Physics DF025

Chapter 14

CHAPTER 14: Kinetic theory of gases

(3 Hours)

Physics DF025

Chapter 14

Learning Outcome:14.1 Ideal gas equations (1 hour)

a. Sketch and interpret

– P-V graph at constant temperature

– V-T graph at constant pressure

– P-T graph at constant volume

of an ideal gas.

nRTPV

At the end of this chapter, students should be able to:

b. Use the ideal gas equation:

Physics DF025

Chapter 14

14.1 Ideal gas equation14.1.1 Boyle’s law• states : “The pressure

of a fixed mass of gas at constant

temperature is inversely proportional to its volume.”

OR

VP

1 if constantT

• The related equations to the Boyle’s law are

constantPV

OR

2211 VPVP

where pressure initial:1Ppressure final:2P volumeinitial:1V

volumefinal:2V

Physics DF025

Chapter 14• Graphs of the Boyle’s law.

P

T1

V

T2

0

a.

T2

T1

P

V

1

0

b.

The pressure of a fixed mass of gass at constant temperature is inversely propertional to its volume.

Physics DF025

Chapter 1414.1.2 Charles’s law

• states : “The volume of a fixed mass of gas at constant pressure is directly proportional to its absolute temperature.”

OR

TV If constantP

The related equations to the Charles’ law are

constantT

V

OR

2

2

1

1

T

V

T

V

where

re temperatuabsolute initial:1Tre temperatuabsolute final:2T

volumeinitial:1V volumefinal:2V

Physics DF025

Chapter 14

Graphs of the Charles’s law.

a.

T(C)

V

0273.15

b. V

T(K)0

Physics DF025

Chapter 1414.1.3 Gay-lussac’s

(pressure) lawstates : “The pressure of a fixed mass of gas at constant volume is directly proportional to its absolute temperature.”

OR

TP If constantV

The related equations to the Gay-lussac’s law are

constantT

P

OR

2

2

1

1

T

P

T

P

where

re temperatuabsolute initial:1Tre temperatuabsolute final:2T

pressure initial:1Ppressure final:2P

Physics DF025

Chapter 14Graphs of the Gay-lussac’s (pressure) law

T(C)

P

0273.15

a. P

T(K)0

b.

Physics DF025

Chapter 14

• An ideal gas is defined as a perfect gas which obeys the three gas laws (Boyle’s, Charles’s and Gay-Lussac’s) exactly.

• Consider an ideal gas in a container changes its pressure P, volume V and temperature T as shown in Figure 14.1.

14.1.4 Equation of state for an ideal gas

1st stage

1P1V1T

2P'V

1T

2P2V2T

2nd stage

Figure 14.1

Physics DF025

Chapter 14– In 1st stage, temperature is

kept at T1 ,

Using Boyle’s law :

– In 2nd stage, pressure is kept

constant at P2 ,

Using Charles’s law :

112 ' VPVP

2

11'P

VPV (1)

2

2

1

'

T

V

T

V

2

12'T

TVV (2)

– Equating eqs. (1) and (2), thus

constantT

PV(3)

Initial Final

OR

2

22

1

11

T

VP

T

VP

Physics DF025

Chapter 14• Consider 1 mole of gas at standard temperature and pressure

(S.T.P.), T = 273.15 K, P = 101.3 kPa and Vm = 0.0224 m3

– From equation (3),

where R is called molar gas constant and its value is the same for all gases.

– Thus

• For n mole of an ideal gas, the equation of state is written as

– where n : the number of mole gas

15.273

0224.0103.101 3m

T

PVR

11 mol K J 31.8 R

RT

PVm

RTPV m where gas mole 1 of volume:mV

nRTPV

Physics DF025

Chapter 14

M

mn

where gas a of mass:m

gas a of massmolar :M

OR

AN

Nn

where molecules ofnumber :N

constant sAvogadro':AN123 mol10026 .

– If the Boltzmann constant, k is defined as

then the equation of state becomes

123

A

K J 1038.1 N

Rk

NkTPV

Physics DF025

Chapter 14

The volume of vessel A is three times of the volume vessel B. The vessels are filled with an ideal gas and are at a steady state. The temperature of vessel A and vessel B are at 300 K and 500 K respectively as shown in Figure 14.2.

If the mass of the gas in the vessel A is m, obtain the mass of the gas in the vessel B in terms of m.

Example 1 :

A B(300 K) (500 K)

Figure 14.2

Solution :

Since the vessels A and B are connected by a narrow tube thus the pressure for both vessels is same, finally i.e.

The system is in the steady state, thus

By applying the equation of state for an ideal gas,

K 500K; 300; ;3 0B0AABA TTmmVV

PPP BA

K 500K; 300 B0BA0A TTTT

nRTPV andM

mn

RTM

mPV

Physics DF025

Chapter 14

Therefore,

Vessel A :

AA

AA RTM

mVP

3003 B RM

mVP

Vessel B :

K 500K; 300; ;3 0B0AABA TTmmVV

RM

mPV

100B (1)

(2) RM

mPV

B

B 500

By equating the eqs. (1) and (2) hence

5B

mm

RM

mR

M

m

B500100

Physics DF025

Chapter 14

Refer to Figure 14.3. Initially A contains 3.00 m3 of an ideal gas at

a temperature of 250 K and a pressure of 5.00 104 Pa, while B contains 7.20 m3 of the same gas at 400 K and 2.00 104 Pa. Calculate the pressure after the connecting tap has been opened

and the system reached equilibrium, assuming that A is kept at

250 K and B is kept at 400 K.

Example 2 :

AB

connecting tap

Figure 14.3

Physics DF025

Chapter 14

Solution :

After the connecting tap has been opened and the system reached equilibrium, thus

By using the equation of state for ideal gas,

PPP 1B1A

1

11

0

00

T

VP

T

VP

1B

1B

1A

1A

0B

0B0B

0A

0A0A

T

V

T

VP

T

VP

T

VP

400

20.7

250

00.3

400

20.71000.2

250

00.31000.5 44

P

K; 250 ;m 00.3 1A0A3

1A0A TTVV;m 20.7Pa; 1000.5 3

1B0B4

0A VVPPa 1000.2K; 400 4

0B1B0B PTT

Pa 1020.3 4P

Physics DF025

Chapter 14

Given R = 8.31 J mol1 K1 and NA = 6.0 1023 mol1

1. A gas has a volume of 60.0 cm3 at 20 C and 900 mmHg. What would its volume be at STP?(Given the atmospheric pressure = 101.3 kPa and the density of mercury = 13600 kg m3)

ANS. : 66.2 cm3

2. Estimate the number of molecules in a flask of volume 5.0 104 m3 which contains oxygen gas at a pressure of 2.0 105 Pa and temperature of 300 K.

ANS. : 2.41 1022 molecules3. A cylinder contains a hydrogen gas of volume 2.40 103 m3 at

17 C and 2.32 106 Pa. Calculatea. the number of molecules of hydrogen in the cylinder,b. the mass of the hydrogen,c. the density of hydrogen under these conditions.(Given the molar mass of hydrogen = 2 g mol1)

ANS. : 1.39 1024 molecules; 4.62 g; 1.93 kg m3

Exercise 14.1 :

Physics DF025

Chapter 14

Learning Outcome:14.2 Kinetic theory of gases (1 hour)

At the end of this chapter, students should be able to:

a. State the assumptions of kinetic theory of gases.

b. Apply the equations of ideal gas,

2

3

1vNmPV

2

3

1vP

in related problems. c. Explain and use root

mean square (rms) speed,

m

kTv 32

of gas molecules.

and pressure ,

Physics DF025

Chapter 1414.2 Kinetic theory of gases• The macroscopic behaviour of an ideal gas can be describe

by using the equation of state but the microscopic behaviour only can be describe by kinetic theory of gases.

14.2.1 Assumption of kinetic theory of gases• All gases are made up of identical atoms or molecules. • All atoms or molecules move randomly and haphazardly.• The volume of the atoms or molecules is negligible when

compared with the volume occupied by the gas.• The intermolecular forces are negligible except during

collisions.• Inter-atomic or molecular collisions are elastic.• The duration of a collision is negligible compared with the

time spent travelling between collisions.• Atoms and molecules move with constant speed between

collisions. Gravity has no effect on molecular motion.

Physics DF025

Chapter 14

• Consider, initially a single molecule moving with a velocity vx towards wall A and after colliding elastically , it moves in the opposite direction with a velocity vx

as shown in Figure 14.5.

• Consider an ideal gas of N molecules are contained in a cubical container of side d as shown in Figure 14.4.

Wall A

Wall B

14.2.2 Force exerted by an ideal gas

Figure 14.4

• Let each molecule of the gas have the mass m and velocity v.

• The velocity, v of each molecule can be resolved into their components i.e. vx, vy and vz.

Wa

ll AWa

ll B

Wa

ll AWa

ll B

Figure 14.5

Physics DF025

Chapter 14Therefore the change in the linear momentum of the molecule is given by

xxx mvmvP

xx mvP 2

The molecule has to travel a distance 2d (from A to B and back to A) before its next collision with wall A. The time taken for this movement is

xv

dt

2

If Fx1 is the magnitude of the average force exerted by a molecule on the wall in the time t, thus by applying Newton’s second law of motion gives

x

xxx

vd

mv

t

PF

2

21

21 xx v

d

mF

For N molecules of the ideal gas,

222

21 ....... xNxxx v

d

mv

d

mv

d

mF

222

21 ....... xNxxx vvv

d

mF

Physics DF025

Chapter 14where vx1 is the x component of velocity of molecule 1, vx2 is the x component of velocity of molecule 2 and so on.

• The mean (average ) value of the square of the velocity in the x direction for N molecules is

N

vvvv xNxx

x

222

212 .......

2

222

21 .......

x

xNxx

vN

vvv

2xx vN

d

mF

2222zyx vvvv

2222zyx vvvv

Thus, the x component for the total force exerted on the wall of the cubical container is

• The magnitude of the velocity v is given by

then

Physics DF025

Chapter 14• Since the velocities of the molecules in the ideal gas are

completely random, there is no preference to one direction or another. Hence

• The total force exerted on the wall in all direction, F is given by

22 3 xvv

222zyx vvv

3

22

vvx

d

vmNF

2

3

2xvN

d

mF

3

2vN

d

mF

where molecule theof velocity squaremean :2 v

Physics DF025

Chapter 14

• From the definition of pressure,

14.2.3 Pressure of an ideal gas

A

FP 2dA where

d

vm

3

NF

2

and

3

2

3

1

d

vNmP

(14.1)

2v

V

Nm

3

1P

Vd 3and

2

3

1vNmPV

where container in the gas idealan of mass:Nm

(14.2)

Physics DF025

Chapter 14• Since the density of the

gas, is given by

hence the equation (15.1) can be written as

V

Nm

wheregas by the pressure :Pgas theofdensity :

molecules gas theof

velocity squaremean :2 v

2

3

1vP (14.3)

• is defined as

• From the equation of state in terms

of Boltzmann constant, k :

• By equating the eqs. (15.4) and (14.2), thus

14.2.4 Root mean square velocity ( vrms)

NkTPV

2

3

1vNmNkT

(14.4)

m

kTv

32

2rms vv

Physics DF025

Chapter 14

• Since

therefore the equation of root mean square velocity of the gas molecules also can be written as

where

re temperatuabsolute :T

gas molecule a of mass:m

(speed) velocity squaremean root : rmsv

gas a of massmolar :M

2

3

1vP

thus P3

v2

P

v3

rms

• Therefore

ORm

kTv

3rms

M

RTv

3rms

Physics DF025

Chapter 14

Eight gas molecules chosen at random are found to have speeds of 1,1,2,2,2,3,4 and 5 m s1. Determine

a. the mean speed of the

molecules,

b. the mean square speed of

the molecules,

c. the root mean square

speed of the molecules.

Example 3 : Solution :

a. The mean speed of the molecules is given by

8N

N

vv

N

ii

1

8

54322211 v

1s m 5.2 v

Physics DF025

Chapter 14Solution :

b. The mean square speed of the molecules is given by

8N

N

vv

N

ii

1

2

2

8

54322211 222222222 v

222 s m 8 v

c. The root mean square speed of the molecules is

2rms vv

8rms v

1rms s m 83.2 v

Physics DF025

Chapter 14

Solution :

a. By using the equation of

state, thus

A cylinder of volume 0.08 m3 contains oxygen gas at a temperature of 280 K and pressure of 90 kPa. Determine

a. the mass of oxygen in the

cylinder,

b. the number of oxygen

molecules in the cylinder,

c. the root mean square

speed of the oxygen

molecules in the cylinder.

Example 4 :

Pa 1090

K; 280;m 08.03

3

P

TV

nRTPV M

mn

RTM

mPV

and

(Given R = 8.31 J mol1 K1, k = 1.38 1023 J K1, molar mass of oxygen, M = 32 g mol1, NA = 6.02 1023 mol1)

Physics DF025

Chapter 14Solution :a.

b. The number of oxygen molecules in the cylinder is

c. The root mean square speed of the oxygen molecules is

1rms s m 467

032.0

28031.83 v

kg 1090.9 2m

AN

N

M

mn

ANM

mN

Pa 1090 K; 280;m 08.0 33 PTV

28031.8032.0

08.01090 3

m

232

1002.6032.0

1090.9

N

molecules 1086.1 24N

M

RTv

3rms

Physics DF025

Chapter 14Exercise 15.2 :

Given R = 8.31 J mol1 K1, Boltzmann constant, k = 1.381023 K1

1. In a period of 1.00 s, 5.00 1023 nitrogen molecules strike a wall with an area of 8.00 cm2. If the molecules move with a speed of 300 m s1 and strike the wall head-on in the elastic collisions, determine the pressure exerted on the wall.

(The mass of one N2 molecule is 4.68 1026 kg)ANS. : 17.6 kPa2. Initially, the r.m.s. speed of an atom of a monatomic ideal gas is

250 m s1. The pressure and volume of the gas are each doubled while the number of moles of the gas is kept constant.

Calculate the final translational r.m.s. speed of the atoms. ANS. : 500 m s1

3. Given that the r.m.s. of a helium atom at a certain temperature is 1350 m s1, determine the r.m.s. speed of an oxygen (O2) molecule at this temperature.

(The molar mass of O2 is 32.0 g mol1 and the molar mass of He is 4.00 g mol1)

ANS. : 477 m s1

Physics DF025

Chapter 14

a. Explain and use translational kinetic energy of gases,

b. State the principle of equipartition of energy.

c. Define degree of freedom

d. State the number of degree of freedom for monoatomic, diatomic, and polyatomic gas molecules.

Learning Outcome:

kTTN

RK

2

3

2

3

Atr

At the end of this chapter, students should be able to:

14.3 Molecular kinetic energy and internal energy

Physics DF025

Chapter 14

e. Explain internal energy of gas and relate the internal energy to the number of degree of freedom,

Learning Outcome:

At the end of this chapter, students should be able to:

14.3 Molecular kinetic energy and internal energy

f. Explain and use internal energy of an ideal gas

Physics DF025

Chapter 1414.3 Molecular kinetic energy and internal energy14.3.1 Translational kinetic energy of molecule

• From equation (14.1), thus

This equation shows that

(14.5)

2

2

1

3

2vm

V

NP

increases

V

N

increases

2vm

2

1P increases () When

2v

V

Nm

3

1P

Physics DF025

Chapter 1414.3.1 Translational kinetic energy of molecule

increases

V

N

increases

2vm

2

1

P increases () When

2

2

1

3

2vmNPV and NkTPV

This equation shows that

Rearrange equation (14.5), thus

Physics DF025

Chapter 14

2

2

1

3

2vmNNkT

and

kTvm2

3

2

1 2

tr2

2

1Kvm

TN

RkTK

Atr 2

3

2

3

where

molecule

a ofenergy kinetic

onal translatiaverage : trK

re temperatuabsolute :TconstantBoltzmann :k

constant gasmolar : Rconstant Avogadro :AN

Physics DF025

Chapter 14

• For N molecules of an ideal gas in the cubical container, the total average (mean) translational kinetic energy, E is given by

NkTE2

3

trNKE

kT

2

3NE

nRTE2

3

OR

Physics DF025

Chapter 14

• States : “the mean (average) kinetic energy of every degrees

of freedom of a molecule is

Therefore

14.3.2 Principle of equipartition of energy

.2

1kT

kTf

K2

freedom of degrees:fwhere

re temperatuabsolute:T

OR

RTf

K2

Mean (average) kinetic energy per molecule

Mean (average) kinetic energy per mole

Physics DF025

Chapter 14

A vessel contains hydrogen gas of 7.50 1017 molecules per unit volume and the root mean square speed of the molecules is 2.50 km s1at a temperature of 30 C. Determine

a. the average translational kinetic energy of a molecule for

hydrogen gas,

b. the pressure of hydrogen gas.

(Given the molar mass of hydrogen gas = 2 g mol1, NA= 6.02 1023 mol1and k = 1.38 1023 J K1)

Example 5 :

K 15.303

;s m 1050.2;1050.7 13rms

17

T

vV

N

kTK2

3tr

15.3031038.12

3 23tr

K

J 1028.6 21tr

K

Solution :

a. The average translational kinetic energy of a molecule is

Physics DF025

Chapter 14

Solution :

b. The pressure of hydrogen gas is given by

2

3

1vm

V

NP

K 15.303 ;s m 1050.2;1050.7 13rms

17 TvV

N

AN

Mm and

2rms

A3

1v

N

M

V

NP

where2

rms2 vv

2323

17 1050.21002.6

002.01050.7

3

1

P

Pa 1019.5 3P

Physics DF025

Chapter 14

• is defined as a number of independent ways in which an atom or molecule can absorb or release or store the energy.

Monatomic gas (e.g. He,Ne,Ar)

• The number of degrees of freedom is 3 i.e. three direction of translational motion where contribute translational kinetic energy as shown in Figure 14.6.

14.3.3 Degree of freedom ( f )

yv

xv

zv

x

y

z

He

Figure 14.6

Physics DF025

Chapter 14Diatomic gas (e.g. H2, O2, N2)

• The number of degrees of freedom is

Polyatomic gas (e.g. H2O, CO2, NH3)

• The number of degrees of freedom is

Translational kinetic energy 3Rotational kinetic energy 2

5Figure 14.7

xv

zv

H Hyv

x

y

z

Figure 14.8

xv

zv

yv

x

y

z

O

HH

6

Translational kinetic energy 3Rotational kinetic energy 3

Physics DF025

Chapter 14

Molecule Example

Degrees of Freedom ( f )Average kinetic

energy per molecule,<K>Translational Rotational Total

• Table 14.1 shows the degrees of freedom for various molecules.

Monatomic

Diatomic

Polyatomic

He

H2

H2O

3 kT2

3

(At temperature of 300 K)

Table 14.1

0 3

3 kT2

52 5

3 kTkT 32

63 6

Physics DF025

Chapter 14

• Degrees of freedom depend on the absolute temperature of the gases.

– For example : Diatomic gas (H2)

– Hydrogen gas have the vibrational kinetic energy (as shown in Figure 14.9) where contribute 2 degrees of freedom which correspond to the kinetic energy and the potential energy associated with vibrations along the bond between the atoms.

H H

vibration

Figure 14.9

3f5f7f

when the temperature,

At 250 KAt 250 – 750 KAt >750 K

Physics DF025

Chapter 14Solution :

By applying the equation of the total translational kinetic energy, thus

A vessel contains an ideal diatomic gas at temperature of 20 C. The total translational kinetic energy of the gas molecules is 3.00 106 J. The mass of the gas is then doubled and the total translational kinetic energy of the molecules becomes 9.00 106 J. Determine the new temperature of the gas.

Example 6 :

; J 1000.9

J; 1000.3;K 15.2936

2

611

E

ET

12 2mm

NkTE2

3

ANM

mN

andwhere

AN

Rk

TN

RN

M

mE

AA2

3

RTM

mE

2

3

Physics DF025

Chapter 14

Solution :

For temperature T1 :

For temperature T2 :

(2)(1) :

11

1 2

3RT

M

mE

J 1000.9J; 1000.3;K 15.293 62

611

EET

12 2mm (1)

22

2 2

3RT

M

mE

21

2

2

2

3RT

M

mE

(2)

1

2

1

2 2

T

T

E

E

15.293

2

1000.3

1000.9 26

6 T

K 4402 T OR C 167

Physics DF025

Chapter 14

• Explain internal energy of gas and relate the internal energy to the number of degree of freedom.

• Explain and use internal energy of an ideal gas

Learning Outcome:

RCC VP

NkTf

U2

V

P

C

C

and

At the end of this chapter, students should be able to:

• Define molar specific heat at constant pressure and volume.

• Use equations,

Physics DF025

Chapter 14

14.3.4 Internal energy of gas and relate the internal energy to the number of degree of freedom.

• is defined as the sum of total kinetic energy and total potential energy of the gas molecules.

• But in ideal gas, the intermolecular forces are assumed to be negligible thus the potential energy of the molecules can be neglected.

KNU

nRT2

fU

NkT2

fU

AN

Rk and

gas theof

energy internal:Uwhere

OR

Thus for N molecules,

Physics DF025

Chapter 14

Monatomic Diatomic Polyatomic

Table 14.2 shows the properties for 1 mole of an ideal gas.

Degrees of freedom, f

Average kinetic energy per

molecule, <K>

Internal energy, U

3 5 6

kT2

3

Table 14.2

kT2

5kTkT 3

2

6

RT2

3 RT2

5RTRT 3

2

6

Physics DF025

Chapter 14

Exercise 14.3 :

Given R = 8.31 J mol1 K1, Boltzmann constant, k = 1.381023 K1

1. One mole of oxygen has a mass of 32 g. Assuming oxygen behaves as an ideal gas, calculate

a. the volume occupied by one mole of oxygen gas

b. the density of oxygen gas

c. the r.m.s. speed of its molecules

d. the average translational kinetic energy of a molecule

at 273 K and pressure of 1.01105 Pa.

ANS. : 2.25102 m3; 1.42 kg m3; 461 m s1; 5.651021 J

Physics DF025

Chapter 14

THE END…Next Chapter…CHAPTER 15 :Thermodynamics