# Gases The Gas Laws Labs 18 Molar Mass of a Volatile Liquid 19 Calcium Carbonate Analysis Molar Volume of

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```					Gases
The Gas Laws

Labs
#18 Molar Mass of a Volatile Liquid
#19 Calcium Carbonate Analysis:
Molar Volume of Carbon Dioxide

2/16/2012                                         1
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

2/16/2012                                              2
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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Pressure: force per unit area (P= f/a)
•          Force generated by                   • Collision frequency
collisions of gas particle             dependent on
w/container walls                       • Gas particle velocity
•        Related to velocity of gas         • Distance to container walls.
particles                       • Changing temperature
•        Total force = sum of forces       changes collision force, as
of all collisions each second
per unit area                     well as collision frequency
•        Force = m x acceleration        • Collision frequency
•          Pressure dependent on                  changed by altering size of
•        Gas particle velocity             container
•        Collision frequency                • Force of collisions is not
affected

2/16/2012                                                                     4
(force of 1
newton exerted
on one square
meter of area)

(29.92” Hg)
(psi)

2/16/2012                                    5
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

2/16/2012                                                   6
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
2/16/2012                                               7
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
2/16/2012                                                      8
•  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
2/16/2012                                         9
2/16/2012   10
• A sample of CH4 is confined in
a water manometer. The
temperature of the system is
30.0 °C and the atmospheric
pressure is 98.70 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 13.534 g/mL).

2/16/2012                            11
• 1) Convert 30.0 mm of H2O to equivalent mm of mercury:
• (30.0 mm) (1.00 g/mL) = (x) (13.534 g/mL) x = 2.21664 mm (I will
carry some guard digits.)
• 2) Convert mmHg to kPa:
• 2.21664 mmHg x (101.325 kPa/760.0 mmHg) = 0.29553 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 + 0.29553 kPa
• x = 98.4045 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.)
• 98.4045 kPa = x + 4.23965 kPa
• x = 94.1648 kPa
• Based on provided data, use three significant figures; so 94.2 kPa.
2/16/2012                                                                 12
Boyle’s Law                    X axis independent variable
Y axis dependent variable

Pressure-Volume Relationship
Temperature/# molecules (n) constant

2/16/2012                                              13
• 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)
2/16/2012                                   14
Pressure applied vs. volume measured

• Shows inverse
proportion
• If pressure doubled,
volume decreased by
½

2/16/2012                                   15
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

2/16/2012                        16
• 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

2/16/2012                                      17
Charles’ Law
Temperature-Volume Relationship
Pressure/n constant

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• Volume of gas
directly proportional
to temperature (T 
V), and extrapolates
to zero at zero
Kelvin (convert
Celsius to Kelvin)

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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
http://www.absorblearning.com/media/attachment.action?quick=10o&att=2629

2/16/2012                                                          20
• If T↓, V↓ and vs.
• V = kT
• P is constant
• k = proportionality
constant
• V/T = constant

V1   V2
             ( P  constant)
T1   T2
2/16/2012                                       21
• A gas at 30oC and 1.00 atm occupies a
volume of 0.842 L. What volume will the
gas occupy at 60.0oC and 1.00 atm?
• V1 = 0.842 atm
• T1 = 30oC = 303 K
• V2 = ?
• T2 = 60.0oC = 333 K
• V1/T1 = V2/T2
• 0.842/303 K = X/333 K = 0.925 L

2/16/2012                                   22
Gay Lussac’s Law
Pressure-Temperature Relationship
Volume/n constant

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•       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

2/16/2012                                            24
(at low pressures)
Volume-Amount Relationship
Temperature/Pressure constant

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•   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

2/16/2012                                                       26
• 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

2/16/2012                                       27
• A 5.20 L sample at 18.0oC and 2.00 atm pressure
contains 0.436 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 = 0.436 mol
• V2 = X
• n2 = 0.436 + 1.27 = 1.70 mol
• V1/n1 = V2/n2
• 5.20 L/0.436 mol = X/1.70 mol = 20.3 L

2/16/2012                                        28
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)

2/16/2012                                               29
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
2/16/2012                                               30
•          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)
•    0.08206 liter ∙ atm/mole ∙ K or 0.08206 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

2/16/2012                                                               31
•     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

2/16/2012                                      32
• A sample containing 0.614 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)(0.08206 L atm/K
mol)(285 K) = 1.11 atm

2/16/2012                                    33
2/16/2012   34
Homework:
Q pp. 232-234, #29, 31-34, 44

2/16/2012                       35
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 22.42
L
•     Allows you to
compare gases at STP
to each other

2/16/2012                       36
• What volume will 1.18 mole of O2 occupy
at STP?
• PV = nRT
• (1atm)(X) = (1.18 mol)(0.08206 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

2/16/2012                                    37
• 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 = 0.341 mol CO2
44.0 g CO2
• PV = nRT
• (1.04 atm)(X) = (0.341 mol)(0.08206 L atm/K
mol)(295K) = 7.94 L

2/16/2012                                          38
• 0.500 L of H2(g) are reacted with 0.600 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 = 0.0223 mol H2/2
22.4 L
• 0.600 L O2 1 mol O2 = 0.0268 mol O2/1
22.4 L
• PV = nRT
• (1atm)(X) = (0.0223 mol)(0.08206 L atm/K
mol)(623 K) = 1.14 L

2/16/2012                                             39
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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

2/16/2012                              41
• 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)(0.08206 L atm/K mol)(307K)
(1.75 atm)
• M.M. = 48.9 g/mol

2/16/2012                                         42
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)/(0.0821 L-
atm/mol-K)(273 K) = 1.78 g/L
• density Ne = 0.900 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

2/16/2012                                                               43
Molar Volume (V/n)
•     V = RT
n P
•     (0.0821 L-atm/mol-K)(273K) = 22.4 L/mol
1.00 atm

2/16/2012                                       44
Homework:
Q pp. 234-235, #52, 56, 58, 60

2/16/2012                        45
Dalton’s Partial Pressure

2/16/2012                   46
•        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 = P A + P B + P C + …

2/16/2012                               47
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•         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

2/16/2012                                                49
• 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(0.08206 L atm/k mol)(319 K)
• n = 0.0917 mol He
• When gases are combined under STP, partial pressure of
He will change while N2 will remain the same since it is
•     PHeV = nRT
•     (X)(4.5 L) = (0.091 mol)(0.08206 L atm/K mol)(273 K)
•     P = 0.457 atm
•     Total pressure = 1.00 + 0.46 = 1.46 atm

2/16/2012                                                          50
• Mole fraction
• 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 = 0.201 mol N2
22.4 L
• XN2 = 0.201 mol = 0.69    1.00 atm = 0.69
0.293 mol          1.457 atm
• X He = 0.0917 mol = 0.31 0.457 atm = 0.31
0.293 mol         1.457 atm

2/16/2012                                                   51
•       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 + 4.00 mol + 1.00 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 = 0.375 x 1,200 torr = 450. torr
• PAr = 0.500 x 1,200 torr = 600. torr
• PNe = 0.125 x 1,200 torr = 150. torr
•       He-1.2 L, 0.63 atm, 16oC          Ne-3.4 L, 2.8 atm, 16oC

2/16/2012                                                   52
•     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

2/16/2012                                        53
•         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 = 0.950 +
0.049 = 0.999

2/16/2012                                                                      54
• 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 = 28.3 torr = 0.036
Pair      783 torr

2/16/2012                                       55
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2/16/2012   57
• 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
(0.0821 L-atm/mol-K)(299K)
• 1.7 mol N2 2 mol NaN3 65.0g NaN3 = 72 g NaN3
3 mol N2        1 mol NaN3

2/16/2012                                                      58
Homework:
Q pg. 235, #64, 66, 68, 70, 72

2/16/2012                        59
Kinetic Molecular Theory of
Gases
Describes the properties of gases

2/16/2012                                       60
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

2/16/2012                                                                      61
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
2/16/2012range of speeds                                62
Average KE, e, related to
At higher temperatures,         root mean square (rms)
greater fraction of molecules   speed u
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

2/16/2012                                                 63
• 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:

2/16/2012                                      64
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J = kg· m2/s2

2/16/2012       67
•           Based on understanding of
pressure, molecular
meaning of gas laws can
be appreciated as follows:

2/16/2012                           68
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

2/16/2012                                           69
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

2/16/2012                                                           70
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
2/16/2012                                                71

•      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

2/16/2012                                        72
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.

2/16/2012                                                             73
Maxwell-Boltzmann distribution
• Probability
distribution that forms
basis of kinetic theory
of gases
• Explains many
fundamental gas
properties, including
pressure and diffusion

2/16/2012                        74
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)

2/16/2012                                                75
• Mean free path
Diffusion                  • Average distance molecule
travels between collisions
• Spread of gas            • Factors affecting it:
• Density (increasing density
molecules                     decreases MFP)
throughout volume           • Radius of molecule (increasing
• Much slower than           size increases collision
frequency, reducing MFP)
average molecular
• Pressure (increasing pressure
speed because of
increases collision frequency,
collision between          reducing MFP)
molecules                • Temperature (increasing T
increases collision frequency,
but does not affect MFP)

2/16/2012                                                 76
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

2/16/2012                                            77
•  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
2/16/2012                                        78
•       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

2/16/2012                                                             79
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Effusion:
Rate of effusion for gas 1     M2

Rate of effusion for gas 2     M1

Diffusion:
Distance traveled by gas 1     M2

Distance traveled by gas 2     M1

2/16/2012                         81
• How many times faster than He would
NO2 gas effuse?
• MNO2 = 46.01 g/mol
• MHe = 4.003 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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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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• 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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Van der Waals Equation

Developed modification of ideal
gas law to deal with nonideal
behavior of real gases

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Real Gases

Must correct ideal gas behavior
when at high pressure (smaller
volume) and low temperature
(attractive forces become important)

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correction for attractive forces   correction for small/finite volume

2
[ Pobs  a (n / V ) ]  V  nb  nRT
                 
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, 0.0821 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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• Calculate the pressure exerted by 0.3000 mol of
He in a 0.2000 L container at -25oC
• Using the ideal gas law
• Using van der Waal’s equation
• PV = nRT
• (X)(0.2000L) = (0.3000 mol)().08206 L atm/K
mol)(248) = 30.55 atm
2
[ Pobs  a (n / V ) ]  V  nb  nRT
• From table 5.3, a = 0.0341 atm L2/mol2 and b =
0.0237 L mol
• (P + 0.0341 x 0.30002/0.20002)(0.2000 – 0.3000 x
0.0237) = 0.3000(0.08206)(248)= 31.60 atm

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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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Homework:
Q pp. 235-236, #74, 76, 78, 80, 82, 84, 86, 93
Do one additional exercise and one challenge problem.
Submit quizzes by email to me:
http://www.cengage.com/chemistry/book_content/0547125321_z
umdahl/ace/launch_ace.html?folder_path=/chemistry/book_con
tent/0547125321_zumdahl/ace&layer=act&src=ch05_ace1.xml
http://www.cengage.com/chemistry/book_content/0547125321_z
umdahl/ace/launch_ace.html?folder_path=/chemistry/book_con
tent/0547125321_zumdahl/ace&layer=act&src=ch05_ace2.xml
http://www.cengage.com/chemistry/book_content/0547125321_z
umdahl/ace/launch_ace.html?folder_path=/chemistry/book_con
tent/0547125321_zumdahl/ace&layer=act&src=ch05_ace3.xml

2/16/2012                                               93

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