1. Chapter 5 Gases Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA

2. The Structure of a Gas
Gases are composed of particles that are flying around very fast in their container(s)
The particles in straight lines until they encounter either the container wall or another particle, then they bounce off
If you were able to take a snapshot of the particles in a gas, you would find that there is a lot of empty space in there
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3. Gases Pushing Gas molecules are constantly in motion
As they move and strike a surface, they push on that surface
push = force
If we could measure the total amount of force exerted by gas molecules hitting the entire surface at any one instant, we would know the pressure the gas is exerting
pressure = force per unit area
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4. The Effect of Gas Pressure
The pressure exerted by a gas can cause some amazing and startling effects
Whenever there is a pressure difference, a gas will flow from an area of high pressure to an area of low pressure
the bigger the difference in pressure, the stronger the flow of the gas
If there is something in the gas’s path, the gas will try to push it along as the gas flows
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5. Air Pressure
The atmosphere exerts a pressure on everything it contacts
the atmosphere goes up about 370 miles, but 80% is in the first 10 miles from the earth’s surface
This is the same pressure that a column of water would exert if it were about 10.3 m high
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6. Atmospheric Pressure Effects
Differences in air pressure result in weather and wind patterns
The higher in the atmosphere you climb, the lower the atmospheric pressure is around you
at the surface the atmospheric pressure is 14.7 psi, but at 10,000 ft it is only 10.0 psi
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7. Pressure Imbalance in the Ear
If there is a difference
in pressure across
the eardrum membrane,
the membrane will be
pushed out – what we
commonly call a
“popped eardrum”
picture from Tro Intro Chem 2nd ed.
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8. The Pressure of a Gas
Gas pressure is a result of the constant movement of the gas molecules and their collisions with the surfaces around them
The pressure of a gas depends on several factors
number of gas particles in a given volume
volume of the container
average speed of the gas particles
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9. Measuring Air Pressure
We measure air pressure with a barometer
Column of mercury supported by air pressure
Force of the air on the surface of the mercury counter balances the force of gravity on the column of mercury
gravity
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10. Practice – What happens to the height of the column of mercury in a mercury barometer as you climb to the top of a mountain?
The height of the column increases because atmospheric pressure decreases with increasing altitude
The height of the column decreases because atmospheric pressure decreases with increasing altitude
The height of the column decreases because atmospheric pressure increases with increasing altitude
The height of the column increases because atmospheric pressure increases with increasing altitude
The height of the column increases because atmospheric pressure decreases with increasing altitude
The height of the column decreases because atmospheric pressure decreases with increasing altitude
The height of the column decreases because atmospheric pressure increases with increasing altitude
The height of the column increases because atmospheric pressure increases with increasing altitude
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11. Common Units of Pressure
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12. because mmHg are smaller than psi, the answer makes sense
Example 5.1: A high-performance bicycle tire has a pressure of 132 psi. What is the pressure in mmHg?
Given:
Find:
132 psi
mmHg
Conceptual Plan:
Relationships:
1 atm = 14.7 psi, 1 atm = 760 mmHg
psi
atm
mmHg
Solution:
Check:
because mmHg are smaller than psi, the answer makes sense
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13. Practice—Convert 45.5 psi into kPa
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14. Practice—Convert 45.5 psi into kPa
Given:
Find:
645.5 psi
kPa
Conceptual Plan:
Relationships:
1 atm = 14.7 psi, 1 atm = kPa
psi
atm
kPa
Solution:
Check:
because kPa are smaller than psi, the answer makes sense
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15. Manometers
The pressure of a gas trapped in a container can be measured with an instrument called a manometer
Manometers are U-shaped tubes, partially filled with a liquid, connected to the gas sample on one side and open to the air on the other
A competition is established between the pressures of the atmosphere and the gas
The difference in the liquid levels is a measure of the difference in pressure between the gas and the atmosphere
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16. Manometer
for this sample, the gas has a larger pressure than the atmosphere, so
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17. Boyle’s Law Pressure of a gas is inversely proportional to its volume
Robert Boyle (1627–1691)
Pressure of a gas is inversely proportional to its volume
constant T and amount of gas
graph P vs V is curve
graph P vs 1/V is straight line
As P increases, V decreases by the same factor
P x V = constant
P1 x V1 = P2 x V2
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18. Boyle’s Experiment Added Hg to a J-tube with air trapped inside
Used length of air column as a measure of volume
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21. Boyle’s Experiment, P x V
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22. Boyle’s Law: A Molecular View
Pressure is caused by the molecules striking the sides of the container
When you decrease the volume of the container with the same number of molecules in the container, more molecules will hit the wall at the same instant
This results in increasing the pressure
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23. Boyle’s Law and Diving
Because water is more dense than air, for each m you dive below the surface, the pressure on your lungs increases atm
at 20 m the total pressure is 3 atm
If your tank contained air at 1 atm of pressure, you would not be able to inhale it into your lungs
you can only generate enough force to overcome about 1.06 atm
Scuba tanks have a regulator so that the air from the tank is delivered at the same pressure as the water surrounding you.
This allows you to take in air even when the outside pressure is large.
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24. Boyle’s Law and Diving
If a diver holds her breath and rises to the surface quickly, the outside pressure drops to 1 atm
According to Boyle’s law, what should happen to the volume of air in the lungs?
Because the pressure is decreasing by a factor of 3, the volume will expand by a factor of 3, causing damage to internal organs. Always Exhale When Rising!!
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25. Example 5. 2: A cylinder with a movable piston has a volume of 7
Example 5.2: A cylinder with a movable piston has a volume of 7.25 L at 4.52 atm. What is the volume at 1.21 atm?
Given:
Find:
V1 =7.25 L, P1 = 4.52 atm, P2 = 1.21 atm
V2, L
Conceptual Plan:
Relationships:
P1 ∙ V1 = P2 ∙ V2
V1, P1, P2 V2
Solution:
Check:
because P and V are inversely proportional, when the pressure decreases ~4x, the volume should increase ~4x, and it does
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26. Practice – A balloon is put in a bell jar and the pressure is reduced from 782 torr to atm. If the volume of the balloon is now 2.78 x 103 mL, what was it originally?
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27. P1 ∙ V1 = P2 ∙ V2 , 1 atm = 760 torr (exactly) V2, P1, P2 V1
A balloon is put in a bell jar and the pressure is reduced from 782 torr to atm. If the volume of the balloon is now 2.78x 103 mL, what was it originally?
Given:
Find:
V2 =2780 mL, P1 = 762 torr, P2 = atm
V1, mL
Conceptual Plan:
Relationships:
P1 ∙ V1 = P2 ∙ V2 , 1 atm = 760 torr (exactly)
V2, P1, P2 V1
Solution:
Check:
because P and V are inversely proportional, when the pressure decreases ~2x, the volume should increase ~2x, and it does
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28. Charles’s Law Volume is directly proportional to temperature
Jacques Charles (1746–1823)
Volume is directly proportional to temperature
constant P and amount of gas
graph of V vs. T is straight line
As T increases, V also increases
Kelvin T = Celsius T + 273
V = constant x T
if T measured in Kelvin
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29. If you plot volume vs. temperature for any gas at constant pressure, the points will all fall on a straight line
If the lines are extrapolated back to a volume of “0,” they all show the same temperature, − °C, called absolute zero
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30. Charles’s Law – A Molecular View
The pressure of gas inside and outside the balloon are the same
At low temperatures, the gas molecules are not moving as fast, so they don’t hit the sides of the balloon as hard – therefore the volume is small
The pressure of gas inside and outside the balloon are the same
At high temperatures, the gas molecules are moving faster, so they hit the sides of the balloon harder – causing the volume to become larger
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31. Example 5. 3: A gas has a volume of 2. 57 L at 0. 00 °C
Example 5.3: A gas has a volume of 2.57 L at 0.00 °C. What was the temperature at 2.80 L?
Given:
Find:
V1 =2.57 L, V2 = 2.80 L, t2 = 0.00 °C
t1, K and °C
Conceptual Plan:
Relationships:
V1, V2, T2 T1
Solution:
Check:
because T and V are directly proportional, when the volume decreases, the temperature should decrease, and it does
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32. Practice – The temperature inside a balloon is raised from 25
Practice – The temperature inside a balloon is raised from 25.0 °C to °C. If the volume of cold air was 10.0 L, what is the volume of hot air?
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33. The temperature inside a balloon is raised from 25. 0 °C to 250. 0 °C
The temperature inside a balloon is raised from 25.0 °C to °C. If the volume of cold air was 10.0 L, what is the volume of hot air?
Given:
Find:
V1 =10.0 L, t1 = 25.0 °C L, t2 = °C
V2, L
Conceptual Plan:
Relationships:
V1, T1, T2 V2
Solution:
Check:
when the temperature increases, the volume should increase, and it does
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34. Avogadro’s Law
Amedeo Avogadro (1776–1856)
Volume directly proportional to the number of gas molecules
V = constant x n
constant P and T
more gas molecules = larger volume
Count number of gas molecules by moles
Equal volumes of gases contain equal numbers of molecules
the gas doesn’t matter
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35. Example 5. 4:A 0. 225 mol sample of He has a volume of 4. 65 L
Example 5.4:A mol sample of He has a volume of 4.65 L. How many moles must be added to give 6.48 L?
Given:
Find:
V1 = 4.65 L, V2 = 6.48 L, n1 = mol
n2, and added moles
Conceptual Plan:
Relationships:
V1, V2, n1 n2
Solution:
Check:
because n and V are directly proportional, when the volume increases, the moles should increase, and they do
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36. Practice — If 1. 00 mole of a gas occupies 22
Practice — If 1.00 mole of a gas occupies 22.4 L at STP, what volume would moles occupy?
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37. Practice — If 1. 00 mole of a gas occupies 22
Practice — If 1.00 mole of a gas occupies 22.4 L at STP, what volume would moles occupy?
Given:
Find:
V1 =22.4 L, n1 = 1.00 mol, n2 = mol V2
Conceptual Plan:
Relationships:
V1, n1, n2 V2
Solution:
Check:
because n and V are directly proportional, when the moles decrease, the volume should decrease, and it does
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38. Ideal Gas Law
By combining the gas laws we can write a general equation
R is called the gas constant
The value of R depends on the units of P and V
we will use and convert P to atm and V to L
The other gas laws are found in the ideal gas law if two variables are kept constant
Allows us to find one of the variables if we know the other three
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39. Example 5.6: How many moles of gas are in a basketball with total pressure 24.3 psi, volume of 3.24 L at 25°C?
Given:
Find:
V = 3.24 L, P = 24.3 psi, t = 25 °C
n, mol
Conceptual Plan:
Relationships:
P, V, T, R n
Solution:
Check:
1 mole at STP occupies 22.4 L, because there is a much smaller volume than 22.4 L, we expect less than 1 mole of gas
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40. Standard Conditions
Because the volume of a gas varies with pressure and temperature, chemists have agreed on a set of conditions to report our measurements so that comparison is easy – we call these standard conditions
STP
Standard pressure = 1 atm
Standard temperature = 273 K
0 °C
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41. Practice – A gas occupies 10. 0 L at 44. 1 psi and 27 °C
Practice – A gas occupies 10.0 L at 44.1 psi and 27 °C. What volume will it occupy at standard conditions?
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42. A gas occupies 10. 0 L at 44. 1 psi and 27 °C
A gas occupies 10.0 L at 44.1 psi and 27 °C. What volume will it occupy at standard conditions?
Given:
Find:
V1 = 10.0L, P1 = 44.1 psi, t1 = 27 °C, P2 = 1.00 atm, t2 = 0 °C
V2, L
Conceptual Plan:
Relationships:
P1, V1, T1, R n
P2, n, T2, R V2
Solution:
Check:
1 mole at STP occupies 22.4 L, because there is more than 1 mole, we expect more than 22.4 L of gas
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43. Practice — Calculate the volume occupied by 637 g of SO2 (MM 64
Practice — Calculate the volume occupied by 637 g of SO2 (MM 64.07) at 6.08 x 104 mmHg and –23 °C
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44. Practice—Calculate the volume occupied by 637 g of SO2 (MM 64.07) at 6.08 x 104 mmHg and –23 °C.
Given:
Find:
mSO2 = 637 g, P = 6.08 x 104 mmHg, t = −23 °C,
V, L
Conceptual Plan:
Relationships: g n
P, n, T, R V
Solution:
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45. Molar Volume
Solving the ideal gas equation for the volume of 1 mol of gas at STP gives 22.4 L
6.022 x 1023 molecules of gas
notice: the gas is immaterial
We call the volume of 1 mole of gas at STP the molar volume
it is important to recognize that one mole measures of different gases have different masses, even though they have the same volume
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46. Molar Volume
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47. Practice — How many liters of STP can be made from the decomposition of g of PbO2? 2 PbO2(s) → 2 PbO(s) + O2(g) (PbO2 = 239.2, O2 = 32.00)
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48. 1 mol O2 = 22.4 L, 1 mol PbO2 = 239.2g, 1 mol O2 ≡ 2 mol PbO2
Practice — How many liters of STP can be made from the decomposition of g of PbO2? 2 PbO2(s) → 2 PbO(s) + O2(g)
Given:
Find:
100.0 g PbO2, 2 PbO2 → 2 PbO + O2
L O2
Conceptual Plan:
Relationships:
1 mol O2 = 22.4 L, 1 mol PbO2 = 239.2g, 1 mol O2 ≡ 2 mol PbO2
L O2
mol PbO2
g PbO2
mol O2
Solution:
because less than 1 mole PbO2, and ½ moles of O2 as PbO2, the number makes sense
Check:
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49. Density at Standard Conditions
Density is the ratio of mass to volume
Density of a gas is generally given in g/L
The mass of 1 mole = molar mass
The volume of 1 mole at STP = 22.4 L
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50. Practice – Calculate the density of N2(g) at STP
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51. Practice – Calculate the density of N2(g) at STP
Given:
Find:
N2,
dN2, g/L MM d
Conceptual Plan:
Relationships:
Solution:
Check:
the units and number are reasonable
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52. Gas Density Density is directly proportional to molar mass
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53. Example 5.7: Calculate the density of N2 at 125°C and 755 mmHg
Given:
Find:
P = 755 mmHg, t = 125 °C,
dN2, g/L
P, MM, T, R d
Conceptual Plan:
Relationships:
Solution:
Check:
because the density of N2 is 1.25 g/L at STP, we expect the density to be lower when the temperature is raised, and it is
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54. Practice — Calculate the density of a gas at 775 torr and 27 °C if 0
Practice — Calculate the density of a gas at 775 torr and 27 °C if moles weighs g
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55. the unit is correct and the value is reasonable
Practice — Calculate the density of a gas at 775 torr and 27 °C if moles weighs g
Given:
Find:
m=9.988g, n=0.250 mol, P=775 mmHg, t=27 °C,
density, g/L
m = 9.988g, n = mol, P = atm, T = 300.K
density, g/L
P, n, T, R V
V, m d
Conceptual Plan:
Relationships:
Solution:
Check:
the unit is correct and the value is reasonable
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56. Molar Mass of a Gas
One of the methods chemists use to determine the molar mass of an unknown substance is to heat a weighed sample until it becomes a gas, measure the temperature, pressure, and volume, and use the ideal gas law
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57. the unit is correct, the value is reasonable
Example 5.8: Calculate the molar mass of a gas with mass g that has a volume of L at 55°C and 886 mmHg
Given:
Find:
m=0.311g, V=0.225 L, P= atm, T=328 K,
molar mass, g/mol
m=0.311g, V=0.225 L, P=886 mmHg, t=55°C,
molar mass, g/mol
Conceptual Plan:
Relationships:
P, V, T, R n
n, m MM
Solution:
Check:
the unit is correct, the value is reasonable
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58. Practice — What is the molar mass of a gas if 12
Practice — What is the molar mass of a gas if 12.0 g occupies 197 L at 3.80 x 102 torr and 127 °C?
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59. the unit is correct and the value is reasonable
Practice — What is the molar mass of a gas if 12.0 g occupies 197 L at 380 torr and 127 °C?
Given:
Find:
m = 12.0g, V = 197 L, P = 0.50 atm, T =400 K,
molar mass, g/mol
m=12.0 g, V= 197 L, P=380 torr, t=127°C,
molar mass, g/mol
P, V, T, R n
n, m MM
Conceptual Plan:
Relationships:
Solution:
Check:
the unit is correct and the value is reasonable
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60. Mixtures of Gases
When gases are mixed together, their molecules behave independent of each other
all the gases in the mixture have the same volume
all completely fill the container  each gas’s volume = the volume of the container
all gases in the mixture are at the same temperature
therefore they have the same average kinetic energy
Therefore, in certain applications, the mixture can be thought of as one gas
even though air is a mixture, we can measure the pressure, volume, and temperature of air as if it were a pure substance
we can calculate the total moles of molecules in an air sample, knowing P, V, and T, even though they are different molecules
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61. Composition of Dry Air
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62. Partial Pressure
The pressure of a single gas in a mixture of gases is called its partial pressure
We can calculate the partial pressure of a gas if
we know what fraction of the mixture it composes and the total pressure
or, we know the number of moles of the gas in a container of known volume and temperature
The sum of the partial pressures of all the gases in the mixture equals the total pressure
Dalton’s Law of Partial Pressures
because the gases behave independently
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63. The partial pressure of each gas in a mixture can be calculated using the ideal gas law
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64. Example 5.9: Determine the mass of Ar in the mixture
PHe=341 mmHg, PNe=112 mmHg, Ptot = 662 mmHg,
V = 1.00 L, T=298 K
massAr, g
PAr = atm, V = 1.00 L, T=298 K
massAr, g
Given:
Find:
Conceptual Plan:
Relationships:
Ptot, PHe, PNe
PAr
PAr, V, T
nAr
mAr
PAr = Ptot – (PHe + PNe)
Solution:
Check:
the units are correct, the value is reasonable
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65. Practice – Find the partial pressure of neon in a mixture with total pressure 3.9 atm, volume 8.7 L, temperature 598 K, and 0.17 moles Xe.
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66. the unit is correct, the value is reasonable
Find the partial pressure of neon in a mixture with total pressure 3.9 atm, volume 8.7 L, temperature 598 K, and 0.17 moles Xe
Given:
Find:
Ptot = 3.9 atm, V = 8.7 L, T = 598 K, Xe = 0.17 mol
PNe, atm
nXe, V, T, R
PXe
Ptot, PXe
PNe
Conceptual Plan:
Relationships:
Solution:
the unit is correct, the value is reasonable
Check:
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67. Mole Fraction
The fraction of the total pressure that a single gas contributes is equal to the fraction of the total number of moles that a single gas contributes
The ratio of the moles of a single component to the total number of moles in the mixture is called the mole fraction, c
for gases, = volume % / 100%
The partial pressure of a gas is equal to the mole fraction of that gas times the total pressure
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68. Ex 5. 10: Find the mole fractions and partial pressures in a 12
Ex 5.10: Find the mole fractions and partial pressures in a 12.5 L tank with 24.2 g He and 4.32 g O2 at 298 K
Given:
Find:
nHe = 6.05 mol, nO2 = mol, V = 12.5 L, T = 298 K
cHe= , cO2= , PHe, atm, PO2, atm, Ptotal, atm
mHe = 24.2 g, mO2 = 4.32 g, V = 12.5 L, T = 298 K
cHe, cO2, PHe, atm, PO2, atm, Ptotal, atm
mgas
ngas
cgas
ntot, V, T, R
Ptot
Conceptual Plan:
Relationships:
cgas, Ptotal
Pgas
Solution:
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69. Practice – Find the mole fraction of neon in a mixture with total pressure 3.9 atm, volume 8.7 L, temperature 598 K, and 0.17 moles Xe
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70. the unit is correct, the value is reasonable
Find the mole fraction of neon in a mixture with total pressure 3.9 atm, volume 8.7 L, temperature 598 K, and 0.17 moles Xe
Given:
Find:
Ptot = 3.9 atm, V = 8.7 L, T = 598 K, Xe = 0.17 mol
PNe, atm
Conceptual Plan:
Relationships:
nXe, V, T, R
PXe
Ptot, PXe
PNe
Ptot, PNe
cNe
Solution:
the unit is correct, the value is reasonable
Check:
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71. Collecting Gases
Gases are often collected by having them displace water from a container
The problem is that because water evaporates, there is also water vapor in the collected gas
The partial pressure of the water vapor, called the vapor pressure, depends only on the temperature
so you can use a table to find out the partial pressure of the water vapor in the gas you collect
if you collect a gas sample with a total pressure of mmHg* at 25 °C, the partial pressure of the water vapor will be mmHg – so the partial pressure of the dry gas will be mmHg
Table 5.4*
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72. Collecting Gas by Water Displacement
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73. Vapor Pressure of Water
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74. Ex 5.11: 1.02 L of O2 collected over water at 293 K with a total pressure of mmHg. Find mass O2.
Given:
Find:
V=1.02 L, PO2= atm, T=293 K
mass O2, g
V=1.02 L, P=755.2 mmHg, T=293 K
mass O2, g
Ptot, PH2O
PO2
PO2,V,T
nO2
gO2
Conceptual Plan:
Relationships:
Solution:
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75. Practice – 0. 12 moles of H2 is collected over water in a 10
Practice – 0.12 moles of H2 is collected over water in a 10.0 L container at 323 K. Find the total pressure.
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76. 12 moles of H2 is collected over water in a 10. 0 L container at 323 K
0.12 moles of H2 is collected over water in a 10.0 L container at 323 K. Find the total pressure.
Given:
Find:
V=10.0 L, nH2 = 0.12 mol, T = 323 K
Ptotal, mmHg
Conceptual Plan:
Relationships:
nH2,V,T
PH2
PH2, PH2O
Ptotal
Solution:
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77. Reactions Involving Gases
The principles of reaction stoichiometry from Chapter 4 can be combined with the gas laws for reactions involving gases
In reactions of gases, the amount of a gas is often given as a volume
instead of moles
as we’ve seen, you must state pressure and temperature
The ideal gas law allows us to convert from the volume of the gas to moles; then we can use the coefficients in the equation as a mole ratio
When gases are at STP, use 1 mol = 22.4 L
P, V, T of Gas A
mole A
mole B
P, V, T of Gas B
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78. Ex 5. 12: What volume of H2 is needed to make 35
Ex 5.12: What volume of H2 is needed to make 35.7 g of CH3OH at 738 mmHg and 355 K? CO(g) + 2 H2(g) → CH3OH(g)
Given:
Find:
mCH3OH = 37.5g, P=738 mmHg, T=355 K
VH2, L
nH2 = mol, P= atm, T=355 K
VH2, L
g CH3OH
mol CH3OH
mol H2
P, n, T, R V
Conceptual Plan:
Relationships:
Solution:
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79. Ex 5. 13: How many grams of H2O form when 1
Ex 5.13: How many grams of H2O form when 1.24 L H2 reacts completely with O2 at STP? O2(g) + 2 H2(g) → 2 H2O(g)
Given:
Find:
VH2 = 1.24 L, P = 1.00 atm, T = 273 K
massH2O, g
g H2O
L H2
mol H2
mol H2O
Concept Plan:
Relationships:
H2O = g/mol, 1 mol = 22.4 STP
2 mol H2O : 2 mol H2
Solution:
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80. Practice – What volume of O2 at 0
Practice – What volume of O2 at atm and 313 K is generated by the thermolysis of 10.0 g of HgO? 2 HgO(s)  2 Hg(l) + O2(g) (MMHgO = g/mol)
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81. What volume of O2 at atm and 313 K is generated by the thermolysis of 10.0 g of HgO? 2 HgO(s)  2 Hg(l) + O2(g)
Given:
Find:
mHgO = 10.0g, P=0.750 atm, T=313 K
VO2, L
nO2 = mol, P=0.750 atm, T=313 K
VO2, L
g HgO
mol HgO
mol O2
P, n, T, R V
Conceptual Plan:
Relationships:
Solution:
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82. Properties of Gases Expand to completely fill their container
Take the shape of their container
Low density
much less than solid or liquid state
Compressible
Mixtures of gases are always homogeneous
Fluid
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83. Kinetic Molecular Theory
The particles of the gas (either atoms or molecules) are constantly moving
The attraction between particles is negligible
When the moving gas particles hit another gas particle or the container, they do not stick; but they bounce off and continue moving in another direction
like billiard balls
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84. Kinetic Molecular Theory
There is a lot of empty space between the gas particles
compared to the size of the particles
The average kinetic energy of the gas particles is directly proportional to the Kelvin temperature
as you raise the temperature of the gas, the average speed of the particles increases
but don’t be fooled into thinking all the gas particles are moving at the same speed!!
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85. Gas Properties Explained – Pressure
Because the gas particles are constantly moving, they strike the sides of the container with a force
The result of many particles in a gas sample exerting forces on the surfaces around them is a constant pressure
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86. Gas Properties Explained – Indefinite Shape and Indefinite Volume
Because the gas
molecules have
enough kinetic
energy to overcome
attractions, they
keep moving around
and spreading out
until they fill the
container.
As a result, gases
take the shape and
the volume of the
container they
are in.
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87. Gas Properties Explained - Compressibility
Because there is a lot of unoccupied space in the structure
of a gas, the gas molecules can be squeezed closer together
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88. Gas Properties Explained – Low Density
Because there is a lot of unoccupied space in the structure of a gas, gases do not have a lot of mass in a given volume; the result is they have low density
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89. Density & Pressure
Pressure is the result of the constant movement of the gas molecules and their collisions with the surfaces around them
When more molecules are added, more molecules hit the container at any one instant, resulting in higher pressure
also higher density
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90. Gas Laws Explained – Boyle’s Law
Boyle’s Law says that the volume of a gas is inversely proportional to the pressure
Decreasing the volume forces the molecules into a smaller space
More molecules will collide with the container at any one instant, increasing the pressure
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91. Gas Laws Explained – Charles’s Law
Charles’s Law says that the volume of a gas is directly proportional to the absolute temperature
Increasing the temperature increases their average speed, causing them to hit the wall harder and more frequently
on average
To keep the pressure constant, the volume must then increase
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92. Gas Laws Explained – Avogadro’s Law
Avogadro’s Law says that the volume of a gas is directly proportional to the number of gas molecules
Increasing the number of gas molecules causes more of them to hit the wall at the same time
To keep the pressure constant, the volume must then increase
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93. Gas Laws Explained – Dalton’s Law of Partial Pressures
Dalton’s Law says that the total pressure of a mixture of gases is the sum of the partial pressures
Kinetic-molecular theory says that the gas molecules are negligibly small and don’t interact
Therefore the molecules behave independently of each other, each gas contributing its own collisions to the container with the same average kinetic energy
Because the average kinetic energy is the same, the total pressure of the collisions is the same
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94. Dalton’s Law & Pressure
Because the gas molecules are not sticking together, each gas molecule contributes its own force to the total force on the side
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95. Kinetic Energy and Molecular Velocities
Average kinetic energy of the gas molecules depends on the average mass and velocity
KE = ½mv2
Gases in the same container have the same temperature, therefore they have the same average kinetic energy
If they have different masses, the only way for them to have the same kinetic energy is to have different average velocities
lighter particles will have a faster average velocity than more massive particles
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96. Molecular Speed vs. Molar Mass
To have the same average kinetic energy, heavier molecules must have a slower average speed
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97. Temperature and Molecular Velocities_
KEavg = ½NAmu2
NA is Avogadro’s number
KEavg = 1.5RT
R is the gas constant in energy units, J/mol∙K
1 J = 1 kg∙m2/s2
Equating and solving we get
NA∙mass = molar mass in kg/mol
As temperature increases, the average velocity increases
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98. Molecular Velocities
All the gas molecules in a sample can travel at different speeds
However, the distribution of speeds follows a statistical pattern called a Boltzman distribution
Ee talk about the “average velocity” of the molecules, but there are different ways to take this kind of average
The method of choice for our average velocity is called the root-mean-square method, where the rms average velocity, urms, is the square root of the average of the sum of the squares of all the molecule velocities
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99. Boltzman Distribution
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100. Temperature vs. Molecular Speed
As the absolute temperature increases, the average velocity increases
the distribution function “spreads out,” resulting in more molecules with faster speeds
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101. Ex 5.14: Calculate the rms velocity of O2 at 25 °C
Given:
Find:
O2, t = 25 °C
urms
Conceptual Plan:
Relationships:
MM, T
urms
Solution:
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102. Practice – Calculate the rms velocity of CH4 (MM 16.04) at 25 °C
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103. Practice – Calculate the rms velocity of CH4 at 25 °C
Given:
Find:
CH4, t = 25 °C
urms
Conceptual Plan:
Relationships:
MM, T
urms
Solution:
The mass of O2 (32.00) is 2x the mass of CH4 (16.04).
The rms of CH4 (681 m/s) is √2x the rms of O2 (482 m/s)
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104. Mean Free Path
Molecules in a gas travel in straight lines until they collide with another molecule or the container
The average distance a molecule travels between collisions is called the mean free path
Mean free path decreases as the pressure increases
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105. Diffusion and Effusion
The process of a collection of molecules spreading out from high concentration to low concentration is called diffusion
The process by which a collection of molecules escapes through a small hole into a vacuum is called effusion
The rates of diffusion and effusion of a gas are both related to its rms average velocity
For gases at the same temperature, this means that the rate of gas movement is inversely proportional to the square root of its molar mass
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106. Effusion
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107. Graham’s Law of Effusion
Thomas Graham (1805–1869)
For two different gases at the same temperature, the ratio of their rates of effusion is given by the following equation:
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108. Ex 5. 15: Calculate the molar mass of a gas that effuses at a rate 0
Ex 5.15: Calculate the molar mass of a gas that effuses at a rate times N2
Given:
Find:
MM, g/mol
Conceptual Plan:
Relationships:
rateA/rateB, MMN2
MMunknown
Solution:
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109. Practice – Calculate the ratio of rate of effusion for oxygen to hydrogen
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110. This means that, on average, the O2 molecules
Practice – Calculate the ratio of rate of effusion for oxygen to hydrogen
Given:
Find:
O2, g/mol; H g/mol
Conceptual Plan:
Relationships:
MMO2, MMH2
rateA/rateB
Solution:
This means that, on average, the O2 molecules
are traveling at ¼ the speed of H2 molecules
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111. Ideal vs. Real Gases
Real gases often do not behave like ideal gases at high pressure or low temperature
Ideal gas laws assume
1. no attractions between gas molecules
2. gas molecules do not take up space
based on the kinetic-molecular theory
At low temperatures and high pressures these assumptions are not valid
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112. Real Gas Behavior
Because real molecules take up space, the molar volume of a real gas is larger than predicted by the ideal gas law at high pressures
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113. The Effect of Molecular Volume
Johannes van der Waals (1837–1923)
At high pressure, the amount of space occupied by the molecules is a significant amount of the total volume
The molecular volume makes the real volume larger than the ideal gas law would predict
van der Waals modified the ideal gas equation to account for the molecular volume
b is called a van der Waals constant and is different for every gas because their molecules are different sizes
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114. Real Gas Behavior
Because real molecules attract each other, the molar volume of a real gas is smaller than predicted by the ideal gas law at low temperatures
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115. The Effect of Intermolecular Attractions
At low temperature, the attractions between the molecules is significant
The intermolecular attractions makes the real pressure less than the ideal gas law would predict
van der Waals modified the ideal gas equation to account for the intermolecular attractions
a is another van der Waals constant and is different for every gas because their molecules have different strengths of attraction
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116. van der Waals’ Equation
Combining the equations to account for molecular volume and intermolecular attractions we get the following equation
used for real gases
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117. Real Gases
A plot of PV/RT vs. P for 1 mole of a gas shows the difference between real and ideal gases
It reveals a curve that shows the PV/RT ratio for a real gas is generally lower than ideal for “low” pressures – meaning the most important factor is the intermolecular attractions
It reveals a curve that shows the PV/RT ratio for a real gas is generally higher than ideal for “high” pressures – meaning the most important factor is the molecular volume
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118. PV/RT Plots
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119. Structure of the Atmosphere
The atmosphere shows several layers, each with its own characteristics
The troposphere is the layer closest to the Earth’s surface
circular mixing due to thermal currents – weather
The stratosphere is the next layer up
less air mixing
The boundary between the troposphere and stratosphere is called the tropopause
The ozone layer is a layer of high O3 concentration located in the stratosphere
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120. Air Pollution
Air pollution is materials added to the atmosphere that would not be present in the air without, or are increased by, man’s activities
though many of the “pollutant” gases have natural sources as well
Pollution added to the troposphere has a direct effect on human health and the materials we use because we come in contact with it
and the air mixing in the troposphere means that we all get a smell of it!
Pollution added to the stratosphere may have indirect effects on human health caused by depletion of ozone
and the lack of mixing and weather in the stratosphere means that pollutants last longer before “washing” out
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121. Pollutant Gases, SOx
SO2 and SO3, oxides of sulfur, come from coal combustion in power plants and metal refining
as well as volcanoes
Lung and eye irritants
Major contributors to acid rain
2 SO2 + O2 + 2 H2O  2 H2SO4
SO3 + H2O  H2SO4
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122. Pollutant Gases, NOx
NO and NO2, oxides of nitrogen, come from burning of fossil fuels in cars, trucks, and power plants
as well as lightning storms
NO2 causes the brown haze seen in some cities
Lung and eye irritants
Strong oxidizers
Major contributors to acid rain
4 NO + 3 O2 + 2 H2O  4 HNO3
4 NO2 + O2 + 2 H2O  4 HNO3
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123. Pollutant Gases, CO
CO comes from incomplete burning of fossil fuels in cars, trucks, and power plants
Adheres to hemoglobin in your red blood cells, depleting your ability to acquire O2
At high levels can cause sensory impairment, stupor, unconsciousness, or death
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124. Pollutant Gases, O3
Ozone pollution comes from other pollutant gases reacting in the presence of sunlight
as well as lightning storms
known as photochemical smog and ground-level ozone
O3 is present in the brown haze seen in some cities
Lung and eye irritant
Strong oxidizer
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125. Major Pollutant Levels
Government regulation has resulted in a decrease in the emission levels for most major pollutants
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126. O3(g) + UV light  O2(g) + O(g)
Stratospheric Ozone
Ozone occurs naturally in the stratosphere
Stratospheric ozone protects the surface of the earth from over-exposure to UV light from the Sun
O3(g) + UV light  O2(g) + O(g)
Normally the reverse reaction occurs quickly, but the energy is not UV light
O2(g) + O(g)  O3(g)
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127. CF2Cl2 + UV light  CF2Cl + Cl
Ozone Depletion
Chlorofluorocarbons became popular as aerosol propellants and refrigerants in the 1960s
CFCs pass through the tropopause into the stratosphere
There, CFCs can be decomposed by UV light, releasing Cl atoms
CF2Cl2 + UV light  CF2Cl + Cl
Cl atoms catalyze O3 decomposition and remove O atoms so that O3 cannot be regenerated
NO2 also catalyzes O3 destruction
Cl + O3  ClO + O2
O3 + UV light  O2 + O
ClO + O  O2 + Cl
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128. Ozone Holes
Satellite data over the past 3 decades reveals a marked drop in ozone concentration over certain regions
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