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# Chapter 17 - Thermometry, Thermal Expansion - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

## NEET : Chapter 17 - Thermometry, Thermal Expansion - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

``` Page 1

17. Thermometry, Thermal
Expansion & Kinetic Theory of Gases
Introductory Exercise 17.1
1. (a)
C F
5
32
9
=
-
for F = 0, C = - ´
5
9
32
= - ° 17 8 . C
(b)
K F -
=
- 273. 5 1
5
32
9
for K = 0,
F = - ´ + = - °
9
5
273 15 32 . 459.67 F
2. (a)
x x
5
2 32
9
=
-
Þ x
x
= -
10
9
17 8 .
Þ  17.8 = -
æ
è
ç
ö
ø
÷
10
9
1 x
Þ  x = ´ = ° 17.8 C 9 160 2 .
(b)
x x
5
2 32
9
=
- /
Þ  x x = -
5
18
17 8 .
Þ   17 8
13
18
. = - x  Þ  x = - ° 24 65 . C
3.
C F -
-
=
-
-
5
99 5
32
212 32
Þ
C F -
=
- 5
94
32
180
Þ
52 5
94
32
180
-
=
- F
Þ F = + ´ = ° 32
180
94
47 122 F
4.
K F -
=
- 273 15
5
32
9
.
Þ  x x - = - 273 15
5
9
17 8 . .
Þ
4
9
255 35 x = . Þ x = 574 54 .
5.
C F
5
32
9
=
-
Þ
9
5
32 x x = -
Þ
4
5
32 x = -
Þ x = - ´ = - °
5
4
32 40 C
6. D D t t =
1
2
a q
= ´ ´ ´ ´
-
1
2
10 86400 30
5
1.2
= ´ ´ = 1.5 1.2 8.64 s 15.55 s  given.
7. As from 0°C to 4°C, den sity of wa ter
in creases so the vol ume of wooden block
above wa ter level in creases and as from
4°C to 10°C den sity of wa ter de creases
so the vol ume of block above wa ter
de cr eases .
8. V g V g
1 1 2 1
r s =
¢
and V g V g
2 2 2 2
r s =
¢
Þ
DV
V
V
V
1
1
1
1
1
1
1 1 = -
¢
= - -
s
r
and
DV
V
2
2
2
2
1 = -
s
r

\
DV
V
V
V
2
2
1
1
-
D
= -
æ
è
ç
ç
ö
ø
÷
÷
- -
æ
è
ç
ç
ö
ø
÷
÷
= - 1 1
2
2
1
1
1
1
2
2
s
r
s
r
s
r
s
r
= -
-
-
s
r
s g
r g
1
1
1 2
1 1
1
1
( )
( )
DT
T
Page 2

17. Thermometry, Thermal
Expansion & Kinetic Theory of Gases
Introductory Exercise 17.1
1. (a)
C F
5
32
9
=
-
for F = 0, C = - ´
5
9
32
= - ° 17 8 . C
(b)
K F -
=
- 273. 5 1
5
32
9
for K = 0,
F = - ´ + = - °
9
5
273 15 32 . 459.67 F
2. (a)
x x
5
2 32
9
=
-
Þ x
x
= -
10
9
17 8 .
Þ  17.8 = -
æ
è
ç
ö
ø
÷
10
9
1 x
Þ  x = ´ = ° 17.8 C 9 160 2 .
(b)
x x
5
2 32
9
=
- /
Þ  x x = -
5
18
17 8 .
Þ   17 8
13
18
. = - x  Þ  x = - ° 24 65 . C
3.
C F -
-
=
-
-
5
99 5
32
212 32
Þ
C F -
=
- 5
94
32
180
Þ
52 5
94
32
180
-
=
- F
Þ F = + ´ = ° 32
180
94
47 122 F
4.
K F -
=
- 273 15
5
32
9
.
Þ  x x - = - 273 15
5
9
17 8 . .
Þ
4
9
255 35 x = . Þ x = 574 54 .
5.
C F
5
32
9
=
-
Þ
9
5
32 x x = -
Þ
4
5
32 x = -
Þ x = - ´ = - °
5
4
32 40 C
6. D D t t =
1
2
a q
= ´ ´ ´ ´
-
1
2
10 86400 30
5
1.2
= ´ ´ = 1.5 1.2 8.64 s 15.55 s  given.
7. As from 0°C to 4°C, den sity of wa ter
in creases so the vol ume of wooden block
above wa ter level in creases and as from
4°C to 10°C den sity of wa ter de creases
so the vol ume of block above wa ter
de cr eases .
8. V g V g
1 1 2 1
r s =
¢
and V g V g
2 2 2 2
r s =
¢
Þ
DV
V
V
V
1
1
1
1
1
1
1 1 = -
¢
= - -
s
r
and
DV
V
2
2
2
2
1 = -
s
r

\
DV
V
V
V
2
2
1
1
-
D
= -
æ
è
ç
ç
ö
ø
÷
÷
- -
æ
è
ç
ç
ö
ø
÷
÷
= - 1 1
2
2
1
1
1
1
2
2
s
r
s
r
s
r
s
r
= -
-
-
s
r
s g
r g
1
1
1 2
1 1
1
1
( )
( )
DT
T
=
-
-
æ
è
ç
ç
ö
ø
÷
÷
D
s
r
g g
g
1
1
2 1
1
1
T
T
= -
-
-
s
r
s g
r g
1
1
1 2
1 1
1
1
( )
( )
DT
T
=
-
-
æ
è
ç
ç
ö
ø
÷
÷
s
r
g g
g
1
1
2 1
1
1
T
T D
9. On cool ing brass con tracts more than
iron ( ) a a
B
>
Fe
such that brass disk gets
loosen from hole of iron.
10. V T µ Þ V kT =  Þ ln ln ln V k T = +
Þ
D D V
V
T
T
=  Þ
D
D
V
V T T
= =
1
g
Introductory Exercise 17.2
1. For ideal gases, pV nRT =
Þ  T
V
nR
p = =
VM
mR
p
Slope =
VM
mR
As slope µ
1
m
Þ m m
2 1
<
2. pV nRT = Þ
p
p
T
T
2
1
2
1
=
= =
360
300
6
5
Þ p p
2 1
6
5
6
5
10 12 = = ´ = atm atm
3. M
mix
=
´ + ´
+
=
+
=
1
4
28
1
4
44
1
4
1
4
7 11
1
2
36
pV nRT
m
M
RT = =
Þ pM
m
V
RT RT = = r Þ r =
pM
RT
\              r =
´ ´ ´
´
-
101 10 36 10
8 31 290
5 3
.
.
=
´
´
=
101 36
8 31 29
15
.
. .
. kg/m
3
4. pV nRT
N
N
RT
A
= =
Þ N
pVN
RT
A
=
=
´ ´ ´ ´ ´
´ ´
´
- -
10 10 10 250 10
10
300
6 3 6
23
13.6
6.02
8.31
=
´ ´
´
´ = ´
13.6 6.02
8.31
8.21
5
6
10 10
15 15
5. pV nRT =
Þ   V
nR
p
T = ×
Slope µ
1
r
Þ p p
1 2
>
6. pV nRT = Þ p nRT
V
= ( )
1
Þ y mx = is a straight line passing
through origin.
Introductory Exercise  17.3
1. Av er age ve loc ity de pends on the
di rec tion of mo tion of gas mol e cules and
as con tainer do not move such that their
net ef fect be comes zero, due to the
rea son that some mol e cules are mov ing
in one di rec tion while other are mov ing
in op po site di rec tion. But in case of
av er age speed only mag ni tudes are in
use which do not cancel each other.
38  |  Thermometry, Thermal Expansion & Kinetic Theory of Gases
Page 3

17. Thermometry, Thermal
Expansion & Kinetic Theory of Gases
Introductory Exercise 17.1
1. (a)
C F
5
32
9
=
-
for F = 0, C = - ´
5
9
32
= - ° 17 8 . C
(b)
K F -
=
- 273. 5 1
5
32
9
for K = 0,
F = - ´ + = - °
9
5
273 15 32 . 459.67 F
2. (a)
x x
5
2 32
9
=
-
Þ x
x
= -
10
9
17 8 .
Þ  17.8 = -
æ
è
ç
ö
ø
÷
10
9
1 x
Þ  x = ´ = ° 17.8 C 9 160 2 .
(b)
x x
5
2 32
9
=
- /
Þ  x x = -
5
18
17 8 .
Þ   17 8
13
18
. = - x  Þ  x = - ° 24 65 . C
3.
C F -
-
=
-
-
5
99 5
32
212 32
Þ
C F -
=
- 5
94
32
180
Þ
52 5
94
32
180
-
=
- F
Þ F = + ´ = ° 32
180
94
47 122 F
4.
K F -
=
- 273 15
5
32
9
.
Þ  x x - = - 273 15
5
9
17 8 . .
Þ
4
9
255 35 x = . Þ x = 574 54 .
5.
C F
5
32
9
=
-
Þ
9
5
32 x x = -
Þ
4
5
32 x = -
Þ x = - ´ = - °
5
4
32 40 C
6. D D t t =
1
2
a q
= ´ ´ ´ ´
-
1
2
10 86400 30
5
1.2
= ´ ´ = 1.5 1.2 8.64 s 15.55 s  given.
7. As from 0°C to 4°C, den sity of wa ter
in creases so the vol ume of wooden block
above wa ter level in creases and as from
4°C to 10°C den sity of wa ter de creases
so the vol ume of block above wa ter
de cr eases .
8. V g V g
1 1 2 1
r s =
¢
and V g V g
2 2 2 2
r s =
¢
Þ
DV
V
V
V
1
1
1
1
1
1
1 1 = -
¢
= - -
s
r
and
DV
V
2
2
2
2
1 = -
s
r

\
DV
V
V
V
2
2
1
1
-
D
= -
æ
è
ç
ç
ö
ø
÷
÷
- -
æ
è
ç
ç
ö
ø
÷
÷
= - 1 1
2
2
1
1
1
1
2
2
s
r
s
r
s
r
s
r
= -
-
-
s
r
s g
r g
1
1
1 2
1 1
1
1
( )
( )
DT
T
=
-
-
æ
è
ç
ç
ö
ø
÷
÷
D
s
r
g g
g
1
1
2 1
1
1
T
T
= -
-
-
s
r
s g
r g
1
1
1 2
1 1
1
1
( )
( )
DT
T
=
-
-
æ
è
ç
ç
ö
ø
÷
÷
s
r
g g
g
1
1
2 1
1
1
T
T D
9. On cool ing brass con tracts more than
iron ( ) a a
B
>
Fe
such that brass disk gets
loosen from hole of iron.
10. V T µ Þ V kT =  Þ ln ln ln V k T = +
Þ
D D V
V
T
T
=  Þ
D
D
V
V T T
= =
1
g
Introductory Exercise 17.2
1. For ideal gases, pV nRT =
Þ  T
V
nR
p = =
VM
mR
p
Slope =
VM
mR
As slope µ
1
m
Þ m m
2 1
<
2. pV nRT = Þ
p
p
T
T
2
1
2
1
=
= =
360
300
6
5
Þ p p
2 1
6
5
6
5
10 12 = = ´ = atm atm
3. M
mix
=
´ + ´
+
=
+
=
1
4
28
1
4
44
1
4
1
4
7 11
1
2
36
pV nRT
m
M
RT = =
Þ pM
m
V
RT RT = = r Þ r =
pM
RT
\              r =
´ ´ ´
´
-
101 10 36 10
8 31 290
5 3
.
.
=
´
´
=
101 36
8 31 29
15
.
. .
. kg/m
3
4. pV nRT
N
N
RT
A
= =
Þ N
pVN
RT
A
=
=
´ ´ ´ ´ ´
´ ´
´
- -
10 10 10 250 10
10
300
6 3 6
23
13.6
6.02
8.31
=
´ ´
´
´ = ´
13.6 6.02
8.31
8.21
5
6
10 10
15 15
5. pV nRT =
Þ   V
nR
p
T = ×
Slope µ
1
r
Þ p p
1 2
>
6. pV nRT = Þ p nRT
V
= ( )
1
Þ y mx = is a straight line passing
through origin.
Introductory Exercise  17.3
1. Av er age ve loc ity de pends on the
di rec tion of mo tion of gas mol e cules and
as con tainer do not move such that their
net ef fect be comes zero, due to the
rea son that some mol e cules are mov ing
in one di rec tion while other are mov ing
in op po site di rec tion. But in case of
av er age speed only mag ni tudes are in
use which do not cancel each other.
38  |  Thermometry, Thermal Expansion & Kinetic Theory of Gases
2.   KE =
3
2
kT = ´
´
´
3
2 6 10
300
23
8.31
J
= ´ ´
-
3
4
10
21
8.31 J
= ´
-
6.21 J 10
21

3. v
RT
M
rms
=
3
,
v
He
8.31
=
´ ´
´
-
3 300
4 10
3
= ´ 1.37 m/s 10
3
v
Ne
8.31
20.2
608.5 =
´ ´
´
=
-
3 300
10
3
m/s
KE 6.21 J = = ´
-
3
2
10
21
kT
4. v
RT
M
rms
=
3

Þ T
Mv
R
= =
´ ´
´
-
rms
8.31
2 3 6
3
4 10 10
3
= 160.45 K
5. r
r r r r
=
+
+
=
- +
- +
n n
n n
n n
n n
1 1 2 2
1 2
2 1 2 2
2 2
1
1
( )
= + - r r r
1 2 2 1
n ( )
Þ  n
2
1
2 1
=
-
-
=
-
-
r r
r r
1.293 1.429
1.251 1.429
= = =
136
178
0.764 76.4% by mass
6.
V
V
p T
p T
p h g
p
2
1
1 2
2 1
0
0
277
= =
+
´
( ) r
=
´ + ´ ´ ´
´ ´
( ) 1.01
1.01
10 40 10 10 293
10 277
5 3
5
=
´
´
=
5.01
1.01
5.25
293
277
Þ V V
2 1
3
105 = = 5.25 cm
7. N nN
A
= = ´ ´
1
18
6 10
23
= ´
1
3
10
23
;
S R = = ´ ´ ´ ´ 4 4 6400 10 10
2 3 2 2
p 3.14 ( )
= ´ 5.14 cm 10
18 2
\
N
S
=
´ ´
10
3 10
23
18
5.14

= ´ 6.5 10
3
molecules/cm
2
(a)   nC nR
V
= =
3
2
35 J/K
Þ n
R
= =
70
3
2.8 mole
(b) U nRT = = ´ =
3
2
35 273 J/K K 9555 J
(c) C C R R
p V
= + = =
5
2
20.8 J/ K mole
8. (a) n C C nR
p V
( ) - = = 29.1 J/K
Þ n =
29.1
8.314
mole = 3.5 mole
(b) C nc n R
V V
= = = ´ ´
3
2
3.5 1.5 8.314
= 43.65 J/K
C nc n R C nR
p p V
= = = +
5
2
= + ´ 43.65 3.5 8.314
= 72.75 J/K
(c) C nc
V V
¢ = = ´ = n R
5
3
72.75 J/K
C nc n R
p p
¢ = = ´
7
2
= + ´ 72.75 3.5 8.314
= 101.85 J/K
10. v
RT
M
rms
=
3
and v
RT
M
av
=
8
p

Here 3
8
>
p
Þ v v
rms av
> ,
i e . ., the statement is true.
Thermometry, Thermal Expansion & Kinetic Theory of Gases   | 39
Page 4

17. Thermometry, Thermal
Expansion & Kinetic Theory of Gases
Introductory Exercise 17.1
1. (a)
C F
5
32
9
=
-
for F = 0, C = - ´
5
9
32
= - ° 17 8 . C
(b)
K F -
=
- 273. 5 1
5
32
9
for K = 0,
F = - ´ + = - °
9
5
273 15 32 . 459.67 F
2. (a)
x x
5
2 32
9
=
-
Þ x
x
= -
10
9
17 8 .
Þ  17.8 = -
æ
è
ç
ö
ø
÷
10
9
1 x
Þ  x = ´ = ° 17.8 C 9 160 2 .
(b)
x x
5
2 32
9
=
- /
Þ  x x = -
5
18
17 8 .
Þ   17 8
13
18
. = - x  Þ  x = - ° 24 65 . C
3.
C F -
-
=
-
-
5
99 5
32
212 32
Þ
C F -
=
- 5
94
32
180
Þ
52 5
94
32
180
-
=
- F
Þ F = + ´ = ° 32
180
94
47 122 F
4.
K F -
=
- 273 15
5
32
9
.
Þ  x x - = - 273 15
5
9
17 8 . .
Þ
4
9
255 35 x = . Þ x = 574 54 .
5.
C F
5
32
9
=
-
Þ
9
5
32 x x = -
Þ
4
5
32 x = -
Þ x = - ´ = - °
5
4
32 40 C
6. D D t t =
1
2
a q
= ´ ´ ´ ´
-
1
2
10 86400 30
5
1.2
= ´ ´ = 1.5 1.2 8.64 s 15.55 s  given.
7. As from 0°C to 4°C, den sity of wa ter
in creases so the vol ume of wooden block
above wa ter level in creases and as from
4°C to 10°C den sity of wa ter de creases
so the vol ume of block above wa ter
de cr eases .
8. V g V g
1 1 2 1
r s =
¢
and V g V g
2 2 2 2
r s =
¢
Þ
DV
V
V
V
1
1
1
1
1
1
1 1 = -
¢
= - -
s
r
and
DV
V
2
2
2
2
1 = -
s
r

\
DV
V
V
V
2
2
1
1
-
D
= -
æ
è
ç
ç
ö
ø
÷
÷
- -
æ
è
ç
ç
ö
ø
÷
÷
= - 1 1
2
2
1
1
1
1
2
2
s
r
s
r
s
r
s
r
= -
-
-
s
r
s g
r g
1
1
1 2
1 1
1
1
( )
( )
DT
T
=
-
-
æ
è
ç
ç
ö
ø
÷
÷
D
s
r
g g
g
1
1
2 1
1
1
T
T
= -
-
-
s
r
s g
r g
1
1
1 2
1 1
1
1
( )
( )
DT
T
=
-
-
æ
è
ç
ç
ö
ø
÷
÷
s
r
g g
g
1
1
2 1
1
1
T
T D
9. On cool ing brass con tracts more than
iron ( ) a a
B
>
Fe
such that brass disk gets
loosen from hole of iron.
10. V T µ Þ V kT =  Þ ln ln ln V k T = +
Þ
D D V
V
T
T
=  Þ
D
D
V
V T T
= =
1
g
Introductory Exercise 17.2
1. For ideal gases, pV nRT =
Þ  T
V
nR
p = =
VM
mR
p
Slope =
VM
mR
As slope µ
1
m
Þ m m
2 1
<
2. pV nRT = Þ
p
p
T
T
2
1
2
1
=
= =
360
300
6
5
Þ p p
2 1
6
5
6
5
10 12 = = ´ = atm atm
3. M
mix
=
´ + ´
+
=
+
=
1
4
28
1
4
44
1
4
1
4
7 11
1
2
36
pV nRT
m
M
RT = =
Þ pM
m
V
RT RT = = r Þ r =
pM
RT
\              r =
´ ´ ´
´
-
101 10 36 10
8 31 290
5 3
.
.
=
´
´
=
101 36
8 31 29
15
.
. .
. kg/m
3
4. pV nRT
N
N
RT
A
= =
Þ N
pVN
RT
A
=
=
´ ´ ´ ´ ´
´ ´
´
- -
10 10 10 250 10
10
300
6 3 6
23
13.6
6.02
8.31
=
´ ´
´
´ = ´
13.6 6.02
8.31
8.21
5
6
10 10
15 15
5. pV nRT =
Þ   V
nR
p
T = ×
Slope µ
1
r
Þ p p
1 2
>
6. pV nRT = Þ p nRT
V
= ( )
1
Þ y mx = is a straight line passing
through origin.
Introductory Exercise  17.3
1. Av er age ve loc ity de pends on the
di rec tion of mo tion of gas mol e cules and
as con tainer do not move such that their
net ef fect be comes zero, due to the
rea son that some mol e cules are mov ing
in one di rec tion while other are mov ing
in op po site di rec tion. But in case of
av er age speed only mag ni tudes are in
use which do not cancel each other.
38  |  Thermometry, Thermal Expansion & Kinetic Theory of Gases
2.   KE =
3
2
kT = ´
´
´
3
2 6 10
300
23
8.31
J
= ´ ´
-
3
4
10
21
8.31 J
= ´
-
6.21 J 10
21

3. v
RT
M
rms
=
3
,
v
He
8.31
=
´ ´
´
-
3 300
4 10
3
= ´ 1.37 m/s 10
3
v
Ne
8.31
20.2
608.5 =
´ ´
´
=
-
3 300
10
3
m/s
KE 6.21 J = = ´
-
3
2
10
21
kT
4. v
RT
M
rms
=
3

Þ T
Mv
R
= =
´ ´
´
-
rms
8.31
2 3 6
3
4 10 10
3
= 160.45 K
5. r
r r r r
=
+
+
=
- +
- +
n n
n n
n n
n n
1 1 2 2
1 2
2 1 2 2
2 2
1
1
( )
= + - r r r
1 2 2 1
n ( )
Þ  n
2
1
2 1
=
-
-
=
-
-
r r
r r
1.293 1.429
1.251 1.429
= = =
136
178
0.764 76.4% by mass
6.
V
V
p T
p T
p h g
p
2
1
1 2
2 1
0
0
277
= =
+
´
( ) r
=
´ + ´ ´ ´
´ ´
( ) 1.01
1.01
10 40 10 10 293
10 277
5 3
5
=
´
´
=
5.01
1.01
5.25
293
277
Þ V V
2 1
3
105 = = 5.25 cm
7. N nN
A
= = ´ ´
1
18
6 10
23
= ´
1
3
10
23
;
S R = = ´ ´ ´ ´ 4 4 6400 10 10
2 3 2 2
p 3.14 ( )
= ´ 5.14 cm 10
18 2
\
N
S
=
´ ´
10
3 10
23
18
5.14

= ´ 6.5 10
3
molecules/cm
2
(a)   nC nR
V
= =
3
2
35 J/K
Þ n
R
= =
70
3
2.8 mole
(b) U nRT = = ´ =
3
2
35 273 J/K K 9555 J
(c) C C R R
p V
= + = =
5
2
20.8 J/ K mole
8. (a) n C C nR
p V
( ) - = = 29.1 J/K
Þ n =
29.1
8.314
mole = 3.5 mole
(b) C nc n R
V V
= = = ´ ´
3
2
3.5 1.5 8.314
= 43.65 J/K
C nc n R C nR
p p V
= = = +
5
2
= + ´ 43.65 3.5 8.314
= 72.75 J/K
(c) C nc
V V
¢ = = ´ = n R
5
3
72.75 J/K
C nc n R
p p
¢ = = ´
7
2
= + ´ 72.75 3.5 8.314
= 101.85 J/K
10. v
RT
M
rms
=
3
and v
RT
M
av
=
8
p

Here 3
8
>
p
Þ v v
rms av
> ,
i e . ., the statement is true.
Thermometry, Thermal Expansion & Kinetic Theory of Gases   | 39
AIEEE Corner
¢ Subjective Questions (Level 1)
1.
C¢
=
-
= =
5
68 32
9
36
9
4
Þ C¢ = ° 20 C ;
K¢ -
=
-
=
273
5
68 32
9
4
Þ K¢ = 293 K
C¢
=
-
= - = -
5
5 32
9
27
9
3
Þ C¢ = - ° 15 C;
K¢ -
=
-
= -
273
5
5 32
9
3
Þ K¢ =258 K
C¢
=
-
= =
5
176 32
9
144
9
16
Þ C¢ = ° 80 C;
K¢ -
=
273
5
16
Þ K¢ =353 K
2.
30
5
32
9
=
¢ - F
Þ F¢ = + = ° 54 32 86 F
= ° 546 R
5
5
32
9
=
¢ - F
Þ F¢ = + = ° 9 32 41 F = ° 501 R
- =
¢ - 20
5
32
9
F
Þ     F¢ = - + = - ° 36 32 41 F
= ° 456 R
3.
x x
5
32
9
=
-
Þ 32
9
5
4
5
= - = - x x x
Þ x = - ´ = - °
5
4
32 40
Þ - ° = - ° 40 40 C F
4.
D D C F
5 9
= Þ D D F C = = ´ = °
9
5
9
5
40 72
\ F F
2 1
72 140 2 = + ° = ° . F
5.
32 20
80 20
0
100 0
-
-
=
¢ -
-
C

Þ
12
60 100
=
¢ C
Þ C¢ =
´
= °
12 100
60
20 C
6.
T
T
p
p
2
1
2
1
160
80
2 = = =   Þ T T
2 1
2 =
\   T
2
2 = ´ = 273.15 K 546.30 K
7. R R
t
= +
0
1 ( ) a q D
Þ 3 50 250 1 100 . . ( ) = + a Þ 1 250 = K
or a = = ´
-
10
250
4 10
3
.
/°C
\   650 250 1 4 10
3
. . ( ) = + ´
-
Dq
Þ    4 10
2
= ´
-
Dq
Þ   Dq = 400   Þ    q
2
400 = °C
Þ    Dq = 400   Þ     q
2
400 = °C
i e . ., boiling point of sulphur is 400°C.
8.
T
T
p
p
2
1
2
1
75 45
75 5
120
80
3
2
= =
+
+
= =
T T
2 1
3
2
3
2
30015 = = ´ . K
= = ° 450225 . K 177.08 C
9. D D g g a a q = - ( )
Br Fe
Þ    D
D
q
g
g a a
= ×
-
1
Br Fe

=
´
´
×
-
-
-
0.01
Br Fe
10
6 10
1
3
2
a a
=
-
-
10
6
3
( ) a a
Br Fe
\    q q
a a
2 1
3
10
6
= +
-
-
( )
Br Fe
= ° +
-
= ° +
´
-
30
10
6
30
100
6
3
C C
0.63
Br Fe
( ) a a
= ° 57.78 C .
10. (a) D D l l = - ´ ´ ´
-
a q
~
88.42 2.4 10 30
5

= 0.064 cm
(b) D D l l = - ( ) a a q
Al St
= - ´ ´
-
88.42 2.4 1. ( ) 2 10 30
5

= 0.032 cm
l l l
S
= + = + D 88.42 0.032 cm
= 88.45 cm
40  |  Thermometry, Thermal Expansion & Kinetic Theory of Gases
Page 5

17. Thermometry, Thermal
Expansion & Kinetic Theory of Gases
Introductory Exercise 17.1
1. (a)
C F
5
32
9
=
-
for F = 0, C = - ´
5
9
32
= - ° 17 8 . C
(b)
K F -
=
- 273. 5 1
5
32
9
for K = 0,
F = - ´ + = - °
9
5
273 15 32 . 459.67 F
2. (a)
x x
5
2 32
9
=
-
Þ x
x
= -
10
9
17 8 .
Þ  17.8 = -
æ
è
ç
ö
ø
÷
10
9
1 x
Þ  x = ´ = ° 17.8 C 9 160 2 .
(b)
x x
5
2 32
9
=
- /
Þ  x x = -
5
18
17 8 .
Þ   17 8
13
18
. = - x  Þ  x = - ° 24 65 . C
3.
C F -
-
=
-
-
5
99 5
32
212 32
Þ
C F -
=
- 5
94
32
180
Þ
52 5
94
32
180
-
=
- F
Þ F = + ´ = ° 32
180
94
47 122 F
4.
K F -
=
- 273 15
5
32
9
.
Þ  x x - = - 273 15
5
9
17 8 . .
Þ
4
9
255 35 x = . Þ x = 574 54 .
5.
C F
5
32
9
=
-
Þ
9
5
32 x x = -
Þ
4
5
32 x = -
Þ x = - ´ = - °
5
4
32 40 C
6. D D t t =
1
2
a q
= ´ ´ ´ ´
-
1
2
10 86400 30
5
1.2
= ´ ´ = 1.5 1.2 8.64 s 15.55 s  given.
7. As from 0°C to 4°C, den sity of wa ter
in creases so the vol ume of wooden block
above wa ter level in creases and as from
4°C to 10°C den sity of wa ter de creases
so the vol ume of block above wa ter
de cr eases .
8. V g V g
1 1 2 1
r s =
¢
and V g V g
2 2 2 2
r s =
¢
Þ
DV
V
V
V
1
1
1
1
1
1
1 1 = -
¢
= - -
s
r
and
DV
V
2
2
2
2
1 = -
s
r

\
DV
V
V
V
2
2
1
1
-
D
= -
æ
è
ç
ç
ö
ø
÷
÷
- -
æ
è
ç
ç
ö
ø
÷
÷
= - 1 1
2
2
1
1
1
1
2
2
s
r
s
r
s
r
s
r
= -
-
-
s
r
s g
r g
1
1
1 2
1 1
1
1
( )
( )
DT
T
=
-
-
æ
è
ç
ç
ö
ø
÷
÷
D
s
r
g g
g
1
1
2 1
1
1
T
T
= -
-
-
s
r
s g
r g
1
1
1 2
1 1
1
1
( )
( )
DT
T
=
-
-
æ
è
ç
ç
ö
ø
÷
÷
s
r
g g
g
1
1
2 1
1
1
T
T D
9. On cool ing brass con tracts more than
iron ( ) a a
B
>
Fe
such that brass disk gets
loosen from hole of iron.
10. V T µ Þ V kT =  Þ ln ln ln V k T = +
Þ
D D V
V
T
T
=  Þ
D
D
V
V T T
= =
1
g
Introductory Exercise 17.2
1. For ideal gases, pV nRT =
Þ  T
V
nR
p = =
VM
mR
p
Slope =
VM
mR
As slope µ
1
m
Þ m m
2 1
<
2. pV nRT = Þ
p
p
T
T
2
1
2
1
=
= =
360
300
6
5
Þ p p
2 1
6
5
6
5
10 12 = = ´ = atm atm
3. M
mix
=
´ + ´
+
=
+
=
1
4
28
1
4
44
1
4
1
4
7 11
1
2
36
pV nRT
m
M
RT = =
Þ pM
m
V
RT RT = = r Þ r =
pM
RT
\              r =
´ ´ ´
´
-
101 10 36 10
8 31 290
5 3
.
.
=
´
´
=
101 36
8 31 29
15
.
. .
. kg/m
3
4. pV nRT
N
N
RT
A
= =
Þ N
pVN
RT
A
=
=
´ ´ ´ ´ ´
´ ´
´
- -
10 10 10 250 10
10
300
6 3 6
23
13.6
6.02
8.31
=
´ ´
´
´ = ´
13.6 6.02
8.31
8.21
5
6
10 10
15 15
5. pV nRT =
Þ   V
nR
p
T = ×
Slope µ
1
r
Þ p p
1 2
>
6. pV nRT = Þ p nRT
V
= ( )
1
Þ y mx = is a straight line passing
through origin.
Introductory Exercise  17.3
1. Av er age ve loc ity de pends on the
di rec tion of mo tion of gas mol e cules and
as con tainer do not move such that their
net ef fect be comes zero, due to the
rea son that some mol e cules are mov ing
in one di rec tion while other are mov ing
in op po site di rec tion. But in case of
av er age speed only mag ni tudes are in
use which do not cancel each other.
38  |  Thermometry, Thermal Expansion & Kinetic Theory of Gases
2.   KE =
3
2
kT = ´
´
´
3
2 6 10
300
23
8.31
J
= ´ ´
-
3
4
10
21
8.31 J
= ´
-
6.21 J 10
21

3. v
RT
M
rms
=
3
,
v
He
8.31
=
´ ´
´
-
3 300
4 10
3
= ´ 1.37 m/s 10
3
v
Ne
8.31
20.2
608.5 =
´ ´
´
=
-
3 300
10
3
m/s
KE 6.21 J = = ´
-
3
2
10
21
kT
4. v
RT
M
rms
=
3

Þ T
Mv
R
= =
´ ´
´
-
rms
8.31
2 3 6
3
4 10 10
3
= 160.45 K
5. r
r r r r
=
+
+
=
- +
- +
n n
n n
n n
n n
1 1 2 2
1 2
2 1 2 2
2 2
1
1
( )
= + - r r r
1 2 2 1
n ( )
Þ  n
2
1
2 1
=
-
-
=
-
-
r r
r r
1.293 1.429
1.251 1.429
= = =
136
178
0.764 76.4% by mass
6.
V
V
p T
p T
p h g
p
2
1
1 2
2 1
0
0
277
= =
+
´
( ) r
=
´ + ´ ´ ´
´ ´
( ) 1.01
1.01
10 40 10 10 293
10 277
5 3
5
=
´
´
=
5.01
1.01
5.25
293
277
Þ V V
2 1
3
105 = = 5.25 cm
7. N nN
A
= = ´ ´
1
18
6 10
23
= ´
1
3
10
23
;
S R = = ´ ´ ´ ´ 4 4 6400 10 10
2 3 2 2
p 3.14 ( )
= ´ 5.14 cm 10
18 2
\
N
S
=
´ ´
10
3 10
23
18
5.14

= ´ 6.5 10
3
molecules/cm
2
(a)   nC nR
V
= =
3
2
35 J/K
Þ n
R
= =
70
3
2.8 mole
(b) U nRT = = ´ =
3
2
35 273 J/K K 9555 J
(c) C C R R
p V
= + = =
5
2
20.8 J/ K mole
8. (a) n C C nR
p V
( ) - = = 29.1 J/K
Þ n =
29.1
8.314
mole = 3.5 mole
(b) C nc n R
V V
= = = ´ ´
3
2
3.5 1.5 8.314
= 43.65 J/K
C nc n R C nR
p p V
= = = +
5
2
= + ´ 43.65 3.5 8.314
= 72.75 J/K
(c) C nc
V V
¢ = = ´ = n R
5
3
72.75 J/K
C nc n R
p p
¢ = = ´
7
2
= + ´ 72.75 3.5 8.314
= 101.85 J/K
10. v
RT
M
rms
=
3
and v
RT
M
av
=
8
p

Here 3
8
>
p
Þ v v
rms av
> ,
i e . ., the statement is true.
Thermometry, Thermal Expansion & Kinetic Theory of Gases   | 39
AIEEE Corner
¢ Subjective Questions (Level 1)
1.
C¢
=
-
= =
5
68 32
9
36
9
4
Þ C¢ = ° 20 C ;
K¢ -
=
-
=
273
5
68 32
9
4
Þ K¢ = 293 K
C¢
=
-
= - = -
5
5 32
9
27
9
3
Þ C¢ = - ° 15 C;
K¢ -
=
-
= -
273
5
5 32
9
3
Þ K¢ =258 K
C¢
=
-
= =
5
176 32
9
144
9
16
Þ C¢ = ° 80 C;
K¢ -
=
273
5
16
Þ K¢ =353 K
2.
30
5
32
9
=
¢ - F
Þ F¢ = + = ° 54 32 86 F
= ° 546 R
5
5
32
9
=
¢ - F
Þ F¢ = + = ° 9 32 41 F = ° 501 R
- =
¢ - 20
5
32
9
F
Þ     F¢ = - + = - ° 36 32 41 F
= ° 456 R
3.
x x
5
32
9
=
-
Þ 32
9
5
4
5
= - = - x x x
Þ x = - ´ = - °
5
4
32 40
Þ - ° = - ° 40 40 C F
4.
D D C F
5 9
= Þ D D F C = = ´ = °
9
5
9
5
40 72
\ F F
2 1
72 140 2 = + ° = ° . F
5.
32 20
80 20
0
100 0
-
-
=
¢ -
-
C

Þ
12
60 100
=
¢ C
Þ C¢ =
´
= °
12 100
60
20 C
6.
T
T
p
p
2
1
2
1
160
80
2 = = =   Þ T T
2 1
2 =
\   T
2
2 = ´ = 273.15 K 546.30 K
7. R R
t
= +
0
1 ( ) a q D
Þ 3 50 250 1 100 . . ( ) = + a Þ 1 250 = K
or a = = ´
-
10
250
4 10
3
.
/°C
\   650 250 1 4 10
3
. . ( ) = + ´
-
Dq
Þ    4 10
2
= ´
-
Dq
Þ   Dq = 400   Þ    q
2
400 = °C
Þ    Dq = 400   Þ     q
2
400 = °C
i e . ., boiling point of sulphur is 400°C.
8.
T
T
p
p
2
1
2
1
75 45
75 5
120
80
3
2
= =
+
+
= =
T T
2 1
3
2
3
2
30015 = = ´ . K
= = ° 450225 . K 177.08 C
9. D D g g a a q = - ( )
Br Fe
Þ    D
D
q
g
g a a
= ×
-
1
Br Fe

=
´
´
×
-
-
-
0.01
Br Fe
10
6 10
1
3
2
a a
=
-
-
10
6
3
( ) a a
Br Fe
\    q q
a a
2 1
3
10
6
= +
-
-
( )
Br Fe
= ° +
-
= ° +
´
-
30
10
6
30
100
6
3
C C
0.63
Br Fe
( ) a a
= ° 57.78 C .
10. (a) D D l l = - ´ ´ ´
-
a q
~
88.42 2.4 10 30
5

= 0.064 cm
(b) D D l l = - ( ) a a q
Al St
= - ´ ´
-
88.42 2.4 1. ( ) 2 10 30
5

= 0.032 cm
l l l
S
= + = + D 88.42 0.032 cm
= 88.45 cm
40  |  Thermometry, Thermal Expansion & Kinetic Theory of Gases
11.
D
D
l
l
´ = ´ 100 100 % % a q
= - ´ ´ ´
-
1.2 10 35 100
5
%
= - 0.042%
12. F YA
l
l
YA = =
D
D a q
= ´ 2 10
11
´ ´ ´ ´ ´
- -
2 10 10 40
6 5
1.2
= ´ ´ = ´ = 4 40 160 192 1.2 N 1.2 N N
13. V g s = - ´
-
( ) 50 45 10
3
kg
= ´
-
5 10
3
kg
V g ¢ ¢ = - ´
-
s ( . ) 50 451 10
3
kg
= ´
-
4.9 kg 10
3
V g
s
l
( ) 1
1
10
3
+
+
= ´
-
g q
s
g q
D
D
4.9
1
1
4 9
5
+
+
=
g q
g q
s
l
D
D
.

Þ 5 5 + = + g q g q
s e
D D 4.9 4.9
g
g q
q q
g
l
s
s
=
+
= +
0.1
4.9 4.9
5 1
49
5 D
D D
g
s
=
´
+ ´ ´
-
1
49 75
5
4 9
12 10
6
.
= ´ + ´
- -
272.1 12.2 10 10
6 6
= ´ °
-
2.84 C 10
4
14. M = + = 14 3 17 g/mole
= ´
-
17 10
3
kg/mole
Þ M =
´
´
-
-
17 10
6033 10
3
23
.
kg/molecule
= ´
-
282 10
26
. kg/molecule
15. n
pV
RT
= =
´ ´
´
=
-
1.52
8.314 298.15
6.13
10 10
6 2

r = = =
´ ´
-
-
m
V
nM
V
6.13 2 10
10
3
2
= 1.23 kg/m
3
r r ¢ =
¢
=
¢
= =
m
V
nM
V
nM
V
16
16
= 19 62
3
. kg/m
16. p p
V
V
2 1
1
2
1
76
6
= = ´ atm
= 12.7 atm
17. V
p V
T
T
p
p
p
T
T
V
2
1 1
1
2
2
1
2
2
1
1
= × = × ×
= ´ ´
1 270
300
500
3
0.5
m = 900
3
m
18.
p V
T
p V
T
1 1
1
2 2
2
=
Þ
mg
A
p A h
mg
A
p Ah
i f
+
æ
è
ç
ö
ø
÷
×
=
+
æ
è
ç
ö
ø
÷
0 0
293 273
Þ h h
f i
= = ´ =
373
293
373
293
4 cm 50.9 cm
19. p p
1 2
= Þ
n
V
n
V L A L A
1
1
2
2 1 2
25 28 40 4
= = =
/ /

Þ
L
L
1
2
25
28
1
10
5
56
= ´ = = 0.089
n
n
1
2
25 28
40 4
25
280
5
56
= = = =
/
/
0.089
20. n n n = +
1 2

Þ p V V p V p V ( )
1 2 1 1 2 2
+ = +
Þ p
p V p V
V V
=
+
+
1 1 2 2
1 2
\ p =
´ + ´
+
1.38 0.11 0.69 0.16
0.11 0.16
MP
a
=
+
=
0.1518 0.1104
0.27
0.2622
0.27
= 0.97 MP
a
21.
pV
T
pV
T
p V
T
p V
T
1 2 1 1
1
1 2
2
+ = +
1
293
600
3
atm
K
cm ´
= +
æ
è
ç
ç
ö
ø
÷
÷
p
1
3 3
400
373
200
273
cm
K
cm
K
Þ p
1
600 293
400
373
200
273
=
+
/
at m
\ p
1
3
2 9 3
2
3 7 3
1
2 7 3
=
+
æ
è
ç
ö
ø
÷
a t m
=
+
=
3
1 5 7
3
. 1.07 2.64
at m
= 1.136 atm
Thermometry, Thermal Expansion & Kinetic Theory of Gases   | 41
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