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Physics DF025
Chapter 14
CHAPTER 14: Kinetic theory of gases
(3 Hours)
Physics DF025
Chapter 14
Learning Outcome:14.1 Ideal gas equations (1 hour)
a. Sketch and interpret
– P-V graph at constant temperature
– V-T graph at constant pressure
– P-T graph at constant volume
of an ideal gas.
nRTPV
At the end of this chapter, students should be able to:
b. Use the ideal gas equation:
Physics DF025
Chapter 14
14.1 Ideal gas equation14.1.1 Boyle’s law• states : “The pressure
of a fixed mass of gas at constant
temperature is inversely proportional to its volume.”
OR
VP
1 if constantT
• The related equations to the Boyle’s law are
constantPV
OR
2211 VPVP
where pressure initial:1Ppressure final:2P volumeinitial:1V
volumefinal:2V
Physics DF025
Chapter 14• Graphs of the Boyle’s law.
P
T1
V
T2
0
a.
T2
T1
P
V
1
0
b.
The pressure of a fixed mass of gass at constant temperature is inversely propertional to its volume.
Physics DF025
Chapter 1414.1.2 Charles’s law
• states : “The volume of a fixed mass of gas at constant pressure is directly proportional to its absolute temperature.”
OR
TV If constantP
The related equations to the Charles’ law are
constantT
V
OR
2
2
1
1
T
V
T
V
where
re temperatuabsolute initial:1Tre temperatuabsolute final:2T
volumeinitial:1V volumefinal:2V
Physics DF025
Chapter 14
Graphs of the Charles’s law.
a.
T(C)
V
0273.15
b. V
T(K)0
Physics DF025
Chapter 1414.1.3 Gay-lussac’s
(pressure) lawstates : “The pressure of a fixed mass of gas at constant volume is directly proportional to its absolute temperature.”
OR
TP If constantV
The related equations to the Gay-lussac’s law are
constantT
P
OR
2
2
1
1
T
P
T
P
where
re temperatuabsolute initial:1Tre temperatuabsolute final:2T
pressure initial:1Ppressure final:2P
Physics DF025
Chapter 14Graphs of the Gay-lussac’s (pressure) law
T(C)
P
0273.15
a. P
T(K)0
b.
Physics DF025
Chapter 14
• An ideal gas is defined as a perfect gas which obeys the three gas laws (Boyle’s, Charles’s and Gay-Lussac’s) exactly.
• Consider an ideal gas in a container changes its pressure P, volume V and temperature T as shown in Figure 14.1.
14.1.4 Equation of state for an ideal gas
1st stage
1P1V1T
2P'V
1T
2P2V2T
2nd stage
Figure 14.1
Physics DF025
Chapter 14– In 1st stage, temperature is
kept at T1 ,
Using Boyle’s law :
– In 2nd stage, pressure is kept
constant at P2 ,
Using Charles’s law :
112 ' VPVP
2
11'P
VPV (1)
2
2
1
'
T
V
T
V
2
12'T
TVV (2)
– Equating eqs. (1) and (2), thus
constantT
PV(3)
Initial Final
OR
2
22
1
11
T
VP
T
VP
Physics DF025
Chapter 14• Consider 1 mole of gas at standard temperature and pressure
(S.T.P.), T = 273.15 K, P = 101.3 kPa and Vm = 0.0224 m3
– From equation (3),
where R is called molar gas constant and its value is the same for all gases.
– Thus
• For n mole of an ideal gas, the equation of state is written as
– where n : the number of mole gas
15.273
0224.0103.101 3m
T
PVR
11 mol K J 31.8 R
RT
PVm
RTPV m where gas mole 1 of volume:mV
nRTPV
Physics DF025
Chapter 14
M
mn
where gas a of mass:m
gas a of massmolar :M
OR
AN
Nn
where molecules ofnumber :N
constant sAvogadro':AN123 mol10026 .
– If the Boltzmann constant, k is defined as
then the equation of state becomes
123
A
K J 1038.1 N
Rk
NkTPV
Physics DF025
Chapter 14
The volume of vessel A is three times of the volume vessel B. The vessels are filled with an ideal gas and are at a steady state. The temperature of vessel A and vessel B are at 300 K and 500 K respectively as shown in Figure 14.2.
If the mass of the gas in the vessel A is m, obtain the mass of the gas in the vessel B in terms of m.
Example 1 :
A B(300 K) (500 K)
Figure 14.2
Solution :
Since the vessels A and B are connected by a narrow tube thus the pressure for both vessels is same, finally i.e.
The system is in the steady state, thus
By applying the equation of state for an ideal gas,
K 500K; 300; ;3 0B0AABA TTmmVV
PPP BA
K 500K; 300 B0BA0A TTTT
nRTPV andM
mn
RTM
mPV
Physics DF025
Chapter 14
Therefore,
Vessel A :
AA
AA RTM
mVP
3003 B RM
mVP
Vessel B :
K 500K; 300; ;3 0B0AABA TTmmVV
RM
mPV
100B (1)
(2) RM
mPV
B
B 500
By equating the eqs. (1) and (2) hence
5B
mm
RM
mR
M
m
B500100
Physics DF025
Chapter 14
Refer to Figure 14.3. Initially A contains 3.00 m3 of an ideal gas at
a temperature of 250 K and a pressure of 5.00 104 Pa, while B contains 7.20 m3 of the same gas at 400 K and 2.00 104 Pa. Calculate the pressure after the connecting tap has been opened
and the system reached equilibrium, assuming that A is kept at
250 K and B is kept at 400 K.
Example 2 :
AB
connecting tap
Figure 14.3
Physics DF025
Chapter 14
Solution :
After the connecting tap has been opened and the system reached equilibrium, thus
By using the equation of state for ideal gas,
PPP 1B1A
1
11
0
00
T
VP
T
VP
1B
1B
1A
1A
0B
0B0B
0A
0A0A
T
V
T
VP
T
VP
T
VP
400
20.7
250
00.3
400
20.71000.2
250
00.31000.5 44
P
K; 250 ;m 00.3 1A0A3
1A0A TTVV;m 20.7Pa; 1000.5 3
1B0B4
0A VVPPa 1000.2K; 400 4
0B1B0B PTT
Pa 1020.3 4P
Physics DF025
Chapter 14
Given R = 8.31 J mol1 K1 and NA = 6.0 1023 mol1
1. A gas has a volume of 60.0 cm3 at 20 C and 900 mmHg. What would its volume be at STP?(Given the atmospheric pressure = 101.3 kPa and the density of mercury = 13600 kg m3)
ANS. : 66.2 cm3
2. Estimate the number of molecules in a flask of volume 5.0 104 m3 which contains oxygen gas at a pressure of 2.0 105 Pa and temperature of 300 K.
ANS. : 2.41 1022 molecules3. A cylinder contains a hydrogen gas of volume 2.40 103 m3 at
17 C and 2.32 106 Pa. Calculatea. the number of molecules of hydrogen in the cylinder,b. the mass of the hydrogen,c. the density of hydrogen under these conditions.(Given the molar mass of hydrogen = 2 g mol1)
ANS. : 1.39 1024 molecules; 4.62 g; 1.93 kg m3
Exercise 14.1 :
Physics DF025
Chapter 14
Learning Outcome:14.2 Kinetic theory of gases (1 hour)
At the end of this chapter, students should be able to:
a. State the assumptions of kinetic theory of gases.
b. Apply the equations of ideal gas,
2
3
1vNmPV
2
3
1vP
in related problems. c. Explain and use root
mean square (rms) speed,
m
kTv 32
of gas molecules.
and pressure ,
Physics DF025
Chapter 1414.2 Kinetic theory of gases• The macroscopic behaviour of an ideal gas can be describe
by using the equation of state but the microscopic behaviour only can be describe by kinetic theory of gases.
14.2.1 Assumption of kinetic theory of gases• All gases are made up of identical atoms or molecules. • All atoms or molecules move randomly and haphazardly.• The volume of the atoms or molecules is negligible when
compared with the volume occupied by the gas.• The intermolecular forces are negligible except during
collisions.• Inter-atomic or molecular collisions are elastic.• The duration of a collision is negligible compared with the
time spent travelling between collisions.• Atoms and molecules move with constant speed between
collisions. Gravity has no effect on molecular motion.
Physics DF025
Chapter 14
• Consider, initially a single molecule moving with a velocity vx towards wall A and after colliding elastically , it moves in the opposite direction with a velocity vx
as shown in Figure 14.5.
• Consider an ideal gas of N molecules are contained in a cubical container of side d as shown in Figure 14.4.
Wall A
Wall B
14.2.2 Force exerted by an ideal gas
Figure 14.4
• Let each molecule of the gas have the mass m and velocity v.
• The velocity, v of each molecule can be resolved into their components i.e. vx, vy and vz.
Wa
ll AWa
ll B
Wa
ll AWa
ll B
Figure 14.5
Physics DF025
Chapter 14Therefore the change in the linear momentum of the molecule is given by
xxx mvmvP
xx mvP 2
The molecule has to travel a distance 2d (from A to B and back to A) before its next collision with wall A. The time taken for this movement is
xv
dt
2
If Fx1 is the magnitude of the average force exerted by a molecule on the wall in the time t, thus by applying Newton’s second law of motion gives
x
xxx
vd
mv
t
PF
2
21
21 xx v
d
mF
For N molecules of the ideal gas,
222
21 ....... xNxxx v
d
mv
d
mv
d
mF
222
21 ....... xNxxx vvv
d
mF
Physics DF025
Chapter 14where vx1 is the x component of velocity of molecule 1, vx2 is the x component of velocity of molecule 2 and so on.
• The mean (average ) value of the square of the velocity in the x direction for N molecules is
N
vvvv xNxx
x
222
212 .......
2
222
21 .......
x
xNxx
vN
vvv
2xx vN
d
mF
2222zyx vvvv
2222zyx vvvv
Thus, the x component for the total force exerted on the wall of the cubical container is
• The magnitude of the velocity v is given by
then
Physics DF025
Chapter 14• Since the velocities of the molecules in the ideal gas are
completely random, there is no preference to one direction or another. Hence
• The total force exerted on the wall in all direction, F is given by
22 3 xvv
222zyx vvv
3
22
vvx
d
vmNF
2
3
2xvN
d
mF
3
2vN
d
mF
where molecule theof velocity squaremean :2 v
Physics DF025
Chapter 14
• From the definition of pressure,
14.2.3 Pressure of an ideal gas
A
FP 2dA where
d
vm
3
NF
2
and
3
2
3
1
d
vNmP
(14.1)
2v
V
Nm
3
1P
Vd 3and
2
3
1vNmPV
where container in the gas idealan of mass:Nm
(14.2)
Physics DF025
Chapter 14• Since the density of the
gas, is given by
hence the equation (15.1) can be written as
V
Nm
wheregas by the pressure :Pgas theofdensity :
molecules gas theof
velocity squaremean :2 v
2
3
1vP (14.3)
• is defined as
• From the equation of state in terms
of Boltzmann constant, k :
• By equating the eqs. (15.4) and (14.2), thus
14.2.4 Root mean square velocity ( vrms)
NkTPV
2
3
1vNmNkT
(14.4)
m
kTv
32
2rms vv
Physics DF025
Chapter 14
• Since
therefore the equation of root mean square velocity of the gas molecules also can be written as
where
re temperatuabsolute :T
gas molecule a of mass:m
(speed) velocity squaremean root : rmsv
gas a of massmolar :M
2
3
1vP
thus P3
v2
P
v3
rms
• Therefore
ORm
kTv
3rms
M
RTv
3rms
Physics DF025
Chapter 14
Eight gas molecules chosen at random are found to have speeds of 1,1,2,2,2,3,4 and 5 m s1. Determine
a. the mean speed of the
molecules,
b. the mean square speed of
the molecules,
c. the root mean square
speed of the molecules.
Example 3 : Solution :
a. The mean speed of the molecules is given by
8N
N
vv
N
ii
1
8
54322211 v
1s m 5.2 v
Physics DF025
Chapter 14Solution :
b. The mean square speed of the molecules is given by
8N
N
vv
N
ii
1
2
2
8
54322211 222222222 v
222 s m 8 v
c. The root mean square speed of the molecules is
2rms vv
8rms v
1rms s m 83.2 v
Physics DF025
Chapter 14
Solution :
a. By using the equation of
state, thus
A cylinder of volume 0.08 m3 contains oxygen gas at a temperature of 280 K and pressure of 90 kPa. Determine
a. the mass of oxygen in the
cylinder,
b. the number of oxygen
molecules in the cylinder,
c. the root mean square
speed of the oxygen
molecules in the cylinder.
Example 4 :
Pa 1090
K; 280;m 08.03
3
P
TV
nRTPV M
mn
RTM
mPV
and
(Given R = 8.31 J mol1 K1, k = 1.38 1023 J K1, molar mass of oxygen, M = 32 g mol1, NA = 6.02 1023 mol1)
Physics DF025
Chapter 14Solution :a.
b. The number of oxygen molecules in the cylinder is
c. The root mean square speed of the oxygen molecules is
1rms s m 467
032.0
28031.83 v
kg 1090.9 2m
AN
N
M
mn
ANM
mN
Pa 1090 K; 280;m 08.0 33 PTV
28031.8032.0
08.01090 3
m
232
1002.6032.0
1090.9
N
molecules 1086.1 24N
M
RTv
3rms
Physics DF025
Chapter 14Exercise 15.2 :
Given R = 8.31 J mol1 K1, Boltzmann constant, k = 1.381023 K1
1. In a period of 1.00 s, 5.00 1023 nitrogen molecules strike a wall with an area of 8.00 cm2. If the molecules move with a speed of 300 m s1 and strike the wall head-on in the elastic collisions, determine the pressure exerted on the wall.
(The mass of one N2 molecule is 4.68 1026 kg)ANS. : 17.6 kPa2. Initially, the r.m.s. speed of an atom of a monatomic ideal gas is
250 m s1. The pressure and volume of the gas are each doubled while the number of moles of the gas is kept constant.
Calculate the final translational r.m.s. speed of the atoms. ANS. : 500 m s1
3. Given that the r.m.s. of a helium atom at a certain temperature is 1350 m s1, determine the r.m.s. speed of an oxygen (O2) molecule at this temperature.
(The molar mass of O2 is 32.0 g mol1 and the molar mass of He is 4.00 g mol1)
ANS. : 477 m s1
Physics DF025
Chapter 14
a. Explain and use translational kinetic energy of gases,
b. State the principle of equipartition of energy.
c. Define degree of freedom
d. State the number of degree of freedom for monoatomic, diatomic, and polyatomic gas molecules.
Learning Outcome:
kTTN
RK
2
3
2
3
Atr
At the end of this chapter, students should be able to:
14.3 Molecular kinetic energy and internal energy
Physics DF025
Chapter 14
e. Explain internal energy of gas and relate the internal energy to the number of degree of freedom,
Learning Outcome:
At the end of this chapter, students should be able to:
14.3 Molecular kinetic energy and internal energy
f. Explain and use internal energy of an ideal gas
Physics DF025
Chapter 1414.3 Molecular kinetic energy and internal energy14.3.1 Translational kinetic energy of molecule
• From equation (14.1), thus
This equation shows that
(14.5)
2
2
1
3
2vm
V
NP
increases
V
N
increases
2vm
2
1P increases () When
2v
V
Nm
3
1P
Physics DF025
Chapter 1414.3.1 Translational kinetic energy of molecule
increases
V
N
increases
2vm
2
1
P increases () When
2
2
1
3
2vmNPV and NkTPV
This equation shows that
Rearrange equation (14.5), thus
Physics DF025
Chapter 14
2
2
1
3
2vmNNkT
and
kTvm2
3
2
1 2
tr2
2
1Kvm
TN
RkTK
Atr 2
3
2
3
where
molecule
a ofenergy kinetic
onal translatiaverage : trK
re temperatuabsolute :TconstantBoltzmann :k
constant gasmolar : Rconstant Avogadro :AN
Physics DF025
Chapter 14
• For N molecules of an ideal gas in the cubical container, the total average (mean) translational kinetic energy, E is given by
NkTE2
3
trNKE
kT
2
3NE
nRTE2
3
OR
Physics DF025
Chapter 14
• States : “the mean (average) kinetic energy of every degrees
of freedom of a molecule is
Therefore
14.3.2 Principle of equipartition of energy
.2
1kT
kTf
K2
freedom of degrees:fwhere
re temperatuabsolute:T
OR
RTf
K2
Mean (average) kinetic energy per molecule
Mean (average) kinetic energy per mole
Physics DF025
Chapter 14
A vessel contains hydrogen gas of 7.50 1017 molecules per unit volume and the root mean square speed of the molecules is 2.50 km s1at a temperature of 30 C. Determine
a. the average translational kinetic energy of a molecule for
hydrogen gas,
b. the pressure of hydrogen gas.
(Given the molar mass of hydrogen gas = 2 g mol1, NA= 6.02 1023 mol1and k = 1.38 1023 J K1)
Example 5 :
K 15.303
;s m 1050.2;1050.7 13rms
17
T
vV
N
kTK2
3tr
15.3031038.12
3 23tr
K
J 1028.6 21tr
K
Solution :
a. The average translational kinetic energy of a molecule is
Physics DF025
Chapter 14
Solution :
b. The pressure of hydrogen gas is given by
2
3
1vm
V
NP
K 15.303 ;s m 1050.2;1050.7 13rms
17 TvV
N
AN
Mm and
2rms
A3
1v
N
M
V
NP
where2
rms2 vv
2323
17 1050.21002.6
002.01050.7
3
1
P
Pa 1019.5 3P
Physics DF025
Chapter 14
• is defined as a number of independent ways in which an atom or molecule can absorb or release or store the energy.
Monatomic gas (e.g. He,Ne,Ar)
• The number of degrees of freedom is 3 i.e. three direction of translational motion where contribute translational kinetic energy as shown in Figure 14.6.
14.3.3 Degree of freedom ( f )
yv
xv
zv
x
y
z
He
Figure 14.6
Physics DF025
Chapter 14Diatomic gas (e.g. H2, O2, N2)
• The number of degrees of freedom is
Polyatomic gas (e.g. H2O, CO2, NH3)
• The number of degrees of freedom is
Translational kinetic energy 3Rotational kinetic energy 2
5Figure 14.7
xv
zv
H Hyv
x
y
z
Figure 14.8
xv
zv
yv
x
y
z
O
HH
6
Translational kinetic energy 3Rotational kinetic energy 3
Physics DF025
Chapter 14
Molecule Example
Degrees of Freedom ( f )Average kinetic
energy per molecule,<K>Translational Rotational Total
• Table 14.1 shows the degrees of freedom for various molecules.
Monatomic
Diatomic
Polyatomic
He
H2
H2O
3 kT2
3
(At temperature of 300 K)
Table 14.1
0 3
3 kT2
52 5
3 kTkT 32
63 6
Physics DF025
Chapter 14
• Degrees of freedom depend on the absolute temperature of the gases.
– For example : Diatomic gas (H2)
– Hydrogen gas have the vibrational kinetic energy (as shown in Figure 14.9) where contribute 2 degrees of freedom which correspond to the kinetic energy and the potential energy associated with vibrations along the bond between the atoms.
H H
vibration
Figure 14.9
3f5f7f
when the temperature,
At 250 KAt 250 – 750 KAt >750 K
Physics DF025
Chapter 14Solution :
By applying the equation of the total translational kinetic energy, thus
A vessel contains an ideal diatomic gas at temperature of 20 C. The total translational kinetic energy of the gas molecules is 3.00 106 J. The mass of the gas is then doubled and the total translational kinetic energy of the molecules becomes 9.00 106 J. Determine the new temperature of the gas.
Example 6 :
; J 1000.9
J; 1000.3;K 15.2936
2
611
E
ET
12 2mm
NkTE2
3
ANM
mN
andwhere
AN
Rk
TN
RN
M
mE
AA2
3
RTM
mE
2
3
Physics DF025
Chapter 14
Solution :
For temperature T1 :
For temperature T2 :
(2)(1) :
11
1 2
3RT
M
mE
J 1000.9J; 1000.3;K 15.293 62
611
EET
12 2mm (1)
22
2 2
3RT
M
mE
21
2
2
2
3RT
M
mE
(2)
1
2
1
2 2
T
T
E
E
15.293
2
1000.3
1000.9 26
6 T
K 4402 T OR C 167
Physics DF025
Chapter 14
• Explain internal energy of gas and relate the internal energy to the number of degree of freedom.
• Explain and use internal energy of an ideal gas
Learning Outcome:
RCC VP
NkTf
U2
V
P
C
C
and
At the end of this chapter, students should be able to:
• Define molar specific heat at constant pressure and volume.
• Use equations,
Physics DF025
Chapter 14
14.3.4 Internal energy of gas and relate the internal energy to the number of degree of freedom.
• is defined as the sum of total kinetic energy and total potential energy of the gas molecules.
• But in ideal gas, the intermolecular forces are assumed to be negligible thus the potential energy of the molecules can be neglected.
KNU
nRT2
fU
NkT2
fU
AN
Rk and
gas theof
energy internal:Uwhere
OR
Thus for N molecules,
Physics DF025
Chapter 14
Monatomic Diatomic Polyatomic
Table 14.2 shows the properties for 1 mole of an ideal gas.
Degrees of freedom, f
Average kinetic energy per
molecule, <K>
Internal energy, U
3 5 6
kT2
3
Table 14.2
kT2
5kTkT 3
2
6
RT2
3 RT2
5RTRT 3
2
6
Physics DF025
Chapter 14
Exercise 14.3 :
Given R = 8.31 J mol1 K1, Boltzmann constant, k = 1.381023 K1
1. One mole of oxygen has a mass of 32 g. Assuming oxygen behaves as an ideal gas, calculate
a. the volume occupied by one mole of oxygen gas
b. the density of oxygen gas
c. the r.m.s. speed of its molecules
d. the average translational kinetic energy of a molecule
at 273 K and pressure of 1.01105 Pa.
ANS. : 2.25102 m3; 1.42 kg m3; 461 m s1; 5.651021 J
Physics DF025
Chapter 14
THE END…Next Chapter…CHAPTER 15 :Thermodynamics