NEET  >  HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases

# HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases - Notes | Study Physics Class 11 - NEET

``` Page 1

24.1
CHAPTER 24
KINETIC THEORY OF GASES
1. Volume of 1 mole of gas
PV = nRT ? V =
P
RT
=
1
273 082 . 0 ?
= 22.38 ˜ 22.4 L = 22.4 × 10
–3
= 2.24 × 10
–2
m
3
2. n =
RT
PV
=
273 082 . 0
10 1 1
3
?
? ?
?
=
4 . 22
10
3 ?
=
22400
1
No of molecules = 6.023 × 10
23
×
22400
1
= 2.688 × 10
19
3. V = 1 cm
3
, T = 0°C, P = 10
–5
mm of Hg
n =
RT
PV
=
RT
V gh ƒ ?
=
273 31 . 8
1 10 980 36 . 1
6
?
? ? ?
?
= 5.874 × 10
–13
No. of moluclues = No × n = 6.023 × 10
23
× 5.874 × 10
–13
= 3.538 × 10
11
4. n =
RT
PV
=
273 082 . 0
10 1 1
3
?
? ?
?
=
4 . 22
10
3 ?
mass =
? ?
4 . 22
32 10
3
?
?
g = 1.428 × 10
–3
g = 1.428 mg
5. Since mass is same
n
1
= n
2
= n
P
1
=
0
V
300 nR ?
, P
2
=
0
V 2
600 nR ?
2
1
P
P
=
600 nR
V 2
V
300 nR
0
0
?
?
?
=
1
1
= 1 : 1
6. V = 250 cc = 250 × 10
–3
P = 10
–3
mm = 10
–3
× 10
–3
m = 10
–6
× 13600 × 10 pascal = 136 × 10
–3
pascal
T = 27°C = 300 K
n =
RT
PV
=
3
3
10
300 3 . 8
250 10 136
?
?
?
?
? ?
=
6
10
300 3 . 8
250 136
?
?
?
?
No. of molecules =
23 6
10 6 10
300 3 . 8
250 136
? ? ?
?
?
?
= 81 × 10
17
˜ 0.8 × 10
15
7. P
1
= 8.0 × 10
5
P
a
, P
2
= 1 × 10
6
P
a
, T
1
= 300 K, T
2
= ?
Since, V
1
= V
2
= V
1
1 1
T
V P
=
2
2 2
T
V P
?
300
V 10 8
5
? ?
=
2
6
T
V 10 1 ? ?
? T
2
=
5
6
10 8
300 10 1
?
? ?
= 375° K
8. m = 2 g, V = 0.02 m
3
= 0.02 × 10
6
cc = 0.02 × 10
3
L, T = 300 K, P = ?
M = 2 g,
PV = nRT ? PV = RT
M
m
? P × 20 = 300 082 . 0
2
2
? ?
? P =
20
300 082 . 0 ?
= 1.23 atm = 1.23 × 10
5
pa ˜ 1.23 × 10
5
pa
9. P =
V
nRT
=
V
RT
M
m
? =
M
RT ƒ
ƒ ? 1.25 × 10
–3
g/cm
3
R ? 8.31 × 10
7
ert/deg/mole
T ? 273 K
? M =
P
RT ƒ
=
76 980 6 . 13
273 10 31 . 8 10 25 . 1
7 3
? ?
? ? ? ?
?
= 0.002796 × 10
4
˜ 28 g/mol
600 K
V 0 2V 0
300 K
Page 2

24.1
CHAPTER 24
KINETIC THEORY OF GASES
1. Volume of 1 mole of gas
PV = nRT ? V =
P
RT
=
1
273 082 . 0 ?
= 22.38 ˜ 22.4 L = 22.4 × 10
–3
= 2.24 × 10
–2
m
3
2. n =
RT
PV
=
273 082 . 0
10 1 1
3
?
? ?
?
=
4 . 22
10
3 ?
=
22400
1
No of molecules = 6.023 × 10
23
×
22400
1
= 2.688 × 10
19
3. V = 1 cm
3
, T = 0°C, P = 10
–5
mm of Hg
n =
RT
PV
=
RT
V gh ƒ ?
=
273 31 . 8
1 10 980 36 . 1
6
?
? ? ?
?
= 5.874 × 10
–13
No. of moluclues = No × n = 6.023 × 10
23
× 5.874 × 10
–13
= 3.538 × 10
11
4. n =
RT
PV
=
273 082 . 0
10 1 1
3
?
? ?
?
=
4 . 22
10
3 ?
mass =
? ?
4 . 22
32 10
3
?
?
g = 1.428 × 10
–3
g = 1.428 mg
5. Since mass is same
n
1
= n
2
= n
P
1
=
0
V
300 nR ?
, P
2
=
0
V 2
600 nR ?
2
1
P
P
=
600 nR
V 2
V
300 nR
0
0
?
?
?
=
1
1
= 1 : 1
6. V = 250 cc = 250 × 10
–3
P = 10
–3
mm = 10
–3
× 10
–3
m = 10
–6
× 13600 × 10 pascal = 136 × 10
–3
pascal
T = 27°C = 300 K
n =
RT
PV
=
3
3
10
300 3 . 8
250 10 136
?
?
?
?
? ?
=
6
10
300 3 . 8
250 136
?
?
?
?
No. of molecules =
23 6
10 6 10
300 3 . 8
250 136
? ? ?
?
?
?
= 81 × 10
17
˜ 0.8 × 10
15
7. P
1
= 8.0 × 10
5
P
a
, P
2
= 1 × 10
6
P
a
, T
1
= 300 K, T
2
= ?
Since, V
1
= V
2
= V
1
1 1
T
V P
=
2
2 2
T
V P
?
300
V 10 8
5
? ?
=
2
6
T
V 10 1 ? ?
? T
2
=
5
6
10 8
300 10 1
?
? ?
= 375° K
8. m = 2 g, V = 0.02 m
3
= 0.02 × 10
6
cc = 0.02 × 10
3
L, T = 300 K, P = ?
M = 2 g,
PV = nRT ? PV = RT
M
m
? P × 20 = 300 082 . 0
2
2
? ?
? P =
20
300 082 . 0 ?
= 1.23 atm = 1.23 × 10
5
pa ˜ 1.23 × 10
5
pa
9. P =
V
nRT
=
V
RT
M
m
? =
M
RT ƒ
ƒ ? 1.25 × 10
–3
g/cm
3
R ? 8.31 × 10
7
ert/deg/mole
T ? 273 K
? M =
P
RT ƒ
=
76 980 6 . 13
273 10 31 . 8 10 25 . 1
7 3
? ?
? ? ? ?
?
= 0.002796 × 10
4
˜ 28 g/mol
600 K
V 0 2V 0
300 K
Kinetic Theory of Gases
24.2
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
? PV = RT
M
m
? PM = RT
V
m
? ƒ =
RT
PM
Kalka ƒ
Simla ƒ
=
M P
RT
RT
M P
Kalka
Kalka
Simla
Simla
?
?
?
=
8 . 9 13600 10 76 288
308 8 . 9 13600 10 72
2
2
? ? ? ?
? ? ? ?
?
?
=
288 76
308 72
?
?
= 1.013
Simla ƒ
Kalka ƒ
=
013 . 1
1
= 0.987
11. n
1
= n
2
= n
P
1
=
V
nRT
, P
2
=
V 3
nRT
2
1
P
P
=
nRT
V 3
V
nRT
? = 3 : 1
12. r.m.s velocity of hydrogen molecules = ?
T = 300 K, R = 8.3, M = 2 g = 2 × 10
–3
Kg
C =
M
RT 3
? C =
3
10 2
300 3 . 8 3
?
?
? ?
= 1932. 6 m/s ˜1930 m/s
Let the temp. at which the C = 2 × 1932.6 is T ?
2 × 1932.6 =
3
10 2
T 3 . 8 3
?
?
? ? ?
? (2 × 1932.6)
2
=
3
10 2
T 3 . 8 3
?
?
? ? ?
?
3 . 8 3
10 2 ) 6 . 1932 2 (
3 2
?
? ? ?
?
= T ?
? T ? = 1199.98 ˜ 1200 K.
13. V
rms
=
ƒ
P 3
P = 10
5
Pa = 1 atm, ƒ =
3
4
10
10 77 . 1
?
?
?
=
4
3 5
10 77 . 1
10 10 3
?
?
?
? ?
= 1301.8 ˜ 1302 m/s.
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
? T =
23
19
10 38 . 1 3
10 6 . 1 04 . 0 2
?
?
? ?
? ? ?
= 0.0309178 × 10
4
= 309.178 ˜ 310 K
15. V
avg
=
M
RT 8
?
=
032 . 0 14 . 3
300 3 . 8 8
?
? ?
T =
Speed
ce tan Dis
=
25 . 445
2 6400000 ?
= 445.25 m/s
=
3600
83 . 28747
km = 7.985 ˜ 8 hrs.
16. M = 4 × 10
–3
Kg
V
avg
=
M
RT 8
?
=
3
10 4 14 . 3
273 3 . 8 8
?
? ?
? ?
= 1201.35
Momentum = M × V
avg
= 6.64 × 10
–27
× 1201.35 = 7.97 × 10
–24
˜ 8 × 10
–24
Kg-m/s.
P T
V V
P T P 2T
3V V
P 1 -
Page 3

24.1
CHAPTER 24
KINETIC THEORY OF GASES
1. Volume of 1 mole of gas
PV = nRT ? V =
P
RT
=
1
273 082 . 0 ?
= 22.38 ˜ 22.4 L = 22.4 × 10
–3
= 2.24 × 10
–2
m
3
2. n =
RT
PV
=
273 082 . 0
10 1 1
3
?
? ?
?
=
4 . 22
10
3 ?
=
22400
1
No of molecules = 6.023 × 10
23
×
22400
1
= 2.688 × 10
19
3. V = 1 cm
3
, T = 0°C, P = 10
–5
mm of Hg
n =
RT
PV
=
RT
V gh ƒ ?
=
273 31 . 8
1 10 980 36 . 1
6
?
? ? ?
?
= 5.874 × 10
–13
No. of moluclues = No × n = 6.023 × 10
23
× 5.874 × 10
–13
= 3.538 × 10
11
4. n =
RT
PV
=
273 082 . 0
10 1 1
3
?
? ?
?
=
4 . 22
10
3 ?
mass =
? ?
4 . 22
32 10
3
?
?
g = 1.428 × 10
–3
g = 1.428 mg
5. Since mass is same
n
1
= n
2
= n
P
1
=
0
V
300 nR ?
, P
2
=
0
V 2
600 nR ?
2
1
P
P
=
600 nR
V 2
V
300 nR
0
0
?
?
?
=
1
1
= 1 : 1
6. V = 250 cc = 250 × 10
–3
P = 10
–3
mm = 10
–3
× 10
–3
m = 10
–6
× 13600 × 10 pascal = 136 × 10
–3
pascal
T = 27°C = 300 K
n =
RT
PV
=
3
3
10
300 3 . 8
250 10 136
?
?
?
?
? ?
=
6
10
300 3 . 8
250 136
?
?
?
?
No. of molecules =
23 6
10 6 10
300 3 . 8
250 136
? ? ?
?
?
?
= 81 × 10
17
˜ 0.8 × 10
15
7. P
1
= 8.0 × 10
5
P
a
, P
2
= 1 × 10
6
P
a
, T
1
= 300 K, T
2
= ?
Since, V
1
= V
2
= V
1
1 1
T
V P
=
2
2 2
T
V P
?
300
V 10 8
5
? ?
=
2
6
T
V 10 1 ? ?
? T
2
=
5
6
10 8
300 10 1
?
? ?
= 375° K
8. m = 2 g, V = 0.02 m
3
= 0.02 × 10
6
cc = 0.02 × 10
3
L, T = 300 K, P = ?
M = 2 g,
PV = nRT ? PV = RT
M
m
? P × 20 = 300 082 . 0
2
2
? ?
? P =
20
300 082 . 0 ?
= 1.23 atm = 1.23 × 10
5
pa ˜ 1.23 × 10
5
pa
9. P =
V
nRT
=
V
RT
M
m
? =
M
RT ƒ
ƒ ? 1.25 × 10
–3
g/cm
3
R ? 8.31 × 10
7
ert/deg/mole
T ? 273 K
? M =
P
RT ƒ
=
76 980 6 . 13
273 10 31 . 8 10 25 . 1
7 3
? ?
? ? ? ?
?
= 0.002796 × 10
4
˜ 28 g/mol
600 K
V 0 2V 0
300 K
Kinetic Theory of Gases
24.2
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
? PV = RT
M
m
? PM = RT
V
m
? ƒ =
RT
PM
Kalka ƒ
Simla ƒ
=
M P
RT
RT
M P
Kalka
Kalka
Simla
Simla
?
?
?
=
8 . 9 13600 10 76 288
308 8 . 9 13600 10 72
2
2
? ? ? ?
? ? ? ?
?
?
=
288 76
308 72
?
?
= 1.013
Simla ƒ
Kalka ƒ
=
013 . 1
1
= 0.987
11. n
1
= n
2
= n
P
1
=
V
nRT
, P
2
=
V 3
nRT
2
1
P
P
=
nRT
V 3
V
nRT
? = 3 : 1
12. r.m.s velocity of hydrogen molecules = ?
T = 300 K, R = 8.3, M = 2 g = 2 × 10
–3
Kg
C =
M
RT 3
? C =
3
10 2
300 3 . 8 3
?
?
? ?
= 1932. 6 m/s ˜1930 m/s
Let the temp. at which the C = 2 × 1932.6 is T ?
2 × 1932.6 =
3
10 2
T 3 . 8 3
?
?
? ? ?
? (2 × 1932.6)
2
=
3
10 2
T 3 . 8 3
?
?
? ? ?
?
3 . 8 3
10 2 ) 6 . 1932 2 (
3 2
?
? ? ?
?
= T ?
? T ? = 1199.98 ˜ 1200 K.
13. V
rms
=
ƒ
P 3
P = 10
5
Pa = 1 atm, ƒ =
3
4
10
10 77 . 1
?
?
?
=
4
3 5
10 77 . 1
10 10 3
?
?
?
? ?
= 1301.8 ˜ 1302 m/s.
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
? T =
23
19
10 38 . 1 3
10 6 . 1 04 . 0 2
?
?
? ?
? ? ?
= 0.0309178 × 10
4
= 309.178 ˜ 310 K
15. V
avg
=
M
RT 8
?
=
032 . 0 14 . 3
300 3 . 8 8
?
? ?
T =
Speed
ce tan Dis
=
25 . 445
2 6400000 ?
= 445.25 m/s
=
3600
83 . 28747
km = 7.985 ˜ 8 hrs.
16. M = 4 × 10
–3
Kg
V
avg
=
M
RT 8
?
=
3
10 4 14 . 3
273 3 . 8 8
?
? ?
? ?
= 1201.35
Momentum = M × V
avg
= 6.64 × 10
–27
× 1201.35 = 7.97 × 10
–24
˜ 8 × 10
–24
Kg-m/s.
P T
V V
P T P 2T
3V V
P 1 -
Kinetic Theory of Gases
24.3
17. V
avg
=
M
RT 8
?
=
032 . 0 14 . 3
300 3 . 8 8
?
? ?
Now,
2
RT 8
1
? ?
=
4
RT 8
2
? ?
2
1
T
T
=
2
1
18. Mean speed of the molecule =
M
RT 8
?
Escape velocity = gr 2
M
RT 8
?
= gr 2 ?
M
RT 8
?
= 2gr
? T =
R 8
M gr 2 ?
=
3 . 8 8
10 2 14 . 3 6400000 8 . 9 2
3
?
? ? ? ? ?
?
= 11863.9 ˜ 11800 m/s.
19. V
avg
=
M
RT 8
?
2 avg
2 avg
N V
H V
=
RT 8
28
2
RT 8 ? ?
?
? ?
=
2
28
= 14 = 3.74
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
1
M
RT 3
=
2
M
RT 8
?
?
1
M
RT 3
=
2
M
RT 8
?
?
2
1
M
M
?
=
8
3
?
2
1
M
M
=
8
3 ?
= 1.1775 ˜ 1.18
21. V
mean
=
M
RT 8
?
=
3
10 2 14 . 3
273 3 . 8 8
?
? ?
? ?
= 1698.96
Total Dist = 1698.96 m
No. of Collisions =
7
10 38 . 1
96 . 1698
?
?
= 1.23 × 10
10
22. P = 1 atm = 10
5
Pascal
T = 300 K, M = 2 g = 2 × 10
–3
Kg
(a) V
avg
=
M
RT 8
?
=
3
10 2 14 . 3
300 3 . 8 8
?
? ?
? ?
= 1781.004 ˜ 1780 m/s
(b) When the molecules strike at an angle 45°,
Force exerted = mV Cos 45° – (–mV Cos 45°) = 2 mV Cos 45° = 2 m V
2
1
= 2 mV
No. of molecules striking per unit area =
Area mv 2
Force
?
=
mV 2
essure Pr
=
23
3
5
10 6
1780 10 2 2
10
?
? ? ?
?
=
31
10
1780 2
3
?
?
= 1.19 × 10
–3
× 10
31
= 1.19 × 10
28
˜ 1.2 × 10
28
23.
1
1 1
T
V P
=
2
2 2
T
V P
P
1
? 200 KPa = 2 × 10
5
pa P
2
= ?
T
1
= 20°C = 293 K T
2
= 40°C = 313 K
V
2
= V
1
+ 2% V
1
=
100
V 102
1
?
?
293
V 10 2
1
5
? ?
=
313 100
V 102 P
1 2
?
? ?
? P
2
=
293 102
313 10 2
7
?
? ?
= 209462 Pa = 209.462 KPa
Page 4

24.1
CHAPTER 24
KINETIC THEORY OF GASES
1. Volume of 1 mole of gas
PV = nRT ? V =
P
RT
=
1
273 082 . 0 ?
= 22.38 ˜ 22.4 L = 22.4 × 10
–3
= 2.24 × 10
–2
m
3
2. n =
RT
PV
=
273 082 . 0
10 1 1
3
?
? ?
?
=
4 . 22
10
3 ?
=
22400
1
No of molecules = 6.023 × 10
23
×
22400
1
= 2.688 × 10
19
3. V = 1 cm
3
, T = 0°C, P = 10
–5
mm of Hg
n =
RT
PV
=
RT
V gh ƒ ?
=
273 31 . 8
1 10 980 36 . 1
6
?
? ? ?
?
= 5.874 × 10
–13
No. of moluclues = No × n = 6.023 × 10
23
× 5.874 × 10
–13
= 3.538 × 10
11
4. n =
RT
PV
=
273 082 . 0
10 1 1
3
?
? ?
?
=
4 . 22
10
3 ?
mass =
? ?
4 . 22
32 10
3
?
?
g = 1.428 × 10
–3
g = 1.428 mg
5. Since mass is same
n
1
= n
2
= n
P
1
=
0
V
300 nR ?
, P
2
=
0
V 2
600 nR ?
2
1
P
P
=
600 nR
V 2
V
300 nR
0
0
?
?
?
=
1
1
= 1 : 1
6. V = 250 cc = 250 × 10
–3
P = 10
–3
mm = 10
–3
× 10
–3
m = 10
–6
× 13600 × 10 pascal = 136 × 10
–3
pascal
T = 27°C = 300 K
n =
RT
PV
=
3
3
10
300 3 . 8
250 10 136
?
?
?
?
? ?
=
6
10
300 3 . 8
250 136
?
?
?
?
No. of molecules =
23 6
10 6 10
300 3 . 8
250 136
? ? ?
?
?
?
= 81 × 10
17
˜ 0.8 × 10
15
7. P
1
= 8.0 × 10
5
P
a
, P
2
= 1 × 10
6
P
a
, T
1
= 300 K, T
2
= ?
Since, V
1
= V
2
= V
1
1 1
T
V P
=
2
2 2
T
V P
?
300
V 10 8
5
? ?
=
2
6
T
V 10 1 ? ?
? T
2
=
5
6
10 8
300 10 1
?
? ?
= 375° K
8. m = 2 g, V = 0.02 m
3
= 0.02 × 10
6
cc = 0.02 × 10
3
L, T = 300 K, P = ?
M = 2 g,
PV = nRT ? PV = RT
M
m
? P × 20 = 300 082 . 0
2
2
? ?
? P =
20
300 082 . 0 ?
= 1.23 atm = 1.23 × 10
5
pa ˜ 1.23 × 10
5
pa
9. P =
V
nRT
=
V
RT
M
m
? =
M
RT ƒ
ƒ ? 1.25 × 10
–3
g/cm
3
R ? 8.31 × 10
7
ert/deg/mole
T ? 273 K
? M =
P
RT ƒ
=
76 980 6 . 13
273 10 31 . 8 10 25 . 1
7 3
? ?
? ? ? ?
?
= 0.002796 × 10
4
˜ 28 g/mol
600 K
V 0 2V 0
300 K
Kinetic Theory of Gases
24.2
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
? PV = RT
M
m
? PM = RT
V
m
? ƒ =
RT
PM
Kalka ƒ
Simla ƒ
=
M P
RT
RT
M P
Kalka
Kalka
Simla
Simla
?
?
?
=
8 . 9 13600 10 76 288
308 8 . 9 13600 10 72
2
2
? ? ? ?
? ? ? ?
?
?
=
288 76
308 72
?
?
= 1.013
Simla ƒ
Kalka ƒ
=
013 . 1
1
= 0.987
11. n
1
= n
2
= n
P
1
=
V
nRT
, P
2
=
V 3
nRT
2
1
P
P
=
nRT
V 3
V
nRT
? = 3 : 1
12. r.m.s velocity of hydrogen molecules = ?
T = 300 K, R = 8.3, M = 2 g = 2 × 10
–3
Kg
C =
M
RT 3
? C =
3
10 2
300 3 . 8 3
?
?
? ?
= 1932. 6 m/s ˜1930 m/s
Let the temp. at which the C = 2 × 1932.6 is T ?
2 × 1932.6 =
3
10 2
T 3 . 8 3
?
?
? ? ?
? (2 × 1932.6)
2
=
3
10 2
T 3 . 8 3
?
?
? ? ?
?
3 . 8 3
10 2 ) 6 . 1932 2 (
3 2
?
? ? ?
?
= T ?
? T ? = 1199.98 ˜ 1200 K.
13. V
rms
=
ƒ
P 3
P = 10
5
Pa = 1 atm, ƒ =
3
4
10
10 77 . 1
?
?
?
=
4
3 5
10 77 . 1
10 10 3
?
?
?
? ?
= 1301.8 ˜ 1302 m/s.
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
? T =
23
19
10 38 . 1 3
10 6 . 1 04 . 0 2
?
?
? ?
? ? ?
= 0.0309178 × 10
4
= 309.178 ˜ 310 K
15. V
avg
=
M
RT 8
?
=
032 . 0 14 . 3
300 3 . 8 8
?
? ?
T =
Speed
ce tan Dis
=
25 . 445
2 6400000 ?
= 445.25 m/s
=
3600
83 . 28747
km = 7.985 ˜ 8 hrs.
16. M = 4 × 10
–3
Kg
V
avg
=
M
RT 8
?
=
3
10 4 14 . 3
273 3 . 8 8
?
? ?
? ?
= 1201.35
Momentum = M × V
avg
= 6.64 × 10
–27
× 1201.35 = 7.97 × 10
–24
˜ 8 × 10
–24
Kg-m/s.
P T
V V
P T P 2T
3V V
P 1 -
Kinetic Theory of Gases
24.3
17. V
avg
=
M
RT 8
?
=
032 . 0 14 . 3
300 3 . 8 8
?
? ?
Now,
2
RT 8
1
? ?
=
4
RT 8
2
? ?
2
1
T
T
=
2
1
18. Mean speed of the molecule =
M
RT 8
?
Escape velocity = gr 2
M
RT 8
?
= gr 2 ?
M
RT 8
?
= 2gr
? T =
R 8
M gr 2 ?
=
3 . 8 8
10 2 14 . 3 6400000 8 . 9 2
3
?
? ? ? ? ?
?
= 11863.9 ˜ 11800 m/s.
19. V
avg
=
M
RT 8
?
2 avg
2 avg
N V
H V
=
RT 8
28
2
RT 8 ? ?
?
? ?
=
2
28
= 14 = 3.74
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
1
M
RT 3
=
2
M
RT 8
?
?
1
M
RT 3
=
2
M
RT 8
?
?
2
1
M
M
?
=
8
3
?
2
1
M
M
=
8
3 ?
= 1.1775 ˜ 1.18
21. V
mean
=
M
RT 8
?
=
3
10 2 14 . 3
273 3 . 8 8
?
? ?
? ?
= 1698.96
Total Dist = 1698.96 m
No. of Collisions =
7
10 38 . 1
96 . 1698
?
?
= 1.23 × 10
10
22. P = 1 atm = 10
5
Pascal
T = 300 K, M = 2 g = 2 × 10
–3
Kg
(a) V
avg
=
M
RT 8
?
=
3
10 2 14 . 3
300 3 . 8 8
?
? ?
? ?
= 1781.004 ˜ 1780 m/s
(b) When the molecules strike at an angle 45°,
Force exerted = mV Cos 45° – (–mV Cos 45°) = 2 mV Cos 45° = 2 m V
2
1
= 2 mV
No. of molecules striking per unit area =
Area mv 2
Force
?
=
mV 2
essure Pr
=
23
3
5
10 6
1780 10 2 2
10
?
? ? ?
?
=
31
10
1780 2
3
?
?
= 1.19 × 10
–3
× 10
31
= 1.19 × 10
28
˜ 1.2 × 10
28
23.
1
1 1
T
V P
=
2
2 2
T
V P
P
1
? 200 KPa = 2 × 10
5
pa P
2
= ?
T
1
= 20°C = 293 K T
2
= 40°C = 313 K
V
2
= V
1
+ 2% V
1
=
100
V 102
1
?
?
293
V 10 2
1
5
? ?
=
313 100
V 102 P
1 2
?
? ?
? P
2
=
293 102
313 10 2
7
?
? ?
= 209462 Pa = 209.462 KPa
Kinetic Theory of Gases
24.4
24. V
1
= 1 × 10
–3
m
3
, P
1
= 1.5 × 10
5
Pa, T
1
= 400 K
P
1
V
1
= n
1
R
1
T
1
? n =
1 1
1 1
T R
V P
=
400 3 . 8
10 1 10 5 . 1
3 5
?
? ? ?
?
? n =
4 3 . 8
5 . 1
?
? m
1
= M
4 3 . 8
5 . 1
?
?
= 32
4 3 . 8
5 . 1
?
?
= 1.4457 ˜ 1.446
P
2
= 1 × 10
5
Pa, V
2
= 1 × 10
–3
m
3
, T
2
= 300 K
P
2
V
2
= n
2
R
2
T
2
? n
2
=
2 2
2 2
T R
V P
=
300 3 . 8
10 10
3 5
?
?
?
=
3 . 8 3
1
?
= 0.040
? m
2
= 0.04 × 32 = 1.285
?m = m
1
– m
2
=1.446 – 1.285 = 0.1608 g ˜ 0.16 g ?
25. P
1
= 10
5
+ ƒgh = 10
5
+ 1000 × 10 × 3.3 = 1.33 × 10
5
pa
P
2
= 10
5
, T
1
= T
2
= T, V
1
=
3
4
?(2 × 10
–3
)
3
V
2
=
3
4
?r
3
, r = ?
1
1 1
T
V P
=
2
2 2
T
V P
?
1
3 3 5
T
) 10 2 (
3
4
10 33 . 1
?
? ? ? ? ? ?
=
2
2 5
T
r
3
4
10 ? ? ?
? 1.33 × 8 × 10
5
× 10
–9
= 10
5
× r
3
? r =
3 3
10 64 . 10
?
? = 2.19 × 10
–3
˜ 2.2 mm
26. P
1
= 2 atm = 2 × 10
5
pa
V
1
= 0.002 m
3
, T
1
= 300 K
P
1
V
1
= n
1
RT
1
? n =
1
1 1
RT
V P
=
300 3 . 8
002 . 0 10 2
5
?
? ?
=
3 3 . 8
4
?
= 0.1606
P
2
= 1 atm = 10
5
pa
V
2
= 0.0005 m
3
, T
2
= 300 K
P
2
V
2
= n
2
RT
2
? n
2
=
2
2 2
RT
V P
=
300 3 . 8
0005 . 0 10
5
?
?
=
10
1
3 . 8 3
5
?
?
= 0.02
?n = moles leaked out = 0.16 – 0.02 = 0.14
27. m = 0.040 g, T = 100°C, M
He
= 4 g
U =
2
3
nRt = RT
M
m
2
3
? ? T ? = ?
Given 12 RT
M
m
2
3
? ? ? = T R
M
m
2
3
? ? ?
? 1.5 × 0.01 × 8.3 × 373 + 12 = 1.5 × 0.01 × 8.3 × T ?
? T ? =
1245 . 0
4385 . 58
= 469.3855 K = 196.3°C ˜ 196°C
28. PV
2
= constant
? P
1
V
1
2
= P
2
V
2
2
?
2
1
1
1
V
V
nRT
? =
2
2
2
2
V
V
nRT
?
? T
1
V
1
= T
2
V
2
= TV = T
1
× 2V ? T
2
=
2
T
Page 5

24.1
CHAPTER 24
KINETIC THEORY OF GASES
1. Volume of 1 mole of gas
PV = nRT ? V =
P
RT
=
1
273 082 . 0 ?
= 22.38 ˜ 22.4 L = 22.4 × 10
–3
= 2.24 × 10
–2
m
3
2. n =
RT
PV
=
273 082 . 0
10 1 1
3
?
? ?
?
=
4 . 22
10
3 ?
=
22400
1
No of molecules = 6.023 × 10
23
×
22400
1
= 2.688 × 10
19
3. V = 1 cm
3
, T = 0°C, P = 10
–5
mm of Hg
n =
RT
PV
=
RT
V gh ƒ ?
=
273 31 . 8
1 10 980 36 . 1
6
?
? ? ?
?
= 5.874 × 10
–13
No. of moluclues = No × n = 6.023 × 10
23
× 5.874 × 10
–13
= 3.538 × 10
11
4. n =
RT
PV
=
273 082 . 0
10 1 1
3
?
? ?
?
=
4 . 22
10
3 ?
mass =
? ?
4 . 22
32 10
3
?
?
g = 1.428 × 10
–3
g = 1.428 mg
5. Since mass is same
n
1
= n
2
= n
P
1
=
0
V
300 nR ?
, P
2
=
0
V 2
600 nR ?
2
1
P
P
=
600 nR
V 2
V
300 nR
0
0
?
?
?
=
1
1
= 1 : 1
6. V = 250 cc = 250 × 10
–3
P = 10
–3
mm = 10
–3
× 10
–3
m = 10
–6
× 13600 × 10 pascal = 136 × 10
–3
pascal
T = 27°C = 300 K
n =
RT
PV
=
3
3
10
300 3 . 8
250 10 136
?
?
?
?
? ?
=
6
10
300 3 . 8
250 136
?
?
?
?
No. of molecules =
23 6
10 6 10
300 3 . 8
250 136
? ? ?
?
?
?
= 81 × 10
17
˜ 0.8 × 10
15
7. P
1
= 8.0 × 10
5
P
a
, P
2
= 1 × 10
6
P
a
, T
1
= 300 K, T
2
= ?
Since, V
1
= V
2
= V
1
1 1
T
V P
=
2
2 2
T
V P
?
300
V 10 8
5
? ?
=
2
6
T
V 10 1 ? ?
? T
2
=
5
6
10 8
300 10 1
?
? ?
= 375° K
8. m = 2 g, V = 0.02 m
3
= 0.02 × 10
6
cc = 0.02 × 10
3
L, T = 300 K, P = ?
M = 2 g,
PV = nRT ? PV = RT
M
m
? P × 20 = 300 082 . 0
2
2
? ?
? P =
20
300 082 . 0 ?
= 1.23 atm = 1.23 × 10
5
pa ˜ 1.23 × 10
5
pa
9. P =
V
nRT
=
V
RT
M
m
? =
M
RT ƒ
ƒ ? 1.25 × 10
–3
g/cm
3
R ? 8.31 × 10
7
ert/deg/mole
T ? 273 K
? M =
P
RT ƒ
=
76 980 6 . 13
273 10 31 . 8 10 25 . 1
7 3
? ?
? ? ? ?
?
= 0.002796 × 10
4
˜ 28 g/mol
600 K
V 0 2V 0
300 K
Kinetic Theory of Gases
24.2
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
? PV = RT
M
m
? PM = RT
V
m
? ƒ =
RT
PM
Kalka ƒ
Simla ƒ
=
M P
RT
RT
M P
Kalka
Kalka
Simla
Simla
?
?
?
=
8 . 9 13600 10 76 288
308 8 . 9 13600 10 72
2
2
? ? ? ?
? ? ? ?
?
?
=
288 76
308 72
?
?
= 1.013
Simla ƒ
Kalka ƒ
=
013 . 1
1
= 0.987
11. n
1
= n
2
= n
P
1
=
V
nRT
, P
2
=
V 3
nRT
2
1
P
P
=
nRT
V 3
V
nRT
? = 3 : 1
12. r.m.s velocity of hydrogen molecules = ?
T = 300 K, R = 8.3, M = 2 g = 2 × 10
–3
Kg
C =
M
RT 3
? C =
3
10 2
300 3 . 8 3
?
?
? ?
= 1932. 6 m/s ˜1930 m/s
Let the temp. at which the C = 2 × 1932.6 is T ?
2 × 1932.6 =
3
10 2
T 3 . 8 3
?
?
? ? ?
? (2 × 1932.6)
2
=
3
10 2
T 3 . 8 3
?
?
? ? ?
?
3 . 8 3
10 2 ) 6 . 1932 2 (
3 2
?
? ? ?
?
= T ?
? T ? = 1199.98 ˜ 1200 K.
13. V
rms
=
ƒ
P 3
P = 10
5
Pa = 1 atm, ƒ =
3
4
10
10 77 . 1
?
?
?
=
4
3 5
10 77 . 1
10 10 3
?
?
?
? ?
= 1301.8 ˜ 1302 m/s.
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
? T =
23
19
10 38 . 1 3
10 6 . 1 04 . 0 2
?
?
? ?
? ? ?
= 0.0309178 × 10
4
= 309.178 ˜ 310 K
15. V
avg
=
M
RT 8
?
=
032 . 0 14 . 3
300 3 . 8 8
?
? ?
T =
Speed
ce tan Dis
=
25 . 445
2 6400000 ?
= 445.25 m/s
=
3600
83 . 28747
km = 7.985 ˜ 8 hrs.
16. M = 4 × 10
–3
Kg
V
avg
=
M
RT 8
?
=
3
10 4 14 . 3
273 3 . 8 8
?
? ?
? ?
= 1201.35
Momentum = M × V
avg
= 6.64 × 10
–27
× 1201.35 = 7.97 × 10
–24
˜ 8 × 10
–24
Kg-m/s.
P T
V V
P T P 2T
3V V
P 1 -
Kinetic Theory of Gases
24.3
17. V
avg
=
M
RT 8
?
=
032 . 0 14 . 3
300 3 . 8 8
?
? ?
Now,
2
RT 8
1
? ?
=
4
RT 8
2
? ?
2
1
T
T
=
2
1
18. Mean speed of the molecule =
M
RT 8
?
Escape velocity = gr 2
M
RT 8
?
= gr 2 ?
M
RT 8
?
= 2gr
? T =
R 8
M gr 2 ?
=
3 . 8 8
10 2 14 . 3 6400000 8 . 9 2
3
?
? ? ? ? ?
?
= 11863.9 ˜ 11800 m/s.
19. V
avg
=
M
RT 8
?
2 avg
2 avg
N V
H V
=
RT 8
28
2
RT 8 ? ?
?
? ?
=
2
28
= 14 = 3.74
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
1
M
RT 3
=
2
M
RT 8
?
?
1
M
RT 3
=
2
M
RT 8
?
?
2
1
M
M
?
=
8
3
?
2
1
M
M
=
8
3 ?
= 1.1775 ˜ 1.18
21. V
mean
=
M
RT 8
?
=
3
10 2 14 . 3
273 3 . 8 8
?
? ?
? ?
= 1698.96
Total Dist = 1698.96 m
No. of Collisions =
7
10 38 . 1
96 . 1698
?
?
= 1.23 × 10
10
22. P = 1 atm = 10
5
Pascal
T = 300 K, M = 2 g = 2 × 10
–3
Kg
(a) V
avg
=
M
RT 8
?
=
3
10 2 14 . 3
300 3 . 8 8
?
? ?
? ?
= 1781.004 ˜ 1780 m/s
(b) When the molecules strike at an angle 45°,
Force exerted = mV Cos 45° – (–mV Cos 45°) = 2 mV Cos 45° = 2 m V
2
1
= 2 mV
No. of molecules striking per unit area =
Area mv 2
Force
?
=
mV 2
essure Pr
=
23
3
5
10 6
1780 10 2 2
10
?
? ? ?
?
=
31
10
1780 2
3
?
?
= 1.19 × 10
–3
× 10
31
= 1.19 × 10
28
˜ 1.2 × 10
28
23.
1
1 1
T
V P
=
2
2 2
T
V P
P
1
? 200 KPa = 2 × 10
5
pa P
2
= ?
T
1
= 20°C = 293 K T
2
= 40°C = 313 K
V
2
= V
1
+ 2% V
1
=
100
V 102
1
?
?
293
V 10 2
1
5
? ?
=
313 100
V 102 P
1 2
?
? ?
? P
2
=
293 102
313 10 2
7
?
? ?
= 209462 Pa = 209.462 KPa
Kinetic Theory of Gases
24.4
24. V
1
= 1 × 10
–3
m
3
, P
1
= 1.5 × 10
5
Pa, T
1
= 400 K
P
1
V
1
= n
1
R
1
T
1
? n =
1 1
1 1
T R
V P
=
400 3 . 8
10 1 10 5 . 1
3 5
?
? ? ?
?
? n =
4 3 . 8
5 . 1
?
? m
1
= M
4 3 . 8
5 . 1
?
?
= 32
4 3 . 8
5 . 1
?
?
= 1.4457 ˜ 1.446
P
2
= 1 × 10
5
Pa, V
2
= 1 × 10
–3
m
3
, T
2
= 300 K
P
2
V
2
= n
2
R
2
T
2
? n
2
=
2 2
2 2
T R
V P
=
300 3 . 8
10 10
3 5
?
?
?
=
3 . 8 3
1
?
= 0.040
? m
2
= 0.04 × 32 = 1.285
?m = m
1
– m
2
=1.446 – 1.285 = 0.1608 g ˜ 0.16 g ?
25. P
1
= 10
5
+ ƒgh = 10
5
+ 1000 × 10 × 3.3 = 1.33 × 10
5
pa
P
2
= 10
5
, T
1
= T
2
= T, V
1
=
3
4
?(2 × 10
–3
)
3
V
2
=
3
4
?r
3
, r = ?
1
1 1
T
V P
=
2
2 2
T
V P
?
1
3 3 5
T
) 10 2 (
3
4
10 33 . 1
?
? ? ? ? ? ?
=
2
2 5
T
r
3
4
10 ? ? ?
? 1.33 × 8 × 10
5
× 10
–9
= 10
5
× r
3
? r =
3 3
10 64 . 10
?
? = 2.19 × 10
–3
˜ 2.2 mm
26. P
1
= 2 atm = 2 × 10
5
pa
V
1
= 0.002 m
3
, T
1
= 300 K
P
1
V
1
= n
1
RT
1
? n =
1
1 1
RT
V P
=
300 3 . 8
002 . 0 10 2
5
?
? ?
=
3 3 . 8
4
?
= 0.1606
P
2
= 1 atm = 10
5
pa
V
2
= 0.0005 m
3
, T
2
= 300 K
P
2
V
2
= n
2
RT
2
? n
2
=
2
2 2
RT
V P
=
300 3 . 8
0005 . 0 10
5
?
?
=
10
1
3 . 8 3
5
?
?
= 0.02
?n = moles leaked out = 0.16 – 0.02 = 0.14
27. m = 0.040 g, T = 100°C, M
He
= 4 g
U =
2
3
nRt = RT
M
m
2
3
? ? T ? = ?
Given 12 RT
M
m
2
3
? ? ? = T R
M
m
2
3
? ? ?
? 1.5 × 0.01 × 8.3 × 373 + 12 = 1.5 × 0.01 × 8.3 × T ?
? T ? =
1245 . 0
4385 . 58
= 469.3855 K = 196.3°C ˜ 196°C
28. PV
2
= constant
? P
1
V
1
2
= P
2
V
2
2
?
2
1
1
1
V
V
nRT
? =
2
2
2
2
V
V
nRT
?
? T
1
V
1
= T
2
V
2
= TV = T
1
× 2V ? T
2
=
2
T
Kinetic Theory of Gases
24.5
29.
2
O
P =
V
RT n
2
o
,
2
H
P =
V
RT n
2
H
2
O
n =
2
O
M
m
=
32
60 . 1
= 0.05
Now, P
mix
= RT
V
n n
2 2
H O
?
?
?
?
?
?
?
? ?
2
H
n =
2
H
M
m
=
28
80 . 2
= 0.1
P
mix
=
166 . 0
300 3 . 8 ) 1 . 0 05 . 0 ( ? ? ?
= 2250 N/m
2
30. P
1
= Atmospheric pressure = 75 × ƒg
V
1
= 100 × A
P
2
= Atmospheric pressure + Mercury pessue = 75ƒg + hgƒg (if h = height of mercury)
V
2
= (100 – h) A
P
1
V
1
= P
2
V
2
? 75ƒg(100A) = (75 + h)ƒg(100 – h)A
? 75 × 100 = (74 + h) (100 – h) ? 7500 = 7500 – 75 h + 100 h – h
2
? h
2
– 25 h = 0 ? h
2
= 25 h ? h = 25 cm
Height of mercury that can be poured = 25 cm
31. Now, Let the final pressure; Volume & Temp be
After connection = P
A
? ? Partial pressure of A
P
B
? ? Partial pressure of B
Now,
T
V 2 P
A
?
?
=
A
A
T
V P ?
Or
T
P
A
?
=
A
A
T 2
P
…(1)
Similarly,
T
P
B
?
=
B
B
T 2
P
…(2)
T
P
T
P
B A
?
?
?
=
B
B
A
A
T 2
P
T 2
P
? =
?
?
?
?
?
?
?
?
?
B
B
A
A
T
P
T
P
2
1
?
T
P
=
?
?
?
?
?
?
?
?
?
B
B
A
A
T
P
T
P
2
1
[ ? P
A
? + P
B
? = P]
32. V = 50 cc = 50 × 10
–6
cm
3
P = 100 KPa = 10
5
Pa M = 28.8 g
(a) PV = nrT
1
? PV =
1
RT
M
m
? m =
1
RT
PMV
=
273 3 . 8
10 50 8 . 28 10
6 5
?
? ? ?
?
=
273 3 . 8
10 8 . 28 50
1
?
? ?
?
= 0.0635 g.
(b) When the vessel is kept on boiling water
PV =
2
RT
M
m
? m =
2
RT
PVM
=
373 3 . 8
10 50 8 . 28 10
6 5
?
? ? ?
?
=
373 3 . 8
10 8 . 28 50
1
?
? ?
?
= 0.0465
(c) When the vessel is closed
P × 50 × 10
–6
= 273 3 . 8
8 . 28
0465 . 0
? ?
? P =
6
10 50 8 . 28
273 3 . 8 0465 . 0
?
? ?
? ?
= 0.07316 × 10
6
Pa ˜ 73 KPa
B A
P
A
: T
A
V
P
B
: T
B
V
```

## Physics Class 11

127 videos|464 docs|210 tests

## Physics Class 11

127 videos|464 docs|210 tests

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