Thermodynamics: JEE Main Previous Year Questions (2021-2026)

 Page 1


JEE Main Previous Year Questions (2021-2026): 
Thermodynamics  
 
(January 2026) 
 
Q1: The mean free path of a molecule of diameter 5 × 10
-10
 m at the temperature 41 
°
C and 
pressure 1.38 × 10
5
 Pa, is given as ________ m.  
(Given k
B
 = 1.38 × 10
-23
 J/K).  
(a) 2v2 × 10
-10
  
(b) 10v2 × 10
-8
  
(c) 2v2 × 10
-8
  
(d) 2 × 10
-8 
Ans: (c) 
Sol:  
The diameter of molecule is, d = 5 × 10
-10
 m  
Temperature, T 41
°
C = (41 + 273) K = 314 K  
Pressure, P = 1.38 × 10
5
 Pa  
Boltzmann constant, k
B
 = 1.38 × 10
-23
 J / K  
The standard formula for the mean free path ? is: 
 
Substituting the values, 
Page 2


JEE Main Previous Year Questions (2021-2026): 
Thermodynamics  
 
(January 2026) 
 
Q1: The mean free path of a molecule of diameter 5 × 10
-10
 m at the temperature 41 
°
C and 
pressure 1.38 × 10
5
 Pa, is given as ________ m.  
(Given k
B
 = 1.38 × 10
-23
 J/K).  
(a) 2v2 × 10
-10
  
(b) 10v2 × 10
-8
  
(c) 2v2 × 10
-8
  
(d) 2 × 10
-8 
Ans: (c) 
Sol:  
The diameter of molecule is, d = 5 × 10
-10
 m  
Temperature, T 41
°
C = (41 + 273) K = 314 K  
Pressure, P = 1.38 × 10
5
 Pa  
Boltzmann constant, k
B
 = 1.38 × 10
-23
 J / K  
The standard formula for the mean free path ? is: 
 
Substituting the values, 
 
Hence, the correct option is (d). 
 
Q2: Which of the following best represents the temperature versus heat supplied graph 
for water, in the range of - 20
°
C to 120
°
C ? 
(a)  
Page 3


JEE Main Previous Year Questions (2021-2026): 
Thermodynamics  
 
(January 2026) 
 
Q1: The mean free path of a molecule of diameter 5 × 10
-10
 m at the temperature 41 
°
C and 
pressure 1.38 × 10
5
 Pa, is given as ________ m.  
(Given k
B
 = 1.38 × 10
-23
 J/K).  
(a) 2v2 × 10
-10
  
(b) 10v2 × 10
-8
  
(c) 2v2 × 10
-8
  
(d) 2 × 10
-8 
Ans: (c) 
Sol:  
The diameter of molecule is, d = 5 × 10
-10
 m  
Temperature, T 41
°
C = (41 + 273) K = 314 K  
Pressure, P = 1.38 × 10
5
 Pa  
Boltzmann constant, k
B
 = 1.38 × 10
-23
 J / K  
The standard formula for the mean free path ? is: 
 
Substituting the values, 
 
Hence, the correct option is (d). 
 
Q2: Which of the following best represents the temperature versus heat supplied graph 
for water, in the range of - 20
°
C to 120
°
C ? 
(a)  
 
(b)  
 
(c)  
Page 4


JEE Main Previous Year Questions (2021-2026): 
Thermodynamics  
 
(January 2026) 
 
Q1: The mean free path of a molecule of diameter 5 × 10
-10
 m at the temperature 41 
°
C and 
pressure 1.38 × 10
5
 Pa, is given as ________ m.  
(Given k
B
 = 1.38 × 10
-23
 J/K).  
(a) 2v2 × 10
-10
  
(b) 10v2 × 10
-8
  
(c) 2v2 × 10
-8
  
(d) 2 × 10
-8 
Ans: (c) 
Sol:  
The diameter of molecule is, d = 5 × 10
-10
 m  
Temperature, T 41
°
C = (41 + 273) K = 314 K  
Pressure, P = 1.38 × 10
5
 Pa  
Boltzmann constant, k
B
 = 1.38 × 10
-23
 J / K  
The standard formula for the mean free path ? is: 
 
Substituting the values, 
 
Hence, the correct option is (d). 
 
Q2: Which of the following best represents the temperature versus heat supplied graph 
for water, in the range of - 20
°
C to 120
°
C ? 
(a)  
 
(b)  
 
(c)  
 
(d)  
 
Ans: (c) 
Sol:  
To determine the correct graph, we must analyse the phase changes and temperature changes 
that occur when heat is supplied to water starting from - 20
°
C.  
Step 1: Heating Ice (- 20
°
C to 0
°
C)  
Initially, the substance is ice. As heat is supplied, the temperature of the ice increases linearly 
from - 20
°
C to 0
°
C. This is represented by a sloping line on the graph.  
Page 5


JEE Main Previous Year Questions (2021-2026): 
Thermodynamics  
 
(January 2026) 
 
Q1: The mean free path of a molecule of diameter 5 × 10
-10
 m at the temperature 41 
°
C and 
pressure 1.38 × 10
5
 Pa, is given as ________ m.  
(Given k
B
 = 1.38 × 10
-23
 J/K).  
(a) 2v2 × 10
-10
  
(b) 10v2 × 10
-8
  
(c) 2v2 × 10
-8
  
(d) 2 × 10
-8 
Ans: (c) 
Sol:  
The diameter of molecule is, d = 5 × 10
-10
 m  
Temperature, T 41
°
C = (41 + 273) K = 314 K  
Pressure, P = 1.38 × 10
5
 Pa  
Boltzmann constant, k
B
 = 1.38 × 10
-23
 J / K  
The standard formula for the mean free path ? is: 
 
Substituting the values, 
 
Hence, the correct option is (d). 
 
Q2: Which of the following best represents the temperature versus heat supplied graph 
for water, in the range of - 20
°
C to 120
°
C ? 
(a)  
 
(b)  
 
(c)  
 
(d)  
 
Ans: (c) 
Sol:  
To determine the correct graph, we must analyse the phase changes and temperature changes 
that occur when heat is supplied to water starting from - 20
°
C.  
Step 1: Heating Ice (- 20
°
C to 0
°
C)  
Initially, the substance is ice. As heat is supplied, the temperature of the ice increases linearly 
from - 20
°
C to 0
°
C. This is represented by a sloping line on the graph.  
 
Step 2: Melting (Phase Change at 0
°
C)  
At 0
°
C, the ice begins to melt into water. During a phase change, the temperature remains 
constant even as heat is added (latent heat of fusion).  
This is represented by a horizontal line at 0
°
C.  
 
Step 3: Heating Liquid Water (0
°
C to 100
°
C)  
Once all the ice has melted, further heat increases the temperature of the liquid water linearly 
from 0
°
C up to its boiling point at 100
°
C. This is another sloping line.  
 
Step 4: Boiling (Phase Change at 100
°
C)  
At 100
°
C, the water begins to turn into steam. The temperature stays constant during this 
process (latent heat of vaporization). This is represented by a second horizontal line at 100
°
C.  
  
Step 5: Heating Steam (100
°
C to 120
°
C)  
After all the water has turned into steam, the temperature of the steam increases from 100
°
C 
toward 120
°
C. This is represented by a final sloping line.  
 
So, the correct graph must show :  
1. A start point below 0
°
C (at - 20
°
C).  
2. A horizontal plateau at 0
°
C (melting).  
3. A horizontal plateau at 100
°
C (boiling).  
Hence, the correct option is (C). 
 
 
Q3: 10 kg of ice at - 10
°
C is added to 100 kg of water to lower its temperature from 25
°
C. 
Consider no heat exchange to surroundings. The decrement to the temperature of water 
is _______
°
C.  
(specific heat of ice = 2100 J / Kg. 
°
C, specific heat of water = 4200 J / Kg. 
°
C, latent heat 
of fusion of ice = 3.36 × 10
5
 J / Kg)  
(a) 15  
(b) 10  
(c) 6.67  
(d) 11.6  
Ans: (b)