Chapter 02 Thermal Properties of Matter (Pp 19-58)

CH 02 THERMAL PROPERTIES OF MATTER 19

CHAPTER 02 THERMAL PROPERTIES OF MATTER

2-1 BOYLE’S LAW

Problem 2-1

A perfect gas undergoes isothermal compression, which

reduces its volume by331020.2 m

−× . The final pressure and

volume of the gas are Torr31078.3 × and

331065.4 m−×

respectively. Calculate the original pressure of the gas.

Solution According to Boyle’s law 2211 VpVp =

×+×

××=

=

−−

−

)1065.4()1020.2(

1065.4)1078.3(

33

33

1

221

V

Vpp

Torrp 31 1057.2 ×=

PaPap 531 1043.3)32.133)(1057.2( ×=×=

PaTorr 32.1331 =Θ Problem 2-2

A sample of air occupies litre0.1 at C025 and atm0.1 . What

pressure is needed to compress it to 3100 cm at this

temperature?


×

××=

=

−

−

6

35

2

112 10100

101)10013.1(

V

Vpp

atmmNorPap 10/10013.1 242 =×=


Problem 2-3

A bubble rises from the bottom of a tall open tank of water

that is at uniform temperature. Just before it bursts at the

surface the bubble is three times its original volume.

Determine the absolute pressure at the bottom of the tank.


×=

=

0

05

2

112

3)10013.1(

V

V

V

Vpp

252 /10039.3 mNorPap ×=


2-2 CHARLES’S LAW

Problem 2-4

To what temperature must a sample of a perfect gas of

volume 3500 cm be cooled from C

035 in order to reduce its

volume to3150 cm ?

Solution According to Charles’s law

f

f

i

i

T

V

T

V=

+=

=

500

150)27335(

i

f

ifV

VTT

CKT f

06.1804.92 −==

Problem 2-5

A given mass of an ideal gas occupies mL50 at C020 . If its

pressure is held constant, what volume does it occupy at a

temperature of C050 ?

Solution According to Charles’s law

f

f

i

i

T

V

T

V=

mLT

TVV

i

f

if 1.5527320

27350)50( =

+

+=

=


2-3 GAY-LUSSAC’S LAW

Problem 2-6

A container of helium gas at STP is sealed and then raised

to a temperature of K730 . What will be its new pressure?

Solution According to Gay-Lussac’s law

.1

1

2

2 constT

p

T

p==

=

1

212

T

Tpp

Now Paatmp 31 10013.11 ×==

KCT 273001 ==

KT 7302 = Therefore

Pap55

2 10709.2273730

)10013.1( ×=

×=

Problem 2-7

An aerosol can of whipped cream is pressurized at kPa440

when it is refrigerated at C03 . The can warms against

temperature in excess of C050 . What is the maximum safe

pressure of the can?

Solution Under these conditions

.1

1

2

2 constT

p

T

p== (Gay-Lussac’s Law)

+

+×=

=

2733

27350)10440( 3

1

212

T

Tpp

Pap 52 10149.5 ×=


Problem 2-8

One mole of oxygen gas is at a pressure of atm6 and a

temperature of C027 . (a) If the gas is heated at constant

volume until the pressure triples, what is the final

temperature? (b) If the gas is heated until both the pressure

and volume are doubled, what is the final temperature?

Solution

(a) Now 2

21

1

1

T

p

T

p= (Gay-Lussac’s Law)

KTp

pT 900)27327)(3(1

1

22 =+=

=

(b) Now 2

22

1

11

T

Vp

T

Vp= (The General Gas Law)

KTV

V

p

pT 1200)27327)(2)(2(1

1

2

1

22 =+=

=

Problem 2-9

A sample of hydrogen gas was found to have a pressure of

kPa125 when the temperature was C023 . What its

pressure be expected to be when the temperature is C011 ?

Solution The relation between pressure and temperature at constant volume is given by

i

i

f

f

T

p

T

p=

PaT

Tpp

i

f

if

53 10183.127323

27311)10125( ×=

+

+×=

=


2-4 THE IDEAL GAS LAW

Problem 2-10

If 300.3 m of a gas initially at STP is placed under a

pressure of atm20.3 , the temperature of the gas rises to

C00.38 . What is the new volume?

Solution According to general gas equation

.2

22

1

11 constT

Vp

T

Vp==

31

22

112 068.1)00.3(

)273)(20.3()38273)(1(

mVTp

TpV =

+

=

=

Problem 2-11

Calculate the volume occupied by one mole of an ideal gas

at STP conditions.

Solution

For an ideal gas TRnVp =

p

TRnV =

Now moln 1= , 11314.8 −−= KmolJR KCT 27300 == , Paatmp 510013.11 ×== Therefore

litresmV 4.221024.210013.1

)273)(314.8)(1( 325

=×=×

= −

litresm33 1011 ×=Θ

Problem 2-12

A mol5.2 sample of an ideal gas at atm00.1 pressure has a

temperature of C020 . What is the volume occupied by the

gas?


Solution For an ideal gas TRnVp =

p

TRnV =

litresmV 60060.010013.1

)27320)(314.8)(5.2( 35 ==

×

+=

Problem 2-13

Calculate the volume occupied by 5 moles of an ideal gas at

100 kPa and 300 K.

Solution According to ideal gas law TRnVp =

33

125.010100

)300)(314.8)(5(m

p

TRnV =

×==

Problem 2-14

The boiling point of helium at one atmosphere is K2.4 .

What is the volume occupied by the helium gas due to

evaporation of g100 of liquid helium at atm1 pressure and

a temperature of K300 ?

Solution

According to ideal gas law TRnVp =

p

TRnV =

Now moln 254

100==

11314.8 −−= KmolJR KT 300=

Paatmp 510013.11 ×== Hence

litresmV 616616.010013.1

)300)(314.8)(25( 35

==×

=


Problem 2-15

Calculate the density of oxygen at S.T.P. using the ideal gas

law. The mass of one mole of oxygen is 32 g.


325

10241.210013.1

)273)(314.8)(1(m

p

TRnV

−×=×

==

As mass of one mole of oxygen )( 2O is equal to g32)16(2 = , therefore

kgkggm 23 102.3103232 −− ×=×== The desired density of oxygen at S.T.P. conditions is

32

2

/428.110241.2

102.3mkg

V

m=

×

×==

−

−

ρ

Problem 2-16

For an ideal gas at temperature K300 and atm1 pressure,

what are dimensions of a cube that contains 1000 particles

of the gas?

Solution Now TkNVp =

3235

23

10090.410013.1

)300)(10381.1)(1000(m

p

TkNV

−−

×=×

×==

If ‘x’ is the side of the cube, then Vx =3

nmmVx 34.010445.3)10090.4( 83/1233/1 ≅×=×== −− Problem 2-17

What is the pressure of an ideal gas if moles5.3 occupy

litres0.2 at a temperature of C0150− ?



V

TRnp =

Pap 63

1079.1)100.2(

)273150)(314.8)(5.3(×=

×

+−=

−

Problem 2-18

A sample of mg255 of neon occupies litres0.3 at K122 .

Calculate the pressure of the gas.

Solution

Now TRnVp =

V

TRnp =

As 20.18 g of neon is equal to 1 mol, therefore

nmolmolgmg =×=×

=×= −−

− 23

3 10264.118.2010255

10255255

Hence

Pap3

3

2

10274.4103

)122)(314.8)10264.1(×=

×

×=

−

−

atmatmp2

5

3

10219.410013.110274.4 −×=

×

×=

Problem 2-19

The temperature in outer space is about K7.2 and the

matter there consists mainly of isolated hydrogen atom with

a density of 3/3.0 matoms . What is the pressure of this gas?

Solution According to ideal gas law TkNVp =


PaTkV

Np 2323 10119.1)7.2)(10381.1)(3.0( −− ×=×=

=

Problem 2-20

A storage tank at S.T.P. contains kg5.18 of nitrogen ( 2N ).

(a) What is the volume of the tank?

(b) What is the pressure if an additional kg0.15 of nitrogen

is added without changing the temperature?

The mass of one mole of nitrogen is g28 .

Solution (a) According to ideal gas law

TRM

mTRnVp

==

Mp

TRmV =

335

8.14)1028)(10013.1(

)273)(314.8)(5.18(mV =

××=

−

(b) Now TRM

mTRnVp

== 1

1

VM

TRmp 1=

Pap 53

10835.1)8.14)(1028(

)273)(314.8)(155.18(×=

×

+=

−

atmatmp 811.110013.1

10835.15

5

=×

×=

Problem 2-21

Estimate the mass of air in a room whose dimensions are

mmm 458 ×× high at S.T.P. The mass of one mole of air

is g29 .

Solution According to ideal gas law


TRM

mTRnVp

==

TR

MVpm =

kgm 207)273)(314.8(

)1029)(458)(10013.1( 35

=××××

=−

Problem 2-22

Consider an ideal gas at C020 and a pressure of MPa00.2

in a 321000.1 m

−× tank. Determine the gram moles of gas

present.


molesTR

Vpn 21.8

)27320)(314.8()1000.1)(1000.2( 26

=+

××==

−

Problem 2-23

An industrial firm supplies compressed air cylinders of

volume 325.0 m filled to a pressure of MPa20 at C017 .

Calculate the contents of the cylinder in (a) moles (mol) (b)

kilograms. The molar mass of air is 1029.0 −molkg .

Solution

(a) TRnVp =

molTR

Vpn

36

10074.2)27317)(314.8(

)25.0)(1020(×=

+

×==

(b) moln310074.2 ×=

kgn 60)029.0)(10074.2( 3 =×= Problem 2-24

Calculate the number of moles in one 3m of an ideal gas at

C020 and atmospheric pressure.



molesTR

Vpn 6.41

)27320)(314.8()1)(10013.1( 5

=+

×==

Problem 2-25

A sample of an ideal gas occupies a volume of litres4 at

C020 with a pressure of atm3 . How many moles are present

in the sample?


TRnVp =

TR

Vpn =

Now PaPaatmp 55 10039.3)10013.1(33 ×=××==

331044 mlitresV−×==

11314.8 −−= KmolJR KKCT 293)27320(200 =+== Hence

molesn 499.0)293)(314.8(

)104)(10039.3( 35

=××

=−

Problem 2-26

A tyre is filled with air at C015 to a gauge pressure of

kPa220 . If the tyre reaches a temperature of C038 , what

fraction of the original air must be removed if the original

pressure of kPa220 is to be maintained?

Solution As volume and pressure remains same, therefore the fractional change in number of moles can be written as


2

21

1

12

)/(

)/()/(

T

TT

TRVp

TRVpTRVp

n

n −=

−=

∆

%4.7074.0)27338(

)27338()27315(−=−=

+

+−+=

∆

n

n

Hence 7.4 % of original air must be removed to maintain the same pressure i.e. kPa220 in the tyre. Problem 2-27

A quantity of ideal gas at C00.12 and a pressure of kPa108

occupy a volume of347.2 m . (a) How many moles of the gas

are present? (b) If the pressure is now raised to kPa316

and the temperature is raised to C00.31 , how much volume

will the gas now occupy? Assume there are no leaks.

Solution (a) According to ideal gas law TRnVp =

molesTR

Vpn 113

)27312)(314.8()47.2)(10108( 3

=+

×==

(b) Now 111 TRnVp =

1

11

p

TRnV =

litresormV 904904.010316

)27331)(314.8)(113( 331 =

×

+=

Problem 2-28

A tank of volume 35.0 m contains oxygen at an absolute

pressure of 26 /105.1 mN× and at a temperature of C020 .

Assume that oxygen behaves like an ideal gas.

(a) How many kilomoles of oxygen are there in the tank?

(b) Find the pressure if the temperature is increased to

C0500 . B.U. B.Sc. 2008S

Solution (a) According to ideal gas law TRnVp =


)27320)(314.8(

)5.0)(105.1( 6

+

×==

TR

Vpn

kilomolesmolesn 308.01008.3 2 =×=

(b) Now .1

1

2

2 constT

p

T

p== for constant V.

)105.1(27320273500 6

11

22 ×

+

+=

= p

T

Tp

262 /1096.3 mNp ×=

Problem 2-29

Oxygen gas having a volume of 31130 cm at C

042 and a

pressure of kPa101 expands until its volume is 31530 cm

and its pressure is kPa106 . Find

(a) the number of moles of oxygen in the system and

(b) its final temperature.

Solution (a) For an ideal gas iii TRnVp =

molTR

Vpn

i

ii 263

1036.4)27342)(314.8(

)101130)(10101( −−

×=+

××==

(b) Now f

ff

i

ii

T

Vp

T

Vp=

i

i

f

i

f

f TV

V

p

pT

=

KT f 6.447)27342(11301530

101106

=+

=

Problem 2-30


Calculate the value of gas constant if one mole of an ideal

gas at S.T.P. conditions occupies litres41.22 .


1135

316.8)2730)(1(

)1041.22)(10013.1( −−−

=+

××== KmolJ

Tn

VpR

Problem 2-31

Write the ideal gas law in terms of the density of the gas.

Solution The ideal gas law is

TRM

mTRnVp

==

where ‘m’ and ‘M’ are mass and molecular weight of the gas respectively. The above equation can be rewritten as

M

TR

M

TR

V

mp

ρ=

=

Problem 2-32

Calculate the density of oxygen at S.T.P. using the ideal gas.

The molecular weight of oxygen is molg /32 .


M

TRp

ρ=

TR

Mp=ρ

335

/428.1)273)(314.8(

)1032)(10013.1(mkg=

××=

−

ρ

Problem 2-33


How many molecules are in ideal gas sample at K350 that

occupies litres5.8 when the pressure is kPa180 ?

Solution According to gas law TkNVp =

Tk

VpN =

moleculesN23

23

33

10165.3)350)(10381.1(

)105.8)(10180(×=

×

××=

−

−

Problem 2-34

An auditorium has dimensions mmm 30200.10 ×× . How

many molecules of air fill the auditorium at C00.20 and a

pressure of atm1 ?


TRN

NTRnVp

A

== AN

Nn =Θ

TR

NVpN A=

)27320)(314.8()10022.6)(0.300.200.10)(10013.1( 235

+

××××=N

291050.1 ×=N molecules

Problem 2-35

A sealed flask of volume 380 cm contains argon gas at a

pressure of kPa10 and a temperature of C027 . Calculate

the number of molecules of argon gas in the vessel.



Tk

VpN =

moleculesN20

23

63

1093.1)27327)(10381.1(

)1080)(1010(×=

+×

××=

−

−

Problem 2-36

A certain vacuum pump is capable of reading the gas

pressure in a sealed container is ×001.0 standard pressure if

the temperature is maintained at K300 . Calculate the

number of molecules per 3

m in the container at this

pressure and temperature.


)300)(10381.1(

)10013.1(001.023

5

−×

××==

Tk

p

V

N

322 /1045.2 mmoleculesV

N×=

Problem 2-37

Calculate the number of molecules / m3 in an ideal gas at

S.T.P.


32523

5

/10687.2)300)(10381.1(

)10013.1(mmolecules

Tk

p

V

N×=

×

×==

−

Problem 2-38 The best vacuum attainable in the laboratory is about

Pa18100.5 −× at K293 . How many molecules are there per

3m in such a vacuum?

Solution

TRN

NTRnVp

A

== AN

Nn =Θ


TR

Np

V

N A=

32318

/1236)293)(314.8(

)10022.6)(100.5(mmolecules

V

N=

××=

−

Problem 2-39

One mole of helium gas is at room temperature ( K300 )

and one atmospheric pressure. Calculate the number of

helium atoms per unit volume.

Solution For an ideal gas

TRN

NTRnVp

A

== AN

Nn =Θ

TR

Np

V

N A=

325235

/1045.2)300)(314.8(

)10022.6)(10013.1(matoms

V

N×=

××=

Problem 2-40

A measured amount of heat is added to mol3102 −× of a

particular ideal gas to change its volume from 30.63 cm to 30.113 cm at constant pressure of atmosphere1 . Calculate

the change in temperature. P.U. B.Sc. 2006

Solution For an ideal gas at constant pressure 11 TRnVp = (1)

22 TRnVp = (2) Subtract Eq.(1) from Eq.(2)

)()( 1212 TTRnVVp −=−

Rn

VVpTTT

)( 1212

−=−=∆

KT 305)314.8)(102(

}10)63113){(10013.11(3

65

=×

×−××=∆

−

−


2- 5 THE INTERNAL ENERGY OF AN IDEAL GAS

Problem 2-41

At what temperature the relation eVTk

00.12

= holds?

Solution

Now JeVTk 1910602.100.1

2−×==

JeV1910602.100.1 −×=Θ

1923

10602.12

)10381.1( −−

×=× T

KT4

23

19

1032.210381.1

)10602.1(2×=

×

×=

−

−

Problem 2-42

What is the average kinetic energy of a molecule of a gas at

K300 ?

Solution The average kinetic energy of a molecule is given by

TkK2

3=><

JK2123 10215.6)300)(10381.1(

2

3 −− ×=×=><

Problem 2-43

What is the total random kinetic energy (in Joules) of the

molecules in one mole of a gas at temperature C027 .

Solution The total random kinetic energy of one mole of a monoatomic gas is given by

TRE2

3=

JE 3741)27327)(314.8(2

3=+=


Problem 2-44

Find the average translational kinetic energy of individual

nitrogen molecule at K1600 , in electron volts.

P.U. B.Sc. 2004

Solution The average translational kinetic energy of a nitrogen molecule is given by

TkK2

3=><

JK2023 103144.3)1600)(10381.1(

2

3 −− ×=×=><

eVK 207.010602.1

103144.319

20

=×

×=><

−

−

JeV1910602.11 −×=Θ

Problem 2-45

What is the total translational kinetic energy of mol2 of

2O molecule at C020 ?

Solution The average translational kinetic energy of a molecule of a gas is given by

TkK2

3=><

JK2123 10069.6)27320)(10381.1(

2

3 −− ×=+×=><

As there are AN2 molecules in mol2 of 2O , therefore total translational kinetic energy is

)10069.6)(10022.6(22 2123 −××=><= KNK ATOTAL

JKTOTAL 7310=

Problem 2-46

The mean kinetic energy of hydrogen molecule at C00 is

J211062.5 −× . Calculate the value of Avogadro’s number.


Solution

The mean kinetic energy is given by

TN

RTkK

A

==><

23

23

2321

10058.6)1062.5(2)273)(314.8(3

23

×=×

=><

=−K

TRN A

Problem 2-47

Calculate the total kinetic energy of the molecules of one

gram of helium gas at C00 .

Solution The mean kinetic energy of a gas molecule is given by

TkK23

=><

As molar mass of helium is 4 g / mole, therefore the number of molecules in one gram of helium gas will be

4AN

N =

Hence the total kinetic energy of molecules of the given gas sample will be

=><= Tk

NKNK A

TOTAL 23

4

TRTNkK ATOTAL 83

)(83

==

JKTOTAL 851)273)(314.8(83

==

Problem 2-48

Calculate the total rotational kinetic energy of all the

molecules in one mole of air at C025 .

Solution The internal energy of air molecules has rotational, vibrational and translational kinetic energies, therefore on average the


rotational kinetic energy will be INTERNALE31

at a given

temperature. Hence for polyatomic gas

INTERNALROTATIONAL EEK31

.).( =

JEK ROTATIONAL 2478)27325)(314.8(31

.).( =+=

Problem 2-49

Calculate the internal energy of one mole of an ideal gas at

C0250 .

Solution The internal energy of monoatomic gas is given by

TRnTkNE INTERNAL 23

23

==

JEINTERNAL 6522)273250)(314.8)(1(23

=+=

Problem 2-50

Calculate the internal energy of an ideal gas of volume 34104.3 m

−× when its pressure is kPa100 .

The internal energy of monoatomic gas is given by

TRnTkNE INTERNAL 23

23

==

VpE INTERNAL 23

= TRnVp =Θ

JEINTERNAL 51)104.3)(10100(23 43 =××= −


Problem 2-51

What is the internal energy of mol50.4 of an ideal diatomic

gas at K600 , assuming that all degrees of freedom are

active?

Solution A diatomic molecule free to translate, rotate and vibrate will have seven degrees of freedom. The internal energy of such diatomic gas will be

TRnE INTERNAL 27

=

JEINTERNAL

410857.7)600)(314.8)(50.4(27

×==

Problem 2-52

Certain excited state of hydrogen atom is found to have

energy of J1810632.1 −× above the lowest (ground) state. At

what temperature would the average translational kinetic

energy be equal to the energy of the excited state?

(Given that KeVk /106.8 5−×= ) P.U. B.Sc. 2007

Solution The average translational kinetic energy is given by

TkK23

=><

Now JK1810632.1 −×=

KJKeVk /)10602.1()106.8(/106.8 1955 −−− ×××=×=

KJk /10378.1 23−×= Hence

T)10378.1(23

10632.1 2318 −− ×=×

KT4

23

18

109.7)10378.1(3)10632.1(2

×=×

×=

−

−


Example 2-53

A mole of an ideal at K300 is subjected to a pressure of

Pa410 and its volume is 3025.0 cm . Calculate

(a) the molar gas constant R

(b) the Boltzmann constant k and

(c) the average translational kinetic energy of a molecule of

the gas.

Solution (a) According to ideal gas law

TRnVp =

115

33.8325

)300)1()025.0)(10( −−==== KmolJ

Tn

VpR

(b) The Boltzmann constant is related to gas constant as ANkR =

12323

1038.110022.6)3/25( −−×=

×== KJ

N

Rk

A

(c) The average translational kinetic energy of gas molecules is given by

TkK23

>=<

JK2123 1021.6)300)(1038.1(

23 −− ×=×>=<


2-6 THE KINETIC THEORY OF GASES

Problem 2-54

Twelve molecules have the following speeds, given in

arbitrary units:

.87,3,5,8,1,4,0,6,4,2,6 and

Calculate (a) the mean speed and (b) the r.m.s. speed.

Solution

(a) 12

873581406426 +++++++++++==

∑N

vvmean

smvmean /5.41254

==

(b) N

vvrms

∑=

2

12873581406426 222222222222 +++++++++++

=rmsv

smvrms /16.512

320==

Problem 2-55

The speeds of ten particles in m/s are

0.60.5,0.4,0.4,0.3,0.3,0.3,0.2,0.1,0 and

Find (a) the average speed (b) the root-mean-square speed

and (c) the most probable speed of these particles.

K.U. B.Sc. 1999, 2006

Solution (a) The average speed is calculated as

N

vvaverage

∑=

100.60.50.40.40.30.30.30.20.10 +++++++++

=averagev


smvaverage /1.31031

==

(b) The mean square speed is defined as

N

vvrms

∑=

2

2/1

2222

222222

)0.6()0.5()0.4()0.4(

)0.3()0.3()0.3()0.2()0.1()0(

101

++++

+++++=rmsv

smvrms /54.310

125==

(c) As the entry 3.0 has the highest frequency in the given data, therefore the most probable speed is

smv prob /0.3=

Problem 2-56

Calculate the root-mean-square speed of hydrogen molecule

at C00 and atm1 pressure assuming hydrogen to be an ideal

gas. Under these conditions for hydrogen 32 /1099.8 mkg−×=ρ . P.U. B.Sc. 2000, 2009

Solution The root-mean square speed is defined as

ρ

pvrms

3=

Now PaormNatmp 25 /10013.11 ×==

32 /1099.8 mkg−×=ρ Therefore

smvrms /18391099.8

)10013.1(32

5

=×

×=

−

Problem 2-57

At C00.44 and atm

21023.1 −× the density of a gas is 35 /1032.1 cmg−× . Find rmsv for the gas molecule.

P.U. B.Sc. 2001, 2008


Solution The root-mean-square speed is defined as

ρ

pvrms

3=

Now atmp 21023.1 −×=

PaPaormNp 3252 1024599.1/)10013.1)(1023.1( ×=××= −

35 /1032.1 cmg−×=ρ

32332

35

/1032.1/)10(

)10)(1032.1(mkgmkg

−

−

−−

×=×

=ρ

Hence

smvrms /5321032.1

)1024599.1(32

3

=×

×=

−

Problem 2-58

Calculate the root mean square velocity of Nitrogen at C00 .

Given that the density of Nitrogen at N.T.P. is 3/25.1 mkg .

B.U. B.Sc. 2002A

Solution

The rmsv is given by

smp

vrms /49325.1

)10013.1(33 5

=×

==ρ

Problem 2-59

A cylindrical container of length cm0.56 and diameter

cm5.12 holds mol350.0 of nitrogen gas at a pressure of

atm05.2 . Find the r.m.s. speed of nitrogen molecule.

Solution The root-mean-square speed of nitrogen molecule is defined as

ρ

pvrms

3= (1)

Now Paatmp )10013.1)(05.2(05.2 5×==


Pap 510077.2 ×= For calculation of density we proceed as under. 1 mol of nitrogen gas has mass kgg 3102828 −×==

0.350 mol of nitrogen gas has mass kg)1028)(350.0( 3−×=

kgm 3108.9 −×= The volume of the cylinder is given by λλ 22 )2/(DrV ππ ==

3332

22

10872.6)100.56(2

105.12mmV

−−−

×=×

×= π

The density is defined as

33

3

/426.110872.6

108.9mkg

V

m=

×

×==

−

−

ρ

Substitute the values of p and ρ in Eq.(1)

smvrms /661426.1

)10077.2(3 5

=×

=

Problem 2-60

What is the average speed and root-mean-square speed of

oxygen molecule at K300 ? Molar mass of oxygen

molekg /032.0= . K.U. B.Sc. 2007

Solution The average speed is defined as

M

TRvaverage

π

8=

smvaverage /5.445)032.0(

)300)(314.8(8==

π

The root mean square velocity is defined as

M

TRvrms

3=


smvrms /6.483032.0

)300)(314.8(3==

Problem 2-61

Root mean square velocity of the gas molecules (of certain

density and pressure) if found to be 1531 −

sm at a

temperature of C044 . Find the molecular mass of the gas.

P.U. B.Sc. 2005

Solution The root mean square velocity is defined as

M

TRvrms

3=

M

TRvrms

32 =

2

3

rmsv

TRM =

2)531(

)27344)(314.8(3 +=M

molgmolkgM /28/1080.2 2 =×= − Problem 2-62

The temperature in interstellar space is K7.2 . Find the

root-mean-square speed of hydrogen molecule at this

temperature. The molar mass of hydrogen is g2 .

Solution

The root-mean-square speed is defined as

M

TRvrms

3=

smvrms /5.183102

)7.2)(314.8(33

=×

=−

Problem 2-63


At what temperature do helium atoms have an r.m.s. speed

equal to the escape speed from the surface of Earth

( skmvESCAPE /2.11= )?


M

TRvrms

3=

M

TRvrms

32 =

R

MvT rms

3

2

=

Now skmvv ESCAPErms /2.11==

smsmvrms /1012.1/102.11 43 ×=×=

molkgmolgM /104/4 3−×==

11314.8 −−= KmolJR Therefore

KT4

324

1001.2)314.8(3

)104()1012.1(×=

××=

−

Problem 2-64

At what temperature, pressure remaining unchanged, will

the speed of hydrogen molecules be double of its value at

S.T.P.?

Solution For a given mass of gas

Tvrms ∝

Let 1v and 2v be the r.m.s. speeds of hydrogen molecules at

1T and 2T respectively, then

11 Tv ∝

22 Tv ∝ Divide second relation by first relation


1

2

1

2

T

T

v

v=

Substituting the given values in above equation

273

2 2

1

1 T

v

v=

273

2 2T=

273

4 2T=

KT 1092)273(42 == Problem 2-65

Calculate the temperature of oxygen molecules to have the

same root mean square speed as that of hydrogen molecules

at C0100 .


M

TRvrms

3=

Now

rmsv of oxygen molecule at 1T = rmsv of oxygen molecule at 2T

2

2

1

1 33

M

TR

M

TR=

2

2

1

1 33

M

TR

M

TR=

22

11 T

M

MT

=

As the molecular weights of oxygen and hydrogen are 32 and 2 respectively, therefore

162

32

2

1 ==M

M


and KKCT 173)273100(10002 =+−=−=

Hence KT 2768)173)(16(1 == Problem 2-66

At what temperature do atoms of helium have an r.m.s.

speed equal to 1.00 % of the speed of light?


M

TRvrms

3=

M

TRvrms

32 =

R

MvT rms

3

2

=

Now

ccofvrms 01.0%00.1 ==

smsmvrms /10998.2/)10998.2)(01.0( 68 ×=×=

molkgmolgM /104/4 3−×== , 11314.8 −−= KmolJR Therefore

KT9

326

10441.1)314.8(3

)104()10998.2(×=

××=

−

Problem 2-67

To increase the r.m.s. speed of a gas by 1 %, by what

percentage must the temperature increase?

Solution


M

TRvrms

3= (1)

Let TT ∆+ be the temperature at which the r.m.s. speed is rmsrms vofv %1+

rmsrmsrms vvv 01.101.0 =+=


Hence

M

TTRvrms

)(301.1

∆+= (2)

Divide Eq.(2) by Eq.(1)

T

TT ∆+=01.1

Square both sides of above equation

T

T

T

TT ∆+=

∆+= 10201.1

0201.010201.1 =−=∆

T

T

01.21000201.0100 =×=×∆

T

T

The desired percentage increase in temperature is 2.01 %. Problem 2-68 At what temperature do the atoms of helium gas have same

root mean square speed as molecules of hydrogen gas at

C027 ? The molar mass of helium is double that of

hydrogen. K.U. B.Sc. 2002

Solution


M

TRvrms

3=

where M is the molar mass. Now

rmsv of helium gas atom = rmsv of hydrogen gas molecule

2

2

1

1 33

M

TR

M

TR=

2

2

1

1

M

T

M

T=

2

2

1

1

M

T

M

T=


=

2

121

M

MTT

CKM

MT

0

2

21 327600

2)27327( ==

+=

Problem 2-69

Calculate the root-mean-square speed of smoke particles of

mass g14102.5 −× in air at C014 and atm1 pressure.

Solution The root-mean-square speed is given by

m

Tkvrms

3=

Now KJk /10381.1 23−×= KKCT 287)27314(140 =+==

kggm 1714 102.5102.5 −− ×=×= Therefore

smvrms /10521.1102.5

)287)(10381.1(3 217

23−

−

−

×=×

×=

Problem 2-70

Calculate the root mean square speed of hydrogen at C0127 .

The mass of hydrogen molecule is kg271034.3 −× .

Solution

The root-mean-square speed is given by

m

Tkvrms

3=

Now KJk /10381.1 23−×= KKCT 400)273127(1270 =+==

kgm 271034.3 −×= Therefore

smvrms /1023.21034.3

)400)(10381.1(3 327

23

×=×

×=

−

−


2-7 THE MEAN FREE PATH

Problem 2-71

What is the average distance between nitrogen molecules at

S.T.P.?

Solution The ideal gas law can be used to calculate the molecular density (i.e. number of molecules per m3) for nitrogen at S.T.P. as under. TkNVp =

Tk

p

V

N=

32523

5

/10687.2)273)(10381.1(

10013.1mmolecules

V

N×=

×

×=

−

Assuming that nitrogen molecule has a spherical shape of diameter‘d’, then

623

434 33

3 ddr

N

V πππ =

==

)/(

663

VNN

Vd

ππ==

3/1

)/(6

=VN

dπ

nmmd 142.410142.4)10687.2(

6 9

3/1

25=×=

×= −

π

the desired distance between nitrogen molecules. Problem 2-72

The mean free path of 2CO molecules at S.T.P. is measured

to be about 8106.5 −× . Estimate the diameter of 2CO

molecule.

Solution The mean free path of a gas molecule is given by


pd

Tk

22 πλ =

λπ p

Tkd

2

2 =

2/1

2

=λπ p

Tkd

md10

2/1

85

23

10868.3)106.5)(10013.1(2

)273)(10381.1( −

−

−

×=

××

×=

π

the desired diameter of a 2CO molecule. Problem 2-73

The molecular diameter of different kinds of gas molecules

can be found experimentally by measuring the rates at

which different diffuse into each other. For nitrogen,

md101015.3 −×= has been reported. What are the mean free

path and average rate of collision for nitrogen at room

temperature ( K300 ) and at atmospheric pressure?

P.U. B.Sc. 2000, 2008

Solution The number of molecules per unit volume is given by

Tk

p

pTkN

N

V

Nn ===

)/(ρ TkNVp =Θ

32523

5

/10445.2)300)(10381.1(

10013.1mmoleculesn ×=

×

×=

−ρ

The mean free path is given by

nd ρπ

λ22

1=

m8

2521010278.9

)10445.2()1015.3(2

1 −

−×=

××=

πλ

The average rate of collision is given by


Rate of collision M

TRvrms 31λλ

==

028.0

)300)(314.8(3

10278.9

18−×

=

1910572.5 −×= s Problem 2-74

Calculate the mean free path of a gas if its diameter at

S.T.P. is 2 Å.


nd ρπ

λ22

1=

pd

Tk

22 πλ =

Tk

pn =ρΘ

m7

5210

23

1009.2)10013.1()102(2

)273)(10381.1( −

−

−

×=××

×=

πλ

Problem 2-75

Calculate the diameter of benzene molecule if there are 325 /1079.2 mmolecules× and mean free path for benzene is

m8102.2 −× .

Solution The mean free path is given by

nd ρπ

λ22

1=

λρπ n

d2

12 =

2/1)2(

1

λρπ n

d =


{ } 2/1825 )102.2)(1079.2(2

1

−××

=

π

d

06.61006.6 10 =×= −md Å

the desired diameter of benzene molecule. Problem 2-76

A cubic box m20.1 on a side is evacuated so the pressure of

air inside is torr610− . Estimate how many molecular

collisions there are per each collision with a wall ( C00 ). The

diameter of air molecule is m10103 −× .


pd

Tk

22 πλ =

m7.70)10333.110()103(2

)273)(10381.1(26210

23

=×××

×=

−−

−

πλ

22 /10333.11 mNtorr ×=Θ The desired number of molecular collisions is

1210697.17.70

20.1 −−×== ma

λ

Problem 2-77

At what frequency would the wavelength of sound be of the

order of mean free path in oxygen at S.T.P.? Take the

diameter of oxygen molecule to be m10103 −× and speed of

sound in oxygen equal to sm /330 .

Solution The number of molecules per unit volume is given by

Tk

p

pTkN

N

V

Nn ===

)/(ρ TkNVp =Θ

32523

5

/10687.2)273)(10381.1(


×

×=

−ρ

The mean free path of a gas molecule is given by


nd ρπ

λ22

1=

m8

2521010305.9

)10687.2()103(2

1 −

−×=

××=

πλ

The desired frequency is

Hzv

f9

810546.3

10307.9

330×=

×==

−λ

Problem 2-78

At standard temperature and pressure (0 0C and 1.00 atm)

the mean free path in helium gas is 285 nm. Determine

(a) the number of molecules per cubic metre and

(b) the effective diameter of the helium atoms.

Solution

(c) The number of molecules per unit volume is given by

Tk

p

pTkN

N

V

Nn ===

)/(ρ TkNVp =Θ

32523

5

/10687.2)273)(10381.1(


×

×=

−ρ

(d) The mean free path is given by

nd ρπ

λ22

1=

λρπ n

d2

12 =

2/1)2(

1

λρπ n

d =

{ } 2/1925 )10285)(10687.2(2

1

−××

=

π

d

nmmd 174.010741.1 10 ≅×= −


ADDITIONAL PROBLEMS (1) Calculate the root-mean-square (r.m.s.) velocity of

nitrogen molecule at S.T.P. conditions. The density

of nitrogen at S.T.P. is 325.1 −mkg .

B.U. B.Sc. 2002A

(2) Calculate the root mean square speed of oxygen

molecule at C027 . The density of oxygen at S.T.P. is

343.1 −mkg .

(3) Calculate the r.m.s. speed of carbon dioxide

molecule at C027 . The molar mass of carbon dioxide

is g44 .

(4) Calculate the r.m.s. speed of the molecules of

nitrogen gas at C010 . The molar mass of nitrogen

molecule is kg028.0 .

(5) Calculate the number of molecules in a flask of

volume 3500 cm containing oxygen at a pressure of

kPa200 and a temperature of C027 .

ANSWERS

(1) sm /493 (2) sm /5.483 (3) sm /412 (4) sm /502 (5) molecules

221042.2 ×

Documents

Chapter 02 Thermal Properties of Matter (Pp 19-58)