Physics 2: Fluid Mechanics and Thermodynamics
Đào Ngọc Hạnh Tâm
Office: A1.503, email: [email protected]
HCMIU, Vietnam National University
Acknowledgment:
Most of these slides are supported by Prof. Phan Bao Ngoc
Chapter 1 Fluid Mechanics
Chapter 2 Heat, Temperature, and the First Law of Thermodynamics
Chapter 3 The Kinetic Theory of Gases
Midterm exam after Lecture 6
Chapter 4 Entropy and the Second Law of Thermodynamics
Final exam
(Chapters 14, 18, 19, 20 of Principles of Physics, Halliday et al.)
Contents of Physics 2
Assignment 1
Assignment 2
3.1. Ideal Gases
3.1.1. Experimental Laws and the Equation of State
3.1.2. Molecular Model of an Ideal Gas
3.3. Mean Free Path
3.4. The Boltzmann Distribution Law and The Distribution
of Molecular Speeds
3.5. The Molar Specific Heats of an Ideal Gas
3.6. The Equipartition of Energy Theorem
3.7. The Adiabatic Expansion of an Ideal Gas
Chapter 3: The Kinetic Theory of Gases
In this chapter, we consider the physics of gases at the microscopic level: A gas consists of atoms that fill their container’s volume and the volume is a result of the freedom of the atoms spread throughout the container.
The temperature is a measure of the kinetic energy of the atoms. The pressure exerted by a gas is produced by the collisions of the atoms with the container’s wall.
The kinetic theory of gases relates the motion of the atoms to the volume, pressure, and temperature of the gas.
Overview
• A truly ideal gas does not exist in nature, but all real gases approach the ideal state at low enough densities:
+ Molecules are far enough apart, so they do not interact with one another.
+ An ideal gas obeys the ideal gas law (see next slides).
3.1. Ideal Gases
123A mol106.02N
3.1. Ideal Gases
NA: Avogadro’s Number
One mole is the number of atoms in a 12g sample of carbon-12
Avogadro’s law Equal volumes of gases, at the same temperature and
pressure, contain the same number of molecules.
One mole contains 6.02 x 1023 elementary units (atoms or molecules)
Number of moles contained in a sample of any substance:
AN
Nn
where N is the number of molecules in the sample
A
samplesample
mN
M
M
Mn
where Msample is the mass of the sample
M is the molar mass (the mass of 1 mol)
m is the mass of one molecule
Problem 1 (p. 530) Find the mass in kilograms of 7.5x1024 atoms of arsenic (As), which has molar mass of 74.9 g/mol.
M
Mn arsenic
(kg) 0.933
(g)9339.741002.6
107.5M
N
NnMM
23
24
A
arsenic
The number of moles n:
3.1.1. Experimental Laws and the Equation of State
Boyle’s Law (Boyle-Mariotte): For a given mass, at constant temperature (isothermal), the pressure times the volume is a constant for an ideal gas.
constantpV
Glenn Research Center/NASA
Robert Boyle (1627-1691)
Charles’s Law: For a given mass, at constant pressure (isobaric), the volume is directly proportional to the temperature.
TconstantV
Glenn Research Center/NASA
Jacques Charles (1746-1823)
Gay-Lussac’s Law: For a given mass, at constant volume (isochoric), the pressure is directly proportional to the temperature.
Tconstantp
Glenn Research Center/NASA
J. L. Gay-Lussac (1778-1850)
Equation of State:
Boyle’s Law: constantpV
Charles’s Law: TconstantV
Gay-Lussac’s Law: Tconstantp
The gas laws of Boyle, Charles and Gay-Lussac can be combined into a single equation of state:
law) gas (ideal nRTpV
where p is the absolute pressure n is the number of moles of gas
T is the temperature (in K) R is a constant, called the gas constant
11Kmol J 13.8R
law) gas (ideal NkTpV
1-23
123
11
A
K J1038.1mol106.02
K mol 8.31J
N
Rk
(N is the number of molecules) kNkNN
NnR A
A
Rewrite the ideal gas law in an alternative form
with the Boltzmann constant k
law) gas (ideal nRTpV
Sample Problem (p. 510)
A cylinder contains 12 L of oxygen at 200C and 15 atm. The temperature is raised to 350C, and the volume is reduced to 8.5 L. What is the final pressure of the gas in atmospheres? Assume that the gas is ideal.
nRTpV Key equation:
At state i: nRTVp iii
At the final state f: nRTVp fff
fi
fiif
i
f
ii
ff
VT
TVpp
T
T
Vp
Vp
We must convert temperatures in C0 to that in K:
K; 293K 20)(273Ti K 083K 35)(273Tf
atm3.22L) K)(8.5 (293
L) K)(12 atm)(308 (15p f
Problem 4 (p. 530) A quantity of ideal gas at 100C and 100 kPa occupies a volume of 3.0 m3. (a) How many moles of the gas are present? (b) If the pressure is now raised to 300 kPa and the temperature is raised to 300C, how much volume does the gas occupy? Assume no leaks.
nRTVp iii
11Kmol J 13.8R RT
pVn (a)
K 28310273T;m 3.0V Pa;10kPa 100p 35
(moles) 27.6128331.8
0.310n
5
(b) At any state i (p, V, T):
12
2112
2
22
1
11
Tp
TVpVnR
T
Vp
T
Vp
; nRTpV
)(m 1.10.3283300
303100
Tp
TVpV 3
12
2112
:K 30330273T2
Problem 5 (p. 530) The best laboratory vacuum has a pressure of about 10-18 atm, or 1.01x10-13 Pa. How many gas molecules are there per cubic centimeter in such a vacuum at 293 K?
; RT
pVn
11Kmol J 13.8R
)(molecules251002.629331.8
101001.1N 23
613
K 293T;m10cm 1V Pa; 101.01p 36313-
NnN AThe number of molecules:
The number of moles:
Work Done by an Ideal Gas at Constant Temperature
V
1constant
V
1nRTp
A process at constant temperature is called an isothermal expansion/compression. The equation of state for n moles:
The work done during an isothermal process:
f
i
f
i
f
i
V
V
V
V
V
VlnV nRTdV
V
nRTpdVW
i
f
V
Vln nRTW
f
i
V
VpdVW
:)(isochoric constantV 1) If 0W
:(isobaric) constantp 2) If Vp)Vp(VW if
Summary
The equations below allows us to calculate work done by the gas for three special cases:
3) If :l)(isotherma constantT
i
f
V
Vln nRTW
3.1.2. Molecular Model for an Ideal Gas
In this model:
1. The molecules obey Newton's laws of motion.
2. The molecules move in all direction with equal probability.
3. There is no interactions between molecules (no collisions between molecules).
4. The molecules undergo elastic collisions with the walls.
a. Pressure, Temperature, and RMS Speed
First, we consider a cubical box of edge length L, containing n moles of an ideal gas. A molecule of mass m and velocity v collide with the shaded wall.
Key question: What is the connection between the pressure p exerted by the gas and the speed of the molecules?
Problem: Let n moles of an ideal gas be confined in a cubical box of volume V. The walls of the box are held at temperature T.
For an elastic collision, the
particle’s momentum (=m.v) along the x axis is reserved and change with an amount:
xxxx 2mv)(mv)mv(Δp
The average rate at which momentum is delivered to the shaded wall by this molecule:
L
mv
2L/v
2mv
Δt
Δp 2
x
x
xx
dt
pd
dt
)vd(m
dt
vdmamF
Recall: L
mvF
2
xx,1
The pressure exerted on the wall by this single molecule:
2
x,1
1L
Fp
For N molecules, the total pressure p:
2
2
Nx,
2
x,2
2
x,1
2
x
L
/Lmv.../Lmv/Lmv
L
Fp
Travel time b/w 2 walls with a speed v
Note: Pressure is the force applied perpendicular to the surface of an object
)v...vv(L
mp 2
Nx,
2
x,2
2
x,13
The average value of the square of the x components of all the molecular speeds:
N
v...vv 2
Nx,
2
x,2
2
x,12
xv
2
x3
A vL
nmNp
gas theof massmolar the:mNM ASince
2
xvV
nMp :LV 3
For any molecule: 2
z
2
y
2
x
2 vvvv
As all molecules move in random directions: 22
x v3
1v
2v3V
nMp
The square root of is called the root-mean-square speed: 2v
rms
2 vv
3V
nMvp
2
rms
Combining with the equation of state: nRTpV
M
3RTvrms
This relationship shows us how the pressure of the gas (a macroscopic quantity) depends on the speed of the molecules (a microscopic quantity)
b. Translational Kinetic Energy
Consider a single molecule of an ideal gas moving around in the box (see Section a) .
2rms
22 mv2
1v
2
1
2
1K mmv
M/m
3RT
2
1
M
3RT
2
1K
m
A2N
3RTK
The Boltzmann constant k: AN
Rk
kT2
3K
22z
2y
2x v
3
1vvv kT
2
1vm
2
1vm
2
1vm
2
1 2z
2y
2x
Kdoes not depend on the mass of the molecule
24. At 273 K and 1.0 x 10-2 atm, the density of a gas is 1.24 x 10-5 g/cm3. (a) Find vrms for the gas molecules. (b) Find the molar mass of the gas and (c) identify the gas (hint: see Table 19-1).
)1(3
M
RTvrms
Root-mean-square speed:
)2(n
VM
V
nM
V
M gas
(1) and (2):
p
V
nRTvrms
33
3235 kg/m1024.1g/cm1024.1
Pa1001.1atm100.1 32 p
m/s494rmsv
(a)
)2(n
VM
Equation of state:
)3(nRTpV
p
RT
n
VM
g/mol28kg/mol028.0 M
From Table 19.1, the gas is nitrogen (N2)
(b)
(c)
Homework: 9, 13, 14 , 18, 20, 23, 25, 27 (p. 531-532)
Chapter 3 The Kinetic Theory of Gases 3.1. Ideal Gases
3.1.1. Experimental Laws and the Equation of State
3.1.2. Molecular Model of an Ideal Gas
3.2. Mean Free Path
3.3. The Boltzmann Distribution Law and
The Distribution of Molecular Speeds
3.4. The Molar Specific Heats of an Ideal Gas
3.5. The Equipartition-of-Energy Theorem
3.6. The Adiabatic Expansion of an Ideal Gas
3.2. Mean Free Path
3.2.1 Concept
•A molecule traveling through a gas changes both speed and direction as it elastically collides with other molecules in its path.
•Between collisions, the molecules moves in a straight line at constant speed.
•The mean free path is the average
distance traversed by a molecule between
collisions.
V
N
1
density
1λ
V
N is the number of molecules per unit volume or the density of molecules
molecules ofnumber theis N
gas theof volume theis V where
To count the number of collisions: We further consider that this single molecule has an equivalent radius of d and all the other molecules are points (see cartoons next slides for an equivalent problem).
Our goal: Estimate of of a single molecule.
Assumptions: + Our molecule is traveling with a constant speed v and all the other molecules are at rest. + All molecules are spheres of diameter d a collision occurs as the centers of 2 molecules come within a distance d.
= 1 collision
d
= 1 collision . d
Equivalent problem
1st collision
2nd collision
3rd collision
Equivalent problem
d
2d
The number of collisions = the number molecules lie in a cylinder of length vt and cross-sectional area d2
If all the molecules are moving:
V
Nd 22
1
Using the equation of state: pV = NkT
pd
kT22
The average time between collisions (the mean free time):
vpd
kT
vt
22
The average time between collisions (the mean free time):
The frequency of collisions:
tf
1
Approximate the air around you as a
collection of nitrogen molecules, each of which has a
diameter of 2.00 10-10 m. Temperature is at 20 oC.
The pressure of the atmosphere is 1.01 x 105 N/m2
How far does a typical molecule move before it collides
with another molecule?
SOLUTION
PROBLEM
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?
SOLUTION
PROBLEM
3.3. The Boltzmann Distribution Law and the Distribution of Molecular Speeds
The Boltzmann distribution law: if the energy is associated with some state or condition of a system is then the frequency with which that state or condition occurs, or the probability of its occurrence is proportional to:
kTe /
constantBoltzmann the:k
Many of the most familiar laws of physical chemistry are special cases of the Boltzmann distribution law:
P(v)dv is the fraction of molecules with speeds in the infinitesimal range (v,v+dv).
1)(0
dvvP
The fraction of molecules with speeds from v1 to v2:
2
1)(frac
v
vdvvP
3.3.1. The distribution of molecular speeds (or the Maxwell speed distribution law):
Let M be the molar mass of the gas, v be the molecular speed, and P(v) be the speed distribution function:
)1(2
4)( 2/22
2/3
kTMvevRT
MvP
Average, RMS, and Most Probable Speeds
)2()(0
dvvvPvThe average speed:
from (1) & (2): M
RTv
8
0
22 )( dvvPvv
M
RTv
32
The root-mean-square speed:
M
RTvvrms
32
The most probable speed is the speed at which P(v) is maximum:
0)(
dv
vdP
M
RTvP
2
3.3.2. The barometric distribution law:
This law gives the number density (h), i.e. number of molecules per unit volume, of an ideal gas of uniform temperature T as a function of height h in the field of the Earth’s gravity.
kThhmgehh
/)0(
0 )()(
where h0 is an arbitrary fixed reference height; m is the mass of a molecule.
Abell 1982 nasa.gov
Homework: 28, 32, 33, 40 (Page. 531-532)
Chapter 3 The Kinetic Theory of Gases 3.1. Ideal Gases
3.1.1. Experimental Laws and the Equation of State
3.1.2. Molecular Model of an Ideal Gas
3.2. Mean Free Path
3.3. The Boltzmann Distribution Law and
The Distribution of Molecular Speeds
3.4. The Molar Specific Heats of an Ideal Gas
3.5. The Equipartition-of-Energy Theorem
3.6. The Adiabatic Expansion of an Ideal Gas
3.4. The Molar Specific Heats of an Ideal Gas
Let’s consider our ideal gas of n moles that is a monatomic gas, which has individual atoms, e.g. helium, argon, neon. For a single atom, the average translational KE:
kTK2
3
The internal energy Eint of the gas (no rotational KE for monatomic gases):
nRTnkTKEN
2
3N
2
3A
1int
Recall: molar specific heat: TCnQ
Þ DEint =3
2nRDT
WQE int
TnRWTnCWQE V 2
3int
R2
3C V
TnCE V int
So, the change in internal energy can be calculated by:
TnRE 2
3int
or
a. Molar specific heat at constant volume:
Consider n moles of an ideal gas at state i: p, T, and fixed V state f: p+p, T+T
TnCQ V
CV is a constant and called the molar specific heat at constant volume.
V is constant, so W=0
11 K mol J5.12
b. Molar specific heat at constant pressure:
TnCQ p
WQE int
Cp is the molar specific heat at constant pressure.
TnRVpW
TnRTnCTnR p 2
3
RRRCp2
5
2
3
RCC Vp
42. What is the internal energy of 2.0 mol of an ideal monatomic gas at 273 K?
TnCE V
1-1- K mol J 5.122
3 RCV
(J)68252735.120.2 E
(kJ)8.6E
Example: (Problem 8, page 530) Suppose 1.8 mol of an ideal gas is taken from a volume of 3.0 m3 to a volume of 1.5 m3 via an isothermal compression at 300C. (a) How much energy is transferred as heat during the compression, and (b) is the transfer to or from the gas?
(a) We have:
An isothermal process: T=constant
Work done by the gas for isotherm:
(b) Q<0: heat transferred from the gas
3.5. The Equipartition-of-Energy Theorem
Every kind of molecule has a certain number f of degrees of freedom. For each degree of freedom in which a molecule can store energy, the average internal energy is per molecule. kT
2
1
Molecule Example
Degrees of freedom
Translational Rotational Total (f)
Monatomic He
Diatomic O2
Polyatomic CH4
3 0 3
3 2 5
3 3 6
RCC
Rf
C
Vp
V
2TnRf
E 2
int
5 degrees of freedom of a diatomic molecule
Six degrees of freedom Technical Aspects of robotics
wac.nsw.edu.au
3.6. The Adiabatic Expansion of an Ideal Gas
What is an adiabatic process ?
constantpV
Vp/CC where
nRTpV
constant1 TV
Proof of the equations above, see p. 526-527
Equation of state:
Free expansions:
0 :Recall WQ
fi TTE 0int
ffii VpVp
Is a process for which Q = 0
constantpV
56. Suppose 1.0L of a gas with =1.30, initially at 285 K and 1.0 atm, is suddenly compressed adiabatically to half its initial volume. Find its final (a) pressure and (b) temperature. (c) If the gas is then cooled to 273 K at constant pressure, what is its final volume?
;
fiVpVp fi if VV
2
1
f
iif
V
Vpp
1
f
iif
V
VTT
f
f
f
f
T
T
V
VpnRTpV
''constant,
Homework: 42, 44, 46, 54, 56, 78 (p. 533-535)
