DC Pandey Solutions: Thermometry, Thermal Expansion

# DC Pandey Solutions: Thermometry, Thermal Expansion | DC Pandey Solutions for NEET Physics PDF Download

``` 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
```

122 docs

## FAQs on DC Pandey Solutions: Thermometry, Thermal Expansion - DC Pandey Solutions for NEET Physics

 1. What is thermometry and how is it related to thermal expansion?
Ans. Thermometry is the science of measuring temperature. It involves the use of various devices and techniques to determine the temperature of an object or a system. Thermal expansion, on the other hand, is the tendency of matter to change in shape, volume, or density in response to a change in temperature. Thermometry and thermal expansion are related because temperature changes can cause objects to expand or contract, leading to changes in their dimensions.
 2. What are the different types of thermometers used in thermometry?
Ans. There are several types of thermometers used in thermometry, including mercury-in-glass thermometers, digital thermometers, infrared thermometers, thermocouples, and resistance temperature detectors (RTDs). Each type of thermometer has its own advantages and limitations, and they are used in different applications based on their accuracy, response time, and measurement range.
 3. How does a mercury-in-glass thermometer work?
Ans. A mercury-in-glass thermometer consists of a glass tube filled with mercury and a calibrated scale. When the temperature changes, the mercury expands or contracts, causing it to rise or fall in the glass tube. The scale on the thermometer allows us to measure the change in height of the mercury column, which corresponds to the temperature of the object being measured.
 4. What is the coefficient of linear expansion and how is it calculated?
Ans. The coefficient of linear expansion is a measure of how much a material expands or contracts in response to a change in temperature. It is denoted by the symbol "α" and is defined as the change in length per unit length per degree Celsius (or Kelvin) change in temperature. The coefficient of linear expansion is calculated using the formula: α = (ΔL / L) / ΔT, where ΔL is the change in length, L is the original length, and ΔT is the change in temperature.
 5. How does thermal expansion affect everyday objects and structures?
Ans. Thermal expansion can have practical implications in everyday life. For example, when a metal object is heated, it expands, which can lead to the loosening of fasteners or the warping of structures. This is why gaps are often left between railway tracks to allow for their expansion in hot weather. Similarly, the expansion and contraction of materials due to temperature changes can also affect the accuracy of measurements in scientific experiments and industrial processes, requiring careful consideration and compensation.

## DC Pandey Solutions for NEET Physics

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