chapter 02 thermal properties of matter (pp 19-58)
Post on 14-Oct-2014
133 Views
Preview:
TRANSCRIPT
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?
Solution According to Boyle’s law 1122 VpVp =
×
××=
=
−
−
6
35
2
112 10100
101)10013.1(
V
Vpp
atmmNorPap 10/10013.1 242 =×=
CH 02 THERMAL PROPERTIES OF MATTER 20
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.
Solution According to Boyle’s law 1122 VpVp =
×=
=
0
05
2
112
3)10013.1(
V
V
V
Vpp
252 /10039.3 mNorPap ×=
CH 02 THERMAL PROPERTIES OF MATTER 21
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( =
+
+=
=
CH 02 THERMAL PROPERTIES OF MATTER 22
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 ×=
CH 02 THERMAL PROPERTIES OF MATTER 23
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( ×=
+
+×=
=
CH 02 THERMAL PROPERTIES OF MATTER 24
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?
CH 02 THERMAL PROPERTIES OF MATTER 25
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
==×
=
CH 02 THERMAL PROPERTIES OF MATTER 26
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.
Solution According to ideal gas law TRnVp =
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− ?
CH 02 THERMAL PROPERTIES OF MATTER 27
Solution According to ideal gas law TRnVp =
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 =
CH 02 THERMAL PROPERTIES OF MATTER 28
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
CH 02 THERMAL PROPERTIES OF MATTER 29
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.
Solution According to ideal gas law TRnVp =
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.
CH 02 THERMAL PROPERTIES OF MATTER 30
Solution According to ideal gas law TRnVp =
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?
Solution According to ideal gas law
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
CH 02 THERMAL PROPERTIES OF MATTER 31
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 =
CH 02 THERMAL PROPERTIES OF MATTER 32
)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
CH 02 THERMAL PROPERTIES OF MATTER 33
Calculate the value of gas constant if one mole of an ideal
gas at S.T.P. conditions occupies litres41.22 .
Solution According to ideal gas law TRnVp =
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 .
Solution According to ideal gas law
M
TRp
ρ=
TR
Mp=ρ
335
/428.1)273)(314.8(
)1032)(10013.1(mkg=
××=
−
ρ
Problem 2-33
CH 02 THERMAL PROPERTIES OF MATTER 34
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 ?
Solution According to ideal gas law
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.
Solution According to ideal gas law TkNVp =
CH 02 THERMAL PROPERTIES OF MATTER 35
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.
Solution According to ideal gas law TkNVp =
)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.
Solution According to ideal gas law TkNVp =
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
CH 02 THERMAL PROPERTIES OF MATTER 36
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
=×
×−××=∆
−
−
CH 02 THERMAL PROPERTIES OF MATTER 37
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=+=
CH 02 THERMAL PROPERTIES OF MATTER 38
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.
CH 02 THERMAL PROPERTIES OF MATTER 39
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
CH 02 THERMAL PROPERTIES OF MATTER 40
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 =××= −
CH 02 THERMAL PROPERTIES OF MATTER 41
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
×=×
×=
−
−
CH 02 THERMAL PROPERTIES OF MATTER 42
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 −− ×=×>=<
CH 02 THERMAL PROPERTIES OF MATTER 43
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
CH 02 THERMAL PROPERTIES OF MATTER 44
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
CH 02 THERMAL PROPERTIES OF MATTER 45
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×==
CH 02 THERMAL PROPERTIES OF MATTER 46
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=
CH 02 THERMAL PROPERTIES OF MATTER 47
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
CH 02 THERMAL PROPERTIES OF MATTER 48
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= )?
Solution The root-mean-square speed is defined as
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
CH 02 THERMAL PROPERTIES OF MATTER 49
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 .
Solution The root-mean-square speed is defined as
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
CH 02 THERMAL PROPERTIES OF MATTER 50
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?
Solution The root-mean-square speed is defined as
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
The root-mean-square speed is defined as
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 =+=
CH 02 THERMAL PROPERTIES OF MATTER 51
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
The root-mean-square speed is defined as
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=
CH 02 THERMAL PROPERTIES OF MATTER 52
=
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
×=×
×=
−
−
CH 02 THERMAL PROPERTIES OF MATTER 53
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
CH 02 THERMAL PROPERTIES OF MATTER 54
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
CH 02 THERMAL PROPERTIES OF MATTER 55
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.
Solution The mean free path of a gas molecule is given by
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 =
CH 02 THERMAL PROPERTIES OF MATTER 56
{ } 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 −× .
Solution The mean free path of a gas molecule is given by
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(
10013.1mmoleculesn ×=
×
×=
−ρ
The mean free path of a gas molecule is given by
CH 02 THERMAL PROPERTIES OF MATTER 57
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(
15013.1mmoleculesn ×=
×
×=
−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 ≅×= −
CH 02 THERMAL PROPERTIES OF MATTER 58
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 ×
top related