Kinetic Theory Practice Questions - DPP for JEE

 Page 1


1. (b) Here, P
1
 = 200kPa
T
1
 = 22°C = 295 K T
2
 = 42°C = 315K
V
2 
?
2. (d) Molar mass of the gas = 4g/mol
Speed of any quantity x
V =  ? 952 = 
? ? = 1.6 = 
Also, ? = 
So, C
P
 =  = 8JK
–1
mol
–1
3. (a) Q
From ; 
Hence .
4. (a)
Page 2


1. (b) Here, P
1
 = 200kPa
T
1
 = 22°C = 295 K T
2
 = 42°C = 315K
V
2 
?
2. (d) Molar mass of the gas = 4g/mol
Speed of any quantity x
V =  ? 952 = 
? ? = 1.6 = 
Also, ? = 
So, C
P
 =  = 8JK
–1
mol
–1
3. (a) Q
From ; 
Hence .
4. (a)
As, TV
?–1
 = K
So, t ? V
? + 1/2
Therefore, q =
5. (a) According to given Vander Waal’s equation
Work done, 
6. (c) As no heat is lost,
Loss of kinetic energy = gain of internal energy of gas
 ? 
?
7. (a) ...(i)
where K
av
 is the average kinetic energy of the proton.
?
T = 
8. (a) Let T be the temperature of the mixture, then
U = U
1
 + U
2
Page 3


1. (b) Here, P
1
 = 200kPa
T
1
 = 22°C = 295 K T
2
 = 42°C = 315K
V
2 
?
2. (d) Molar mass of the gas = 4g/mol
Speed of any quantity x
V =  ? 952 = 
? ? = 1.6 = 
Also, ? = 
So, C
P
 =  = 8JK
–1
mol
–1
3. (a) Q
From ; 
Hence .
4. (a)
As, TV
?–1
 = K
So, t ? V
? + 1/2
Therefore, q =
5. (a) According to given Vander Waal’s equation
Work done, 
6. (c) As no heat is lost,
Loss of kinetic energy = gain of internal energy of gas
 ? 
?
7. (a) ...(i)
where K
av
 is the average kinetic energy of the proton.
?
T = 
8. (a) Let T be the temperature of the mixture, then
U = U
1
 + U
2
? 
= 
?   (2 + 4)T = 2T
0
 + 8T
0
  (  n
1
 = 2, n
2
 = 4)
?  T = 
9. (c) Given
.......(i)
.......(ii)
From Equation (i) & (ii),
 
 
 
Hence C
P
 = 1.6 C
V
 = 1.6 × 8.33 × 10
3
C
P
 = 1.33 × 10
4
10. (c)  or  
i.e.,  versus m graph is straight line passing through origin with
slope R/M, i.e. the slope depends on molecular mass of the gas M
and is different for different gases.
11. (a) V = 22.4 litre = 22.4 × 10
–3
 m
3
, J = 4200 J/kcal
by ideal gas equation for one mole of a gas,
Page 4


1. (b) Here, P
1
 = 200kPa
T
1
 = 22°C = 295 K T
2
 = 42°C = 315K
V
2 
?
2. (d) Molar mass of the gas = 4g/mol
Speed of any quantity x
V =  ? 952 = 
? ? = 1.6 = 
Also, ? = 
So, C
P
 =  = 8JK
–1
mol
–1
3. (a) Q
From ; 
Hence .
4. (a)
As, TV
?–1
 = K
So, t ? V
? + 1/2
Therefore, q =
5. (a) According to given Vander Waal’s equation
Work done, 
6. (c) As no heat is lost,
Loss of kinetic energy = gain of internal energy of gas
 ? 
?
7. (a) ...(i)
where K
av
 is the average kinetic energy of the proton.
?
T = 
8. (a) Let T be the temperature of the mixture, then
U = U
1
 + U
2
? 
= 
?   (2 + 4)T = 2T
0
 + 8T
0
  (  n
1
 = 2, n
2
 = 4)
?  T = 
9. (c) Given
.......(i)
.......(ii)
From Equation (i) & (ii),
 
 
 
Hence C
P
 = 1.6 C
V
 = 1.6 × 8.33 × 10
3
C
P
 = 1.33 × 10
4
10. (c)  or  
i.e.,  versus m graph is straight line passing through origin with
slope R/M, i.e. the slope depends on molecular mass of the gas M
and is different for different gases.
11. (a) V = 22.4 litre = 22.4 × 10
–3
 m
3
, J = 4200 J/kcal
by ideal gas equation for one mole of a gas,
= 1.979 kcal/kmol K
12. (b)
 and 
(C
p
)
mix 
= 3R = 3 × 2 = 6 cal/mol.°C
? Amount of heat needed to raise the temperature from 0°C to
100°C
13. (c) P-V diagram of the gas is a straight line passing through origin.
Hence P ? V or PV
–1
 = constant
Molar heat capacity in the process PV
x
 = constant is
 Here  (For diatomic gas)
14. (a) Number of moles of first gas = 
Number of moles of second gas = 
Number of moles of third gas = 
If there is no loss of energy then
P
1
V
1
 + P
2
V
2
 + P
3
V
3
 = PV
 = 
Page 5


1. (b) Here, P
1
 = 200kPa
T
1
 = 22°C = 295 K T
2
 = 42°C = 315K
V
2 
?
2. (d) Molar mass of the gas = 4g/mol
Speed of any quantity x
V =  ? 952 = 
? ? = 1.6 = 
Also, ? = 
So, C
P
 =  = 8JK
–1
mol
–1
3. (a) Q
From ; 
Hence .
4. (a)
As, TV
?–1
 = K
So, t ? V
? + 1/2
Therefore, q =
5. (a) According to given Vander Waal’s equation
Work done, 
6. (c) As no heat is lost,
Loss of kinetic energy = gain of internal energy of gas
 ? 
?
7. (a) ...(i)
where K
av
 is the average kinetic energy of the proton.
?
T = 
8. (a) Let T be the temperature of the mixture, then
U = U
1
 + U
2
? 
= 
?   (2 + 4)T = 2T
0
 + 8T
0
  (  n
1
 = 2, n
2
 = 4)
?  T = 
9. (c) Given
.......(i)
.......(ii)
From Equation (i) & (ii),
 
 
 
Hence C
P
 = 1.6 C
V
 = 1.6 × 8.33 × 10
3
C
P
 = 1.33 × 10
4
10. (c)  or  
i.e.,  versus m graph is straight line passing through origin with
slope R/M, i.e. the slope depends on molecular mass of the gas M
and is different for different gases.
11. (a) V = 22.4 litre = 22.4 × 10
–3
 m
3
, J = 4200 J/kcal
by ideal gas equation for one mole of a gas,
= 1.979 kcal/kmol K
12. (b)
 and 
(C
p
)
mix 
= 3R = 3 × 2 = 6 cal/mol.°C
? Amount of heat needed to raise the temperature from 0°C to
100°C
13. (c) P-V diagram of the gas is a straight line passing through origin.
Hence P ? V or PV
–1
 = constant
Molar heat capacity in the process PV
x
 = constant is
 Here  (For diatomic gas)
14. (a) Number of moles of first gas = 
Number of moles of second gas = 
Number of moles of third gas = 
If there is no loss of energy then
P
1
V
1
 + P
2
V
2
 + P
3
V
3
 = PV
 = 
? T
mix
 = 
15. (d) We know that  P V = n R T = (m/M) R T
where M = Molecular weight.
Now ...(1)
where d = density of the gas
... (2)
where R = k N
A
, k is Boltzmann constant.
But = m = mass of each molecule  so
16. (d)
Initially, PV = 
Finally,  (if x g gas leaks out)
Hence,   ?  x = 1 gram
17. (a)
or 
Hence the gas is diatomic.
18. (c)