Phys2 ch4-kineticsgas

•• PROGRAM OFPROGRAM OF
“PHYSICS”“PHYSICS”Lecturer: Dr. DO Xuan Hoi
Room 413
E-mail : dxhoi@hcmiu.edu.vn

PHYSICS 2PHYSICS 2
(FLUID MECHANICS AND THERMAL(FLUID MECHANICS AND THERMAL
PHYSICS)PHYSICS)
02 credits (30 periods)
Chapter 1 Fluid Mechanics
Chapter 2 Heat, Temperature and the Zeroth
Law of Thermodynamics
Chapter 3 Heat, Work and the First Law of
Thermodynamics
Chapter 4 The Kinetic Theory of Gases
Chapter 5 Entropy and the Second Law of
Thermodynamics

References :References :
Halliday D., Resnick R. and Walker, J. (2005),Halliday D., Resnick R. and Walker, J. (2005),
Fundamentals of Physics, Extended seventh edition.Fundamentals of Physics, Extended seventh edition.
John Willey and Sons, Inc.John Willey and Sons, Inc.
Alonso M. and Finn E.J. (1992). Physics, Addison-WesleyAlonso M. and Finn E.J. (1992). Physics, Addison-Wesley
Publishing CompanyPublishing Company
Hecht, E. (2000). Physics. Calculus, Second Edition.Hecht, E. (2000). Physics. Calculus, Second Edition.
Brooks/Cole.Brooks/Cole.
Faughn/Serway (2006), Serway’s College Physics,Faughn/Serway (2006), Serway’s College Physics,
Brooks/Cole.Brooks/Cole.
Roger Muncaster (1994), A-Level Physics, StanleyRoger Muncaster (1994), A-Level Physics, Stanley
Thornes.Thornes.

http://ocw.mit.edu/OcwWeb/Physics/index.htm
http://www.opensourcephysics.org/index.html
http://hyperphysics.phy-astr.gsu.edu/hbase/HFram
http://www.practicalphysics.org/go/Default.html
http://www.msm.cam.ac.uk/
http://www.iop.org/index.html
.
.
.

CHAPTER 4
The Kinetic Theory of Gases
• Ideal Gases, Experimental Laws and the
Equation of State
• Molecular Model of an Ideal Gas The
Equipartition
of Energy
The Boltzmann Distribution Law
The Distribution of Molecular Speeds
Mean Free Path
• The Molar Specific Heats of an Ideal Gas
• Adiabatic Expansion of an Ideal Gas

1. Ideal Gases, Experimental Laws and the
Equation of State
1.1 Notions
► Properties of gases
 A gas does not have a fixed volume or pressure
 In a container, the gas expands to fill the container
► Ideal gas
 Collection of atoms or molecules that move randomly
 Molecules exert no long-range force on one another
 Molecules occupy a negligible fraction of the
volume of their container
► Most gases at room temperature
and pressure behave approximately
as an ideal gas

1.2
Moles
► It’s convenient to express the amount of gas in
a given volume in terms of the number of
moles, n
mass
n
molar mass
=
► One mole is the amount of the substance that
contains as many particles as there are atoms
in 12 g of carbon-12

1.3 Avogadro’s Hypothesis
“Equal volumes of gas at the same temperature and
pressure contain the same numbers of molecules”
 Corollary: At standard temperature and pressure, one
mole quantities of all gases contain the same number of
molecules
 This number is called NA
 Can also look at the total number of particles: AN nN=
The number of particles in a mole is called
Avogadro’s Number
NA=6.02 x 1023
particles / mole
The mass of an individual atom :
atom
A
molar mass
m
N
=

The Hope diamond (44.5 carats) is almost pure
carbon and the Rosser Reeves (138 carats) is primarily
aluminum oxide (Al2O3). One carat is equivalent to a mass of
0.200 g. Determine (a) the number of carbon atoms in the
Hope diamond and (b) the number of Al2O3 molecules in the
ruby Rosser Reeves.
SOLUTION
(44.8 )(0.200 / ) 8.90Hopem carats g carat g= =
8.90
0.741
12.011 /
Hope
Hope
m g
n mol
mass per mole g mol
= = =
The mass of the Hope diamond :(a)
The number of moles in the Hope diamond :
(0.741 )H AN mol N=
The number of carbon atoms in the Hope diamond :
PROBLEM 1
23 23
(0.741 )(6.022 10 / ) 4.46 10mol atoms mol atoms= × = ×

ruby Rosser Reeves.
SOLUTION
(138 )(0.200 / ) 27.6Rm carats g carat g= =
27.6
0.271
101.9612 /
R
R
m g
n mol
mass per mole g mol
= = =
The mass of the Rosser Reeves :(b)
Molecular mass :
The number of moles in the Rosser Reeves :
PROBLEM 1
2(26.9815 ) 3(15.9994 ) 101.9612Rm u u u= + = 101.9612 /g mol=

(b)
The number of Al2O3 molecules in the Rosser Reeves :
(0.271 )R AN mol N=
23 23
(0.271 )(6.022 10 / ) 1.63 10mol atoms mol molecules= × = ×
ruby Rosser Reeves.
SOLUTION
PROBLEM 1

1.4 Experimental Laws
• Boyle’s Law
Experiment :
Conclusion :
When the gas is kept at a constant
temperature, its pressure is
inversely proportional to its volume
(Boyle’s law)
PV const=

• Charles’ Law
Experiment :
Conclusion :
At a constant pressure, the
temperature is directly proportional to
the volume
(Charles’ law)
V CT=
( C : constant )

• Gay-Lussac’s Law
Experiment :
Conclusion :
At a constant volume, the temperature
is directly proportional to the pressure
(Gay-Lussac’ law)
T CP=
( C : constant )

1.5 Equation of State for an Ideal Gas
Gay-Lussac’ law : T CP=V = constant →
T = const →Boyle’s law : PV const=
Charles’ law : V CT=P = const →
• The number of moles n of a substance of mass m
(g) :
m
n
M
= (M : molar mass-g/mol)
→ Equation of state for an ideal
gas :
PV = nRT (Ideal gas law)
T : absolute temperature in kelvins
R : a universal constant that is the same for all
gases
R =8.315 J/mol.K

→ Definition of an Ideal Gas :
“An ideal gas is one for which PV/nT is constant
at
all pressures”
AN nN=• Total number of molecules :
PV
R
nT
=
With Boltzmann’s constant :
A
N
PV = RT
N A
R
= nT
N
23
23 1
8.315 / .
1.38 10 /
6.22 10
J mol K
k J K
mol
−
−
= ×
×B
A
R
= =
N
BPV = Nk T→ Ideal gas law :

PV = CT
Ideal gas law for an quantity of
gas:

TestTest
An ideal gas is confined to a container with constant
volume. The number of moles is constant. By what
factor will the pressure change if the absolute
temperature triples?
a. 1/9
b. 1/3
c. 3.0
d. 9.0

An ideal gas occupies a volume of
100cm3
at 20°C and 100 Pa.
(a) Find the number of moles of gas in the container
SOLUTION
PV
n
RT
=
The number of moles of gas :(a)
PROBLEM 2
4 3
6(100 )(10 )
4.10 10
(8.315 / )(293 )
Pa m
mol
J mol K
−
−
= = ×
(b) How many molecules are in the container?
The number molecules in the
container : 6
(4.10 10 ) AN mol N−
= ×
6 23
18
(4.10 10 )(6.022 10 / )
2.47 10
mol atoms mol
molecules
−
= × ×
= ×
(b)

A certain scuba tank is designed to hold 66
ft3
of air when it is at atmospheric pressure at 22°C. When this
volume of air is compressed to an absolute pressure of
3 000 lb/in.2
and stored in a 10-L (0.35-ft3
) tank, the air
becomes so hot that the tank must be allowed to cool before it
can be used.
(a) If the air does not cool, what is its temperature? (Assume
that the air behaves like an ideal gas.)
PROBLEM 3
SCUBA (Self-Contained Underwater
Breathing Apparatus))
The number of moles n remains constant :
1 1 2 2
1 2
;
PV PV
n R
T T
= =
2 3
2 3
(3000 / . )(0.35 )
(295 ) 319
(14.7 / . )(66 )
lb in ft
K K
lb in ft
= =
2 2
2 1
1 1
=
PV
T T
PV
(a)

A certain scuba tank is designed to hold 66
ft3
of air when it is at atmospheric pressure at 22°C. When this
volume of air is compressed to an absolute pressure of
3 000 lb/in.2
and stored in a 10-L (0.35-ft3
) tank, the air
becomes so hot that the tank must be allowed to cool before it
can be used.
(b) What is the air temperature in degrees Celsius and in
degrees Fahrenheit?
PROBLEM 3
45.9°C;
115°F.
(b)

A sculpa consists of a 0.0150 m3
tank filled
with compressed air at a pressure of 2.02×107
Pa. Assume that
air is consumed at a rate of 0.0300 m3
per minute and that the
temperature is the same at all depths, determine how long the
diver can stay under seawater at a depth of
(a) 10.0 m and (b) 30.0 m
The density of seawater is ρ = 1025 kg/m3
.
PROBLEM 4
SOLUTION
2 1P P ghρ= +
5 3 2
1.01 10 (1025 / )(9.80 / )(10.0 )Pa kg m m s m= × +
1 1
2
2
PV
V
P
=
(a)
5
2.01 10 Pa= ×
3
1.51 m=
5 3
5
(2.02 10 )(0.0150 )
(1.01 10 )
Pa m
Pa
×
=
×
The volume available for breathing :
3 3 3
1.51 0.0150 1.50m m m− =

A sculpa consists of a 0.0150 m3
tank filled
with compressed air at a pressure of 2.02×107
Pa. Assume that
air is consumed at a rate of 0.0300 m3
per minute and that the
temperature is the same at all depths, determine how long the
diver can stay under seawater at a depth of
(a) 10.0 m and (b) 30.0 m
The density of seawater is ρ = 1025 kg/m3
.
PROBLEM 4
SOLUTION
(a)
3
3
1.50
50.0 min
0.0300 / min
m
t
m
= =
The compressed air will last for :
(b) 24.6 mint =
The deeper dive must have a shorter duration

A spray can containing a propellant gas at
twice atmospheric pressure (202 kPa) and having a volume of
125 cm3
is at 22°C. It is then tossed into an open fire. When
the temperature of the gas in the can reaches 195°C, what is
the pressure inside the can? Assume any change in the
volume of the can is negligible.
PROBLEM 5
The number of moles n remains constant :
1 1 2 2
1 2
PV PV
n R
T T
= =
(468 )
(202 ) 320
(295 )
K
kPa kPa
K
= =2
2 1
1
T
P P
T
=
SOLUTION
Because the initial and final volumes
of the gas are assumed to be equal :
1 2
1 2
;
P P
T T
=

An ideal gas at 20.0O
C at a pressure of 1.50
×105
Pa when has a number of moles of 6.16×10-2
mol.
SOLUTION
nRT
V
P
=
The volume :(a)
PROBLEM 6
2
5
(6.16 10 )(8.315 / )(293 )
(1.50 10 )
mol J mol K
Pa
−
×
=
×
(a) Find the volume of the gas.
2
5
(6.16 10 )(8.315 / )(293 )
(1.50 10 )
mol J mol K
Pa
−
×
=
×
3 3
1.00 10 1.00m L−
= × =

An ideal gas at 20.0O
C at a pressure of 1.50
×105
Pa when has a number of moles of 6.16×10-2
mol.
SOLUTION
nRT
V
P
=
The volume :(a)
PROBLEM 6
2
5
(6.16 10 )(8.315 / )(293 )
(1.50 10 )
mol J mol K
Pa
−
×
=
×
(b) The gas expands to twice its original volume, while the
pressure falls to atmospheric pressure. Find the final
temperature.
(b)
3 3
1.00 10 1.00m L−
= × =
1 1 2 2
1 2
;
PV PV
n R
T T
= =
5
5
(1.01 10 )(2.00 )
(293 )
(1.50 10 )(1.00 )
Pa L
K
Pa L
×
=
×
2 2
2 1
1 1
PV
T T
PV
= 395 K=

A beachcomber finds a corked bottle
containing a message. The air in the bottle is at the
atmospheric pressure and a temperature of 30.0O
C. The cork
has the cross-sectional area of 2.30 cm3
. The beachcomber
places the bottle over a fire, figuring the increased pressure
will pushout the cork. At a temperature of 99o
C the cork is
ejected from the bottle
PROBLEM
7
1 1 2 1
1 2
;
PV PV
n R
T T
= =
5 5(372 )
(1.01 10 ) 1.24 10
(303 )
K
Pa Pa
K
= × = ×2
2 1
1
T
P P
T
=
(a)
What was the the pressure in the bottle just before the
cork left it ?
(a)
SOLUTION
Message in a bottle found 24 years later - Yahoo!7

A beachcomber finds a corked bottle
containing a message. The air in the bottle is at the
atmospheric pressure and a temperature of 30.0O
C. The cork
has the cross-sectional area of 2.30 cm3
. The beachcomber
places the bottle over a fire, figuring the increased pressure
will pushout the cork. At a temperature of 99o
C the cork is
ejected from the bottle
PROBLEM
7
0 ;F =∑
5 5 4 2
(1.24 10 1.01 10 )(2.30 10 )Pa Pa m−
= × − × ×
1 0in out fricP A P A F− − =(b)
What force of friction held the cork in place?(b)
SOLUTION
5.29 N=
( )fric in outF P P A= −

A room of volume 60.0 m3
contains air
having an equivalent molar mass of 29.0 g/mol. If the
temperature of the room is raised from 17.0°C to 37.0°C, what
mass of air (in kilograms) will leave the room? Assume that the
air pressure in the room is maintained at 101 kPa.
PROBLEM
8
m
PV n RT RT
µ
= =
3 5 3
(29.0 10 / )(1.01 10 ) 60.0 1 1
(8.31 / . ) 290 310
kg mol Pa m
J mol K K K
−
 × × ×
= − ÷
 
1 2
1 2
1 1PV
m m
R T T
µ  
− = − ÷
 
SOLUTION
4.70 kg=

2 Molecular Model of an Ideal Gas
2.1 Assumptions of the molecular model of an ideal
gas
• A container with volume V contains a very large number N of
identical molecules, each with mass m.
• The molecules behave as point particles; their size is small in
comparison to the average distance between particles and to the
dimensions of the container.
• The molecules are in constant
motion; they obey Newton's laws
of motion. Each molecule collides
occasionally with a wall of the
container. These collisions are
perfectly elastic.
• The container walls are perfectly
rigid and infinitely massive and do
not move.
A particle
having a
brownian
motion inside
a polymer like
network
Brownian
motion

2.2 Collisions and Gas Pressure
• Consider a cubical box with sides of length
d containing an ideal gas. The molecule
shown moves with velocity v.
• Consider the collision of one molecule
moving with a velocity v toward the right-hand
face of the box
• Elastic collision with the wall → Its x component
of momentum is reversed, while its y component
remains unchanged :
• The average force exerted on the molecule :
• The average force exerted by the molecule on the wall :
∆ = − − = −( ) 2x x x xp mv mv mv
− − −
= = =
∆
2
1
2 2
2 /
x x x
x
mv mv mv
F
t d v d
−
− = − =
2 2
1
x xmv mv
F
d d

• The total force F exerted by all the
molecules on the wall :
• The average value of the square of the
velocity in the x direction for N molecules :
• The total pressure exerted on the wall:
( )= + +2 2
1 2 ...x x
m
F v v
d
+ + +
=
2 2 2
2 1 2 ...x x xN
x
v v v
v
N
= 2
x
Nm
F v
d
= + +2 2 2 2
;x y zv v v v = + +2 2 2 2
;x y zv v v v =2 2
3 ;xv v
 
=  ÷
 ÷
 
2
3
N mv
F
d
   
= = = = ÷  ÷
   
2 2
2 3
1 1
;
3 3
F F N N
P mv mv
A Vd d
  
=  ÷ ÷
  
22 1
3 2
N
P mv
V

• The equation of state for an ideal gas :
Temperature is a direct measure of average
molecular kinetic energy
The average translational kinetic energy per molecule is
Each degree of freedom contributes to the energy
of a system:
(the theorem of equipartition of energy)
 
=  ÷
 
22 1
3 2
T mv
k
  
=  ÷ ÷
  
22 1
3 2
N
P mv
V
=PV NkT
=21 3
2 2
mv kT
3
2
kT
=2 21
3xv v =21 1
;
2 2xmv kT =21 1
;
2 2ymv kT =21 1
2 2zmv kT
1
2
kT

• The total translational kinetic energy of N molecules of gas
: The number of moles of gas
: Boltzmann’s constant
=21 3
2 2
mv kT
 
= = = ÷
 
21 3 3
2 2 2transE N mv NkT nRT
=
A
N
n
N
=
A
R
k
N
• Assume: Ideal gas is a monatomic gas (which has
individual atoms rather than molecules: helium, neon, or
argon) and the internal energy Eint of ideal gas is simply the
sum of the translational kinetic energies of its atoms
= = =int
3 3
2 2transE E NkT nRT

• The root-mean-square (rms) speed of the molecules :
2 21 1 3
;
2 2 2rms Bmv mv k T= = = =
3 3
rms
kT RT
v
m M
M is the molar mass in kilograms per mole : M = mNA
= 2
rmsv v

Five gas molecules chosen at random are
found to have speeds of 500, 600,700, 800, and 900 m/s.
Find the rms speed. Is it the same as the average speed?
SOLUTION
PROBLEM 9
In general, vrms and vav are not the same.

A tank used for filling helium balloons has a
volume of 0.300 m3
and contains 2.00 mol of helium gas at
20.0°C. Assuming that the helium behaves like an ideal gas,
(a) what is the total translational kinetic energy of the
molecules of the gas?
SOLUTIO
N
PROBLEM 10
(a)

A tank used for filling helium balloons has a
volume of 0.300 m3
and contains 2.00 mol of helium gas at
20.0°C. Assuming that the helium behaves like an ideal gas,
(b) What is the average kinetic energy per molecule?
(c) Using the fact that the molar mass of helium is
4.00×103
kg/mol, determine the rms speed of the atoms
at 20.0°C.
SOLUTIO
N
PROBLEM
10
(b)
(c)

(a) What is the average translational kinetic
energy of a molecule of an ideal gas at a temperature of
27°C ?
(b) What is the total random translational kinetic energy of
the molecules in 1 mole of this gas?
(c) What is the root-mean-square speed of oxygen molecules
at this temperature ?
SOLUTIO
N
PROBLEM 11
(a)
(b)

(a) What is the average translational kinetic
energy of a molecule of an ideal gas at a temperature of
27°C ?
(b) What is the total random translational kinetic energy of
the molecules in 1 mole of this gas?
(c) What is the root-mean-square speed of oxygen molecules
at this temperature ?
SOLUTIO
N
PROBLEM 11
(c)

(a) A deuteron, 2
1H, is the nucleus of a
hydrogen isotope and consists of one proton and one
neutron. The plasma of deuterons in a nuclear fusion reactor
must be heated to about 300 million K. What is the rms
speed of the deuterons? Is this a significant fraction of the
speed of light (c = 3.0 x 108
m/s) ?
(b) What would the temperature of the plasma be if the
deuterons had an rms speed equal to 0.10c ?
SOLUTIO
N
PROBLEM 12

2.3 The Boltzmann Distribution Law
The Maxwell–Boltzmann distribution
function
• Consider the distribution of molecules in our atmosphere :
Determine how the number of molecules per unit volume
varies with altitude
V Vmgn V mgn Ady= =
VdP mgn dy= −
Consider an atmospheric layer of
thickness dy and cross-sectional
area A, having N particles. The air is
in static equilibrium :
( )PA P dP A mgN− + =
where nV is the number density.
;BPV Nk T=
Law of Exponential Atmospheres
• From the equation of state : ;V BP n k T= B VdP k Tdn=
;V B Vmgn dy k Tdn− = ;V
V B
dn mg
dy
n k T
= −
0 0
;
Vn y
V
V Bn
dn mg
dy
n k T
= −∫ ∫

0 0
;
Vn y
V
V Bn
dn mg
dy
n k T
= −∫ ∫ 0 0
ln Vn y
V n
B
mg
n y
k T
= −      
0ln ln ;V
B
mg
n n y
k T
− = −
0
ln ;V
B
n mg
y
n k T
= −
0
expV
B
n mg
y
n k T
 
= − ÷
 
0
ln ;V
B
n mg
y
n k T
= −
0
expV
B
n mg
y
n k T
 
= − ÷
 
/
0
Bmgy k T
Vn n e−
=
/
0
BU k T
Vn n e−
=
The Boltzmann distribution law : the probability of
finding the molecules in a particular energy state varies
exponentially as the negative of the energy divided by kBT.

What is the number density of air at an
altitude of 11.0 km (the cruising altitude of a commercial
jetliner) compared with its number density at sea level?
Assume that the air temperature at this height is the same as
that at the ground, 20°C.
SOLUTIO
N
PROBLEM
13
/
0
Bmgy k T
Vn n e−
=The Boltzmann distribution law :
Assume an average molecular mass of :
−
= × 26
28.9 4.80 10u kg

Density of the number of molecules with speeds between v
and dv :
The Maxwell–Boltzmann distribution function
θ θ ϕ− −
µ =
2 2
/ 2 / 2 2
( ) sinmv kT mv kT
VN v dV e dV e v dv d d
r
π π
θ θ ϕ−
µ ∫ ∫
2
2
/ 2 2
0 0
( ) sinmv kT
VN v dv e v dv d d
π −
µ
2
/ 2 2
( ) 4 mv kT
VN v dv e v dv
π −
=
2
/ 2 2
( ) 4 mv kT
VN v dv A e v dv
With : =∫ ( )VN v dv N
π −
=∫
2
/ 2 2
4 mv kT
A e v dv N

Poisson's Integral Formula:
π
+∞
−
=∫
2
0
1
2
ax
e dx
a
and dv is
π
π
− 
=  ÷
 
2
3 / 2
2 / 2
4
2
mv kT
V
m
N N v e
kT

and dv is
The rms speed :
The average speed:
The most probable speed:
π
π
− 
=  ÷
 
2
3 / 2
2 / 2
4
2
mv kT
V
m
N N v e
kT
= = =2
3 / 1.73 /rmsv v kT m kT m
π= =8 / 1.60 /v kT m kT m
= =2 / 1.41 /mpv kT m kT m
> >rms mpv v v

Definition: The average value of v n
:PROOF:
∞
= ∫
0
1n n
vv v N dv
N
π
π
∞
−
  
 ÷=  ÷ ÷  
∫
2
3/ 2
2 / 2
0
1
4
2
mv kTm
v v N v e dv
N kT
π
π
∞
−
  
 ÷= = ÷ ÷  
∫
2
3/ 2
2 2 2 / 2
0
1
4 3 /
2
mv kTm
v v N v e dv kT m
N kT
The average speed:
π= =8 / 1.60 /v kT m kT m
The mean square speed:
→ = = =2
π
π
−
  
= = ÷ ÷ ÷  
2
3/2
2 /2
0 ; 4 0 ;
2
mv kTvdN d m
N v e
dv dv kT
= 2 /mpv kT m

For diatomic carbon dioxide gas ( CO2 , molar
mass 44.0 g/mol) at T = 300 K, calculate
(a) the most probable speed vmp;
(b) the average speed vav;
(c) the root-mean-square speed vrms.
SOLUTIO
N
PROBLEM 14
The rms speed :
The average speed:
= = =2
π= =8 / 1.60 /v kT m kT m
= =2 / 1.41 /mpv kT m kT m

At what temperature is the root-mean-square
speed of nitrogen molecules equal to the root-mean-square
speed of hydrogen molecules at 20.00
C?
SOLUTIO
N
PROBLEM 15
The rms speed : = =2
3 /rmsv v kT m
A N2 molecule has more mass so N2 gas must be at a
higher temperature to have the same v rms .

2.4 The mean free path
• A molecule moving through a gas
collides with other molecules in a random
fashion.
Notion of the mean free path
• Between collisions, the molecules move with constant
speed along straight lines. The average distance between
collisions is called the mean free path.

The mean free path for a gas molecule
• Consider N spherical molecules with radius r in a volume V.
Suppose only one molecule is moving.
• When it collides with another molecule,
the distance between centers is 2r.
• In a short time dt a molecule with speed v
travels a distance vdt ; during this time it
collides with any molecule that is in the
cylindrical volume of radius 2r and length vdt.
• The volume of the cylinder : π 2
4 r vdt
The number of the molecules with centers in this cylinder :
The number of collisions per unit time :
π= 2
(4 )
N
dN r vdt
V
π= 2
(4 )
dN N
r v
dt V
When all the molecules move at once : π= 2
2(4 )
dN N
r v
dt V

• The average time between collisions (the mean free time)
• The mean free path (the average distance
traveled between collisions) is
• For the ideal-gas :
→
=PV NkT
ππ
= = 2
2
1
2(4 )2(4 )
mean
V
t
N r v Nr v
V
λ
π
= = 2
4 2
mean
V
vt
r N
λ
π
= = 2
4 2
mean
kT
vt
r P

Approximate the air around you as a
collection of nitrogen molecules, each of which has a diameter
of 2.00 × 10-10
m.
How far does a typical molecule move before it collides with
another molecule?
SOLUTIO
N
PROBLEM 16
Assume that the gas is ideal:
The mean free path:

A cubical cage 1.25 m on each side contains
2500 angry bees, each flying randomly at 1.10 m/s. We
can model these insects as spheres 1.50 cm in diameter. On
the average, (a) how far does a typical bee travel between
collisions,
(b) what is the average time between collisions,
and (c) how many collisions per second does a bee make?
SOLUTIO
N
PROBLEM 17

3. The Molar Specific Heats of an ldeal Gas
• Constant volume: = ∆VQ nC T
CV : the molar specific heat at constant volume
= ∆PQ nC T
• Constant pressure:
CP : the molar specific heat at constant pressure
First law of thermodynamics:
∆ = − = ∆ − = ∆int
3
0
2VE Q W nC T nR T
=
3
2VC R =int VE nC T→

C : molar specific heat of Various Gases
=
3
2VC R
Gas constant: R = 8.315 J/mol.K
=
5
2VC R
≈
7
2VC R

C : molar specific heat of Various
Gases
=
3
2VC R
=
5
2VC R
=
7
2VC R
• monatomic molecules:
• diatomic molecules:
(not vibration)
• polyatomic molecules:
f : degree of freedom (the number of independent
coordinates to specify the motion of a molecule)
=
2V
f
C R

V = const → dW = 0
PdQ nC dT=
VdQ nC dT=
• If the heat capacity is measured under constant- volume
conditions: the molar heat capacity CV at constant volume
First law → dU = dQ = nCVdT
• By definition :
dW PdV nRdT= =
(Ideal gas : PV = nRT)
First law : dQ = dU + dW PnC dT dU nRdT= +
VnC dT nRdT= +
P VC C R= +
Relating Cp and Cv for an Ideal Gas

The total work done by the gas as its volume changes from
V1 to Vf : f
i
V
V
W PdV= ∫
Ideal gas : PV nRT=
f
i
V
V
nRT
W dV
V
= ∫
Isothermal process: T const=
;
f
i
V
V
dV
W nRT
V
= ∫ ln f
i
V
W nRT
V
=
Work done by an ideal gas at constant temperature

Also : i i f fPV PV=
ln f
i
V
W nRT
V
=
ln i
f
P
W nRT
P
=
:f iV V> 0W >
When a system expands : work is positive.
When a system is compressed, its volume decreases and
it does negative work on its surroundings

• Work done by an ideal gas at constant
volume
= =∫ 0
f
i
V
V
W PdV
• Work done by an ideal gas at constant
pressure
= = = − = ∆∫ 0 ( )
f
i
V
f i
V
W PdV P V V P V

PROBLEM 18 A bubble of 5.00 mol of helium is submerged at
a certain depth in liquid water when the water (and thus the
helium) undergoes a temperature increase of 20.00
C at
constant pressure. As a result, the bubble expands. The helium
is monatomic and ideal.
a) How much energy is added to the helium as heat during the
increase and expansion?
SOLUTIO
N

C at
(b) What is the change in the internal energy of the helium
during the temperature increase?
SOLUTIO
N

C at
(b) What is the change in the internal energy of the helium
during the temperature increase?
(c) How much work is done by the helium as it expands
against the pressure of the surrounding water during the
temperature increase?
SOLUTIO
N

For adiabatic process : no energy is transferred by heat
between the gas and its surroundings: dQ = 0
dU = dQ – dW = -dW
= P
V
C
C
γ• Definition of the Ratio of Heat Capacities :
The Ratio of Heat Capacities
4 Adiabatic Expansion of an Ideal Gas

= − = − ;dU dQ dW dW
• For ideal gas : = + =;PV nRT PdV VdP nRdT
From : R = CP - CV :
=PV constγ
= −VnC dT PdV
+ = −
V
R
PdV VdP PdV
C
−
+ = − P V
V
C C
PdV VdP PdV
C
Divide by PV :
−
+ = − P V
V
C CdV dP dV
V P C V
= −(1 )
dV
V
γ
+ = 0 ;
dP dV
P V
γ + =ln lnP V constγ
=i i f fPV PVγ γ

• For ideal gas : =PV nRT
=PV constγ
=i i f fPV PVγ γ
−
= =1nRT
V nRTV const
V
γ γ
− −
=1 1
i i f fTV T Vγ γ−
=1
TV constγ

PROBLEM 19 One mole of oxygen (assume it to be an ideal
gas) expands at a constant temperature of 310 K from an initial
volume 12 L to a final volume of 19 L.
a/ How much work is done by the gas during the expansion?
SOLUTIO
N

b/ What would be the final temperature if the gas had
expanded adiabatically to this same final volume? Oxygen
(O2 is diatomic and here has rotation but not oscillation.)
SOLUTIO
N

b/ What would be the final temperature if the gas had
expanded adiabatically to this same final volume? Oxygen
(O2 is diatomic and here has rotation but not oscillation.)
c/ What would be the final temperature and pressure if,
instead, the gas had expanded freely to the new volume,
from an initial pressure of.2.0 Pa?
SOLUTIO
N
The temperature does not change in a free expansion:

PROBLEM 20 Air at 20.0°C in the cylinder of a diesel engine is
compressed from an initial pressure of 1.00 atm and volume
of 800.0 cm3
to a volume of 60.0 cm3
. Assume that air
behaves as an ideal gas with γ = 1.40 and that the compression
is adiabatic. Find the final pressure and temperature of the air.
SOLUTIO
N

PROBLEM 21 A typical dorm room or bedroom contains about
2500 moles of air. Find the change in the internal energy of this
much air when it is cooled from 23.9°C to 11.6°C at a constant
pressure of 1.00 atm.
Treat the air as an ideal gas with γ = 1.400.
SOLUTIO
N

PROBLEM 22 The compression ratio of a diesel engine is 15 to
1; this means that air in the cylinders is compressed to 1/15 of
its initial volume (Fig). If the initial pressure is 1.01 × 105
Pa and
the initial temperature is 27°C (300 K), (a) find the final
pressure and the temperature after compression. Air is mostly a
mixture of diatomic oxygen and nitrogen; treat it as an ideal gas
with γ= 1.40.
SOLUTIO
N(a)

The compression ratio of a diesel engine is 15 to
1; this means that air in the cylinders is compressed to 1/15 of
its initial volume (Fig). If the initial pressure is 1.01 × 105
Pa and
the initial temperature is 27°C (300 K),(b) how much work
does the gas do during the compression if the initial volume of
the cylinder is 1.00 L? Assume that CV for air is 20.8 J/mol.K
and γ = 1.40.
SOLUTIO
N(b)
PROBLEM 22

Two moles of carbon monoxide (CO) start at a
pressure of 1.2 atm and a volume of 30 liters. The gas is then
compressed adiabatically to 1/3 this volume. Assume that the
gas may be treated as ideal. What is the change in the internal
energy of the gas? Does the internal energy increase or
decrease? Does the temperature of the gas increase or
decrease during this process? Explain.
SOLUTIO
N
PROBLEM 23

Two moles of carbon monoxide (CO) start at a
pressure of 1.2 atm and a volume of 30 liters. The gas is then
compressed adiabatically to 1/3 this volume. Assume that the
gas may be treated as ideal. What is the change in the internal
energy of the gas? Does the internal energy increase or
decrease? Does the temperature of the gas increase or
decrease during this process? Explain.
SOLUTIO
N
The internal energy increases because work is done on the gas
(ΔU > 0) and Q = 0.
The temperature increases because the internal energy has
increased.
PROBLEM 23

On a warm summer day, a large mass of air
(atmospheric pressure 1.01 × 105
Pa) is heated by the ground
to a temperature of 26.0°C and then begins to rise through the
cooler surrounding air. (This can be treated as an adiabatic
process). Calculate the temperature of the air mass when it has
risen to a level at which atmospheric pressure is only 0.850 ×
105
Pa. Assume that air is an ideal gas, with γ = 1.40.
SOLUTIO
N
PROBLEM 24

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