All Courses
Get the App Signup / Login
Sign in Open App
NEET Exam  >  NEET Notes  >  Physics Class 11  >  HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases

HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases | Physics Class 11 - NEET PDF Download

Download, print and study this document offline
Please wait while the PDF view is loading
 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
Read More
97 videos|378 docs|103 tests
Join Course for Free

Top Courses for NEET

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.
Related Exams
About this Document
1.6K Views
4.66/5 Rating
Nov 21, 2024 Last updated
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.
Introduction of HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases in English is available as part of our Physics Class 11 for NEET & HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases in Hindi for Physics Class 11 course. Download more important topics related with notes, lectures and mock test series for NEET Exam by signing up for free. NEET: HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases | Physics Class 11 - NEET
Description
Full syllabus notes, lecture & questions for HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases | Physics Class 11 - NEET - NEET | Plus excerises question with solution to help you revise complete syllabus for Physics Class 11 | Best notes, free PDF download
Information about HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases
In this doc you can find the meaning of HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases defined & explained in the simplest way possible. Besides explaining types of HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases theory, EduRev gives you an ample number of questions to practice HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases tests, examples and also practice NEET tests
97 videos|378 docs|103 tests
Join Course for Free
Download as PDF
Explore Courses for NEET exam

Top Courses for NEET

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

mock tests for examination

,

ppt

,

shortcuts and tricks

,

Previous Year Questions with Solutions

,

video lectures

,

Sample Paper

,

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

,

practice quizzes

,

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

,

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

,

Semester Notes

,

Objective type Questions

,

pdf

,

Important questions

,

past year papers

,

Summary

,

study material

,

Viva Questions

,

Free

,

Extra Questions

,

MCQs

,

Exam

;
Additional Information about HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases for NEET Preparation

HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases Free PDF Download

The HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases 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 HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases now and kickstart your journey towards success in the NEET exam.

Importance of HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases

The importance of HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases 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.

HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases Notes

HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases Notes offer in-depth insights into the specific topic to help you master it with ease. This comprehensive document covers all aspects related to HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases. 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, HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases Notes on EduRev are your ultimate resource for success.

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

The "HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases 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 HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases on the App

Students of NEET can study HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases alongwith tests & analysis from the EduRev app, which will help them while preparing for their exam. Apart from the HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases, 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 HC Verma Solutions: Chapter 24 - Kinetic Theory of Gases is prepared as per the latest NEET syllabus.
© EduRev
Education Revolution
Follow Us
Signup to see your scores go up
within 7 days!
Takes less than 10 seconds to signup