Concepts of Physics Solution Manual Volume 2

228 Pages·2020·4.576 MB·English

by  HC Verma

CHAPTER – 23
HEAT AND TEMPERATURE
EXERCISES
1. Ice point = 20° (L ) L = 32°
0 1
Steam point = 80° (L )
100
L L 3220
T = 1 0 100 = 100 = 20°C
L L 8020
100 0
2. P = 1.500 × 104Pa
tr
P = 2.050 × 104Pa
We know, For constant volume gas Thermometer
P 2.050104
T = 273.16K = 273.16 = 373.31
P 1.500104
tr
3. Pressure Measured at M.P = 2.2 × Pressure at Triple Point
P 2.2P
T = 273.16 = tr 273.16 = 600.952 K 601 K
P P
tr tr
4. P = 40 × 103Pa, P = ?
tr
P
T = 100°C = 373 K, T = 273.16K
P
tr
TP 37349103
P = tr = = 54620 Pa = 5.42 × 103pa ≈ 55 K Pa
273.16 273.16
5. P = 70 K Pa, P = ?
1 2
T = 273 K, T = 373K
1 2
P 70103 70273.16103
T = 1 273.16 273 = 273.16 P
tr
P P 273
tr tr
P P 273 37370103
T = 2 273.16 373 = 2 P = = 95.6 K Pa
2 P 70273.16103 2 273
tr
6. P = P = 80 cm of Hg
ice point 0°
P = P 90 cm of Hg
steam point 100°
P = 100 cm
0
PP 80100
t = 0 100 = 100 = 200°C
P P 90100
100 0
V
7. T= T T = 273,
VV 0 0
V = 1800 CC, V= 200 CC
1800
T= 273 = 307.125 307
1600
8. R = 86; R = 80; R = 90
t 0° 100°
R R 8680
t = t 0 100 = 100 = 60°C
R R 9080
100 0
9. R at ice point (R ) = 20
0
R at steam point (R ) = 27.5
100
R at Zinc point (R )= 50
420
R = R (1+ + 2)
 0
R = R + R +R 2
100 0 0 0
R R
 100 0 = + 2
R
0
23.1
23.Heat and Temperature
27.520
 = × 100 + × 10000
20
7.5
 = 100 + 10000 
20
50R
R = R (1+ + 2)  0 = + 2
420 0
R
0
5020 3
 = 420 × + 176400 ×   420 + 176400 
20 2
7.5 3
 = 100 + 10000    420 + 176400 
20 2
10. L = ?, L = 10 m, = 1 × 10–5/°C, t= 35
1 0
L = L (1 + t) = 10(1 +10–5× 35) = 10 + 35 × 10–4= 10.0035m
1 0
11. t = 20°C, t = 10°C, L = 1cm = 0.01 m, L =?
1 2 1 2
 = 1.1 × 10–5/°C
steel
L = L (1 +  T) = 0.01(1 + 101 × 10–5× 10) = 0.01 + 0.01 × 1.1 × 10–4
2 1 steel
= 104× 10–6+ 1.1 ×10–6= 10–6(10000 + 1.1) = 10001.1
=1.00011× 10–2m = 1.00011 cm
12. L = 12 cm, = 11 × 10–5/°C
0
tw = 18°C ts = 48°C
Lw = L (1 + tw) = 12 (1 + 11 × 10–5× 18) = 12.002376 m
0
Ls = L (1 + ts) = 12 (1 + 11 × 10–5× 48) = 12.006336 m
0
L12.006336 –12.002376 = 0.00396 m 0.4cm
13. d = 2 cm = 2 × 10–2
1
t = 0°C, t = 100°C
1 2
 = 2.3 × 10–5/°C
al
d = d (1 + t) = 2 × 10–2(1 + 2.3 × 10–5102)
2 1
= 0.02 + 0.000046 = 0.020046 m = 2.0046 cm
14. L = L at 20°C  = 2.3 × 10–5/°C
st Al Al
So, Lo (1 – × 20) = Lo (1 – × 20)  = 1.1 × 10–5/°C
st st Al AI st
Lo (1 20) 12.310520 0.99954
(a)  st = Al = = = 0.999
Lo (1 20) 11.110520 0.99978
Al st
Lo (1 40) 12.310520 0.99954
(b)  40st = AI = = = 0.999
Lo (1 40) 11.110520 0.99978
40Al st
Lo 12.310510 0.999771.00092
= Al  = = 1.0002496 ≈1.00025
Lo 273 1.00044
st
Lo (1 100) 0.999771.00092
100Al = Al = = 1.00096
Lo (1 100) 1.00011
100St st
15. (a) Length at 16°C = L
L = ? T =16°C, T = 46°C
1 2
= 1.1 × 10–5/°C
L = L= L × 1.1 × 10–5 × 30
L  L 
% of error =  100% =  100% = 1.1 × 10–5× 30 × 100% = 0.033%
 L   2 
(b) T = 6°C
2
L  L 
% of error =  100% =  100% = –1.1 × 10–5× 10 × 100 = –0.011%
 L   L 
23.2
23.Heat and Temperature
16. T = 20°C, L = 0.055mm = 0.55 × 10–3m
1
t = ?  = 11 × 10–6/°C
2 st
We know,
L = L T
0
In our case,
0.055 × 10–3= 1 × 1.1 I 10–6× (T +T )
1 2
0.055 = 11 × 10–3× 20 ± 11 × 10–3× T
2
T = 20 + 5 = 25°C or 20 –5 = 15°C
2
The expt. Can be performed from 15 to 25°C
17. ƒ =0.098 g/m3, ƒ = 1 g/m3
0°C 4°C
ƒ 1 1
ƒ = 4C 0.998 = 1 + 4=
0°C
1T 14 0.998
1
4 + = 1 = 0.0005 ≈ 5 × 10–4
0.998
As density decreases = –5 × 10-4
18. Iron rod Aluminium rod
L L
Fe Al
 = 12 × 10–8 /°C  = 23 × 10–8 /°C
Fe Al
Since the difference in length is independent of temp. Hence the different always remains constant.
L = L (1 +  × T) …(1)
Fe Fe Fe
L = L (1 +  × T) …(2)
Al Al Al
L –L = L –L + L ×  × T –L ×  × T
Fe Al Fe Al Fe Fe Al Al
L  23
Fe = Al = = 23 : 12
L  12
Al Fe
19. g = 9.8 m/s2, g = 9.788 m/s2
1 2
l l l (1T)
T = 2 1 T = 2 2 = 2 1
1 2
g g g
1 2
 = 12 × 10–6/°C
Steel
T = 20°C T = ?
1 2
T = T
1 2
l l (1T) l l (1T)
 2 1 = 2 1  1 = 1
g g g g
1 2 1 2
1 112106T 9.788
 =  = 1+ 12 × 10–6× T
9.8 9.788 9.8
9.788 0.00122
 1 = 12 × 10–6T T =
9.8 12106
T –20 = –101.6 T = –101.6 + 20 = –81.6 ≈ –82°C 
2 2
20. Given
d = 2.005 cm, d = 2.000 cm
St Al
S= 11 × 10–6/°C Al= 23 × 10–6/°C Steel
ds = 2.005 (1+  T) (where T is change in temp.)
s
ds = 2.005 + 2.005 × 11 × 10–6T
Aluminium
d = 2(1+  T) = 2 + 2 × 23 × 10–6T
Al Al
The two will slip i.e the steel ball with fall when both the
diameters become equal.
So,
2.005 + 2.005 × 11 × 10–6T = 2 + 2 × 23 × 10–6T
(46 –22.055)10-6× T = 0.005
0.005106
T = = 208.81
23.945
23.3
23.Heat and Temperature
Now T = T –T = T –10°C [T = 10°C given]
2 1 2 1
T = T + T = 208.81 + 10 = 281.81
2 1
21. The final length of aluminium should be equal to final length of glass.
Let the initial length o faluminium = l
l(1 – T) = 20(1 – )
Al 0
l(1 –24 × 10–6× 40) = 20 (1 –9 × 10–6× 40)
l(1 –0.00096) = 20 (1 –0.00036)
200.99964
l = = 20.012 cm
0.99904
Let initial breadth of aluminium = b
b(1 – T) = 30(1 – )
Al 0
30(1910640) 300.99964
b = = = 30.018 cm
(12410640) 0.99904
22. V = 1000 CC, T = 20°C
g 1
V = ?  = 1.8 × 10–4/°C
Hg Hg
 = 9 × 10–6/°C
g
T remains constant
Volume of remaining space = V –V
g Hg
Now
V = V (1 +  T) …(1)
g g g
V = V (1 +  T) …(2)
Hg Hg Hg
Subtracting (2) from (1)
V –V = V –V + V  T –V  T
g Hg g Hg g g Hg Hg
 Vg = Hg  1000 = 1.8104
V  V 9106
Hg g Hg
9103
V = = 500 CC.
HG 1.8104
23. Volume of water = 500cm3
Area of cross section of can = 125m2
Final Volume of water
= 500(1 + ) = 500[1 + 3.2 × 10–4× (80 –10)] = 511.2 cm3
The aluminium vessel expands in its length only so area expansion of base cab be neglected.
Increase in volume of water = 11.2 cm3
Considering a cylinder of volume = 11.2 cm3
11.2
Height of water increased = = 0.089 cm
125
24. V = 10 × 10× 10 = 1000 CC
0
T = 10°C, V –V = 1.6 cm3
HG g
 = 6.5 × 10–6/°C,  =?,  = 3 × 6.5 × 10–6/°C
g Hg g
V = v (1 +  T) …(1)
Hg HG Hg
V = v (1 +  T) …(2)
g g g
V –V = V –V + V  T –V  T
Hg g Hg g Hg Hg g g
1.6 = 1000 ×  × 10 –1000 × 6.5 × 3 × 10–6× 10
Hg
1.66.33102
 = = 1.789 × 10–41.8 × 10–4/°C
Hg
10000
25. ƒ = 880 Kg/m3, ƒ = 900 Kg/m3
 b
T = 0°C,  = 1.2 × 10–3/°C,
1 
 = 1.5 × 10–3/°C
b
The sphere begins t sink when,
(mg) = displaced water
sphere
23.4
23.Heat and Temperature
Vƒ g = Vƒ g
 b
ƒ ƒ
   b 
1  1 
 b
880 900
 =
11.2103 11.5103
880 + 880 × 1.5 × 10–3() = 900 + 900 × 1.2 × 10–3()
(880 × 1.5 × 10–3– 900 × 1.2 × 10–3) () = 20
(1320 –1080) × 10–3() = 20
= 83.3°C ≈ 83°C
26. L = 100°C
A longitudinal strain develops if and only if, there is an opposition to the expansion.

Since there is no opposition in this case, hence the longitudinal stain here = Zero. 1 m
27.  = 20°C,  = 50°C
1 2
 = 1.2 × 10–5/°C
steel
Longitudinal stain = ?
L L
Stain = = = 
L L
= 1.2 × 10–5× (50 –20) = 3.6 ×10–4
28. A = 0.5mm2 = 0.5 × 10–6m2
T = 20°C, T = 0°C
1 2
 = 1.2 × 10–5/°C, Y = 2 × 2 × 1011N/m2
s
Decrease in length due to compression = L …(1)
Stress F L FL
Y = =  L = …(2)
Strain A L AY
Tension is developed due to (1) & (2)
Equating them,
FL
L= F = AY
AY
= 1.2 × 10-5× (20 –0) × 0.5 × 10–52 × 1011 = 24 N
29.  = 20°C,  = 100°C
1 2
A = 2mm2= 2 × 10–6m2
 = 12 × 10–6/°C, Y = 2 × 1011N/m2
steel steel
Force exerted on the clamps = ?
F
 
A YL YLA
= Y F = L = = YA
Strain L L
= 2 × 1011× 2 × 10–6× 12 × 10–6× 80 = 384 N
30. Let the final length of the system at system of temp. 0°C = ℓ

Initial length of the system = ℓ
0
When temp. changes by . Steel
Strain of the system =  0 Aluminium
1 
 Steel
total stress of system
But the total strain of the system =
total young's modulusof of system
Now, total stress = Stress due to two steel rod + Stress due to Aluminium
=   +  ds +  at = 2%  + 2 Aℓ 
s s s al s
Now young’modulus of system =  +  +  = 2 + 
s s al s al
23.5
23.Heat and Temperature
2    
Strain of system = s s s al
2 
s al
  2    
  0 = s s s al
 2 
0 s al
1  2  
 ℓ= ℓ0  al al s s 
 al 2s 
31. The ball tries to expand its volume. But it is kept in the same volume. So it is kept at a constant volume.
So the stress arises
P V
= B P = B = B ×
V V
 
 v 
= B × 3= 1.6 × 1011× 10–6× 3 × 12 × 10–6× (120 –20) = 57.6 × 1975.8 × 108pa. 
32. Given
 = Moment of Inertia at 0°C
0
=Coefficient of linear expansion
To prove, =  = (1 + 2)
0
Let the temp. change to from 0°C
T = 
Let ‘R’ be the radius of Gyration,
Now, R= R (1 + ),  = MR2 where M is the mass.
0
Now, = MR2= MR2(1 + )2 = MR2(1 + 2)
[By binomial expansion or neglecting 22 which given a very small value.]
So, =  (1 + 2) (proved)
0
33. Let the initial m.. at 0°C be 
0

T = 2
K
=  (1 + 2) (from above question)
0
 (12)  (125)  (110)
At 5°C, T = 2 0 = 2 0 = 2 0
1
K K K
 (1245)  (190)
At 45°C, T = 2 0 = 2 0
2
K K
T 190 1902.4105 1.00216
2 = =
T 110 1102.4105 1.00024
1
T 
% change =  2 1100 = 0.0959% = 9.6 × 10 –2%
 
T1 
34. T = 20°C, T = 50°C, T = 30°C
1 2
= 1.2 × 105/°C
remains constant
V V
(I) = (II) =
R R
Now, R= R(1 + ) = R + R × 1.2 × 10–5× 30 = 1.00036R
From (I) and (II)
V V V
 =
R R 1.00036R
V= 1.00036 V
(1.00036VV)
% change = × 100 = 0.00036 × 100 = 3.6 × 10–2
V

23.6
CHAPTER 24
KINETIC THEORY OF GASES
1. Volume of 1 mole of gas
RT 0.082273
PV = nRT V = = = 22.38 ≈ 22.4 L = 22.4 × 10–3= 2.24 × 10–2m3
P 1
PV 11103 103 1
2. n = = = =
RT 0.082273 22.4 22400
1
No of molecules = 6.023 × 1023× = 2.688 × 1019
22400
3. V = 1 cm3, T = 0°C, P = 10–5mm of Hg
PV ƒghV 1.369801061
n = = = = 5.874 × 10–13
RT RT 8.31273
No. of moluclues = No× n = 6.023 × 1023× 5.874 × 10–13= 3.538 × 1011
PV 11103 103
4. n = = =
RT 0.082273 22.4
 
10332
mass = g = 1.428 × 10–3g = 1.428 mg
22.4
5. Since mass is same
n = n = n
1 2
nR300 nR600
P = , P =
1 2
V0 2V0 2V0 V0
P1 = nR300 2V0 = 1 = 1 : 1 600 K 300 K
P V nR600 1
2 0
6. V = 250 cc = 250 × 10–3
P = 10–3mm = 10–3× 10–3m = 10–6× 13600 × 10 pascal = 136 × 10–3pascal
T = 27°C = 300 K
PV 136103250 136250
n = = 103 = 106
RT 8.3300 8.3300
136250
No. of molecules = 10661023 = 81 × 1017≈ 0.8 × 1015
8.3300
7. P = 8.0 × 105P , P = 1 × 106P , T = 300 K, T = ?
1 a 2 a 1 2
Since, V = V = V
1 2
PV P V 8105V 1106V 1106300
1 1 = 2 2  = T = = 375° K
T T 300 T 2 8105
1 2 2
8. m = 2 g, V = 0.02 m3=0.02 × 106cc = 0.02 × 103L, T = 300 K, P = ?
M = 2 g,
m 2
PV = nRT PV = RT P × 20 = 0.082300
M 2
0.082300
P = = 1.23 atm = 1.23 × 105pa ≈ 1.23 × 105pa
20
nRT m RT ƒRT
9. P = =  =
V M V M
ƒ 1.25 × 10–3g/cm3
R 8.31 × 107ert/deg/mole
T 273 K
ƒRT 1.251038.31107273
M = = = 0.002796 × 104≈ 28 g/mol
P 13.698076
24.1
Kinetic Theory of Gases
10. T at Simla = 15°C = 15 + 273 = 288 K
P at Simla = 72 cm = 72 × 10–2× 13600 ×9.8
T at Kalka = 35°C = 35 + 273 = 308 K
P at Kalka = 76 cm = 76 × 10–2× 13600 × 9.8
PV = nRT
m m PM
PV = RT PM = RT ƒ =
M V RT
ƒSimla P M RT
= Simla  Kalka
ƒKalka RT P M
Simla Kalka
72102136009.8308 72308
= = = 1.013
28876102136009.8 76288
ƒKalka 1
= = 0.987
ƒSimla 1.013
11. n = n = n
1 2
nRT nRT
P1= , P2= V V 3V V
V 3V
P nRT 3V
1 =  = 3 : 1 PT PT P2T P1 -
P V nRT
2
12. r.m.s velocity of hydrogen molecules = ?
T = 300 K, R = 8.3, M = 2 g = 2 × 10–3Kg
3RT 38.3300
C = C = = 1932. 6 m/s ≈1930 m/s
M 2103
Let the temp. at which the C = 2 × 1932.6 is T
38.3T 38.3T
2 × 1932.6 = (2 × 1932.6)2=
2103 2103
(21932.6)22103
 = T
38.3
T= 1199.98 ≈ 1200 K.
3P 1.77104
13. V = P = 105Pa = 1 atm, ƒ =
rms ƒ 103
3105103
= = 1301.8 ≈ 1302 m/s.
1.77104
14. Agv. K.E. = 3/2 KT
3/2 KT = 0.04 × 1.6 × 10–19
(3/2) × 1.38 × 10–23× T = 0.04 × 1.6 × 10–19
20.041.61019
T = = 0.0309178 × 104= 309.178 ≈ 310 K
31.381023
8RT 88.3300
15. V = =
avg
M 3.140.032
Distance 64000002
T = = = 445.25 m/s
Speed 445.25
28747.83
= km = 7.985 ≈ 8 hrs.
3600
16. M = 4 × 10–3Kg
8RT 88.3273
V = = = 1201.35
avg M 3.144103
Momentum = M × V = 6.64 × 10–27× 1201.35 = 7.97 × 10–24≈ 8 × 10–24Kg-m/s.
avg
24.2
Kinetic Theory of Gases
8RT 88.3300
17. V = =
avg
M 3.140.032
8RT 8RT T 1
Now, 1 = 2 1 =
2 4 T 2
2
8RT
18. Mean speed of the molecule =
M
Escape velocity = 2gr
8RT 8RT
= 2gr  = 2gr
M M
2grM 29.864000003.142103
T = = = 11863.9 ≈ 11800 m/s.
8R 88.3
8RT
19. V =
avg
M
VavgH2 = 8RT  28 = 28 = 14 = 3.74
V N 2 8RT 2
avg 2
20. The left side of the container has a gas, let having molecular wt. M
1
Right part has Mol. wt = M
2
Temperature of both left and right chambers are equal as the separating wall is diathermic
3RT 8RT 3RT 8RT M 3 M 3
=  =  1 =  1 = = 1.1775 ≈ 1.18
M M M M M 8 M 8
1 2 1 2 2 2
8RT 88.3273
21. V = = = 1698.96
mean M 3.142103
Total Dist = 1698.96 m
1698.96
No. of Collisions = = 1.23 × 1010
1.38107
22. P = 1 atm = 105Pascal
T = 300 K, M = 2 g = 2 × 10–3Kg
8RT 88.3300
(a) V = = = 1781.004 ≈ 1780 m/s
avg M 3.142103
(b) When the molecules strike at an angle 45°,
1
Force exerted = mV Cos 45° –(–mV Cos 45°) = 2 mV Cos 45° = 2 m V = 2mV
2
Force Pressure
No. of molecules striking per unit area= =
2mvArea 2mV
105 3
= = 1031 = 1.19 × 10–3× 1031= 1.19 × 1028≈ 1.2 × 1028
221031780 21780
61023
PV P V
23. 1 1 = 2 2
T T
1 2
P 200 KPa = 2 × 105pa P = ?
1 2
T = 20°C = 293 K T = 40°C = 313 K
1 2
102V
V = V + 2% V = 1
2 1 1
100
2105V P 102V 2107313
 1 = 2 1 P = = 209462 Pa = 209.462 KPa
2
293 100313 102293
24.3
Kinetic Theory of Gases
24. V = 1 × 10–3m3, P = 1.5 × 105Pa, T = 400 K
1 1 1
P V = n R T
1 1 1 1 1
PV 1.51051103 1.5
n = 1 1 = n =
R T 8.3400 8.34
1 1
1.5 1.5
m = M = 32 = 1.4457 ≈ 1.446
1
8.34 8.34
P = 1 × 105Pa, V = 1 × 10–3m3, T = 300 K
2 2 2
P V = n R T
2 2 2 2 2
P V 105103 1
n = 2 2 = = = 0.040
2
R T 8.3300 38.3
2 2
m = 0.04 × 32 = 1.285
2
m = m –m =1.446 –1.285 = 0.1608 g ≈ 0.16 g
1 2
25. P = 105 + ƒgh = 105+ 1000 × 10 × 3.3 = 1.33 × 105pa
1
4
P = 105, T = T = T, V = (2 × 10–3)3
2 1 2 1
3
4
V = r3, r = ?
2
3
PV P V
1 1 = 2 2
T T
1 2
4 4
1.33105 (2103)3 105 r2
3 3
 =
T T
1 2
1.33 × 8 × 105× 10–9= 105× r3 r = 310.64103 = 2.19 × 10–3≈ 2.2 mm
26. P = 2 atm = 2 × 105pa
1
V = 0.002 m3, T = 300 K
1 1
P V = n RT
1 1 1 1
PV 21050.002 4
n = 1 1= = = 0.1606
RT 8.3300 8.33
1
P = 1 atm = 105pa
2
V = 0.0005 m3, T = 300 K
2 2
P V = n RT
2 2 2 2
P V 1050.0005 5 1
n = 2 2 = =  = 0.02
2
RT 8.3300 38.3 10
2
n = moles leaked out = 0.16 –0.02 = 0.14
27. m = 0.040 g, T = 100°C, M = 4 g
He
3 3 m
U = nRt=  RT T= ?
2 2 M
3 m 3 m
Given  RT12 =  RT
2 M 2 M
1.5 × 0.01 × 8.3 × 373 + 12 = 1.5 × 0.01 × 8.3 × T
58.4385
T= = 469.3855 K = 196.3°C ≈ 196°C
0.1245
28. PV2= constant
P V 2= P V 2
1 1 2 2
nRT nRT
 1V2 = 2 V 2
V 1 V 2
1 2
T
T V = T V = TV = T × 2V T =
1 1 2 2 1 2
2
24.4