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 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)
Adding (1) & (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
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FAQs on HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases - Physics Class 11 - NEET

1. What is the kinetic theory of gases?
Ans. The kinetic theory of gases is a fundamental theory that explains the behavior of gases based on the motion of their particles. It states that gases consist of tiny particles (atoms or molecules) that are in constant random motion and have negligible volume compared to the volume of the container they are in.
2. How does the kinetic theory of gases explain the pressure exerted by a gas?
Ans. According to the kinetic theory of gases, the pressure exerted by a gas is a result of the collisions between the gas particles and the walls of the container. As the gas particles move randomly, they collide with the walls, exerting a force on them. The pressure is the average force per unit area exerted by these collisions.
3. What is the relationship between temperature and the average kinetic energy of gas particles?
Ans. According to the kinetic theory of gases, the average kinetic energy of gas particles is directly proportional to the temperature of the gas. As the temperature increases, the particles gain more kinetic energy and move faster. Conversely, as the temperature decreases, the particles have lower kinetic energy and move slower.
4. How does the kinetic theory of gases explain the expansion of gases?
Ans. The kinetic theory of gases explains the expansion of gases by the concept of increased particle motion. When gases are heated, their particles gain kinetic energy and move with higher velocities. This increased motion leads to an increase in the pressure exerted by the gas, causing expansion to occur in order to occupy a larger volume.
5. What are the assumptions of the kinetic theory of gases?
Ans. The kinetic theory of gases is based on the following assumptions: 1. Gas particles are considered to be point masses with negligible volume. 2. The motion of gas particles is random and follows a straight-line path until they collide with other particles or the walls of the container. 3. Gas particle collisions are elastic, meaning that no energy is lost during collisions. 4. The average kinetic energy of gas particles is directly proportional to the absolute temperature of the gas. 5. There are no intermolecular forces between gas particles.
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Document Description: HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases for NEET 2024 is part of Physics Class 11 preparation. The notes and questions for HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases have been prepared according to the NEET exam syllabus. Information about HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases covers topics like and HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases Example, for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases.
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