thermodynamics lecture series email: [email protected] hotmail.com applied sciences education...

Thermodynamics Lecture Series

email: [email protected]://www3.uitm.edu.my/staff/drjj/

Applied Sciences Education Research Group (ASERG)

Faculty of Applied SciencesUniversiti Teknologi MARA

Kinetic Theory of Gases – Kinetic Theory of Gases – Microscopic ThermodynamicsMicroscopic Thermodynamics

Reference: Chap 20 Halliday & Resnick Fundamental of Physics 6th edition

Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

2

Review – Steam Power PlantReview – Steam Power Plant

Pum

p

Boiler

Turbin

e

Condenser

High T Res., TH

Furnace

qin = qH

in

out

Low T Res., TL

Water from river

A Schematic diagram for a Steam Power Plant

qout = qL

Working fluid:

Water

qin - qout = out - in

qin - qout = net,out


3

Review - Steam Power PlantReview - Steam Power Plant

Steam Power Plant

High T Res., TH

Furnace

qin = qH

net,out

Low T Res., TL

Water from river

An Energy-Flow diagram for a SPP

qout = qL

Working fluid:

WaterPurpose:

Produce work,

Wout, out


4

Review - Steam Power PlantReview - Steam Power Plant

Thermal Efficiency for steam power plants

in

out,net

qnputi equiredr

output desired

in

out,net

q

in

outin

q

qq

in

out

q

q1

H

L

q

q1

H

Lrev T

T1 For real engines, need

to find qL and qH.


5

Entropy Balance –Steady-flow device

Review - Entropy BalanceReview - Entropy Balance

Heat exchanger

Qin1

2

4

3, Hot water inlet

Cold water Inlet Out

Case 1 – blue border

Case 2 – red border


6



outinoutin WWQQ

kW ,)()( inletexit mm

Heat exchanger: energy balance;

kW,hmhmhmhm0 11223344

Assume kemass = 0, pemass = 0

where 34 mm

1 2

4

3Qin

Case 1

kW ,hhmhhm 212344

12 mm


7



Heat exchanger: energy balance;

Assume kemass = 0, pemass = 0

where

kW ,mmQQ 1122outin

1 2

4

3Qin

kW ,hhm0Q 122in

Case 1

Case 2

kW ,hhmhhm 212344

34 mm

12 mm


8



Heat exchanger:

K

kW,smsmsmsm00S 11223344gen

Entropy Balance

where

1 2

4

3Qin

Case 1

34 mm

12 mm

K

kW ,ssmssmS 122344gen


9



K

kW,smsm

T

Q

T

QS 1122

in

in

out

outgen

Heat exchanger:

Entropy Balance

where

1 2

4

3Qin

Case 2

34 mm

12 mm

K

kW ,ssm

T

Q0S 122

in

ingen

Introduction - Introduction - ObjectivesObjectives

1. State terminologies and their relations among each other for ideal gases.

2. Write the ideal gas equation in terms of the universal gas constant and in terms the Boltzmann constant.

3. Derive and obtain the relationship between pressure and root mean square speed of molecules.

4. Obtain the relationship of rms speed and gas temperature

Objectives:Objectives:


11

Microscopic Variables

New Way of Looking at GasesNew Way of Looking at Gases

Classical ThermodynamicsProperties are macroscopic measurables:

P,V,T,UNo inclusion of atomic behaviour Did not discuss about the origin of P,T or

explain V.

T = 30 C

P = 4.246 kPa

T = 30 C

P = 4.246 kPa

H2O:Sat.

liquid


12

Microscopic Variables-Molecular Approach


Kinetic Theory of GasesPressure exerted by gas related to

molecules colliding with wallsT and U related to kinetic energies of

moleculesV filled by gas relate to freedom of

motion of molecules.Must look at same number of molecules

when measure size of samples

High density


13



Kinetic Theory of Gases: SizesMole: the number of atoms contained in 12 g

sample of carbon-12Avogadro’s number:

NA =6.02 x 1023 atoms/molNumber of moles is

N is the ratio of number of molecules with respect to NA

High density

AN

nN


14



Kinetic Theory of Gases: SizesNumber of moles is

N is the ratio of sample mass to the molar mass, M (kg/kmol) or molecular mass m (kg/atoms) High density

A

samplesample

A mN

M

M

M

N

nN

AmNM Where the molar mass is related to the molecular mass by Avogadro number


15

Ideal Gases


Low density (mass in 1 m3) gases.Molecules are further apart

Real gases satisfying condition PPgasgas << P<< Pcritcrit; T; Tgasgas >> T >> Tcritcrit , have low density and can be treated as ideal gases

High density

Low density Molecules far apart


16

Ideal Gases


Equation of StateEquation of State - P--T behaviour

PP=RT=RT (energy contained by 1 kg mass) where is the specific volume in m3/kg, RR is gas constant, kJ/kgK, TT is absolute temp in Kelvin.

High density

Low density

Molecules far apart


17

Ideal Gases


Equation of StateEquation of State - P--T behaviourPP=RT=RT , since = V/Msam then, P(V/

Msam)=RT. So, PV=MPV=MsamsamRTRT, in kPam3=kJ.

Total energy of a system.

Low density

High density


18

Ideal Gases


Equation of StateEquation of State - P--T behaviourPV =MPV =MsamsamRTRT =NMRT=N(MR)T But

RRuu=MR=MR. Hence, can also write PV = PV = NRNRuuTT where

NN is no of kilomoles, kmol,MM is molar mass in kg/kmole ,RR is a gas constant andRRuu is universal gas constant;

RRuu=MR= 8.314 kJ/kmol=MR= 8.314 kJ/kmolKKLow density

High density


19

Ideal Gases


Equation of StateEquation of State - P--T behaviourPV =NRPV =NRuuTT =NkNAT=(n/NA)(kNA)T.

Hence, can also write PV = nkTPV = nkT whereNN is no of kilomoles, kmol,nn is no of molecules,kk is Boltzmann constant; RRuu = 8.314 kJ/kmol = 8.314 kJ/kmolK = K = kNA

k = Rk = Ruu / N / NAA = 1.38 x 10 = 1.38 x 10-23-23 J/K J/KLow density

High density


20

Pressure, Temperature and Root Mean Square Speed


How is the pressure How is the pressure P that an ideal gas P that an ideal gas of N moles confined of N moles confined to a cubical box of to a cubical box of volume V and held volume V and held at temperature T, at temperature T, related to the related to the speeds of the speeds of the molecules??molecules??

y

m

L

L

L

v

z

x

NormalNormal

To wallTo wall

Before collisionBefore collision


21



Assume elastic collision, Assume elastic collision, then after collide with right then after collide with right wall, only x component of wall, only x component of velocity will change. Then velocity will change. Then momentum change is:momentum change is:

y

Ms

L

L

L

v

z

x

NormalNormal

To wallTo wall

After collisionAfter collision

ifx ppp

xx mvmv

xmv2


22



So momentum change So momentum change received by the wall is:received by the wall is:

NormalNormal

To wallTo wall

y

m

L

L

L

v

z

x

After collisionAfter collisionxv

Lt

2

xx mvp 2 The time to hit the right The time to hit the right

wall again iswall again is


23



So average rate of So average rate of momentum transfer momentum transfer received by the wall due received by the wall due to 1 molecule is:to 1 molecule is:

NormalNormal

To wallTo wall

y

m

L

L

L

v

z

x

After collisionAfter collision

x

xx

vL

mv

t

p

/2

2

xxx FL

mv

t

p

2


24



The total force along x is the sum due to collision by all N The total force along x is the sum due to collision by all N molecules with different speeds. The pressure on the wall is the molecules with different speeds. The pressure on the wall is the force exerted for each unit area and is then:force exerted for each unit area and is then:

2

222

21

2

/..//

L

LmvLmvLmv

L

FP xnxxx

222

213

.. xnxx vvvL

mP


25



The total force along x is the sum due to collision by all n The total force along x is the sum due to collision by all n molecules with different speeds. The pressure on the wall is molecules with different speeds. The pressure on the wall is then:then:

222

213

.. xnxx vvvL

mP

But there are n velocities representing n molecules and so we But there are n velocities representing n molecules and so we can represent the different speeds by an average speed. Note can represent the different speeds by an average speed. Note also that also that N = n/NN = n/NAA. So, . So, n =NNn =NNAA. Then the pressure on the wall . Then the pressure on the wall is:is:


26



avgxA v

L

mNNP 2

3

But there are N velocities representing N molecules and so we But there are N velocities representing N molecules and so we can represent the different speeds by and average speed. Note can represent the different speeds by and average speed. Note also that also that N = n/NN = n/NAA. So, . So, n =NNn =NNAA. Then the pressure on the wall . Then the pressure on the wall is:is:

But But mNmNAA is the molar mass, M is the molar mass, M of the gas mass of 1 mol and of the gas mass of 1 mol and LL33 is the volume of the box. So, is the volume of the box. So,

avgxvV

NMP 2


27



avgxA v

L

mNNP 2

3

But But mNmNAA is the molar mass, M is the molar mass, M of the gas mass of 1 mol and of the gas mass of 1 mol and LL33 is the volume of the box. So, is the volume of the box. So,

avgxvV

NMP 2

Then the pressure is:Then the pressure is:

In the 3D box each molecule In the 3D box each molecule has speed along x,y and z has speed along x,y and z direction.direction.

2222zyx vvvv


28



Since there are many molecules in the box each moving with Since there are many molecules in the box each moving with different velocities and in random directions, the average different velocities and in random directions, the average square of velocity components are equal.square of velocity components are equal.

222zyx vvv

avgvV

NMP 2

3

Finally,Finally,

Then,Then,2222xxx vvvv

3

22 v

vx HenceHence


29



The square root of the average of the square of the velocity is The square root of the average of the square of the velocity is called root-mean-square speed of the molecules. It means called root-mean-square speed of the molecules. It means square each speed, find the mean, then take its square root.square each speed, find the mean, then take its square root.

avgrms vv 2

V

NMvP rms

3

2

Hence, the pressure is:Hence, the pressure is:

So,So, avgrms vv 22


30



The square root of the average of the square of the velocity is The square root of the average of the square of the velocity is called root-mean-square speed of the molecules. It means called root-mean-square speed of the molecules. It means square each speed, find the mean, then take its square root.square each speed, find the mean, then take its square root.

avgrms vv 2

V

NMvP rms

3

2

Hence, the pressure is:Hence, the pressure is:

SoSo,, avgrms vv 22

The rms speed can be determined The rms speed can be determined

If P,T is known. Using If P,T is known. Using PV=NRPV=NRuuTT

3

2rms

u

NMvTNR


31



M

TRv urms

32

The root mean square is then:The root mean square is then:

M

TRv urms

3

Since the square of the root mean square of the velocity is:Since the square of the root mean square of the velocity is:


32



Gas

(Values taken at T=300K)

Molar mass, M

(10-3 kg/kmol)

rms,

(m/s)

Hydrogen (H2) 2.02 1920

Helium (He) 4.0 1370

Water vapor (H2O) 18.0 645

Nitrogen (N2) 28.0 517

Oxygen(O2) 32.0 483

Carbon dioxide (CO2) 44.0 412

Sulphur Dioxide (SO2) 64.1 342


33

Temperature-Translational kinetic Energy


Consider a molecule in the box which are colliding with other Consider a molecule in the box which are colliding with other molecules and changes speed after collision. It moves with molecules and changes speed after collision. It moves with translational kinetic energy at any instanttranslational kinetic energy at any instant

2

2mvKE

But the average translational kinetic But the average translational kinetic energy is over a period of time is:energy is over a period of time is:

222

222 rmsavg

avg

avg vm

vmmv

KE


34

Temperature-Translational kinetic Energy


Substitute the rms speed in terms of T, then:Substitute the rms speed in terms of T, then:

A

u

A

uuavg N

TR

Nm

TRm

M

TRmKE

2

3

2

3

2

3

Note that the molar mass Note that the molar mass M=mNM=mNAA. Note also that . Note also that RRuu = = kNkNAA. Hence the average translational kinetic energy is:. Hence the average translational kinetic energy is:

kTN

TRKE

A

uavg 2

3

2

3

Regardless of mass , all ideal Regardless of mass , all ideal gas molecules at temperature T gas molecules at temperature T have the same avg. have the same avg. translational KE.translational KE.

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