JEE Main Previous Year Questions (2026): Thermodynamics

 Page 1


JEE Main Previous Year Questions 
(2025): Thermodynamics 
Q1: A liquid when kept inside a thermally insulated closed vessel at ????
°
?? was 
mechanically stirred from outside. What will be the correct option for the following 
thermodynamic parameters? 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. ?U = 0, q < 0, w > 0 
B. ?U > 0, q = 0, w > 0 
C. ??? = 0, ?? = 0, ?? = 0 
D. ??? < 0, ?? = 0, ?? > 0 
Ans: B 
Solution: 
Let's analyze the situation step by step: 
System: The liquid inside a thermally insulated (adiabatic) and closed vessel. 
Process: The liquid is mechanically stirred from outside. 
Heat Exchange (?? ) 
Because the vessel is thermally insulated, there is no heat exchange between the system and 
surroundings. Hence, 
?? = 0. 
Work ( ?? ) 
The external mechanical stirring does work on the system by agitating the liquid. Work done on 
the system is conventionally taken as positive: 
?? > 0. 
Change in Internal Energy ( ?? ?? ) 
From the First Law of Thermodynamics, 
??? = ?? + ?? . 
Since ?? = 0 and ?? > 0, we get 
??? = ?? > 0. 
Hence: 
??? > 0, ?? = 0, ?? > 0. 
Page 2


JEE Main Previous Year Questions 
(2025): Thermodynamics 
Q1: A liquid when kept inside a thermally insulated closed vessel at ????
°
?? was 
mechanically stirred from outside. What will be the correct option for the following 
thermodynamic parameters? 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. ?U = 0, q < 0, w > 0 
B. ?U > 0, q = 0, w > 0 
C. ??? = 0, ?? = 0, ?? = 0 
D. ??? < 0, ?? = 0, ?? > 0 
Ans: B 
Solution: 
Let's analyze the situation step by step: 
System: The liquid inside a thermally insulated (adiabatic) and closed vessel. 
Process: The liquid is mechanically stirred from outside. 
Heat Exchange (?? ) 
Because the vessel is thermally insulated, there is no heat exchange between the system and 
surroundings. Hence, 
?? = 0. 
Work ( ?? ) 
The external mechanical stirring does work on the system by agitating the liquid. Work done on 
the system is conventionally taken as positive: 
?? > 0. 
Change in Internal Energy ( ?? ?? ) 
From the First Law of Thermodynamics, 
??? = ?? + ?? . 
Since ?? = 0 and ?? > 0, we get 
??? = ?? > 0. 
Hence: 
??? > 0, ?? = 0, ?? > 0. 
Ans: Option B is correct. 
 
Q2: Match List - I with List - II. 
 List - I (Partial Derivatives)  List - II (Thermodynamic Quantity) 
(A) 
(
?? ?? ?? ?? )
?? 
(I) Cp 
(B) 
(
?? ?? ?? ?? )
?? 
(II) -S 
(C) 
(
?? ?? ?? ?? )
?? 
(III) Cv 
(D) 
(
?? ?? ?? ?? )
?? 
(IV) V 
Choose the correct answer from the options given below : 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. (A)-(I), (B)-(II), (C)-(IV), (D)-(III) 
B. (A)-(II), (B)-(III), (C)-(I), (D)-(IV) 
C. (A)-(II), (B)-(I), (C)-(IV), (D)-(III) 
D. (A)-(II), (B)-(I), (C)-(III), (D)-(IV) 
Ans: C 
Solution: 
We want to match each partial derivative in List-I to the appropriate thermodynamic quantity in 
List-II: 
1. (
????
????
)
?? 
A well-known thermodynamic identity is: 
(
????
????
)
?? = -?? . 
Hence, 
(?? ) ? -?? (II). 
2. (
????
????
)
?? 
By definition of the heat capacity at constant pressure, 
(
????
????
)
?? = ?? ?? . 
Hence, 
Page 3


JEE Main Previous Year Questions 
(2025): Thermodynamics 
Q1: A liquid when kept inside a thermally insulated closed vessel at ????
°
?? was 
mechanically stirred from outside. What will be the correct option for the following 
thermodynamic parameters? 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. ?U = 0, q < 0, w > 0 
B. ?U > 0, q = 0, w > 0 
C. ??? = 0, ?? = 0, ?? = 0 
D. ??? < 0, ?? = 0, ?? > 0 
Ans: B 
Solution: 
Let's analyze the situation step by step: 
System: The liquid inside a thermally insulated (adiabatic) and closed vessel. 
Process: The liquid is mechanically stirred from outside. 
Heat Exchange (?? ) 
Because the vessel is thermally insulated, there is no heat exchange between the system and 
surroundings. Hence, 
?? = 0. 
Work ( ?? ) 
The external mechanical stirring does work on the system by agitating the liquid. Work done on 
the system is conventionally taken as positive: 
?? > 0. 
Change in Internal Energy ( ?? ?? ) 
From the First Law of Thermodynamics, 
??? = ?? + ?? . 
Since ?? = 0 and ?? > 0, we get 
??? = ?? > 0. 
Hence: 
??? > 0, ?? = 0, ?? > 0. 
Ans: Option B is correct. 
 
Q2: Match List - I with List - II. 
 List - I (Partial Derivatives)  List - II (Thermodynamic Quantity) 
(A) 
(
?? ?? ?? ?? )
?? 
(I) Cp 
(B) 
(
?? ?? ?? ?? )
?? 
(II) -S 
(C) 
(
?? ?? ?? ?? )
?? 
(III) Cv 
(D) 
(
?? ?? ?? ?? )
?? 
(IV) V 
Choose the correct answer from the options given below : 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. (A)-(I), (B)-(II), (C)-(IV), (D)-(III) 
B. (A)-(II), (B)-(III), (C)-(I), (D)-(IV) 
C. (A)-(II), (B)-(I), (C)-(IV), (D)-(III) 
D. (A)-(II), (B)-(I), (C)-(III), (D)-(IV) 
Ans: C 
Solution: 
We want to match each partial derivative in List-I to the appropriate thermodynamic quantity in 
List-II: 
1. (
????
????
)
?? 
A well-known thermodynamic identity is: 
(
????
????
)
?? = -?? . 
Hence, 
(?? ) ? -?? (II). 
2. (
????
????
)
?? 
By definition of the heat capacity at constant pressure, 
(
????
????
)
?? = ?? ?? . 
Hence, 
(?? ) ? ?? ??  (I). 
3. (
????
????
)
?? 
A standard thermodynamic relation (for a single-component system) is: 
(
????
????
)
?? = ?? . 
Hence, 
(?? ) ? ?? (IV). 
4. (
????
????
)
?? 
By definition of the heat capacity at constant volume, 
(
????
????
)
?? = ?? ?? . 
Hence, 
(?? ) ? ?? ??  (III). 
Final Matching 
(?? ) ? (II), 
(?? ) ? (I), 
(?? ) ? (IV), 
(?? ) ? (III). 
Which corresponds to: 
Option C: (A)-(II), (B)-(I), (C)-(IV), (D)-(III). 
Q3: Ice at -?? °
?? is heated to become vapor with temperature of ??????
°
?? at atmospheric 
pressure. The entropy change associated with this process can be obtained from 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. ?
268 K
273 K
?C
p,m
dT+
?H
m
, fusion 
T
f
+
?H
m, vaporisation 
T
b
+ ?
273 K
373 K
?C
p,m
dT+ ?
373 K
383 K
?C
p,m
dT 
B. ?
268 K
383 K
?C
p
dT+
?H
melting 
273
+
?H
boiling 
373
 
C. ?
268 K
383 K
?C
p
dT+
q
rev
T
 
D. ?
268 K
273 K
?
C
p,m
T
dT+
?H
m
, fusion 
T
f
+
?H
m, vaporisation 
373
 T
b
+ ?
273 K
?
C
p,m
dT
?? + ?
373 K
383 K
?
C
p,m
dT
T
 
Ans: D 
Solution: 
Page 4


JEE Main Previous Year Questions 
(2025): Thermodynamics 
Q1: A liquid when kept inside a thermally insulated closed vessel at ????
°
?? was 
mechanically stirred from outside. What will be the correct option for the following 
thermodynamic parameters? 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. ?U = 0, q < 0, w > 0 
B. ?U > 0, q = 0, w > 0 
C. ??? = 0, ?? = 0, ?? = 0 
D. ??? < 0, ?? = 0, ?? > 0 
Ans: B 
Solution: 
Let's analyze the situation step by step: 
System: The liquid inside a thermally insulated (adiabatic) and closed vessel. 
Process: The liquid is mechanically stirred from outside. 
Heat Exchange (?? ) 
Because the vessel is thermally insulated, there is no heat exchange between the system and 
surroundings. Hence, 
?? = 0. 
Work ( ?? ) 
The external mechanical stirring does work on the system by agitating the liquid. Work done on 
the system is conventionally taken as positive: 
?? > 0. 
Change in Internal Energy ( ?? ?? ) 
From the First Law of Thermodynamics, 
??? = ?? + ?? . 
Since ?? = 0 and ?? > 0, we get 
??? = ?? > 0. 
Hence: 
??? > 0, ?? = 0, ?? > 0. 
Ans: Option B is correct. 
 
Q2: Match List - I with List - II. 
 List - I (Partial Derivatives)  List - II (Thermodynamic Quantity) 
(A) 
(
?? ?? ?? ?? )
?? 
(I) Cp 
(B) 
(
?? ?? ?? ?? )
?? 
(II) -S 
(C) 
(
?? ?? ?? ?? )
?? 
(III) Cv 
(D) 
(
?? ?? ?? ?? )
?? 
(IV) V 
Choose the correct answer from the options given below : 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. (A)-(I), (B)-(II), (C)-(IV), (D)-(III) 
B. (A)-(II), (B)-(III), (C)-(I), (D)-(IV) 
C. (A)-(II), (B)-(I), (C)-(IV), (D)-(III) 
D. (A)-(II), (B)-(I), (C)-(III), (D)-(IV) 
Ans: C 
Solution: 
We want to match each partial derivative in List-I to the appropriate thermodynamic quantity in 
List-II: 
1. (
????
????
)
?? 
A well-known thermodynamic identity is: 
(
????
????
)
?? = -?? . 
Hence, 
(?? ) ? -?? (II). 
2. (
????
????
)
?? 
By definition of the heat capacity at constant pressure, 
(
????
????
)
?? = ?? ?? . 
Hence, 
(?? ) ? ?? ??  (I). 
3. (
????
????
)
?? 
A standard thermodynamic relation (for a single-component system) is: 
(
????
????
)
?? = ?? . 
Hence, 
(?? ) ? ?? (IV). 
4. (
????
????
)
?? 
By definition of the heat capacity at constant volume, 
(
????
????
)
?? = ?? ?? . 
Hence, 
(?? ) ? ?? ??  (III). 
Final Matching 
(?? ) ? (II), 
(?? ) ? (I), 
(?? ) ? (IV), 
(?? ) ? (III). 
Which corresponds to: 
Option C: (A)-(II), (B)-(I), (C)-(IV), (D)-(III). 
Q3: Ice at -?? °
?? is heated to become vapor with temperature of ??????
°
?? at atmospheric 
pressure. The entropy change associated with this process can be obtained from 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. ?
268 K
273 K
?C
p,m
dT+
?H
m
, fusion 
T
f
+
?H
m, vaporisation 
T
b
+ ?
273 K
373 K
?C
p,m
dT+ ?
373 K
383 K
?C
p,m
dT 
B. ?
268 K
383 K
?C
p
dT+
?H
melting 
273
+
?H
boiling 
373
 
C. ?
268 K
383 K
?C
p
dT+
q
rev
T
 
D. ?
268 K
273 K
?
C
p,m
T
dT+
?H
m
, fusion 
T
f
+
?H
m, vaporisation 
373
 T
b
+ ?
273 K
?
C
p,m
dT
?? + ?
373 K
383 K
?
C
p,m
dT
T
 
Ans: D 
Solution: 
 
 
?S
overall 
= ?S
1
+ ?S
2
+ ?S
3
+ ?S
4
+ ?S
5
 
?S
2
=
?H
m
 fusion 
273
 T
f
= 273
'
K
'
 
?S
3
= ?
273
373
?
C
p,m
dT
T
 
?S
4
=
?H
m vaporisation 
373
 T
b
= 373
'
K
'
 
?S
5
= ?
373
383
?
C
p,m
dT
T
 
Answer = (2) 
Q4: The effect of temperature on spontaneity of reactions are represented as : 
 ???? ???? Temperature Spontaneity 
(A) + - any T Non spontaneous 
(B) + + low T spontaneous 
(C) - - low T Non spontaneous 
(D) - + any T spontaneous 
The incorrect combinations are : 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. (A) and (C) only 
B. (B) and (D) only 
C. (A) and (D) only 
D. (B) and (C) only 
Ans: D 
Solution: 
Let's analyze each combination using the Gibbs free energy equation: 
??? = ??? - ?? ??? 
A reaction is spontaneous when ??? < 0. 
Now, consider each case: 
(A) ??? > 0 and ??? < 0 
Page 5


JEE Main Previous Year Questions 
(2025): Thermodynamics 
Q1: A liquid when kept inside a thermally insulated closed vessel at ????
°
?? was 
mechanically stirred from outside. What will be the correct option for the following 
thermodynamic parameters? 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. ?U = 0, q < 0, w > 0 
B. ?U > 0, q = 0, w > 0 
C. ??? = 0, ?? = 0, ?? = 0 
D. ??? < 0, ?? = 0, ?? > 0 
Ans: B 
Solution: 
Let's analyze the situation step by step: 
System: The liquid inside a thermally insulated (adiabatic) and closed vessel. 
Process: The liquid is mechanically stirred from outside. 
Heat Exchange (?? ) 
Because the vessel is thermally insulated, there is no heat exchange between the system and 
surroundings. Hence, 
?? = 0. 
Work ( ?? ) 
The external mechanical stirring does work on the system by agitating the liquid. Work done on 
the system is conventionally taken as positive: 
?? > 0. 
Change in Internal Energy ( ?? ?? ) 
From the First Law of Thermodynamics, 
??? = ?? + ?? . 
Since ?? = 0 and ?? > 0, we get 
??? = ?? > 0. 
Hence: 
??? > 0, ?? = 0, ?? > 0. 
Ans: Option B is correct. 
 
Q2: Match List - I with List - II. 
 List - I (Partial Derivatives)  List - II (Thermodynamic Quantity) 
(A) 
(
?? ?? ?? ?? )
?? 
(I) Cp 
(B) 
(
?? ?? ?? ?? )
?? 
(II) -S 
(C) 
(
?? ?? ?? ?? )
?? 
(III) Cv 
(D) 
(
?? ?? ?? ?? )
?? 
(IV) V 
Choose the correct answer from the options given below : 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. (A)-(I), (B)-(II), (C)-(IV), (D)-(III) 
B. (A)-(II), (B)-(III), (C)-(I), (D)-(IV) 
C. (A)-(II), (B)-(I), (C)-(IV), (D)-(III) 
D. (A)-(II), (B)-(I), (C)-(III), (D)-(IV) 
Ans: C 
Solution: 
We want to match each partial derivative in List-I to the appropriate thermodynamic quantity in 
List-II: 
1. (
????
????
)
?? 
A well-known thermodynamic identity is: 
(
????
????
)
?? = -?? . 
Hence, 
(?? ) ? -?? (II). 
2. (
????
????
)
?? 
By definition of the heat capacity at constant pressure, 
(
????
????
)
?? = ?? ?? . 
Hence, 
(?? ) ? ?? ??  (I). 
3. (
????
????
)
?? 
A standard thermodynamic relation (for a single-component system) is: 
(
????
????
)
?? = ?? . 
Hence, 
(?? ) ? ?? (IV). 
4. (
????
????
)
?? 
By definition of the heat capacity at constant volume, 
(
????
????
)
?? = ?? ?? . 
Hence, 
(?? ) ? ?? ??  (III). 
Final Matching 
(?? ) ? (II), 
(?? ) ? (I), 
(?? ) ? (IV), 
(?? ) ? (III). 
Which corresponds to: 
Option C: (A)-(II), (B)-(I), (C)-(IV), (D)-(III). 
Q3: Ice at -?? °
?? is heated to become vapor with temperature of ??????
°
?? at atmospheric 
pressure. The entropy change associated with this process can be obtained from 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. ?
268 K
273 K
?C
p,m
dT+
?H
m
, fusion 
T
f
+
?H
m, vaporisation 
T
b
+ ?
273 K
373 K
?C
p,m
dT+ ?
373 K
383 K
?C
p,m
dT 
B. ?
268 K
383 K
?C
p
dT+
?H
melting 
273
+
?H
boiling 
373
 
C. ?
268 K
383 K
?C
p
dT+
q
rev
T
 
D. ?
268 K
273 K
?
C
p,m
T
dT+
?H
m
, fusion 
T
f
+
?H
m, vaporisation 
373
 T
b
+ ?
273 K
?
C
p,m
dT
?? + ?
373 K
383 K
?
C
p,m
dT
T
 
Ans: D 
Solution: 
 
 
?S
overall 
= ?S
1
+ ?S
2
+ ?S
3
+ ?S
4
+ ?S
5
 
?S
2
=
?H
m
 fusion 
273
 T
f
= 273
'
K
'
 
?S
3
= ?
273
373
?
C
p,m
dT
T
 
?S
4
=
?H
m vaporisation 
373
 T
b
= 373
'
K
'
 
?S
5
= ?
373
383
?
C
p,m
dT
T
 
Answer = (2) 
Q4: The effect of temperature on spontaneity of reactions are represented as : 
 ???? ???? Temperature Spontaneity 
(A) + - any T Non spontaneous 
(B) + + low T spontaneous 
(C) - - low T Non spontaneous 
(D) - + any T spontaneous 
The incorrect combinations are : 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. (A) and (C) only 
B. (B) and (D) only 
C. (A) and (D) only 
D. (B) and (C) only 
Ans: D 
Solution: 
Let's analyze each combination using the Gibbs free energy equation: 
??? = ??? - ?? ??? 
A reaction is spontaneous when ??? < 0. 
Now, consider each case: 
(A) ??? > 0 and ??? < 0 
Here, ??? = +( positive ) - ?? (-( positive )) = positive +?? ( positive ) 
This is positive for any temperature. 
Thus, the reaction is non-spontaneous for any T , which is correct. 
(B) ??? > 0 and ??? > 0 
Now, ??? = +( positive ) - ?? ( positive ). 
For ??? to be negative, the temperature has to be high enough such that ?? ??? > ??? . 
The table incorrectly states that the reaction is spontaneous at low T . 
(C) ??? < 0 and ??? < 0 
In this case, ??? = -( positive ) - ?? (-( positive )) = -( positive ) + ?? ( positive ). 
At very low temperatures, the term ?? ??? is small, making ??? negative (spontaneous). 
The table mistakenly indicates that the reaction is non-spontaneous at low T . 
(D) ??? < 0 and ??? > 0 
Then, ??? = -( positive ) - ?? ( positive ). 
This expression is negative at all temperatures, so the reaction is correctly marked as 
spontaneous. 
Thus, the incorrect combinations are: 
(B): It should be spontaneous at high T , not low T . 
(C): It should be spontaneous at low T, not non-spontaneous. 
Therefore, the incorrect combinations are (B) and (C) only. 
The correct answer is Option D. 
Q5: Let us consider an endothermic reaction which is non-spontaneous at the freezing 
point of water. However, the reaction is spontaneous at boiling point of water. 
Choose the correct option. 
JEE Main 2025 (Online) 24th January Morning Shift 
Options: 
A. Both ?H and ?S are (-ve) 
B. ?H is (+ve ) but ?S is (-ve) 
C. ?H is ( -ve ) but ?S is ( +ve ) 
D. Both ?H and ?S are ( +ve ) 
Ans: D 
Solution: 
Reaction is spontaneous at relatively high temperature and non-spontaneous at low 
temperature ?G = ?H - T?SIt is only possible when ?H and ?S both are positive. 
Option (1)