Thermodynamics Practice Questions - DPP for JEE

 Page 1


1. (a) U = a + bPV ......(1)
In adiabatic change,
dU = – dW =  = 
? 
or ......(2)
where a is the constant of integration.
Comparing (1) and (2), we get
2. (d) For path ab : 
By using 
For path ca :
  ...(i)
  ...(ii)
Also 
= 0.48 × 8.31 × (300 – 1000) = –2792.16J  ...(iii)
On solving equations (i), (ii) and (iii)
Page 2


1. (a) U = a + bPV ......(1)
In adiabatic change,
dU = – dW =  = 
? 
or ......(2)
where a is the constant of integration.
Comparing (1) and (2), we get
2. (d) For path ab : 
By using 
For path ca :
  ...(i)
  ...(ii)
Also 
= 0.48 × 8.31 × (300 – 1000) = –2792.16J  ...(iii)
On solving equations (i), (ii) and (iii)
3. (a) Here Q = 0 and W = 0. Therefore from first law of
thermodynamics ?U = Q + W = 0
 Internal energy of the system with partition = Internal energy of
the system without partition.
But 
 
4. (a) As, 
But 
So,
or [As PV = u RT]
Therefore, T ? 
5. (c) Coefficient of performance of a refrigerator,
Page 3


1. (a) U = a + bPV ......(1)
In adiabatic change,
dU = – dW =  = 
? 
or ......(2)
where a is the constant of integration.
Comparing (1) and (2), we get
2. (d) For path ab : 
By using 
For path ca :
  ...(i)
  ...(ii)
Also 
= 0.48 × 8.31 × (300 – 1000) = –2792.16J  ...(iii)
On solving equations (i), (ii) and (iii)
3. (a) Here Q = 0 and W = 0. Therefore from first law of
thermodynamics ?U = Q + W = 0
 Internal energy of the system with partition = Internal energy of
the system without partition.
But 
 
4. (a) As, 
But 
So,
or [As PV = u RT]
Therefore, T ? 
5. (c) Coefficient of performance of a refrigerator,
ß =   (Where Q
2
 is heat removed)
Given: T
2
 = 4°C = 4 + 273 = 277 k
T
1
 = 30°C = 30 + 273 = 303 k
? ß = 
? W = 236.5 joule
Power  P =  =  = 236.5 watt.
6. (a) Change in internal energy do not depend upon the path followed
by the process. It only depends on initial and final states i.e.,    ?U
1
= ?U
2
7. (b)
We have,  
For diatomic gas, 
 
Now, efficiency = 
Page 4


1. (a) U = a + bPV ......(1)
In adiabatic change,
dU = – dW =  = 
? 
or ......(2)
where a is the constant of integration.
Comparing (1) and (2), we get
2. (d) For path ab : 
By using 
For path ca :
  ...(i)
  ...(ii)
Also 
= 0.48 × 8.31 × (300 – 1000) = –2792.16J  ...(iii)
On solving equations (i), (ii) and (iii)
3. (a) Here Q = 0 and W = 0. Therefore from first law of
thermodynamics ?U = Q + W = 0
 Internal energy of the system with partition = Internal energy of
the system without partition.
But 
 
4. (a) As, 
But 
So,
or [As PV = u RT]
Therefore, T ? 
5. (c) Coefficient of performance of a refrigerator,
ß =   (Where Q
2
 is heat removed)
Given: T
2
 = 4°C = 4 + 273 = 277 k
T
1
 = 30°C = 30 + 273 = 303 k
? ß = 
? W = 236.5 joule
Power  P =  =  = 236.5 watt.
6. (a) Change in internal energy do not depend upon the path followed
by the process. It only depends on initial and final states i.e.,    ?U
1
= ?U
2
7. (b)
We have,  
For diatomic gas, 
 
Now, efficiency = 
=  =  = 
8. (c) The equation for the line is
P = 
[slope = , c = 3P
0
]
PV
0
 + P
0
V = 3P
0
V
0
...(i)
But PV =  nRT 
? P = ...(ii)
From (i) & (ii) V
0
 + P
0
V = 3P
0
V
0
? nRT V
0
 + P
0
V
2
 = 3P
0
V
0
V ...(iii)
For temperature to be maximum  = 0
Differentiating e.q. (iii) by ‘V’ we get
nRV
0
+ P
0
(2V) = 3P
0
V
0
? nRV
0
 =  3P
0
V
0 
– 2 P
0
V
 = = 0
Page 5


1. (a) U = a + bPV ......(1)
In adiabatic change,
dU = – dW =  = 
? 
or ......(2)
where a is the constant of integration.
Comparing (1) and (2), we get
2. (d) For path ab : 
By using 
For path ca :
  ...(i)
  ...(ii)
Also 
= 0.48 × 8.31 × (300 – 1000) = –2792.16J  ...(iii)
On solving equations (i), (ii) and (iii)
3. (a) Here Q = 0 and W = 0. Therefore from first law of
thermodynamics ?U = Q + W = 0
 Internal energy of the system with partition = Internal energy of
the system without partition.
But 
 
4. (a) As, 
But 
So,
or [As PV = u RT]
Therefore, T ? 
5. (c) Coefficient of performance of a refrigerator,
ß =   (Where Q
2
 is heat removed)
Given: T
2
 = 4°C = 4 + 273 = 277 k
T
1
 = 30°C = 30 + 273 = 303 k
? ß = 
? W = 236.5 joule
Power  P =  =  = 236.5 watt.
6. (a) Change in internal energy do not depend upon the path followed
by the process. It only depends on initial and final states i.e.,    ?U
1
= ?U
2
7. (b)
We have,  
For diatomic gas, 
 
Now, efficiency = 
=  =  = 
8. (c) The equation for the line is
P = 
[slope = , c = 3P
0
]
PV
0
 + P
0
V = 3P
0
V
0
...(i)
But PV =  nRT 
? P = ...(ii)
From (i) & (ii) V
0
 + P
0
V = 3P
0
V
0
? nRT V
0
 + P
0
V
2
 = 3P
0
V
0
V ...(iii)
For temperature to be maximum  = 0
Differentiating e.q. (iii) by ‘V’ we get
nRV
0
+ P
0
(2V) = 3P
0
V
0
? nRV
0
 =  3P
0
V
0 
– 2 P
0
V
 = = 0
V = [From (i)]
? T
max
 = [From (iii)]
9. (a) V
1
 = = 1000 cm
3
,  P
1
 = 72 cm of Hg.
V
2
 = 900 cm
3
, P
2
 = ?
 The process is isothermal
 P
1
V
1
 = P
2
V
2
72 × 1000 = P
2
 × 900
P
2 
= 80 cm of Hg
 Stress = P
2
 – P
1
 = 80 – 72 = 8 cm of Hg.
10. (d)
and 
We know that = constant
or , 
11. (a) Initial and final condition is same for all process
 ?U
1
 = ?U
2
 = ?U
3
from first law of thermodynamics
   ?Q = ?U + ?W
Work done
  ?W
1
 > ?W
2
 > ?W
3
 (Area of P.V. graph)
So ?Q
1
 > ?Q
2
 > ?Q
3
12. (b) According to first law of thermodynamics,
?Q = heat absorbed by gas
?W = work done by gas.
–20J = ?U – 8J
?U = –12J = U
Final
 – U
initial