lecture 3. the kinetic molecular theory of gases
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Lecture 3. The Kinetic Molecular Theory of Gases. THE IDEAL GAS LAW:. A purely empirical law – solely the consequence of experimental observations Explains the behavior of gases over a limited range of conditions - PowerPoint PPT PresentationTRANSCRIPT
Lecture 3. The Kinetic Molecular Theory of Gases
THE IDEAL GAS LAW:
• A purely empirical law – solely the consequence of experimental observations
• Explains the behavior of gases over a limited range of conditions
• A macroscopic explanation. Say nothing about the microscopic behavior of the atoms or molecules that make up the gas.
THE KINETIC MOLECULAR THEORY:
• Starts with a set of assumptions about the microscopic behavior of matter at the atomic level.
• Assumes that the constituent particles (atoms or molecules) obey the laws of classical physics
• Accounts for the random behavior of the particles with statistics.• Offers an explanation of the macroscopic behavior of gases• Predicts experimental phenomena that haven’t been observed
(Maxwell-Boltzmann Speed distribution)
The Four Postulates of the Kinetic Theory
• A gas consists of large number of identical molecules separated by distances that are so large compared to their size.
• The gas molecules are constantly moving in random directions with a distribution of speeds.
• The molecules exert no forces on one another between collisions, so they move in straight lines with constant velocities.
• The collision of molecules with the walls of the container are elastic: no energy is lost during a collision. Collisions with the walls of the container are the source of pressure.
The Ideal Gas in a Cube with sides, L
Cartesian Coordinates for a Gas Particles
Cartesian Components of a Particle’s velocity
n.informatio ldirectiona
no conveys andquantity scalar a is that Note
axes.Cartesian
principal three thealong velocity theof components
are and , and particle theof velocity theis Where
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v
vvvv
vvvv
zyx
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Components of velocity along the X axis
Calculation of the Force Exerted on a Container by Collision of a Single Particle
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and for simalarly and
frequency collision that theRecalling
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Thus
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zyxtot
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Calculation of the Pressure in Terms of Microscopic Properties of the Gas Particles
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sample gasgiven ain particles ofnumber theSince
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The Pressure of a Gas:
• is directly proportional to the kinetic energy.• or to the root mean square velocity of the molecule.
•The larger the velocity , more frequent collisions, greater change in momentum
tran
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The Kinetic Theory Relates the Kinetic Energy of the Particles to Temperature
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Kinetic Energy and Temperature:
• Two ideal gases at the same temperature will have the same average kinetic energy.
TkE Btrans 2
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Mean Square Speed and Temperature
By use of the facts that (a) PV and nRT have units of energy, and (b) the square of the component of velocity of a gas particle striking the wall is on average one third of the mean square speed, the following expression may be derived:
m
Tkv
B
rms
3
where Kb is the Boltzman constant (1.38 x 10-23), T is the temperature in Kelvin, and M is the molar mass expressed in kg/mol (to make the speed come out in units of m/s).
root mean square velocity
Sample Problem
Calculate the Kinetic Energy of (a) a Hydrogen Molecule traveling at 1.57 x 103 m/sec, at 300 K.
Mass =
KE =
Sample Problem
Calculate the kinetic Kinetic Energies for (b) CH4 and (c) CO2 at 200 K
(b) For Methane, CH4 , v = 5.57 x 102 m/sKE =
(c) For Carbon Dioxide, CO2 , v = 3.37 x 102 m/sKE =
Kinetic Energy and Temperature
Finally, we obtain the relationship of kinetic energy of one gas phase particle to temperature:
TN
RE
Atrans
2
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Note that: (a) Temperature is a measure of the molecular motion, and (b) the kinetic energy is only a function of temperature. At the same temperature, all gases have the same average kinetic energy.
Path of One Gas Particle
Molecular Speed Distribution
Thus far we have discussed the random nature of molecular motion in terms of the average (root mean square) speed. But how is this speed distributed? The kinetic theory predicts the distribution function for the molecular speeds. Below we show the distribution of molecular speeds for N2 gas at three temperatures.
The Mathematical Description of the Maxwell-Boltzmann Speed Distribution
K in etemperatur T
and J/K, 101.38066 constant sBoltzman' k
kg in particle of mass m ,s/m in peedsv :where
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m4)v(f
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B2
Features of Speed Distribution
• The most probable speed is at the peak of the curve.
• The most probable speed increases as the temperature increases.
• The distribution broadens as the temperature increases.
Relationship between molar mass and molecular speed
Features of Speed Distribution
• The most probable speed increases as the molecular mass decreases
• The distribution broadens as the molecular mass decreases.
Calculation of Molecular Speeds and Kinetic Energies
Molecule H2 CH4 CO2
MolecularMass (g/mol)
2.016 16.04 44.01
Kinetic Energy (J/molecule)
6.213 x 10 - 21 6.213 x 10 - 21 6.213 x 10 - 21
Velocity (m/s)
1,926 683.8 412.4
T = 300 K
The Three Measures of the Speed of a Typical Particle
M
RT8c
M
RT2c
M
RT3c
avg
mp
rms
Various Ways to Summarize the ‘Mean’ Speed.
Sample Problem: Calculate the Kinetic Energy of (a) a Hydrogen Molecule traveling at 1.57 x 103 m/sec, at
300 K.Mass =
KE =
Sample Problem: Kinetic Energies for (b) CH4 and (c) CO2 at 200 K
(b) For Methane, CH4 , v = 5.57 x 102 m/sKE =
(c) For Carbon Dioxide, CO2 , v = 3.37 x 102 m/s
KE =
Note
• At a given temperature, all gases have the same molecular kinetic energy distributions,
and
• the same average molecular kinetic energy.
EffusionEffusion is the process whereby a gas escapes from its container through a tiny hole into an evacuated space. According to the kinetic theory a lighter gas effuses faster because the most probable speed of its molecules is higher. Therefore more molecules escape through the tiny hole in unit time.
This is made quantitative in Graham's Law of effusion: The rate of effusion of a gas is inversely proportional to the square root of its molar mass.
1
2
2
1
M
M
r
r
M
1 effusionofRate
The Process of Effusion
Effusion CalculationProblem 15-2: Calculate the ratio of the effusion rates of ammonia and hydrochloric acid.Approach: The effusion rate is inversely proportional to square root of molecular mass, so we find the molar ratio of each substance and take its square root. The inverse of the ratio of the square roots is the effusion rate ratio. Numerical Solution:
HCl = NH3 =
RateNH3 =
Diffusion
• The movement of one gas through another by thermal random motion.
• Diffusion is a very slow process in air because the mean free path is very short (for N2 at STP it is 6.6x10-8 m). Given the nitrogen molecule’s high velocity, the collision frequency is very high also (7.7x109 collisions/s).
• Diffusion also follows Graham's law:
M
1diffusionofRate
Diffusion of agas particlethrough aspace filledwith otherparticles
NH3(g) + HCl(g) = NH4Cl(s)
HCl = 36.46 g/mol NH3 = 17.03 g/mol
Problem 15-3: Relative Diffusion Rate of NH3 compared to HCl:
RateNH3 =
The inverse relation betweendiffusion rate andmolar mass.
NH3(g) + HCl(g) NH4Cl(s)
Due to it’s lightmass, ammonia travels 1.46 timesas fast as hydrogen chloride
Relative Diffusion Rates of NH3 and HCl
Gaseous Diffusion Separation of Uranium 235 / 238
235UF6 vs 238UF6
Separation Factor =
after Two runs
after approximately 2000 runs235UF6 is > 99% Purity !
Y - 12 Plant at Oak Ridge National Lab