GASES AND KINETICMOLECULAR THEORY

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• GASES AND KINETIC-MOLECULAR THEORY

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Pressure

• Pressure is force per unit area. It is produced
  by the collision of gas particles against each
  other and with the walls of the container.
• lb/in2 (psi) in U.S.
• N/m2

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Pressure

• Atmospheric pressure is measured using a
  barometer.
• Definitions of standard pressure
• 760 mm Hg
• 760 torr
• 1 atmosphere
• 101.3 kPa

Hg density 13.6 g/mL
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Boyles Law The Volume-Pressure Relationship

• V ? 1/P or
• V k (1/P) or PV k
• P1V1 k1 for one sample of a gas.
• P2V2 k2 for a second sample of a gas.
• k1 k2 for the same sample of a gas at the same
  T.
• Thus we can write Boyles Law mathematically as
  P1V1 P2V2

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Boyles Law The Volume-Pressure Relationship

• Example 1 At 25oC a sample of He has a volume of
  4.00 x 102 mL under a pressure of 7.60 x 102
  torr. What volume would it occupy under a
  pressure of 2.00 atm at the same T?

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Boyles Law The Volume-Pressure Relationship

• Notice that in Boyles law we can use any
  pressure or volume units as long as we
  consistently use the same units for both P1 and
  P2 or V1 and V2.
• Use your intuition to help you decide if the
  volume will go up or down as the pressure is
  changed and vice versa.

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Charles Law The Volume-Temperature
Relationship The Absolute Temperature Scale
absolute zero -273.15 0C
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Charles Law The Volume-Temperature
Relationship The Absolute Temperature Scale

• Charless law states that the volume of a gas is
  directly proportional to the absolute temperature
  at constant pressure.
• Gas laws must use the Kelvin scale to be correct.
• Relationship between Kelvin and centigrade.

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Charles Law The Volume-Temperature
Relationship The Absolute Temperature Scale

• Mathematical form of Charles law.

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Charles Law The Volume-Temperature
Relationship The Absolute Temperature Scale

• Example 2 A sample of hydrogen, H2, occupies
  1.00 x 102 mL at 25.0oC and 1.00 atm. What
  volume would it occupy at 50.0oC under the same
  pressure?
• T1 25 273 298
• T2 50 273 323

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Gay-Lussacs Law The Pressure-Temperature
Relationship

• At constant Volume, the pressure and the
  Temperature (in K) follow a direct
  proportionality.
• P ? T or P1 T1
• P2 T2

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Standard Temperature and Pressure

• Standard temperature and pressure is given the
  symbol STP.
• It is a reference point for some gas
  calculations.
• Standard P ? 1.00000 atm or 101.3 kPa
• Standard T ? 273.15 K or 0.00oC

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The Combined Gas Law Equation

• Boyles, Charles and Gay Lussacs Laws combined
  into one statement is called the combined gas law
  equation.
• Useful when the V, T, and P of a gas are changing.

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The Combined Gas Law Equation

• Example 3 A sample of nitrogen gas, N2, occupies
  7.50 x 102 mL at 75.00C under a pressure of 8.10
  x 102 torr. What volume would it occupy at STP?

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The Combined Gas Law Equation

• Example 4 A sample of methane, CH4, occupies
  2.60 x 102 mL at 32oC under a pressure of 0.500
  atm. At what temperature would it occupy 5.00 x
  102 mL under a pressure of 1.20 x 103 torr?
• You do it!

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The Combined Gas Law Equation
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Avogadros Law and theStandard Molar Volume
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Avogadros Hypothesis

• Avogadros Hypothesis states that at the same
  temperature and pressure, equal volumes of two
  gases contain the same number of molecules (or
  moles) of gas.

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Avogadros Hypothesis

• If we set the temperature and pressure for any
  gas to be STP, then one mole of that gas has a
  volume called the standard molar volume.
• The standard molar volume is 22.4 L at STP.
• This is another way to measure moles.
• For gases, the volume is proportional to the
  number of moles.
• 11.2 L of a gas at STP 0.500 mole
• 44.8 L ? moles

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Avogadros Law and theStandard Molar Volume

• Example 5 One mole of a gas occupies 36.5 L and
  its density is 1.36 g/L at a given temperature
  and pressure. (a) What is its molar mass? (b)
  What is its density at STP?

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Avogadros Law

• The volume of a gas maintained at constant
  pressure and temperature is directly proportional
  to the number of moles of the gas.
• n1 n2
• V1 V2

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Summary of Gas LawsThe Ideal Gas Law

• Boyles Law - V ? 1/P (at constant T n)
• Charles Law V ? T (at constant P n)
• Avogadros Law V ? n (at constant T P)
• Combine these three laws into one statement
• V ? nT/P
• Convert the proportionality into an equality.
• V nRT/P
• This provides the Ideal Gas Law.
• PV nRT
• R is a proportionality constant called the
  universal gas constant.

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Summary of Gas LawsThe Ideal Gas Law

• We must determine the value of R.
• Recognize that for one mole of a gas at 1.00 atm,
  and 273 K (STP), the volume is 22.4 L.
• Use these values in the ideal gas law.

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Summary of Gas LawsThe Ideal Gas Law

• R has other values if the units are changed.
• R 8.314 J/mol K
• Use this value in thermodynamics.
• R 8.314 kg m2/s2 K mol
• Use this later in this chapter for gas
  velocities.
• R 8.314 dm3 kPa/K mol
• This is R in all metric units.
• R 1.987 cal/K mol
• This the value of R in calories rather than J.

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Summary of Gas LawsThe Ideal Gas Law

• Example 6 What volume would 50.0 g of ethane,
  C2H6, occupy at 1.40 x 102 oC under a pressure of
  1.82 x 103 torr?
• To use the ideal gas law correctly, it is very
  important that all of your values be in the
  correct units!
• T 140 273 413 K
• P 1820 torr (1 atm/760 torr) 2.39 atm
• 50 g (1 mol/30 g) 1.67 mol

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Summary of Gas LawsThe Ideal Gas Law
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Summary of Gas LawsThe Ideal Gas Law

• Example 7 Calculate the number of moles in, and
  the mass of, an 8.96 L sample of methane, CH4,
  measured at standard conditions.
• You do it!

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Summary of Gas LawsThe Ideal Gas Law
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Summary of Gas LawsThe Ideal Gas Law

• Example 8 Calculate the pressure exerted by 50.0
  g of ethane, C2H6, in a 25.0 L container at
  25.0oC.
• You do it!

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Determination of Molecular Weights and Molecular
Formulas of Gaseous Substances

• Example 9 A compound that contains only carbon
  and hydrogen is 80.0 carbon and 20.0 hydrogen
  by mass. At STP, 546 mL of the gas has a mass of
  0.732 g . What is the molecular (true) formula
  for the compound?
• 100 g of compound contains 80 g of C and 20 g of
  H.

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Determination of Molecular Weights and Molecular
Formulas of Gaseous Substances
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Determination of Molecular Weights and Molecular
Formulas of Gaseous Substances
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Determination of Molecular Weights and Molecular
Formulas of Gaseous Substances

• Example 10 A 1.74 g sample of a compound that
  contains only carbon and hydrogen contains 1.44 g
  of carbon and 0.300 g of hydrogen. At STP 101 mL
  of the gas has a mass of 0.262 gram. What is its
  molecular formula?
• You do it!

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Determination of Molecular Weights and Molecular
Formulas of Gaseous Substances
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Determination of Molecular Weights and Molecular
Formulas of Gaseous Substances
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Calculations Involving Many Variables
Sometimes you will not be given values for the
necessary variables directly. For
example Calculate the pressure a 0.150 M gas
exerts at 30C. PV nRT solving for
P P nRT / V But n/V is molarity, so P M RT P
0.15 mol/L 0.0821 atmL/molK293K P 0.0123
atm
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Calculations Involving Many Variables
Another example What is the density of CCl4
vapor at 0.900 atm and 125C? PV nRT and
n m/MM Substituting n in the ideal gas
law PV mRT/MM Reorganizing P MM mRT/V
but d m/V P MM dRT solving for d we
get d P MM/RT d 0.900atm 153.8 g/mol
0.0821 atmL/molK
398 K d 4.24 g/L
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Calculations Involving Many Variables
Another example Calculate the average molar mass
of dry air if it has a density of 1.17 g/L at
21C and 0.95 atm. PV nRT and n
m/MM Substituting n in the ideal gas law PV
mRT/MM Reorganizing P MM mRT/V but d
m/V P MM dRT solving for MM we get MM
dRT / P MM 1.17 g/L 0.0821 atmL/molK 294K
29.7 g/mol 0.95
atm
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Daltons Law of Partial Pressures

• Daltons law states that the pressure exerted by
  a mixture of gases is the sum of the partial
  pressures of the individual gases.
• Ptotal PA PB PC .....

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Daltons Law of Partial Pressures

• Example 11 If 1.00 x 102 mL of hydrogen,
  measured at 25.0 oC and 3.00 atm pressure, and
  1.00 x 102 mL of oxygen, measured at 25.0 oC and
  2.00 atm pressure, were forced into one of the
  containers at 25.0 oC, what would be the pressure
  of the mixture of gases?

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Daltons Law of Partial Pressures

• Vapor Pressure is the pressure exerted by a
  substances vapor over the substances liquid at
  equilibrium.
• Water vapor must be taken into account when
  collecting a gas over water.

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Collecting a gas over H2O
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Collecting a gas over H2O
Ammonium nitrite, NH4NO2, decomposes upon heating
to form nitrogen gas NH4NO2(s) ? N2(g)
2 H2O(l) When a sample of NH4NO2 is decomposed
in a test tube, 511 mL of N2 gas is collected
over water at 26C and 745 torr total pressure.
How many grams of NH4NO2 were decomposed?
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Collecting a gas over H2O
Ammonium nitrite, NH4NO2, decomposes upon heating
to form nitrogen gas NH4NO2(s) ? N2(g)
2 H2O(l) When a sample of NH4NO2 is decomposed in
a test tube, 511 mL of N2 gas is collected over
water at 26C and 745 torr total pressure. How
many grams of NH4NO2 were decomposed?
From Appendix B PH2O at 26C is 25.21 torr PN2
PT PH2O 745 25.21 720 torr or 0.947
atm Calculating nN2 nN2 PV/RT 0.947 atm
0.511 L 0.0197 moles
0.0821 (atm L/mol K) 299K 0.0197 mol
N2 1 mol NH4NO2 64 g NH4NO2 1.26 g
1 mol N2 1 mol
NH4NO2
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Daltons Law of Partial Pressures

• Example 12 A sample of hydrogen was collected by
  displacement of water at 25.0 oC. The
  atmospheric pressure was 748 torr. What pressure
  would the dry hydrogen exert in the same
  container?

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Daltons Law of Partial Pressures

• Example 13 A sample of oxygen was collected by
  displacement of water. The oxygen occupied 742
  mL at 27.0 oC. The barometric pressure was 753
  torr. What volume would the dry oxygen occupy at
  STP?
• You do it!

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Daltons Law of Partial Pressures
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Mole fractions
We know PT P1 P2 P3 And P1 n1RT/V P2
n2RT/V P3 n3RT/V Substituting P1, P2 and
P3 in the Daltons law and factoring RT/V we
get PT (n1n2n3..) RT/V nT
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Mole fractions
P1 n1RT/V and PT nTRT/V Dividing P1/PT we
get P1 n1 RT/V P1
n1 PT nT RT/V PT
nT
mole fraction Solving for P1 P1 ?1
PT
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Mole fractions
The pressure exerted by nitrogen gas in our
atmosphere at 760 torr is PN2 ?N2 PN2 PN2
0.78 760 torr 590 torr 78 of the
atmosphere
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Mass-Volume Relationships in Reactions Involving
Gases

• In this section we are looking at reaction
  stoichiometry just including gases in the
  calculations.

• 2 mol KClO3 yields 2 mol KCl and 3 mol O2
• 2(122.6g) yields 2 (74.6g) and 3
  (32.0g)
• Those 3 moles of O2 can also be thought of as
• 3(22.4L) or 67.2 L at STP

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Mass-Volume Relationships in Reactions Involving
Gases

• Example 14 What volume of oxygen measured at
  STP, can be produced by the thermal decomposition
  of 120.0 g of KClO3?
• You do it!

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Mass-Volume Relationships in Reactions Involving
Gases
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The Kinetic-Molecular Theory

• The basic assumptions of kinetic-molecular theory
  are
• Postulate 1
• Gases consist of discrete large number of
  molecules (or atoms) that are in continuous
  random motion.
• Proof - Gases are easily compressible.

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The Kinetic-Molecular Theory

• Postulate 2The combined volume of all the
  molecules of the gas is negligible relative to
  the total volume in which the gas is contained.

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The Kinetic-Molecular Theory

• Postulate 3 Attractive and repulsive forces
  between molecules are negligible.
• Intermolecular forces (usually London dispersion
  forces) are very weak in gases due to their
  constant motion and large occupied volume.

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The Kinetic-Molecular Theory

• Postulate 4
• Energy can be transferred between molecules
  during collisions, but the average kinetic energy
  of the molecules does not change with time, as
  long as the temperature of the gas remains
  constant.
• Gas molecules have elastic collisions with
  themselves and the container.
• Proof - A sealed, confined gas exhibits no
  pressure drop over time.

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The Kinetic-Molecular Theory

• Postulate 5
• The kinetic energy of the molecules is
  proportional to the absolute temperature.
• At any given temperature the molecules of all
  gases have the same average kinetic energy.
• Proof - Brownian motion increases as temperature
  increases.

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The Kinetic-Molecular Theory

• The kinetic energy of the molecules is
  proportional to the absolute temperature. The
  kinetic energy of the molecules is proportional
  to the absolute temperature.
• Displayed in a Maxwellian distribution.

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Root-Means-Square or rms (u)

• Individual molecules move at varying speeds.
• rms is the speed of a molecule possessing average
  kinetic energy. It depends on the Temperature.
• KE ½ mu2
• KE of He at 298 K KE of Xe at 298 K
• This means that the molecule of He must be moving
  at higher rms (u) since the massis lower than the
  Xe molecules.
• KEHe KEXe
• (½ mu2)He (1/2 mu2)Xe
• since mHe lt mXe then u2He gt u2Xe

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Root-Means-Square or rms (u)
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The Kinetic-Molecular Theory

• The root-mean square velocity of gases is a very
  close approximation to the average gas velocity.
• Calculating the root-mean square velocity is
  simple

• To calculate this correctly
• The value of R 8.314 kg m2/s2 K mol
• And M must be in kg/mol.

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Root-Means-Square or rms (u)
Calculate the rms of a nitrogen molecule at
25C T 25C 273 298 K R 8.314 kg m2/s2
mol K MM 28.0 g/mol 28.0x10-3 kg/mol U v3
(8.314 kg m2/s2 mol K) (298 K) 28.0x10-3
kg/mol U 5.15x102 m/s
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The Kinetic-Molecular Theory

• The gas laws that we have looked at earlier in
  this chapter are proofs that kinetic-molecular
  theory is the basis of gaseous behavior.
• Boyles Law
• P ? 1/V
• As the V increases the molecular collisions with
  container walls decrease and the P decreases.
• Daltons Law
• Ptotal PA PB PC .....
• Because gases have few intermolecular
  attractions, their pressures are independent of
  other gases in the container.
• Charles Law
• V ? T
• An increase in temperature raises the molecular
  velocities, thus the V increases to keep the P
  constant.

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The Kinetic-Molecular Theory
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The Kinetic-Molecular Theory

• Example 18 What is the root mean square velocity
  of He atoms at room T, 25.0oC?
• You do it!

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The Kinetic-Molecular Theory
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Diffusion and Effusion of Gases

• Diffusion is the intermingling of gases. One gas
  spread throughout another substance or in space.
• Effusion is the escape of gases through tiny
  holes into an evacuated space.

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Effusion of Gases
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Diffusion and Effusion of Gases

• This is a demonstration of diffusion.

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Diffusion and Effusion of Gases

• The rate of effusion is inversely proportional to
  the square roots of the molecular weights or
  densities.

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Diffusion and Effusion of Gases

• Example 15 Calculate the ratio of the rate of
  effusion of He to that of sulfur dioxide, SO2, at
  the same temperature and pressure.

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Diffusion and Effusion of Gases

• Example 16 A sample of hydrogen, H2, was found
  to effuse through a pinhole 5.2 times as rapidly
  as the same volume of unknown gas (at the same
  temperature and pressure). What is the molecular
  weight of the unknown gas?
• You do it!

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Diffusion and Effusion of Gases

• An unknown gas composed of diatomic molecules
  effuses at a rate of 0.355 times that of O2 at
  the same temperature. Calculate the MM and
  identify the gas.
• runk 0.355 v MMO2
• rO2 Mmunk
• (0.355)2 32 / MMunk MMunk 254 g/mol
• Since the gas is diatomic, the atomic mass is 127
  so the element is iodine I2

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Real Gases Deviations from Ideality

• Real gases behave ideally at ordinary
  temperatures and pressures.
• At low temperatures and high pressures real gases
  do not behave ideally.
• The reasons for the deviations from ideality are
• The molecules are very close to one another, thus
  their volume is important.
• The molecular interactions also become important.

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Real GasesDeviations from Ideality

• van der Waals equation accounts for the behavior
  of real gases at low temperatures and high
  pressures.

• The van der Waals constants a and b take into
  account two things
• a accounts for intermolecular attraction
• b accounts for volume of gas molecules
• At large volumes a and b are relatively small and
  van der Waals equation reduces to ideal gas law
  at high temperatures and low pressures.

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Real GasesDeviations from Ideality

• What are the intermolecular forces in gases that
  cause them to deviate from ideality?
• For nonpolar gases the attractive forces are
  London Forces
• For polar gases the attractive forces are
  dipole-dipole attractions or hydrogen bonds.

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Real GasesDeviations from Ideality

• Example 19 Calculate the pressure exerted by
  84.0 g of ammonia, NH3, in a 5.00 L container at
  200. oC using the ideal gas law.
• You do it!

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Real GasesDeviations from Ideality

• Example 20 Solve Example 19 using the van der
  Waals equation.

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Real GasesDeviations from Ideality