deviations from ideal behavior 1 mole of ideal gas pv = nrt n = pv rt = 1.0 5.8 repulsive forces...
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Deviations from Ideal Behavior
1 mole of ideal gas
PV = nRT
n = PVRT
=1.0
5.8
Repulsive Forces
Attractive Forces
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Maxwell distribution :In 1859, James Clerk Maxwell (1831 – 1879) worked out a formula for
the most probable distribution of speeds in a gas .
Molecules in a gas sample
move at a variety of speeds . Speed of
each molecule constantly changing
due to countless collisions(about 1
billion per second for each molecule).
At low temperature most molecules
move close to the average speed , at
higher temperature there is greater
distribution of speeds.
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Maxwell distribution of speeds :
Just for your knowledge do not memorize
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Boltzman Distribution. The behaviour of the gas molecules under the action of gravity.
(Harcourt school Publishers)
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The Kinetic Theory
• The ideal gas law is an empirical law gives a macroscopic explanation of gas behavior.
• Kinetic Theory starts with a set of assumptions about the microscopic behavior of matter at the atomic level.
• Supposes that the constituent particles (atoms) of the gas obey the laws of classical physics.
• Accounts for the random behavior of the particles with statistics, thereby establishing a new branch of physics - statistical mechanics.
• Predicts experimental phenomena that haven't been observed. (Maxwell-Boltzmann Speed Distribution)
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An ideal gas molecule in a cube
of sides L .
J. B. Callis,Washington University
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Calculating force exerted on a container by Collision of a single particle
J. B. Callis,Washington University
axes..Cartesian three thealong velocity theof components
are and , and particle theof speed theis Where
2222
zyx
zyx
uuuu
uuuu
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2222 2222
,
and for same theand 2
/1frequency collision The
)(2
uL
m
L
mu
L
mu
L
muF
Thus
FFL
umuF
tL
u
mumumumu
t
mu
t
ummaF
zyxtot
zyx
xx
x
xxxx
J. B. Callis,Washington University
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Calculation of the Pressure in Terms of Microscopic Properties of the Gas Particles
V
umnN
V
umnNP
N
nnN
moleculeV
um
L
um
L
Lum
Area
FP
uL
mF
A
A
A
A
Tot
Tot
tot
22
2
3
2
2
2
2
21
3
2
3
number, sAvogadro' is and moles
ofnumber theis where, as expressed becan
sample gasgiven ain particles ofnumber theSince
)1(336
/2
2
J. B. Callis,Washington University
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Kinetic Theory Relates the Kinetic Energy of the Particles to Temperature
nRTPV
Thus
RTKE
KEn
PV
V
nKEP
umNKE
avg
avgavg
Aavg
2
3
3
2or
3
2
2
1 2
J. B. Callis,Washington University
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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
RTurms
3
where R is the gas constant (8.31 J/mol K) ,T is the temperature in kelvin,M is the molar mass expressed in kg/mol (to make the speed come out in units of m/s).
J. B. Callis,Washington University
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Relationship of kinetic energy of one gas particle to temperature:
TN
RE
Ak
2
3
Conclusion:•Temperature is a measure of the molecular motion.• At the same temperature, all gases have the same average kinetic energy.
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Gas diffusion and effusionGas diffusion and effusion
Graham’s law governs effusion Graham’s law governs effusion and diffusion of gas and diffusion of gas molecules. molecules. KE=1/2 mv2
Thomas Graham, 1805-Thomas Graham, 1805-1869. Professor in 1869. Professor in Glasgow and LondonGlasgow and London..
M of AM of B
Rate for B
Rate for A
M
1diffusionofRate
Rate of effusion is inversely Rate of effusion is inversely proportional to its molar massproportional to its molar mass..
Rate of effusion is inversely Rate of effusion is inverselyproportional to its molar massproportional to its molar mass..
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Diffusion• Diffusion is 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).
• Effusion 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.
J. B. Callis,Washington University
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GAS DIFFUSION AND GAS DIFFUSION AND EFFUSIONEFFUSION
• diffusiondiffusion is the gradual is the gradual mixing of molecules mixing of molecules of different gases.of different gases.
• effusioneffusion is the is the movement of molecules movement of molecules through a small hole through a small hole into an empty container.into an empty container.
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The Process of Effusion J. B. Callis,Washington University
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The inverse relation between
diffusion rate andmolar mass.
NH3(g) + HCl(g) NH4Cl(s)
Due to it’s lightmass, ammonia
travels 1.46 timesas fast as
hydrogen chloride
J. B. Callis,Washington University
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Relative Diffusion Rates of NH3 and HCl
J. B. Callis,Washington University
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GAS DIFFUSION AND GAS DIFFUSION AND EFFUSIONEFFUSION
Molecules effuse thru holes in a rubber Molecules effuse thru holes in a rubber balloon, at a rate (= moles/time) that is balloon, at a rate (= moles/time) that is proportional to Tproportional to T
• inversely proportional to M.inversely proportional to M.
QuestionQuestion If you have 2 ballons flled with He & OIf you have 2 ballons flled with He & O2 2 Left
overnight at same T ,Which will effuse more?Which will effuse more?
He effuses more rapidly than OHe effuses more rapidly than O22 at same at same
T.T.
HeHe
OO22
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Examples
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Calculating Molecular Speeds :
Question :
Calculate the average speed of O2 in air at 20 oC .
1660 km/h! (1.6km=1mile).
Question :
What is the r.m.s. speed of SO2 atoms at 25°C?
urms = 340.78 ms-1
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Calculation of Molecular Speeds and Kinetic Energies , T = 300 K
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
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Problem 15-1: Calculate the Kinetic Energy of )a( a Hydrogen Molecule
traveling at 1.57 x 103 m/sec, at 300 K.
Mass =
KE =
KE =
KE =
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Problem 15-1 Kinetic Energies for (b) CH4 and (c) CO2 at 200 K
(b) For Methane, CH4 , u = 5.57 x 102 m/s
KE =
(c) For Carbon Dioxide, CO2 , u = 3.37 x 102 m/s
KE =
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Note
• At a given temperature, all gases have the same molecular kinetic energy distributions,
and
• the same average molecular kinetic energy.
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NH3(g) + HCl(g) = NH4Cl(s)
NH3(g) + HCl(g) = NH4Cl(s)
HCl = 36.46 g/mol NH3 = 17.03 g/mol
Problem
Relative Diffusion Rate of NH3 compared to HCl:
RateNH3 =