kinetic theory of gases i
DESCRIPTION
Kinetic Theory of Gases I. Ideal Gas. The number of molecules is large. The average separation between molecules is large. Molecules moves randomly. Molecules obeys Newton’s Law. Molecules collide elastically with each other and with the wall. Consists of identical molecules. - PowerPoint PPT PresentationTRANSCRIPT
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Kinetic Theory of Gases I
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Ideal GasThe number of molecules is large
The average separation between molecules is large
Molecules moves randomly
Molecules obeys Newton’s Law
Molecules collide elastically with each other and with the wall
Consists of identical molecules
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The Ideal Gas Law
PV nRTn: the number of moles in the ideal gas
n N
NA
total number of molecules
Avogadro’s number: the number ofatoms, molecules, etc, in a mole ofa substance: NA=6.02 x 1023/mol.
R: the Gas Constant: R = 8.31 J/mol · K
in K
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Pressure and TemperaturePressure: Results from collisions of molecules on the surface
P F
APressure:
Force
Area
Force: F dp
dtRate of momentumgiven to the surface
Momentum: momentum given by each collisiontimes the number of collisions in time dt
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Only molecules moving toward the surface hitthe surface. Assuming the surface is normal to the x axis, half the molecules of speed vx movetoward the surface.
Only those close enough to the surface hit it in time dt, those within the distance vxdt
The number of collisions hitting an area A in time dt is 1
2
N
V
Avx dt
The momentum given by each collision to the surface 2mvx
Average density
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dp 2mvx 1
2
N
V
Avxdt
Momentum in time dt:
Force: F dp
dt 2mvx 1
2
N
V
Avx
Pressure: P F
A
N
Vmvx
2
Not all molecules have the same average vx2
P N
Vmvx
2
vx
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vx2 =
1
3v2
1
3vx
2 vy2 vz
2
Pressure: P 1
3
N
Vmv2
2
3
N
V
1
2mv2
Average Translational Kinetic Energy:
K 1
2mv2
1
2mvrms
2
vx2 =
1
3v2
1
3 vrms
2
vrms v2 vx
2 + vy2 + vz
2
3
is the root-mean-square speed vrms
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Pressure: P 2
3N
VK
PV 2
3N KFrom and PV nRT
Temperature: K 3
2nRT
N
3
2kBT
Boltzmann constant: kB R
NA1.3810 23 J/K
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PV 1
3N mvrms
2From
and PV nRT N
NART
vrms 3RT
M
Avogadro’s number
N nNA
M mN A
Molar mass
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Pressure Density x Kinetic Energy
Temperature Kinetic Energy
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Internal Energy
For monatomic gas: the internal energy = sumof the kinetic energy of all molecules:
Eint N K nNA 3
2kBT
3
2nRT
Eint 3
2nRT T
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(d) mass of gas n
pV nRT nConstant temperaturep
T
nR
V nConstant volume
V
T
nR
p nConstant pressure
HRW 16P (5th ed.). Consider a given mass of an ideal gas. Compare curves representing constant-pressure, constant volume, and isothermal processes on (a) a p-V diagram, (b) a p-T diagram, and (c) a V-T diagram. (d) How do these curves depend on the mass of gas?
p p
V
constant pressure
isothermal
constant volume
T
constant pressure
constant volume
isothermal V
T
constant volume
constant pressure
isothermal
PV nRT
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(b) TB pBVB
nR1.8 103 K
(c) TC pCVC
nR6.0 102 K
(d) Cyclic process ∆Eint = 0
Q = W = Enclosed Area= 0.5 x 2m2 x 5x103Pa = 5.0 x 103 J
(a) n pAVA
RTA1.5 mol.
HRW 18P (5th ed.). A sample of an ideal gas is taken through the cyclic process abca shown in the figure; at point a, T = 200 K. (a) How many moles of gas are in the sample? What are (b) the temperature of the gas at point b, (c) the temperature of the gas at point c, and (d) the net heat added to the gas during the cycle?
Volume (m3)
a
b
c
1.0 3.0Pr
essu
re (
kN/m
2 )
2.5
7.5
PV nRTEint Q W
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(a)
vrms 3RT
M
3 8.31 J/molK 293 K 28.0 10-3 kg/mol
511 m/s
for 0.5 vrms T 0.52T 73.3K = -200C
(b) Since vrms Tv rms2
vrms2
T
T
for 2 vrms T 22T 1.17 103K = 899C
HRW 30E (5th ed.).(a) Compute the root-mean-square speed of a nitrogen molecule at 20.0 ˚C. At what temperatures will the root-mean-square speed be (b) half that value and (c) twice that value?
vrms 3RT
M
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(a)K 3
2kBT 3
21.3810 23 J/K 1600K
3.3110 20 J
(b) 1 eV = 1.60 x 10-19 J
K 3.3110 20 J
1.60 10 19 J/eV0.21 eV
HRW 34E (5th ed.). What is the average translational kinetic energy of nitrogen molecules at 1600K, (a) in joules and (b) in electron-volts?
K 3
2kBT
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Kinetic Theory of Gases II
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Mean Free Path
Molecules collide elastically with other molecules
Mean Free Path : average distance between
two consecutive collisions
1
2d2N / V
the more moleculesthe more collisions
the bigger the moleculesthe more collisions
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Molar Specific Heat
Definition:
For constant volume: Q nCVT
For constant pressure: Q nCpT
The 1st Law of Thermodynamics:
Eint 3
2nRT Q W (Monatomic)
Q cmT
Eint Q W
Eint 3
2nRT
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Constant Volume
W PdV 0
3
2nRT Q W
(Monatomic)
Q nCVT
3
2nRT nCVT
CV 3
2R
Eint nCVT
Eint 3
2nRT
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Constant Pressure
W PV nRT
3
2nRT Q W
(Monatomic)
Q nCpT
3
2nRT nCpT nRT
CV Cp R
Cp
CV
Cp 5
2R
5
3
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dEint dW pdV
Adiabatic Process
(Q=0)
dEint dQ dW
pV nRT
1st Law
Ideal Gas Law
Eint nCVT
nCVdT
pdV Vdp nRdT nR pdV
nCV
Divide by pV:
dV
V
dp
p
Cp CV
CV
dV
V(1 )
dV
V
Cp CV R
Cp
CV
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dV
V
dp
p(1 )
dV
V
dp
p
dV
V0
ln p lnV ln(pV ) = const.
pV = const.
pV nRTIdeal Gas Law
(nRT
V)V = const.
TV 1 = const.
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Equipartition of Energy
The internal energy of non-monatomic molecules includes also vibrational androtational energies besides the translational energy.
Each degree of freedom has associated with
it an energy of per molecules.1
2kBT
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Monatomic Gases
3 translational degrees of freedom:
Eint 3
2kBT nNA
3
2nRT
CV 1
ndEint
dT
3
2R
Eint nCVT
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Diatomic Gases
3 translational degrees of freedom
Eint 5
2nRT CV
5
2R
Eint nCVT
2 rotational degrees of freedom
2 vibrational degrees of freedom
HOWEVER, different DOFs require differenttemperatures to excite. At room temperature,only the first two kinds are excited:
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(a) Constant pressure: W = p∆V
Eint Q W Q pV
20.9 1.0 105 Pa 100 cm3 50cm3 10 6 m3/cm3 15.9 J
HRW 63P (5th ed.). Let 20.9 J of heat be added to a particular ideal gas. As a result, its volume changes from 50.0 cm3 to 100 cm3 while the pressure remains constant at 1.00 atm. (a) By how much did the internal energy of the gas change? If the quantity of gas present is 2.00x10-3 mol, find the molar specific heat at (b) constant pressure and (c) constant volume.
Eint Q W
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pV nRT(b) Cp Q
nT Q
n pV / nR R
p
Q
V
8.31 J/molK 20.9J 1.0 105 Pa 50 10 6 cm3 34.4 J/molK
(c) CV Cp R
34.4 J/molK 8.31 J/molK = 26.1 J/molK
HRW 63P (5th ed.). Let 20.9 J of heat be added to a particular ideal gas. As a result, its volume changes from 50.0 cm3 to 100 cm3 while the pressure remains constant at 1.00 atm. (a) By how much did the internal energy of the gas change? If the quantity of gas present is 2.00x10-3 mol, find the molar specific heat at (b) constant pressure and (c) constant volume.
Q nCpT
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ln / 5
3ln /
i f
f i
p p
V V
(a) Adiabatic piVi p f Vf
pi / p f Vf / Vi
Monatomic
HRW 81P (5th ed.). An ideal gas experiences an adiabatic compression from p =1.0 atm, V =1.0x106 L, T = 0.0 ˚C to p =1.0 x 105 atm, V =1.0x103 L. (a) Is the gas monatomic, diatomic, or polyatomic? (b) What is its final temperature? (c) How many moles of gas are present? (d) What is the total translational kinetic energy per mole before and after the compression? (e) What is the ratio of the squares of the rms speeds before and after the compression?
pV = const.
Cp
CV
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(b)piVi
Ti
p f Vf
Tf
pV nRT
Tf pf Vf
piViTi
1.0 105 atm 1.0 103 L 273 K
1.0 atm 1.0 106 L 2.7 104 K
HRW 81P (5th ed.). An ideal gas experiences an adiabatic compression from p =1.0 atm, V =1.0x106 L, T = 0.0 ˚C to p =1.0 x 105 atm, V =1.0x103 L. (a) Is the gas monatomic, diatomic, or polyatomic? (b) What is its final temperature? (c) How many moles of gas are present? (d) What is the total translational kinetic energy per mole before and after the compression? (e) What is the ratio of the squares of the rms speeds before and after the compression?
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n piVi
RTi4.5104 mol.(c) (Pay attention to the units)
K 3
2nRT
N
Ki 3
2RTi 3.4 103 J
(d) For N/n = 1
K f 3
2RTf 3.4 105 J
HRW 81P (5th ed.). An ideal gas experiences an adiabatic compression from p =1.0 atm, V =1.0x106 L, T = 0.0 ˚C to p =1.0 x 105 atm, V =1.0x103 L. (a) Is the gas monatomic, diatomic, or polyatomic? (b) What is its final temperature? (c) How many moles of gas are present? (d) What is the total translational kinetic energy per mole before and after the compression? (e) What is the ratio of the squares of the rms speeds before and after the compression?
pV nRT
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K 1
2mvrms
2 3
2kBT
(e)vrms,i
2
vrms, f2
Ti
Tf0.01
HRW 81P (5th ed.). An ideal gas experiences an adiabatic compression from p =1.0 atm, V =1.0x106 L, T = 0.0 ˚C to p =1.0 x 105 atm, V =1.0x103 L. (a) Is the gas monatomic, diatomic, or polyatomic? (b) What is its final temperature? (c) How many moles of gas are present? (d) What is the total translational kinetic energy per mole before and after the compression? (e) What is the ratio of the squares of the rms speeds before and after the compression?