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Heat and Thermodynamics
Sahraei
Physics Department
Razi University
http://www.razi.ac.ir/sahraei
Lecture 12
specific heat capacity of Solid
Thermometer
Insulator
Heater
Solid
AV
Electrical Method
Experimental Determination of Heat Capacities
The heat capacities of real gases are measured by the electrical method.
Comparison among different s.h.c.
higher s.h.c. difficult to raise its own temperature
Substance specific heat capacity (J/kg .oC)
Water
Glass
Copper
4
1. All ideal gases:
(a) Cv is a function of θ only.
(b) Cp is a function of θ only, and is greater than Cv.
(c) Cp - Cv is not a function of θ, but equal to R.
(d) the ratio Cp / Cv = 𝛾 is a function of θ only, and is greater than 1.
These experimental results indicate that the molar gas constantR = 8.315J/mol.K is a natural unit with which to express the molar heatcapacity of a gas.It is a very interesting consequence of theory that the universal gasconstant is also the natural unit for solids.
5
Gases Molecul
e
Structur
e
cv cp 𝛾
He , Ne , Ar
Monato
mic
gases
3/2 R 5/2 R 5/3
CO, N2, H2
diatomi
c gases
5/2 R 7/2 R 7/5
H2O, CO2
SO2 , Cl2 ,
CH4
Polyato
-mic
gases
vary
with the
temper
at-ure
the
variati
-on
being
differe
nt for
each
gas.
6
In the remainder of this book, we shall specify not the molar heat capacities themselves but the ratios cv/R and cp/R.
The behavior of hydrogen gas H2 is quite exceptional
For all other diatomic gases, Cp/R may always be written
7( )
2
pcf
R
7
Quasi-Static Adiabatic Process
VdPdCdQ
PdVdCdQ
P
V
When an ideal gas undergoes a quasi-static adiabatic process, thepressure,volume, and temperature change in a manner that isdescribed by a relation between P and V, θ and V, or P and θ.
In order to derive the relation between P and V, we start with Eqs.
In an adiabatic process, dQ=0, so
p
vPdV
VdP C
C d
d
Dividing the first equation by the second, we obtain
8
and denoting the ratio of the heat capacities by the symbol 𝛾 we have
dP dV
P V ln ln lnP V c
*PV c
nRPV 1 .V const
A family of curves representing quasi-static adiabatic processes maybe plotted on a PV diagram by assigning different values to theconstant in Eq. *.
9
1( )S
PcV
V
P
V
The slope of any adiabatic curve is
Quasi-static isothermal processes are represented by a family of equilateral hyperbolas obtained by assigning different values to θ in the equation
PV = nRθ
Since( )
P P
V V
PV c
it follows that an adiabatic curve has a steeper negative slope than
does an isothermal curve at the same point, since 𝛾 > 1.
10
The PVθ surface for the ideal gas and its projection onto a PV diagram. (Isotherms are shown as dashed curves, and adiabatics as full curves.)
it may be seen that the adiabatic curves cut across the isotherms.
11
Ruchhardt's Method of Measuring 𝛾
A small positive displacement causes anincreasein volume which is very smallcompared with the equilibrium volume Vandwhich, therefore, can be denoted bydV, where
dV = yA
Similarly, a small positive displacementcauses a decrease in pressure which isverysmall compared with the equilibriumpressure P and which, therefore, can bedenoted by dP, where dP is a negativequantity. The resultant force F
12
Notice that, when y is positive, dP is negative and, therefore, F isnegative; that is, F is a restoring force.
Now, as the ball oscillates fairly rapidly, the variations of P and Vareadiabatic, because there is not enough time for appreciable heattransfer.
Therefore, we may assume that the changes of P and V represent anapproximately quasi-static adiabatic process, and we may write:
PV c
, /dV yA dP F A
Hooke'slaw
13
This is precisely the condition for simple harmonic motion, for which
the period is:
Ruchhardt's method involves errors due to three simplifying assumptions:
(1) that the gas is ideal; (2) that there is no friction; (3) that volume changes are strictly adiabatic.
It is estimated that the second assumption is responsible for the largest error, amounting to about 3 percent.
14
Velocity of a Longitudinal
Wave A is the cross-sectional area of the cylinder
force A(P +P)
constant velocity wo
initial uncompressed
length is w
uncompressed volume
V =Aw
If ρ is the density of the normal or uncompressed gas, the mass of the
free body is ρAw.
15
At time ,
The entire compressed column has a velocity Wo equal to that of the piston.Therefore,
The free body is acted on by a force A( P +P) to the right and a force AP tothe left. Therefore,
AwAw
16
From Newton's second law, the unbalanced force is equal to the rate ofchange of momentum,
has undergone a compression (-V)= Awo
The "uncompressed free body" of volume V=Aw
the velocity of a longitudinal wave
0 0Aw wV
V Aw w
17
isothermal compressibilityby Newton
It was shown later by Laplace that the expression is really the adiabatic compressibility.
1 V
V P
1
( )
wV
V P
18
2/
the center of a compression
the center of a rarefaction
Then, the heat conducted a distance in the time (time for the wave to travel the distance ) is given by
2/
2w/2/
θθ θ
where K is the thermal conductivity of the medium
/ 2 2KAK
w wA
19
The mass of material between the compression and rarefaction is
<<Vwc
K2
Heat necessary to rise temperature
of mass by A /2 2vA c
(adiabatic cond 2
ition)v
KA
wA c
20
The usual range of wavelengths of compressional waves is from a few centi-meters to a few hundred centimeters.
Let us compare these values with
<<Vwc
K2
V
wc
K2
Taking a gas like air as a typical case, we have, roughly,
3
0.0237 / .
1
331 /
0.716
.29 /
/ .v
K W m K
kg
w m s
c k k K
m
J g
V
wc
K2= 155 nm
Therefore, we conclude that, in view of theproperties of ordinary matter, the volume changeswhich take place under the influence of a longitu-dinal wave at ordinary frequencies are adiabatic,not isothermal.
21
sPVV
w
1
/
1
Returnin g now to Eq. for the velocity of a longitudinal wave and
s
sP
V
V
1 adiabatic compressibility
The adiabatic compressibility can be calculated for the ideal gas using Eq. ; thus,
( )S
V V
P P
PV c
1
s
w
1 1( )s S
V
V P P
22
Since the density is
Where M is the molar mass and the v is the molar volume,
1
s
w
Pvw
M
M
v
Rw
M
Equation allows us to calculate 𝛾 from experimental measurements of w and θ.
23
Rw
M
2
1.40Mw
R
For example, the speed of sound in air at 0°C is about 331 m/s Therefore, using the values
8.31 / . , 2
331 / , 273
8.96
/
R kJ kmo
K
l K M kg km
w m
o
s
l
24
The Microscopic Point of View
We have emphasized that the point of view of classical thermodynamics isentirely macroscopic.
Systems are described with the aid of their gross, or large-scale, properties.
v
v
UC
f
iv vQ C d
Equation is true for all hydrostatic systems,whether solid, liquid, or gas.
It enables one to calculate Cv of a hydrostaticsystem, provided that one knows the internalenergy as a function of θ and V.
Eq. may be calculated once the Cv of theparticular system under considerationisknown as a function of θ.
But there is nothing in classicalthermodynamics that provides detailedinformation concerning U or Cv.
25
To obtain detailed information concerning the thermodynamiccoordinates and thermal properties of systems without having toresort to experimental measurements, we require calculations basedon the properties and behavior of the particles of the system.
There are two such microscopic theo-ries: one is called
kinetic theory, and the other is
statistical mechanics
Both theories deal with particles, their internal and externalmotion, their collisionswith one another and with any existing walls,and their forces of interaction.
Making use of the laws of mechanics and statistics, kinetic theoryconcerns itself with the average motion of atoms and theircollisions with walls andother objects in order to calculate theequation of state for the ideal gas.
26
Statistical mechanics avoids the mechanical aspects of particles anddeals with the energy aspects of aggregates or ensembles ofparticles. It relies heavily on statistics and quantum mechanics.
KINETIC THEORY OF THE IDEAL GAS
1. Any small sample of gas consists of an enormous number of particles N.
M denotes the molar mass
then the number of moles n is given by
Any one chemical species, all atoms are identical.
m the mass of each atom,
mN total mass
27
The number of particles per mole of gas is called Avogadro's number NA,where
2. The atoms of an ideal gas are supposed to resemble small hard spheres thatare in perpetual random motion.
The diameter of an atom is of theorder of 2 or 3 x 10-10 m.
Under standard conditions, the average distance between atoms is about 50 times their diameter.
3. The atoms of an ideal gas are assumed to exert no forces ofattraction or repulsion on other atoms except when they collide withone another and with a wall.
Between collisions, they therefore move with uniform rectilinearmotion.
28
Ideal vs. Non-Ideal Gases
xx
29
4. The portion of a wall with which an atom collides is considered to besmooth, and the collision is assumed to be perfectly elastic.
2 xiv
30
5. When there is no external field of force, the atoms are distributed uniformlythroughout a container.
The number density N/ V is assumed constant, sothat in any small element of volume dV there are dN atoms, where
3 191 10 cm atom
6 3 1310 10 cm atom
31
7. Not all atoms have the same speed. A few atoms at any moment move slowly and a few move very rapidly, so that speeds may be considered to cover the range from zero to the speed of light.
dNw represents the number of atoms
w and w + dw with speeds between
it is assumed that dNw remains constant at equilibrium, even though the atoms are perpetually colliding and changing their speeds.
6. There is no preferred direction for the velocity of any atom, so that at any moment there are as many atoms moving in one direction as in another.
32
0 0r 0 2 <
33
Since the velocity vectors of the atoms of gas have no preferred direction,consider an arbitrary velocity vector w directed from the point 0 in Fig. to the elementary area dA'.
The calculation of this quantity involves the concept of a solid angle.
The area dA' on the surfaceof this sphere, formed by two circles of latitude differing by d8 and two circlesof longitude differing by dθhas the magnitude
The solid angle dΩ, formed by lines radiating from 0 and touching the edge ofdA', is by definition
0 0r 0 2 <
34
Since the largest area on the surface of the sphere is that of the entire sphere 4πr2
the maximum solid angle is 4π sr (steradians).
Ω
35
The fraction of atoms with velocity vectors in the neighborhood of w will have speeds between
w & w + dw
and directions within the solid angle dΩ about w.
dNw the number of atoms with speeds between w and w + dw,
then the fraction of these atoms whose directions lie within the solid angle dΩ
dΩ/4πso that the number of atoms within the speed range dw, in the θrange of dθ and the 𝜑 range of d𝜑, is given by
an equation expressing the fact that atomic velocities have no preferred direction.
36
Now consider this group of atoms approaching a small area dA of the wallof the containing vessel. Many of these atoms will undergo collisions along theway, but if we consider only those members of the group that lie within the cylinder (Fig.) whose side is of length
wd, where d is such a short time
atoms within this cylinder will collide with dA.
then all the The volume of the cylinder dV is:
37
Therefore the number of atoms within the speed range dw, in the θrange of dθ and the 𝜑 range of d𝜑, is given by
V total volume of the container
dV /V fraction of the atoms
striking dA in time d is expressed as
which expresses the fact that atoms have no preferred location
38
39
The change in momentum per unit time and per unit area due to
collisions from all directions is the pressure dPw exerted by the wall on
the dNw gas atoms.
Reversing the sign of the momentum change, we get the pressure dPw
exerted by the dNw atoms on the wall:
the total pressure due to atoms of all speeds is given by
40
The average of the square of the atomic speeds <w2 > is defined to be
so that we have
22 1( )
3 2PV N m w < PV nR
22 1( )
3 2N m w nR<
21 3 3
2 2 2A
Rm w k
N <
41
It is worthwhile, at this point, to compare the symbolism used in ourtreatment of kinetic theory with that used in thermodynamics. This is shown in Table .
42
The equation of state for the ideal monatomic gas has a simple form, namely,
21 3
2 2m w k<
22 1( )
3 2PV N m w <
2 3( )( )3 2
PV N k Nk
Equipartition of Energy
3
2E kT 2 2 21 3
2 2x y zm v v v k T
Deduced from kinetic theory of gases:Mean kinetic energy of one particle in a gas
i.e.,
222
zyx vvv and, by symmetry,
each component of energy, is equal to 1/2kT.
This is a statement of the law of equipartition of energy.
Each degree of freedom contributes ½kT to the energy of a system, where possible degrees of freedom are those associated with translation, rotation and vibration of molecules
This tells us that the internal energy of an ideal gas depends only on the temperature
Internal energy is divided equally among the degrees of freedom:
½kT per degree of freedom per molecule.
The number of degrees of freedom increases with the complexity of the molecule.
translational motionvibrational motionrotational motion
Real gases are mostly polyatomic, and store internal energy (Degrees of freedom include) in:
free particles translational kinetic energy (x, y, z components)
< ½ mvx2 > + < ½ mvy
2 > + < ½ mvz 2 > = 3(½ kT)
v
monatomic gas (He, Ne, Ar, …)
3 translational modes (x, y, z)
At room temperature, for most gases:
translation
52
U k T
+ 2 rotational modes (wx, wz)rotation
vibration
Diatomic molecules (N2, O2, CO, …)
3 translational modes (x, y, z)
The water rotations are shown below
non-linear molecules (H2O, NH3, …)
3 translational modes (x, y, z)
+ 3 rotational modes (wx, wy, wz)
kTU 3
Non-linear molecules have 3N - 6 vibrations, while linear molecules have 3N - 5 vibrations. Water,
for example, has 3(3) - 6 = 3 vibrations.
The vibrations of water are shown below.
Rreviev
• The heat capacity ratio or adiabatic index or ratio ofspecific heats, is the ratio of the heat capacity at constantpressure (CP) to heat capacity at constant volume (CV). It issometimes also known as the isentropic expansion factor andis denoted by γ (gamma).
where, C is the heat capacity or the specific heat capacity of agas, suffix P and V refer to constant pressure and constantvolume conditions respectively.
V
P
C
C
Ideal gas relations
Furthermore, the heat capacities can be expressed in termsof heat capacity ratio ( γ ) and the gas constant ( R ):
1
RCP
1
RCV
RCC VP
Relation with degrees of freedom
The heat capacity ratio ( γ ) for an ideal gas can be related to the
degrees of freedom ( f ) of a molecule by:
f
f 2
67.13
5
4.15
7
Thus we observe that for a monatomic gas, with three degrees of
freedom:
while for a diatomic gas, with five degrees of freedom (at
room temperature):
• The terrestrial air is primarily made up of diatomicgasses
• (~78% nitrogen (N2)
• and ~21% oxygen (O2))
• and, at standard conditions it can be considered to be an ideal gas. A diatomic molecule has five degrees of freedom (three translational and two rotational
degrees of freedom). This results in a value of
4.15
7
Ratio of Specific Heats
for some common gases
Gas Ratio of Specific Heats
Carbon Dioxide 1.3
Helium 1.66
Hydrogen 1.41
Methane or Natural Gas 1.31
Nitrogen 1.4
Oxygen 1.4
Standard Air 1.4
• Ideal gas law
– The state of an amount of gas is determined by its pressure, volume, and temperature according to the equation:
where
P is the absolute pressure of the gas,
V is the volume of the gas,
n is the number of moles of gas,
R is the universal gas constant,
T is the absolute temperature.
The value of the ideal gas constant, R, is found to be as follows.
R = 8.314472J·mol−1·K−1
nRTPV
Calculations
Process Constant Equation
Isobaric process Pressure V/T=constant
Isochoric process Volume P/T=constant
Isothermal process Temperature PV=constant
Isentropic process
(Reversible adiabatic
process)
Entropy
PVγ=constant
Pγ-1/Tγ=constant
TVγ-1=constant
56
درواستاتيكيهيسيستميكبرايدرروبيايستاوارفرايندبيانگرزيرروابطازيككدام؟نيست
(الف)
(ب)
(ج)
( د)
ctePV
cteP 1
cteV 1
P
V
V
P
s
ctePV
01 dVPVdPV
V
P
V
P
s
57
براي.ودشمدلبنديطبيعتدردرروبيصورتبهاستممكنگازيكدرصوتانتشارچهوتصسرعتشود،برابردوثابتچگاليدرفشاروقتيبعديسهالايدهگازيك
كند؟ميتغييري
.دشوبرابر مي1/4( الف)
.شودبرابر مي2( ب)
.شودبرابر مي7/0( ج)
.كندتغيير نمي( د)
ctePV
01 dVPVdPV
P
V
P
V
s
PP
V
V s
s
11
00 22
1w
PPw
s
58
تراكمضريبمعكوسحجميمدول)باشدآبحجميمدولاگرشود؟ميمنتشرآندرHz262فركانسباموجيكموجيطولچهبا.(استدرروبي
Bw :سرعت موج طولي
m45.5262
1430
m/s143010
1004.23
9
w
w
Pa1004.2 9B
m10/9( الف)
m8/32( ب)
m4/45( ج)
m4/16( د)
3gr/cm1 :چگالي حجمي آب
طول موج:
59
60
61
62
63
64
65
66
67
68
69
dPشارهيكدرارتفاعتغييراتوفشارتغييراتبينرابطه = - gdyآندركهاستمولكوليوزنباسيالحجميجرمM،gوثقلشتابyاستدرياسطحازارتفاع
.(استبابرابرyارتفاعدرمطلقدماي)لايهنايدرراهواانتقالفرايندهايوكاملگازرازمينجواستراتوسفرلايهاگر-(الف
.وريدآبدسترازمينسطحازارتفاعوفشارتغييراتبينرابطهشود،فرضهمدما(ثابت)شوندفرضدرروبيفرايندهايزمينمجاورهايلايهدراگر-(ب
.بيابيدرادماييتغييراتوفشارتغييراتبينرابطهكنيدمحاسبهارتفاعتغييراترابهدماتغييراتقبلقسمتهاياز-(ج
1P
dyP+dP
P
ydy dp
70
(ب)جواب قسمت
yR
Mg
eP
دررودر انبساط بي:
(الف)جواب قسمت
gdyR
PMdP
R
PM
V
mR
M
mnRPV
gdydP
d
P
dPctePcteP
1
11
´
71
(ج)جواب قسمت gdy
R
PMdP
(ب)از قسمت
(الف)از قسمت
d
P
dP
1
1
R
Mg
dy
dتغييرات دما رابه تغييرات ارتفاع
72
73
74
75
76
آنبازوييك.آوريمميدرLشكلبهرايكنواختداخليقطربابازايشيشهلولهيكچگاليبهمايعيداخلبهرا
صورتبههوادرLطولبهديگربازويوبريمميفرو’.چرخدميعموديبازويمحورحولثابتايزاويهسرعتبالوله.ماندميافقي:بااستبرابررودميبالاعموديبازويدرمايعكهارتفاعيكنيدثابت
w
g
eP
y
R
ML
2
0
22
1P0 فشار جو =
M = وزن مولكولي هوا
g = شتاب گرانشي
77
چگالي هوا
dxxA
dFdP 2w
R
PM
V
mR
M
mnRPV
xdxR
PM
A
dFdP 2w
dxxAdF
xdmdF
gyPP
2
2
0
;
w
w
Adxdm )(
78
xdxR
PM
A
dFdP 2w
:با انتگرال گيري از طرفين اين رابطه
w
g
eP
y
R
ML
2
0
22
1
LP
P
xdxR
PMdP
0
20
w
gyPP از طرفي0
79
80
81
82
83
v
vx
vy
v
vx
vy
A
vxdt