NEET Exam > NEET Notes > DC Pandey Solutions for NEET Physics > DC Pandey Solutions: Thermometry, Thermal Expansion
DC Pandey Solutions: Thermometry, Thermal Expansion | DC Pandey Solutions for NEET Physics PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
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
Read More
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.
About this Document
|
4.81/5 Rating |
|
Nov 23, 2024 Last updated |
Document Description: DC Pandey Solutions: Thermometry, Thermal Expansion for NEET 2024 is part of DC Pandey Solutions for NEET Physics preparation.
The notes and questions for DC Pandey Solutions: Thermometry, Thermal Expansion have been prepared according to the NEET exam syllabus. Information about DC Pandey Solutions: Thermometry, Thermal Expansion covers topics
like and DC Pandey Solutions: Thermometry, Thermal Expansion Example, for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for DC Pandey Solutions: Thermometry, Thermal Expansion.
Introduction of DC Pandey Solutions: Thermometry, Thermal Expansion in English is available as part of our DC Pandey Solutions for NEET Physics
for NEET & DC Pandey Solutions: Thermometry, Thermal Expansion in Hindi for DC Pandey Solutions for NEET Physics course.
Download more important topics related with notes, lectures and mock test series for NEET
Exam by signing up for free. NEET: DC Pandey Solutions: Thermometry, Thermal Expansion | DC Pandey Solutions for NEET Physics
Description
Full syllabus notes, lecture & questions for DC Pandey Solutions: Thermometry, Thermal Expansion | DC Pandey Solutions for NEET Physics - NEET | Plus excerises question with solution to help you revise complete syllabus for DC Pandey Solutions for NEET Physics | Best notes, free PDF download
Information about DC Pandey Solutions: Thermometry, Thermal Expansion
In this doc you can find the meaning of DC Pandey Solutions: Thermometry, Thermal Expansion defined & explained in the simplest way possible. Besides explaining types of
DC Pandey Solutions: Thermometry, Thermal Expansion theory, EduRev gives you an ample number of questions to practice DC Pandey Solutions: Thermometry, Thermal Expansion tests, examples and also practice NEET
tests
|
Explore Courses for NEET exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Viva Questions
,Extra Questions
,Semester Notes
,Previous Year Questions with Solutions
,ppt
,DC Pandey Solutions: Thermometry
,shortcuts and tricks
,Thermal Expansion | DC Pandey Solutions for NEET Physics
,Important questions
,MCQs
,Exam
,Thermal Expansion | DC Pandey Solutions for NEET Physics
,Objective type Questions
,mock tests for examination
,Sample Paper
,Summary
,Thermal Expansion | DC Pandey Solutions for NEET Physics
,video lectures
,study material
,practice quizzes
,DC Pandey Solutions: Thermometry
,past year papers
,DC Pandey Solutions: Thermometry
,Free
;
Additional Information about DC Pandey Solutions: Thermometry, Thermal Expansion for NEET Preparation
DC Pandey Solutions: Thermometry, Thermal Expansion Free PDF Download
The DC Pandey Solutions: Thermometry, Thermal Expansion is an invaluable resource that delves deep into the core of the NEET exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the DC Pandey Solutions: Thermometry, Thermal Expansion now and kickstart your journey towards success in the NEET exam.
Importance of DC Pandey Solutions: Thermometry, Thermal Expansion
The importance of DC Pandey Solutions: Thermometry, Thermal Expansion cannot be overstated, especially for NEET aspirants.
This document holds the key to success in the NEET exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
DC Pandey Solutions: Thermometry, Thermal Expansion Notes
DC Pandey Solutions: Thermometry, Thermal Expansion Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to DC Pandey Solutions: Thermometry, Thermal Expansion.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, DC Pandey Solutions: Thermometry, Thermal Expansion Notes on EduRev are your ultimate resource for success.
DC Pandey Solutions: Thermometry, Thermal Expansion NEET Questions
The "DC Pandey Solutions: Thermometry, Thermal Expansion NEET Questions" guide is a valuable resource for all aspiring students preparing for the
NEET exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study DC Pandey Solutions: Thermometry, Thermal Expansion on the App
Students of NEET can study DC Pandey Solutions: Thermometry, Thermal Expansion alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the DC Pandey Solutions: Thermometry, Thermal Expansion,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of DC Pandey Solutions: Thermometry, Thermal Expansion is prepared as per the latest NEET syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup