1. Gases The Gas Laws Labs #18 Molar Mass of a Volatile Liquid
#19 Calcium Carbonate Analysis: Molar Volume of Carbon Dioxide
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2. Gases have general characteristics
Expansion: Expand indefinitely to fill the space available to them
Indefinite shape: Fill all parts of container evenly, so have no definite shape of their own
Compressibility: Most compressible of states of matter
Mixing: Two or more gases will mix evenly and completely when confined to same container
Low density: Have much lower densities than liquids and solids (typically about 1/1000 those of liquids/solids)
Pressure: Exert pressure on their surroundings
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3. Vapors
Gas phase at temperature where same substance can also exist in liquid or solid state
Below critical temperature of substance (vapor can be condensed by increasing pressure without reducing temperature)
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4. Pressure: force per unit area (P= f/a)
Force generated by collisions of gas particle w/container walls
Related to velocity of gas particles
Total force = sum of forces of all collisions each second per unit area
Force = m x acceleration
Pressure dependent on
Gas particle velocity
Collision frequency
Collision frequency dependent on
Gas particle velocity
Distance to container walls.
Changing temperature changes collision force, as well as collision frequency
Collision frequency changed by altering size of container
Force of collisions is not affected
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5. (force of 1 newton exerted on one square meter of area)
(29.92” Hg)
(psi)
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6. Manometers Used to measure pressure of enclosed gas
U-tube partially filled with liquid, typically Hg
One end connected to container of gas being measured
Other end sealed with vacuum existing above liquid, or open to atmosphere
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7. If left to atmosphere, it measures atmospheric pressure-barometer
Closed-end manometer
Pressure is just difference between two levels (in mm of Hg)-indicates pressure of system attached to apparatus
Gas connected to one arm
Space above Hg in other arm is vacuum
Liquid in tube falls to height (directly proportional to pressure exerted by gas in the tube)
Since pressure of gas causes liquid levels to be different in height, it is this difference (h) that is measure of gas pressure in container
Pgas = Ph
If left to atmosphere, it measures atmospheric pressure-barometer
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8. Open-end manometer Used to measure pressure of gas in container
Difference in Hg levels indicates pressure difference in gas pressure and atmospheric pressure
Atmospheric pressure pushes mercury in one direction
Gas in container pushes it in the other direction
Two ends connected to gases at different pressures
Closed end to gas in bulb (gas filled)
Open end to atmosphere
If pressure of gas higher than atmospheric, Hg level lower in arm connected to gas (Pgas = Pbarometric + h)
If pressure of gas lower than atmospheric, Hg level higher in arm connected to gas (Pgas = Pbarometric - h)
If levels are equal, gas at atmospheric pressure
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9. density fluid If fluid other than mercury is used:
Difference in heights of liquid levels inversely proportional to density of liquid and represents the pressure
Greater density of liquid, smaller difference in height
High density of mercury (13.6 g/mL) allows relatively small monometers to be built
Readings must be corrected for relative densities of fluid used and of mercury (mm Hg = mm fluid)
density fluid
density Hg
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11. A sample of CH4 is confined in a water manometer
A sample of CH4 is confined in a water manometer. The temperature of the system is 30.0 °C and the atmospheric pressure is kPa. What is the pressure of the methane gas, if the height of the water in the manometer is 30.0 mm higher on the confined gas side of the manometer than on the open to the atmosphere side. (Density of Hg is g/mL).
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12. 1) Convert 30.0 mm of H2O to equivalent mm of mercury:
(30.0 mm) (1.00 g/mL) = (x) ( g/mL) x = mm (I will carry some guard digits.)
2) Convert mmHg to kPa:
mmHg x ( kPa/760.0 mmHg) = kPa
3) Determine pressure of enclosed wet CH4:
At point A in the above graphic, we know this: Patmo. press. = Pwet CH4 + Pthe 30.0 mm water column
98.70 kPa = x kPa
x = kPa
4) Determine pressure of dry CH4:
From Dalton's Law, we know this: Pwet CH4 = Pdry CH4 + Pwater vapor
(Water's vapor pressure at 30.0 °C is 31.8 mmHg. Convert it to kPa.)
kPa = x kPa
x = kPa
Based on provided data, use three significant figures; so 94.2 kPa.
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13. Pressure-Volume Relationship Temperature/# molecules (n) constant
Boyle’s Law
X axis independent variable Y axis dependent variable
Pressure-Volume Relationship
Temperature/# molecules (n) constant
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14. Inversely proportional (one goes up, other goes down) V  1/P or
P = k/T where K is constant
PV = constant
P1V1 = P2V2
Gas that strictly obeys Boyle’s Law is ideal gas (holds precisely for gases at very low temperatures)
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15. Pressure applied vs. volume measured
Shows inverse proportion
If pressure doubled, volume decreased by ½
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16. Pressure vs. inverse of volume
Plot of V (P) against 1/P (1/V) gives straight line
Graph of equation P = k1 x 1/V
Same relationship holds whether gas is being expanded or contracted
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17. A gas which has a pressure of 1. 3 atm occupies a volume of 27 L
A gas which has a pressure of 1.3 atm occupies a volume of 27 L. What volume will the gas occupy if the pressure is increased to 3.9 atm at constant pressure?
P1 = 1.3 atm
V1 = 27 L
P2 = 3.9 atm
V2 = ?
P1V1 = P2V2
1.3 atm (27 L) = 3.9 atm (X) = 9.0 L
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18. Temperature-Volume Relationship Pressure/n constant
Charles’ Law
Temperature-Volume Relationship
Pressure/n constant
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19. Volume of gas directly proportional to temperature (T  V), and extrapolates to zero at zero Kelvin (convert Celsius to Kelvin)
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20. Used to determine absolute zero
From extrapolated line, determine T at which ideal gas would have zero volume
Since ideal gases have infinitely small atoms, only contribution to volume of gas is pressure exerted by moving atoms bumping against walls of container
If no volume, then no kinetic energy left
Absolute zero is T at which all KE has been removed
Does not mean all energy has been removed, merely all KE
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21. V = kT If T↓, V↓ and vs. P is constant k = proportionality constant
V/T = constant
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22. A gas at 30oC and 1. 00 atm occupies a volume of 0. 842 L
A gas at 30oC and 1.00 atm occupies a volume of L. What volume will the gas occupy at 60.0oC and 1.00 atm?
V1 = atm
T1 = 30oC = 303 K
V2 = ?
T2 = 60.0oC = 333 K
V1/T1 = V2/T2
0.842/303 K = X/333 K = L
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23. Pressure-Temperature Relationship Volume/n constant
Gay Lussac’s Law
Pressure-Temperature Relationship
Volume/n constant
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24. Direct relationship between pressure and temperature (P  T)
When two gases react, do so in volume ratios always expressed as small whole numbers
When H burns in O, volume of H consumed is always exactly 2x volume of O
Direct relationship between pressure and temperature (P  T)
P/T = constant
Pi/Ti = Pf/Tf
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25. Avogadro’s Law (at low pressures)
Volume-Amount Relationship
Temperature/Pressure constant
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26. Equal volumes of gases, measured at same temperature and pressure, contain equal numbers of molecules
Avogadro's law predicts directly proportional relation between # moles of gas and its volume
Helped establish formulas of simple molecules when distinction between atoms and molecules was not clearly understood, particularly existence of diatomic molecules
Once shown that equal volumes of hydrogen and oxygen do not combine in manner depicted in (1), became clear that these elements exist as diatomic molecules and that formula of water must be H2O rather than HO as previously thought
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27. V of gas proportional to # of moles present (V  n)
At constant T/P, V of container must increase as moles of gas increase
V/n = constant
Vi/ni = Vf/nf
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28. A 5. 20 L sample at 18. 0oC and 2. 00 atm pressure contains 0
A 5.20 L sample at 18.0oC and 2.00 atm pressure contains moles of a gas. If we add an additional 1.27 moles of the gas at the same temperature and pressure, what will the total volume occupied by the gas be?
V1 = 5.20 L
n1 = mol
V2 = X
n2 = = 1.70 mol
V1/n1 = V2/n2
5.20 L/0.436 mol = X/1.70 mol = 20.3 L
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29. Ideal Gas Law (Holds closely at P < 1 atm)
Generalization applicable to most gases, at pressures up to about 10 atm, and at temperatures above 0°C.
Ideal gas’s behavior agrees with that predicted by ideal gas law.
Ideal Gas Law (Holds closely at P < 1 atm)
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30. If proportionality constant called R
Formulated from combination of Boyle’s law, Charles’s law, Gay-Lussac’s law, and Avogadro’s principle
Combined
If proportionality constant called R
Rearrange to form ideal gas equation
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31. Experimentally observed relationship between these properties is called the ideal gas law: PV = nRT
Pressure (P) in atm
Volume (V) in L
Absolute temperature (T) in K (Charles’ Law)
Amount (number of moles, n)
R (universal ideal gas constant)
liter ∙ atm/mole ∙ K or L·atm·K–1·mol–1
8.31 liter ∙ kPa/mole ∙ K
8.31 J/mole ∙ K
8.31 V ∙ C/mole ∙ K
8.31 x 10-7 g ∙ cm2/sec2 ∙ mole ∙ K
6.24 x 104 L ∙ mm Hg/mol ∙ K
1.99 cal/mol ∙ K
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32. You can use ideal gas law equation for all problems
Given 3 of 4 variables and calculate 4th
PiVi= niRTi
PfVf nfRTf
Cancel all constants and R, make appropriate substitutions from given data to perform calculation
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33. P (12.9 L) = (0.614 mol)(0.08206 L atm/K mol)(285 K) = 1.11 atm
A sample containing moles of a gas at 12.0oC occupies a volume of 12.9 L. What pressure does the gas exert?
PV = nRT
P (12.9 L) = (0.614 mol)( L atm/K mol)(285 K) = 1.11 atm
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34. 3/25/2017

35. Homework: Read 5.1-5.3, pp. 189-202 Q pp. 232-234, #29, 31-34, 44
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36. Standard Temperature and Pressure (STP)
STP = 0°C (273K) and 1.00 atm pressure (760 mm Hg)
One mole of gas at STP will occupy L
Allows you to compare gases at STP to each other
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37. What volume will 1.18 mole of O2 occupy at STP? PV = nRT
(1atm)(X) = (1.18 mol)( L atm/K mol)(273 K) = 26.4 L
Alternate way:
At STP, 1 mole occupies 22.4 L
Vi/ni = Vf/nf
22.4 L/1 mol = X/1.18 mol = 26.4 L
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38. A sample containing 15.0 g of dry ice, CO2(s), is put into a balloon and allowed to sublime according to the following equation: CO2(s)  CO2(g) How big will the balloon be (what is the volume of the balloon) at 22oC and 1.04 atm after all of the dry ice has sublimed?
15.0 g CO2 1 mol CO2 = mol CO2
44.0 g CO2
PV = nRT
(1.04 atm)(X) = (0.341 mol)( L atm/K mol)(295K) = 7.94 L
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39. Find limiting reactant:
0.500 L of H2(g) are reacted with L of O2(g) according to the equation 2H2(g) + O2(g)  2H2O(g). What volume will the H2O occupy at 1.00 atm and 350oC?
Find limiting reactant:
0.500 L H2 1 mol H2 = mol H2/2
22.4 L
0.600 L O mol O2 = mol O2/1
PV = nRT
(1atm)(X) = ( mol)( L atm/K mol)(623 K) = 1.14 L
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40. 3/25/2017

41. Molar Mass
Gives density which can be determined if P, T and molar mass are known
Density directly proportional to molar mass
Density increases as gas pressure increases
Density decreases as temperature increases
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42. A gas at 34. 0oC and 1. 75 atm has a density of 3. 40 g/L
A gas at 34.0oC and 1.75 atm has a density of 3.40 g/L. Calculate the molar mass (M.M.) of the gas.
M.M. = dRT P
M.M. = (3.40 g/L)( L atm/K mol)(307K)
(1.75 atm)
M.M. = 48.9 g/mol
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43. Example What are the expected densities of argon, neon and air at STP?
PV = nRT
PV = g (RT)
molar mass
Density = g/V = (Molar mass)P/RT
density Ar = (39.95 g/mol)(1.00 atm)/( L-atm/mol-K)(273 K) = 1.78 g/L
density Ne = g/L
density air = 1.28 g/L
molar mass = (0.80)(28 gN2/mol) + (0/20)(32 g O2/mol) = 28.8 g/mol
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44. Molar Volume (V/n) V = RT n P (0.0821 L-atm/mol-K)(273K) = 22.4 L/mol
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45. Homework: Read 5.4, pp. 203-206 Q pp. 234-235, #52, 56, 58, 60
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46. Dalton’s Partial Pressure
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47. When two gases are mixed together, gas particles tend to act independently of each other
Each component gas of mixture of gases uniformly fills containing vessel
Each component exerts same pressure as it would if it occupied that volume alone
Total pressure of mixture is sum of individual pressures, called partial pressures, of each component
Pt = PA + PB + PC + …
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48. 3/25/2017

49. Since each component obeys ideal gas law (PV=nRT) and has same T and V, it follows that partial pressure of each gas in container is directly proportional to # moles of gas present
Pi/Pt = ni/nt
Pi = ni/nt x Pt
Pi = XiPt
Xi = mole fraction of gas component i
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50. A volume of 2. 0 L of He at 46oC and 1
A volume of 2.0 L of He at 46oC and 1.2 atm pressure was added to a vessel that contained 4.5 L of N2 at STP. What is the total pressure and partial pressure of each gas at STP after the He is added?
Find # moles of He at original conditions-will lead us to finding partial pressure of He at STP.
PV = nRT
(1.2 atm)(2.0 L) = n( L atm/k mol)(319 K)
n = mol He
When gases are combined under STP, partial pressure of He will change while N2 will remain the same since it is already at STP.
PHeV = nRT
(X)(4.5 L) = (0.091 mol)( L atm/K mol)(273 K)
P = atm
Total pressure = = 1.46 atm
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51. Mole fraction 4.5 L N2 1 mol = 0.201 mol N2 22.4 L
Xi = ni = Pi
ntotal Ptotal
You can use either the number of moles or the pressure of each component of your system to evaluate the mole fraction
4.5 L N2 1 mol = mol N2
22.4 L
XN2 = mol = atm = 0.69
0.293 mol atm
X He = mol = atm = 0.31
0.293 mol atm
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52. Calculate the mole fraction of each gas.
A mixture of gases consists of 3.00 moles of helium, 4.00 moles of argon, and 1.00 moles of neon. The total pressure of the mixture is 1,200 torr.
Calculate the mole fraction of each gas.
nt = 3.00 mol mol mol = 8.00 mol
Xi = ni/nt
XHe = 3.00 mol/8.00 mol = 0.375
XAr = 4.00 mol/8.00 mol = 0.500
XNe = 1.00 mol/8.00 mol = 0.125
Calculate the partial pressure of each gas.
Pi = XiPi
PHe = x 1,200 torr = 450. torr
PAr = x 1,200 torr = 600. torr
PNe = x 1,200 torr = 150. torr
He-1.2 L, 0.63 atm, 16oC Ne-3.4 L, 2.8 atm, 16oC
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53. Whenever gases are collected by displacement of water, total gas pressure is sum of partial pressure of collected gas and partial pressure of water vapor
Pt = Pgas + PH2O
Consequently, partial pressure of collected gas is Pgas = Pt – PH2O, where partial pressure of water depends on temperature and corresponds to vapor pressure of water
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54. What are the mole fractions of oxygen and water vapor?
When oxygen gas is collected over water at 30°C and the total pressure is 645 mm Hg:
What is the partial pressure of oxygen? Given the vapor pressure of water at 30°C is 31.8 mm Hg.
Pt = PO2 + PH2O
PO2 = Pt – PH2O = 645 mm Hg – 31.8 mm Hg PO2 = 613 mm Hg
What are the mole fractions of oxygen and water vapor?
Recall that the partial pressures of O2 and H2O are related to their mole fractions, PO2 = XO2Pt      PH2O = XH2OPt
X O2 = 613/645 = 0.950, and XH2O = 31.8/645 = 0.049
Note also that the sum of the mole fractions is 1.0, within the number of significant figures, given: X O2 + XH2O = = 0.999
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55. The vapor pressure of water in air at 28oC is 28. 3 torr
The vapor pressure of water in air at 28oC is 28.3 torr. Calculate the mole fraction of water in a sample of air at 28oC and 1.03 atm pressure.
XH2O = PH2O = torr = 0.036
Pair torr
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56. 3/25/2017

57. 3/25/2017

58. Gas data  mol N2  mol NaN3  g NaN3
The safety air bags in cars are inflated by nitrogen gas generated by the rapid decomposition of sodium azide. If an air bag has a volume of 36 L and is to be filled with nitrogen gas at a pressure of 1.15 atm at a temperature of 26.0oC, how many grams of NaN3 must be decomposed?
2NaN3(s)  2Na(s) + 3N2(g)
Gas data  mol N2  mol NaN3  g NaN3
n = PV/RT = (1.15 atm)(36 L) = 1.7 mol N2
( L-atm/mol-K)(299K)
1.7 mol N2 2 mol NaN g NaN3 = 72 g NaN3
3 mol N mol NaN3
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59. Homework: Read 5.5, pp. 206-211 Q pg. 235, #64, 66, 68, 70, 72
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60. Kinetic Molecular Theory of Gases
Describes the properties of gases
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61. Main assumptions that explain behavior of ideal gas:
Gases composed of tiny, invisible molecules that are widely separated from one another in empty space
Gas molecule behave as independent particles
Volume occupied by molecule considered negligible
Gas molecules are in constant motion, continuous, random and straight-line motion
Collisions of particles with walls of container cause pressure exerted by gas
Molecules collide with one another, but collisions are perfectly elastic (no net loss of energy-exert no force on each other)
Attractive forces between atoms and/or molecules in gas are negligible
Average kinetic energy of collection of gas particles assumed to be directly proportional to Kelvin temperature of gas
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62. Temperature is measure of average kinetic energy of gas
Equal #s of molecules of any gas have same average kinetic energy at same temperature
Greater temperature, greater fraction of molecules moving at higher speeds (higher average kinetic energy of gas molecules)
Individual molecules move at varying speeds
Momentum conserved in each collision, but one colliding molecule might be deflected off at high speed while another is nearly stopped
Result is that molecules at any instant have wide range of speeds
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63. Average KE, e, related to root mean square (rms) speed u
At higher temperatures, greater fraction of molecules moving at higher speeds
4 molecules in gas sample have speeds of 3.0, 4.5, 5.2, and 8.3 m/s.
Average speed
rms
Because mass of molecules does not increase, rms speed of molecules must increase with increasing temperature
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64. Average kinetic energy of single gas molecule
KE = ½ mv2
m = mass of molecule (kg)
v = speed of molecule (meters/sec)
KE measured in joules
Average translational kinetic energy of any kind of molecule in ideal gas is given by:
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65. 3/25/2017

66. 3/25/2017

67. J = kg· m2/s2
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68. Based on understanding of pressure, molecular meaning of gas laws can be appreciated as follows:
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69. Boyle’s Law: inverse relationship between P & V:
Kinetic molecular theory agrees
If V of container decreased, gas particles strike walls of container more frequently, increasing pressure of gas
If V of container increased, fewer impacts of gas molecules with wall per second, decreasing pressure
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70. Charles’ Law: direct relationship between T & V:
Increasing temperature increases average kinetic energy and speed of gas molecules
Increases force of each collision w/container wall
Increases frequency of collisions w/container wall
Both increase pressure
If pressure of gas is to remain constant, volume must increase to decrease frequency of collisions with container walls
If one of the walls is a movable piston, then the volume of the container will expand until balanced by the external force
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71. Gay-Lussac‘s Law: direct relationship between T & P
Increase in temperature increases kinetic energy of gas particles
Force of each collision increases
Frequency of collisions with container walls increases
When gas is heated in container with fixed volume, gas molecules impact more forcefully with wall, increasing pressure
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72. Avogadro’s Law
As more gas molecules are added to the container, # of impacts per second with wall increases, pressure increases correspondingly
If V of container is not fixed, then V will increase
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73. Graham’s Law of Effusion
Experiments show that molecules of a gas do not all move at same speed but are distributed over a range.
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74. Maxwell-Boltzmann distribution
Probability distribution that forms basis of kinetic theory of gases
Explains many fundamental gas properties, including pressure and diffusion
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75. Highest point on curve marks most probable speed-some move with speeds much less than most probable speed, while others move with speeds much greater than this speed
As temp. increases, curve flattens out (more molecules have higher kinetic energies and therefore higher average speeds)
(speed)
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76. Diffusion Mean free path Spread of gas molecules throughout volume
Average distance molecule travels between collisions
Factors affecting it:
Density (increasing density decreases MFP)
Radius of molecule (increasing size increases collision frequency, reducing MFP)
Pressure (increasing pressure increases collision frequency, reducing MFP)
Temperature (increasing T increases collision frequency, but does not affect MFP)
Spread of gas molecules throughout volume
Much slower than average molecular speed because of collision between molecules
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77. Effusion Escape of gas through small opening into vacuum
Rate (in mol/s) proportional to average speed u
More collisions gas has w/walls of container, higher probability it will hit pinhole and go through it
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78. All gases expand to fill container
Average kinetic energy has same value for all gases at same temperature
Gas molecules with higher molar masses will have slower average speeds, while lighter molecules will have higher average speeds
Rates two gases effuse through a pinhole in a container are inversely related to the square roots of the molecular masses of gas particles
Use equation to find relative rates of diffusion of gases, whether evacuated or not)
All gases expand to fill container
Heavier gases diffuse more slowly than lighter ones
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79. KE = cT = ½ mv2
KE-average kinetic energy
c = constant that is same for all gases
T = temperature in Kelvin
M = mass of gas
v2 = average of the square of the velocities of the gas molecules
Since cT will be the same for all gases at the same temperature, the average KE of any two gases at the same T will be the same
KE1 = KE2
½ m1V21 = ½ m2v22
root mean square velocity = vrms = √v2
√m1/m2 = vrms2/vrms1
Higher the T, higher vrms
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80. 3/25/2017

81. Effusion:
Diffusion:
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82. How many times faster than He would NO2 gas effuse? MNO2 = 46.01 g/mol
MHe = g/mol
√MNO2/MHe = rateHe/RateNO2
√46.01/4.003 = 3.390
So He would effuse 3.39 times faster as NO2
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83. Real Gases (Nonideal Gases)
Ideal gas law works very well for most gases, however, it does not work well for gases under high pressures or gases at very low temperature
Conditions used to condense gases
Generalized that any gas close to its BP (condensation point) will deviate significantly from ideal gas law
Kinetic molecular theory makes two assumptions
Gases have no volume
They exhibit no attractive or repulsive forces
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84. As a real gas is cooled and/or compressed, distances between particles decreases dramatically, and these real volumes and forces can no longer be ignored
Cooling gas decreases average KE of its molecules
All gases when cooled enough will condense to liquid
Suggests that intermolecular forces of attraction exist between molecules of real gases
Condensation occurs when average KE is not great enough for molecules to break away from intermolecular attractive forces
When they stick together, there are fewer particles bouncing around and creating P, so real P in nonideal situation will be smaller than P predicted by ideal gas equation
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85. 3/25/2017

86. Van der Waals Equation
Developed modification of ideal gas law to deal with nonideal behavior of real gases
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87. Real Gases
Must correct ideal gas behavior when at high pressure (smaller volume) and low temperature (attractive forces become important)
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88.   corrected pressure corrected volume Pideal Videal
correction for attractive forces
correction for small/finite volume  
corrected pressure
corrected volume
Pideal
Videal
PidealVeffective = nRT
Van der Waals (real gases)
P = pressure of gas (atm)
V = volume of gas (L)
n = #moles of gas (mol)
T = absolute temperature (K)
R = gas constant, L-atm/mol-K
a = constant, different for each gas, that takes into account the attractive forces between molecules
b = constant, different for each gas, that takes into account the volume of each molecule
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89. Calculate the pressure exerted by 0. 3000 mol of He in a 0
Calculate the pressure exerted by mol of He in a L container at -25oC
Using the ideal gas law
Using van der Waal’s equation
PV = nRT
(X)(0.2000L) = ( mol)() L atm/K mol)(248) = atm
From table 5.3, a = atm L2/mol2 and b = L mol
(P x / )( – x ) = ( )(248)= atm
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90. 3/25/2017

91. Pneumatic trough used for collecting gas samples over water
A gas (H2, O2) which is not soluble (or only slightly soluble) in water can be collected over water
Pneumatic trough w/bottle full of water submerged
Gas from reaction bubbles into bottle, displacing water
Pressure of gas inside collecting bottle determined to solve ideal gas equation
When gases are collected over water, there is some water vapor collected
Pressure of water vapor depends on T only and is obtained from reference table
Pressure of gas that was generated is calculated from Dalton’s law of partial pressure Pgas = Patm - Pwater
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92. 3/25/2017

93. Homework:
Read , pp Q pp , #74, 76, 78, 80, 82, 84, 86, 93 Do one additional exercise and one challenge problem. Submit quizzes by to me:
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