Kinetic Theory Practice Questions - DPP for NEET

 Page 1


1. (a) Closed vessel i.e., volume is constant
11
22
250
0.4 1
100
PT PT
TK
PTT
PP
Þ = Þ = Þ=
+ æö
+
ç÷
èø
2. (c)
11
22 22
(273 27) 300
2
VT V
VT
VT V TT
+
µÞ=Þ==
2
600 327 TKC Þ ==
o
3. (c) At low pressure and high temperature real gases
behaves like ideal gases.
4. (d)
A AAB
B BBA
N PVT
PV NkT
N PVT
= Þ =´
(2)4
1
2
4
A
B
N PVT
V
N
PT
´´
Þ==
´´
5. (d)
PV mrT =
Since ,, PVr® remains same
Hence
12
2 12
1 13 (273 52) 325
(273 27) 300
mT
m
T m Tm
+
¥Þ=Þ==
+
2
12 m gm Þ=
i.e., mass released =13gm - 12gm = 1gm
6. (c)
12
,
PV PV
RT RT
m = m=
12
() 2
'2
RT PV RT
PP
V RTV
m +m
= = ´=
7. (d)
31
rms rms
kT
vv
m m
= =µ
8. (a)
9. (a)
22
3 3 3 8.3 300
(1920)
rms
rms
RT RT
v MM
M
v
´´
= Þ = \=
3
2102 kg gm
-
=´ =Þ Gas is hydrogen.
10. (d) r .m.s velocity does not depend upon pressure.
11. (c) Average velocity of gas molecule is
81
uu
= Þµ
p
aa
RT
vv
M M
4
2
1
He H
HeH
M C
CM
<>
Þ = ==
<>
H
C Þ<>
2
He
C =<>
12. (b)
rms a mp
v vv
u
>>
13. (a)
1 1 22
12
12
12
57
11
35
57
11
11 3 35
1.5
11
2
57 11
11
35
mix
´´
+
m g mg
æ öæö
+ --
ç ÷ç÷
g - g-
è øèø
g = = ==
mm
+ +
g - g- æ öæö
--
ç ÷ç÷
è øèø
14. (c) We know that
PV
PV
RR
CCJ
J CC
- = Þ=
-
1.98 ,
PV
cal
CC
gm mol K
-=
--
8.32
J
R
gm mol K
=
--
8.32
4.20
1.98
J \==
J / cal
15. (b)
33
= = Þµ
r
rms rms
PPVP
vv
mm
1 12
2 21
Þ =´
v Pm
v Pm
0
20
2
/2
2
2
Þ = ´ Þ=
P vm
PP
v Pm
16. (d)
2
3
PE =
17. (a) For one gm mole; average kinetic energy =
3
2
RT
18. (c) Average kinetic energy µ Temperature
111
21
222
2
2
ETT E
TT
E T ET
Þ = Þ = Þ=
2
2(273 20) 586 313 T KC = + ==
o
19. (d) Kinetic energy per gm mole 
2
f
E RT =
If nothing is said about gas then we should calculate
Page 2


1. (a) Closed vessel i.e., volume is constant
11
22
250
0.4 1
100
PT PT
TK
PTT
PP
Þ = Þ = Þ=
+ æö
+
ç÷
èø
2. (c)
11
22 22
(273 27) 300
2
VT V
VT
VT V TT
+
µÞ=Þ==
2
600 327 TKC Þ ==
o
3. (c) At low pressure and high temperature real gases
behaves like ideal gases.
4. (d)
A AAB
B BBA
N PVT
PV NkT
N PVT
= Þ =´
(2)4
1
2
4
A
B
N PVT
V
N
PT
´´
Þ==
´´
5. (d)
PV mrT =
Since ,, PVr® remains same
Hence
12
2 12
1 13 (273 52) 325
(273 27) 300
mT
m
T m Tm
+
¥Þ=Þ==
+
2
12 m gm Þ=
i.e., mass released =13gm - 12gm = 1gm
6. (c)
12
,
PV PV
RT RT
m = m=
12
() 2
'2
RT PV RT
PP
V RTV
m +m
= = ´=
7. (d)
31
rms rms
kT
vv
m m
= =µ
8. (a)
9. (a)
22
3 3 3 8.3 300
(1920)
rms
rms
RT RT
v MM
M
v
´´
= Þ = \=
3
2102 kg gm
-
=´ =Þ Gas is hydrogen.
10. (d) r .m.s velocity does not depend upon pressure.
11. (c) Average velocity of gas molecule is
81
uu
= Þµ
p
aa
RT
vv
M M
4
2
1
He H
HeH
M C
CM
<>
Þ = ==
<>
H
C Þ<>
2
He
C =<>
12. (b)
rms a mp
v vv
u
>>
13. (a)
1 1 22
12
12
12
57
11
35
57
11
11 3 35
1.5
11
2
57 11
11
35
mix
´´
+
m g mg
æ öæö
+ --
ç ÷ç÷
g - g-
è øèø
g = = ==
mm
+ +
g - g- æ öæö
--
ç ÷ç÷
è øèø
14. (c) We know that
PV
PV
RR
CCJ
J CC
- = Þ=
-
1.98 ,
PV
cal
CC
gm mol K
-=
--
8.32
J
R
gm mol K
=
--
8.32
4.20
1.98
J \==
J / cal
15. (b)
33
= = Þµ
r
rms rms
PPVP
vv
mm
1 12
2 21
Þ =´
v Pm
v Pm
0
20
2
/2
2
2
Þ = ´ Þ=
P vm
PP
v Pm
16. (d)
2
3
PE =
17. (a) For one gm mole; average kinetic energy =
3
2
RT
18. (c) Average kinetic energy µ Temperature
111
21
222
2
2
ETT E
TT
E T ET
Þ = Þ = Þ=
2
2(273 20) 586 313 T KC = + ==
o
19. (d) Kinetic energy per gm mole 
2
f
E RT =
If nothing is said about gas then we should calculate
DPP/ P 26
75
the translational kinetic energy i.e.,
3
33
8.31 (273 0) 3.4 10
22
Trans
ERTJ = = ´ ´ + =´
20. (b) ()
PP
Q CT D =D m
2 (35 30) 7
PP
cal
CC
moleK
Þ ´ ´ - Þ=
-
PV
C CR -= Q
7 25
VP
cal
C CR
mole kelvin
Þ = - = -=
-
()
VV
Q CT \D =D m
2 5 (35 30) 50 cal = ´´ -=
21. (a) Average kinetic energy per molecule per degree of
freedom 
1
2
kT =
. Since both the gases are diatomic
and at same temperature (300 K), both will have the
same number of rotational degree of freedom i.e. two.
Therefore, both the gases will have the same average
rotational kinetic energy per molecule
1
2
2
kT kT
æö
=´=
ç÷
èø
.
Thus 
1
2
1
1
E
E
=
22. (a)
Coefficient of volume expansion at constant pressure is
1
273
 for all gases. The average transnational K.E. is same
for molecule of all gases and for each molecules it is 
3
kT
2
Mean free path 
2
kT
?
2pdP
= (as P decreases, l increases)
23. (b) 
rms P rms
3RT 2RT
v ,v 0.816v
MM
= ==
rms P rms
8RT
v 0.92 v v v v
pM
= = Þ <<
Further 
2 22
avrms PP
1 133
E mv m v mv
2 224
= ==
24. (d) According to problem mass of gases are equal so
number of moles will not be equal i.e. 
AB
µµ ¹
From ideal gas equation 
A A BB
AB
P V PV
PV µRT
µµ
= Þ=
[As temperature of the container are equal]
From this relation it is clear that if 
AB
PP = then
AA
AB
BB
V µ
1i.e.VV
V µ
= ¹¹
Similarly if 
AB
VV = then 
A
AB
BB
µ PA
1i.e.P P.
P µ
= ¹¹
25. (b) n
1
C
v
 (T – T
0
) + n
2
 C
v
 (T – 2T
0
) = 0
0
3
2
TT =
0
3
2
if
f
i
PT
PP
T
==
26. (c)
10
( –)
vf
Q nCTT D=
00
0 0 00
0
333
–
2 228
PV
R T T PV
RT
æö
= ´´=
ç÷
èø
27. (c) Let DV is change in volume in any compartment then
0
00
1
0
–
2
2
f
f
V
PV
PV
n
RT RT
æö
D
ç÷
èø
== and
0
00
2
0
2 2
2
f
f
V
PV
PV
n
RT RT
æö
+D
ç÷
èø
== 0 V ÞD=
28. (b) Internal energy of an ideal gas does not depend upon
volume of the gas, because there are no forces of
attration/repulsion amongest the molecular of an ideal
gas.
Also internal energy of an ideal gas depends on
temperature.
29. (b) Helium is a monoatomic gas, while oxygen is diatomic.
Therefore, the heat given to helium will be totally used
up in increasing the translational kinetic energy of its
molecules; whereas the heat given to oxygen will be
used up in increasing the translational kinetic energy
of the molecule and also in increasing the kinetic energy
of rotation and vibration. Hence there will be a greater
rise in the temperature of helium.
30. (d) Maxwell speed distribution graph is asymmetric graph,
because it has a long “tail” that extends to infinity .
Also v
rms
 depends upon nature of the gas and it’s
temperature.
v
rms
v
mp
v
av
v
dN
dv