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Chapter 10- Gases Sections 10.1 - 10.4 Homework assignment #1
Characteristics of Gases ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Our atmosphere is a mixture of  N 2  (78%) O 2  (21%) Ar (0.9%) others
 
[object Object]
[object Object],Pressure ,[object Object],P  = F A
Units of Pressure ,[object Object],[object Object],[object Object],[object Object]
Units of Pressure ,[object Object],[object Object],[object Object],[object Object]
 
[object Object]
Manometer ,[object Object]
What is the pressure of the gas in the bulb? ,[object Object],[object Object],[object Object],[object Object],[object Object]
What is the pressure of the gas in the bulb? ,[object Object],[object Object],[object Object],[object Object],[object Object]
Standard Pressure ,[object Object],[object Object],[object Object],[object Object],[object Object]
Boyle’s Law ,[object Object]
Boyle’s Law
As  P  and  V  are inversely proportional ,[object Object],[object Object],[object Object],[object Object],PV  =  k
 
 
[object Object]
A gas initially at 2.0 atm is in an adjustable volume container of 10. L in volume.  If the pressure is decreased to 0.50 atm, what is the new volume?   ,[object Object],[object Object],[object Object],[object Object]
Correct Answer: P V 1 constant   Thus, PV = constant and  P 1 V 1 =P 2 V 2 2.00 atm(10. L) = 0.50 atm ( V final ) V final  = 2.00 atm(10. L)/0.50 atm = 40. L ,[object Object],[object Object],[object Object],[object Object]
Charles’s Law ,[object Object],A plot of  V  versus  T  will be a straight line. ,[object Object],V T =  k
 
[object Object]
Assuming pressure is held constant, to what volume will a balloon initially at 1.0 L change if its temperature is decreased from 300 K to 75 K?   ,[object Object],[object Object],[object Object],[object Object]
Correct Answer: T V   constant Thus, T/V = constant  and T 1 /V 1 =T 2 /V 2 1.0 L/300 K = ( V final )/75 K V final  = 75 K/(1.0 L)300 K  = 0.25 L ,[object Object],[object Object],[object Object],[object Object]
Avogadro’s Law ,[object Object],[object Object],V  =  kn
Ideal-Gas Equation ,[object Object],[object Object],[object Object],[object Object],[object Object],V    nT P
Ideal-Gas Equation ,[object Object]
Ideal-Gas Equation ,[object Object],then becomes or PV  =  nRT nT P V    nT P V  =  R
Ideal-Gas Equation
At standard temperature and pressure, how many moles of gas are present in a box with a volume of 112 L?   ,[object Object],[object Object],[object Object],[object Object]
Correct Answer:       L 22.41 atm 1.000 K 273.15 K L·atm/mol· 0.08206 mol 1     P nRT V nRT PV Thus, at STP 22.41  L   =  112  L 1.00 mol  n  n  = 5.00 moles ,[object Object],[object Object],[object Object],[object Object]
If 250 mL of NO is placed in a flask with O 2 , what volume of O 2  is needed for complete reaction? ,[object Object],[object Object],[object Object],[object Object],[object Object],2 NO   ( g )  + O 2  ( g )  2 NO 2  ( g )
If 250 mL of NO is placed in a flask with O 2 , what volume of O 2  is needed for complete reaction? ,[object Object],[object Object],[object Object],[object Object],[object Object],2 NO   ( g )  + O 2  ( g )  2 NO 2  ( g )
Section 10.4 - 10.6 Second homework assignment
Densities of Gases ,[object Object],n V P RT =
[object Object],[object Object],Densities of Gases n        =  m
[object Object],[object Object],Densities of Gases ,[object Object],n        =  m P  RT m V =
Densities of Gases ,[object Object],[object Object],[object Object],P  RT m V = d  =
 
[object Object]
If an equal mass of each gas is put into a separate balloon, which will have the greatest volume?  Assume that they are all the same temperature and pressure. ,[object Object],[object Object],[object Object],[object Object],[object Object]
If an equal mass of each gas is put into a separate balloon, which will have the greatest volume?  Assume that they are all the same temperature and pressure. ,[object Object],[object Object],[object Object],[object Object],[object Object]
Molecular Mass ,[object Object],Becomes P  RT d  = dRT P    =
For two gases under identical conditions, moles are proportional to volume. Stoichiometry can be done with volumes of gases under identical conditions. 2H 2 (g)   + O 2 (g)     2H 2 O (g) 2L   1L 2L 44.8L   22.4L  44.8L
 
N 2 ( g )  + 3 H 2 ( g )   2 NH 3 ( g ) At STP, 16 L of N 2  and 48 L of H 2  are mixed.  Assuming all the reactants are consumed, how many L of NH 3  will be produced? ,[object Object],[object Object],[object Object],[object Object],[object Object]
Correct Answer: According to Avogadro’s law,  mole ratios in the chemical  equation will be volume ratios  under identical  conditions.  Because the reactants are in a  stoichiometric 3:1 volume ratio, the  product will have stoichiometric  equivalence.  Thus, (16 L N 2 )(2 mol NH 3 /1 mol N 2 ) = 32 L NH 3 ) (constant  T P, n V  ,[object Object],[object Object],[object Object],[object Object],[object Object]
Dalton’s Law of Partial Pressures ,[object Object],[object Object],[object Object]
 
[object Object]
Solution  Because each gas behaves independently, we can use the ideal-gas equation to calculate the pressure that each would exert if the other were not present. The total pressure is the sum of these two partial pressures. SAMPLE EXERCISE 10.10  Applying Dalton’s Law of Partial Pressures A gaseous mixture made from 6.00 g O 2  and 9.00 g CH 4  is placed in a 15.0-L vessel at 0°C. What is the partial pressure of each gas, and what is the total pressure in the vessel? Solve:  We must first convert the mass of each gas to moles: We can now use the ideal-gas equation to calculate the partial pressure of each gas: According to Dalton’s law (Equation 10.12), the total pressure in the vessel is the sum of the partial pressures:
Partial Pressures ,[object Object],[object Object]
Mole Fraction For gas mixtures it is useful to define the mole fraction (X)  X 1 = n 1 /n total The mole fraction or mole percent can be used to determine the partial pressures of gases in a mix IF the total pressure is known.
A container holds a mixture of oxygen, neon, and helium gases whose partial pressures are 150 torr, 300 torr, and 450 torr, respectively. The mole fraction of neon is   ,[object Object],[object Object],[object Object],[object Object],total i P P i  
Correct Answer: X i  =  P i / P total   X i  = (300 torr)/(150 + 300 + 450) torr X i  = 300 torr/900 torr = 0.33 total i P P i   ,[object Object],[object Object],[object Object],[object Object]
Section10.7 - 10.9 Homework due Thursday Gas Stoich Lab on Thursday Quiz 10 Friday Gas Stoich Lab due Monday
Ideal Gas  LAW  (PV=nRT) -  A short statement that  describes   what happens (how gases behave) Kinetic Molecular  THEORY -  Explains  why the gas laws are  observed.
Kinetic-Molecular Theory ,[object Object]
Main Tenets of Kinetic-Molecular Theory ,[object Object]
Main Tenets of Kinetic-Molecular Theory ,[object Object],[object Object]
Main Tenets of Kinetic-Molecular Theory ,[object Object]
Main Tenets of Kinetic-Molecular Theory ,[object Object],[object Object]
 
[object Object]
Gas pressure is caused by gas molecules colliding with a surface.
Gas pressure is caused by gas molecules colliding with a surface. Gas Pressure increases when the frequency of collisions or the force of collisions increases.
Gas pressure is caused by gas molecules colliding with a surface. Gas Pressure increases when the frequency of collisions or the force of collisions increases. Collision frequency increases when temperature increases because molecules move faster.
Gas pressure is caused by gas molecules colliding with a surface. Gas Pressure increases when the frequency of collisions or the force of collisions increases. Collision frequency increases when temperature increases because molecules move faster.  Collision frequency also increases when the amount of gas is increased or when the volume of a gas is decreased.
Gas pressure is caused by gas molecules colliding with a surface. Gas Pressure increases when the frequency of collisions or the force of collisions increases. Collision frequency increases when temperature increases because molecules move faster.  Collision frequency also increases when the amount of gas is increased or when the volume of a gas is decreased. The force of collisions is increased with rising temperature.
Using KMT to explain gas laws Avogadro’s- More particles make more collisions, increasing pressure.
Using KMT to explain gas laws Avogadro’s- More particles make more collisions, increasing pressure.  Boyle’s- More volume means less collisions because molecules must travel farther. Less collisions create less pressure.
Using KMT to explain gas laws Avogadro’s- More particles make more collisions, increasing pressure.  Boyle’s- More volume means less collisions because molecules must travel farther. Less collisions create less pressure. Gay-Lussac’s- Higher temp means faster moving particles. Particles collide more often and with more force. More collisions create more pressure.
Using KMT to explain gas laws Avogadro’s- More particles make more collisions, increasing pressure.  Boyle’s- More volume means less collisions because molecules must travel farther. Less collisions create less pressure. Gay-Lussac’s- Higher temp means faster moving particles. Particles collide more often and with more force. More collisions create more pressure. Charles’- Higher temp means faster moving particles. Particles collide more often and with more force. To maintain constant pressure, volume must increase.
A sample of He gas initially at STP is compressed to a smaller volume at constant temperature.  What effect does this have on the rms speed of the atoms? ,[object Object],[object Object],[object Object]
Correct Answer: The rms speed is directly proportional to the square root of the temperature, which does not change in this example. ,[object Object],[object Object],[object Object]
KE = ½  m  v 2 At the same temperature two gases have equal KE.
KE = ½  m  v 2 At the same temperature two gases have equal KE. Heavier gases must have slower velocity to keep KE constant at a given temperature. Heavier gases move slower
Arrange the gases according to increasing molecular speed.   ,[object Object],[object Object],[object Object],[object Object],[object Object],He (25°C)  He (100°C)  Ne (25°C)  Ne (0°C)
Arrange the gases according to increasing molecular speed.   ,[object Object],[object Object],[object Object],[object Object],[object Object],He (25°C)  He (100°C)  Ne (25°C)  Ne (0°C)
Effusion ,[object Object]
Effusion ,[object Object],Compare effusion rates of two gases r 1 /r 2  =   M 2 /M 1 r = rate of effusion M = molar mass
An unknown gas effuses at half the rate of helium.  This gas is likely to be which of the following? ,[object Object],[object Object],[object Object],[object Object],[object Object]
Correct Answer: ( r 1 /r 2 ) 2  = M 2 /M 1 M 2 = ( r 1 / r 2 ) 2 M 1 M 2 = (2/1) 2 (4.0 g/mol) = 16.0 g/mol Therefore it could be CH 4 ,[object Object],[object Object],[object Object],[object Object],[object Object]
Diffusion ,[object Object],[object Object]
Diffusion occurs slower than effusion because the  mean free path  of molecules is reduced. Diffusion
 
[object Object]
[object Object],Increasing temperature has no effect on mean free path.
SAMPLE EXERCISE 10.15  Applying Graham’s Law  An unknown gas composed of homonuclear diatomic molecules effuses at a rate that is only 0.355 times that of  O 2  at the same temperature. Calculate the molar mass of the unknown, and identify it. We can use Graham’s law of effusion to determine the molar mass of the unknown gas.  If we let  r x  and  represent the rate of effusion and molar mass of the unknown gas, Graham’s Law can be written as follows: From the information given, Thus,
Because we are told that the unknown gas is composed of homonuclear diatomic molecules, it must be an element. The molar mass must represent twice the atomic weight of the atoms in the unknown gas. We conclude that the unknown gas is I 2 . SAMPLE EXERCISE 10.15   continued We now solve for the unknown molar mass,
Real Gases ,[object Object]
Real Gases ,[object Object]
Deviations from Ideal Behavior ,[object Object]
Deviations from Ideal Behavior ,[object Object]
Deviations from Ideal Behavior ,[object Object]
Deviations from Ideal Behavior ,[object Object]
Deviations from Ideal Behavior ,[object Object]
 
[object Object],[object Object]
Ideal Gases ,[object Object]
Ideal Gases ,[object Object],[object Object]
 
[object Object]
Corrections for Nonideal Behavior ,[object Object],[object Object]
The van der Waals Equation ) ( V   −   nb ) =  nRT n 2 a V 2 ( P  +

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Chapter 10 Lecture- Gases

  • 1. Chapter 10- Gases Sections 10.1 - 10.4 Homework assignment #1
  • 2.
  • 3. Our atmosphere is a mixture of N 2 (78%) O 2 (21%) Ar (0.9%) others
  • 4.  
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  • 36.
  • 37. Section 10.4 - 10.6 Second homework assignment
  • 38.
  • 39.
  • 40.
  • 41.
  • 42.  
  • 43.
  • 44.
  • 45.
  • 46.
  • 47. For two gases under identical conditions, moles are proportional to volume. Stoichiometry can be done with volumes of gases under identical conditions. 2H 2 (g) + O 2 (g)  2H 2 O (g) 2L 1L 2L 44.8L 22.4L 44.8L
  • 48.  
  • 49.
  • 50.
  • 51.
  • 52.  
  • 53.
  • 54. Solution Because each gas behaves independently, we can use the ideal-gas equation to calculate the pressure that each would exert if the other were not present. The total pressure is the sum of these two partial pressures. SAMPLE EXERCISE 10.10 Applying Dalton’s Law of Partial Pressures A gaseous mixture made from 6.00 g O 2 and 9.00 g CH 4 is placed in a 15.0-L vessel at 0°C. What is the partial pressure of each gas, and what is the total pressure in the vessel? Solve:  We must first convert the mass of each gas to moles: We can now use the ideal-gas equation to calculate the partial pressure of each gas: According to Dalton’s law (Equation 10.12), the total pressure in the vessel is the sum of the partial pressures:
  • 55.
  • 56. Mole Fraction For gas mixtures it is useful to define the mole fraction (X) X 1 = n 1 /n total The mole fraction or mole percent can be used to determine the partial pressures of gases in a mix IF the total pressure is known.
  • 57.
  • 58.
  • 59. Section10.7 - 10.9 Homework due Thursday Gas Stoich Lab on Thursday Quiz 10 Friday Gas Stoich Lab due Monday
  • 60. Ideal Gas LAW (PV=nRT) - A short statement that describes what happens (how gases behave) Kinetic Molecular THEORY - Explains why the gas laws are observed.
  • 61.
  • 62.
  • 63.
  • 64.
  • 65.
  • 66.  
  • 67.
  • 68. Gas pressure is caused by gas molecules colliding with a surface.
  • 69. Gas pressure is caused by gas molecules colliding with a surface. Gas Pressure increases when the frequency of collisions or the force of collisions increases.
  • 70. Gas pressure is caused by gas molecules colliding with a surface. Gas Pressure increases when the frequency of collisions or the force of collisions increases. Collision frequency increases when temperature increases because molecules move faster.
  • 71. Gas pressure is caused by gas molecules colliding with a surface. Gas Pressure increases when the frequency of collisions or the force of collisions increases. Collision frequency increases when temperature increases because molecules move faster. Collision frequency also increases when the amount of gas is increased or when the volume of a gas is decreased.
  • 72. Gas pressure is caused by gas molecules colliding with a surface. Gas Pressure increases when the frequency of collisions or the force of collisions increases. Collision frequency increases when temperature increases because molecules move faster. Collision frequency also increases when the amount of gas is increased or when the volume of a gas is decreased. The force of collisions is increased with rising temperature.
  • 73. Using KMT to explain gas laws Avogadro’s- More particles make more collisions, increasing pressure.
  • 74. Using KMT to explain gas laws Avogadro’s- More particles make more collisions, increasing pressure. Boyle’s- More volume means less collisions because molecules must travel farther. Less collisions create less pressure.
  • 75. Using KMT to explain gas laws Avogadro’s- More particles make more collisions, increasing pressure. Boyle’s- More volume means less collisions because molecules must travel farther. Less collisions create less pressure. Gay-Lussac’s- Higher temp means faster moving particles. Particles collide more often and with more force. More collisions create more pressure.
  • 76. Using KMT to explain gas laws Avogadro’s- More particles make more collisions, increasing pressure. Boyle’s- More volume means less collisions because molecules must travel farther. Less collisions create less pressure. Gay-Lussac’s- Higher temp means faster moving particles. Particles collide more often and with more force. More collisions create more pressure. Charles’- Higher temp means faster moving particles. Particles collide more often and with more force. To maintain constant pressure, volume must increase.
  • 77.
  • 78.
  • 79. KE = ½ m v 2 At the same temperature two gases have equal KE.
  • 80. KE = ½ m v 2 At the same temperature two gases have equal KE. Heavier gases must have slower velocity to keep KE constant at a given temperature. Heavier gases move slower
  • 81.
  • 82.
  • 83.
  • 84.
  • 85.
  • 86.
  • 87.
  • 88. Diffusion occurs slower than effusion because the mean free path of molecules is reduced. Diffusion
  • 89.  
  • 90.
  • 91.
  • 92. SAMPLE EXERCISE 10.15 Applying Graham’s Law An unknown gas composed of homonuclear diatomic molecules effuses at a rate that is only 0.355 times that of O 2 at the same temperature. Calculate the molar mass of the unknown, and identify it. We can use Graham’s law of effusion to determine the molar mass of the unknown gas. If we let r x and represent the rate of effusion and molar mass of the unknown gas, Graham’s Law can be written as follows: From the information given, Thus,
  • 93. Because we are told that the unknown gas is composed of homonuclear diatomic molecules, it must be an element. The molar mass must represent twice the atomic weight of the atoms in the unknown gas. We conclude that the unknown gas is I 2 . SAMPLE EXERCISE 10.15 continued We now solve for the unknown molar mass,
  • 94.
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  • 100.
  • 101.  
  • 102.
  • 103.
  • 104.
  • 105.  
  • 106.
  • 107.
  • 108. The van der Waals Equation ) ( V − nb ) = nRT n 2 a V 2 ( P +