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Heat and Thermodynamics Sahraei Physics Department Razi University http://www.razi.ac.ir/sahraei Lecture 12

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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

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

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

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

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

xdmdF

gyPP

2

2

0

;

w

w

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