2) Decreases.C. Pressure 2) Decreases when V changesfrom 12.0 L to 24.0 LD. Volume 1) Increases when T changes from15.0 °C to 45.0°C
38 Combined Gas LawThe combined gas law uses Boyle’s Law, Charles’ Law, and Gay-Lussac’s Law (n is constant).P1 V1 = P2 V2T1 T2
39 Combined Gas Law Calculation A sample of helium gas has a volume of L, a pressure of atm and a temperature of 29°C. At what temperature (°C) will the helium have a volume of 90.0 mL and a pressure of 3.20 atm (n constant)?1. Set up Data TableConditions 1 Conditions 2P1 = atm P2 = atmV1 = L (180 mL) V2 = 90.0 mLT1 = 29°C = 302 K T2 = ??
40 Combined Gas Law Calculation (continued) 2. Solve for T2 P1 V1 = P2 V2T1 T2T2 = T1 P2V2P1V1T2 = 302 K x atm x mL = K0.800 atm mLT2 = 604 K – = °C
41 Learning CheckA gas has a volume of 675 mL at 35°C and atm pressure. What is the volume(mL) of the gas at –95°C and a pressure of 802 mm Hg (n constant)?
42 Solution Data Table T1 = 308 K T2 = -95°C + 273 = 178K V1 = 675 mL V2 = ???P1 = 646 mm Hg P2 = 802 mm HgSolve for T2V2 = V1 P1 T2P2T1V2 = mL x 646 mm Hg x 178K = mL mm Hg x 308K
43 Avogadro's Law: Volume and Moles The volume of a gas is directly related to the number of moles of gas when T and P are constant. V1 = V2 n n2
44 Learning CheckIf 0.75 mole of helium gas occupies a volume of 1.5 L, what volume will 1.2 moles of helium occupy at the same temperature and pressure?1) L2) 1.8 L3) 2.4 L
45 Solution 3) 2.4 L Conditions 1 Conditions 2 V1 = 1.5 L V2 = ??? n1 = mole He n2 = 1.2 moles HeV2 = V1n n1 V2 = 1.5 L x moles He = L0.75 mole He
46 STPThe volumes of gases can be compared when they have the same conditions of temperature and pressure (STP).Standard temperature (T)0°C or 273 KStandard pressure (P)1 atm (760 mm Hg)
47 Molar Volume At STP, 1 mole of a gas occupies a volume of 22.4 L. The volume of one mole of a gas is called the molar volume.
48 Molar Volume as a Conversion Factor The molar volume at STP can be used to form conversion factors L and mole 1 mole L
49 Learning Check A. What is the volume at STP of 4.00 g of CH4? 1) L 2) L 3) 44.8 LB. How many grams of He are present in 8.00 L of gas at STP?1) g 2) g 3) 1.43 g
50 Solution 1) 5.60 L 4.00 g CH4 x 1 mole CH4 x 22.4 L (STP) = 5.60 L 16.0 g CH mole CH4B. 3) 1.43 g8.00 L x 1 mole He x g He = g He22.4 L mole He
51 Ideal Gas LawThe relationship between the four properties (P, V, n, and T) of gases can be written equal to a constant R.PV = RnTRearranging this expression gives the expression called the ideal gas law.PV = nRT
52 Universal Gas Constant, R The universal gas constant, R, can be calculated using the molar volume of a gas at STP.At STP (273 K and 1.00 atm), 1 mole of a gas occupies 22.4 L.P VR = PV = (1.00 atm)(22.4 L) nT (1 mole) (273K)n T= L atmmole KNote there are four units associated with R.
53 Learning CheckAnother value for the universal gas constant is obtained using mm Hg for the STP pressure. What is the value of R when a pressure of 760 mm Hg is placed in the R value expression?
54 Solution What is the value of R when the STP value for P is 760 mmHg? R = PV = (760 mm Hg) (22.4 L)nT (1 mole) (273K)= L mm Hg mole K
55 Learning CheckDinitrogen oxide (N2O), laughing gas, is used by dentists as an anesthetic. If a 20.0 L tank of laughing gas contains 2.8 moles N2O at 23°C, what is the pressure (mm Hg) in the tank?
56 Solution1. Adjust the units of the given properties to match the units of R.V = 20.0 L, T = 296 K, n = 2.8 moles, P = ?2. Rearrange the ideal gas law for P.P = nRTVP = (2.8 moles)(62.4 L mm Hg)(296 K)(20.0 L) (mole K)= 2.6 x 103 mm Hg
57 Learning CheckA cylinder contains 5.0 L of O2 at 20.0°C and 0.85 atm. How many grams of oxygen are in the cylinder?
58 Solution 1. Determine the given properties. P = 0.85 atm, V = 5.0 L, T = 293 K, n (or g =?)2. Rearrange the ideal gas law for n (moles).n = PVRT= (0.85 atm)(5.0 L)(mole K) = mole O2(0.0821atm L)(293 K)3. Convert moles to grams using molar mass.= mole O2 x g O2 = 5.8 g O21 mole O2
59 Molar Mass of a GasWhat is the molar mass of a gas if g of the gasoccupy 215 mL at atm and 30.0°C?1. Solve for the moles (n) of gas.n = PV = (0.813 atm) (0.215 L) RT ( L atm/mole K)(303K)= mole2. Set up the molar mass relationship.Molar mass = g = g = g/mole mole mole
60 Gases in EquationsThe amounts of gases reacted or produced in a chemical reaction can be calculated using the ideal gas law and mole factors.Problem:What volume (L) of Cl2 gas at 1.2 atm and 27°C is needed to completely react with 1.5 g of aluminum?2Al(s) Cl2 (g) AlCl3(s)
61 Gases in Equations (continued) 2Al(s) + 3Cl2 (g) AlCl3(s)1.5 g ? L 1.2 atm, 300KCalculate the moles of Cl2 needed.1.5 g Al x 1 mole Al x 3 moles Cl2 = mole Cl227.0 g Al moles Al2. Place the moles Cl2 in the ideal gas equation.V = nRT = (0.083 mole Cl2)( Latm/moleK)(300K)P atm= 1.7 L Cl2
62 Learning CheckWhat volume (L) of O2 at 24°C and atm are needed to react with 28.0 g NH3?4NH3(g) + 5O2(g) NO(g) + 6H2O(g)
63 Solution 1. Calculate the moles of O2 needed. 28.0 g NH3 x 1 mole NH3 x 5 mole O217.0 g NH mole NH3= 2.06 mole O22. Place the moles O2 in the ideal gas equation.V = nRT = (2.06 moles)( L atm/moleK)(297K) P atm= L O2
64 Partial PressureIn a mixture of gases, the partial pressure of each gas is the pressure that gas would exert if it were by itself in the container.
65 Dalton’s Law of Partial Pressures The total pressure exerted by a gas mixture is the sum of the partial pressures of the gases in that mixture.PT = P1 + P
66 Partial PressuresThe total pressure of a gas mixture depends on the total number of gas particles, not on the types of particles.
67 Total PressureFor example, at STP, one mole of gas particles in a volume of 22.4 L will exert the same pressure as one mole of a mixture of gas particles in 22.4 L.V = 22.4 L1.0 mole N20.4 mole O20.6 mole He1.0 mole0.5 mole O20.3 mole He0.2 mole Ar1.0 mole1.0 atm1.0 atm1.0 atm
68 Learning CheckA scuba tank contains O2 with a pressure of atm and He at 855 mm Hg. What is the total pressure in mm Hg in the tank?
69 Solution 1. Convert the pressure in atm to mm Hg 0.450 atm x 760 mm Hg = 342 mm Hg = PO 1 atm2. Calculate the sum of the partial pressures.Ptotal = PO + PHe2Ptotal = 342 mm Hg mm Hg= mm Hg
71 Health Note: Scuba Diving When a scuba diver makes a deep dive, the increased pressure causes more N2 (g) to dissolve in the blood.If a diver rises too fast, the dissolved N2 will form bubbles in the blood, a dangerous and painful condition called "the bends."Helium, which does not dissolve in the blood, is mixed with O2 to prepare breathing mixtures for deep descents.
72 Learning CheckFor a deep dive, some scuba divers are using a mixture of helium and oxygen gases with a pressure of atm. If the oxygen has a partial pressure of 1280 mm Hg, what is the partial pressure of the helium?1) 520 mm Hg2) mm Hg3) mm Hg
73 Solution 4800 mm Hg PTotal = 8.00 atm x 760 mm Hg = 6080 mm Hg 1 atm PTotal = PO PHe2PHe = PTotal - POPHe = mm Hg mm Hg= mm Hg