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

 Page 1


JEE Main Previous Year Questions (2021-2026): 
Thermodynamics  
 
(January 2026) 
 
Q1:  The plot of log 1 0 K vs 1/T gives a straight line. The intercept and slope respectively 
are (where K is the equilibrium constant). 
A:  
B:  
C:  
D:  
Answer: C 
Explanation:
 
 
Q2: 20.0 dm³ of an ideal gas 'X' at 600 K and 0.5 MPa undergoes isothermal reversible 
expansion until the pressure of the gas is 0.2 MPa. 
Which of the following option is correct? 
(Given: log 2 = 0.3010 and log 5 = 0.6989) 
A: w = -9.1 kJ, ?U = 0, ?H = 0, q = 9.1 kJ 
B: w = 9.1 J, ?U = 9.1 J, ?H = 0; q = 0 
C: w = -3.9 kJ, ?U = 0, ?H = 0; q = 3.9 kJ 
D: w = +4.1 kJ, ?U = 0, ?H = 0; q = -4.1 kJ 
Answer: A 
Explanation: 
For an ideal gas undergoing isothermal reversible expansion: 
Temperature is constant ? ?U = 0 (for ideal gas, U depends only on T) 
Hence ?H = 0 (since H also depends only on T for ideal gas) 
Step 1: Given data (convert units) 
Page 2


JEE Main Previous Year Questions (2021-2026): 
Thermodynamics  
 
(January 2026) 
 
Q1:  The plot of log 1 0 K vs 1/T gives a straight line. The intercept and slope respectively 
are (where K is the equilibrium constant). 
A:  
B:  
C:  
D:  
Answer: C 
Explanation:
 
 
Q2: 20.0 dm³ of an ideal gas 'X' at 600 K and 0.5 MPa undergoes isothermal reversible 
expansion until the pressure of the gas is 0.2 MPa. 
Which of the following option is correct? 
(Given: log 2 = 0.3010 and log 5 = 0.6989) 
A: w = -9.1 kJ, ?U = 0, ?H = 0, q = 9.1 kJ 
B: w = 9.1 J, ?U = 9.1 J, ?H = 0; q = 0 
C: w = -3.9 kJ, ?U = 0, ?H = 0; q = 3.9 kJ 
D: w = +4.1 kJ, ?U = 0, ?H = 0; q = -4.1 kJ 
Answer: A 
Explanation: 
For an ideal gas undergoing isothermal reversible expansion: 
Temperature is constant ? ?U = 0 (for ideal gas, U depends only on T) 
Hence ?H = 0 (since H also depends only on T for ideal gas) 
Step 1: Given data (convert units) 
V 1 = 20.0 dm³ = 20.0 × 10 ?³ m³ = 0.020 m³ 
P 1 = 0.5 MPa = 0.5 × 10 6 Pa = 5.0 × 10 5 Pa 
P 2 = 0.2 MPa 
T = 600 K 
Step 2: Find number of moles using PV = nRT 
 
Step 3: Work done in reversible isothermal expansion 
For reversible isothermal process: 
 
Convert to base-10 log form (NCERT style): 
 
Now, 
P 1 / P 2 = 0.5 / 0.2 = 2.5 
Given: 
log(2.5) = log(5/2) = log 5 - log 2 = 0.6989 - 0.3010 = 0.3979 
So, 
w = -2.303(2.0)(8.314)(600)(0.3979) 
First calculate nRT: 
nRT = 2.0 × 8.314 × 600 ˜ 9976.8 J 
Then, 
w ˜ -2.303 × 9976.8 × 0.3979 ˜ -9130 J ˜ -9.1 kJ 
Step 4: Heat and energy changes 
For isothermal ideal gas: 
?U = 0 
First law (NCERT sign convention): 
?U = q + w 
So, 
0 = q + (-9.1) ? q = +9.1 kJ 
Also, 
?H = 0 
Correct option A: w = -9.1 kJ, ?U = 0, ?H = 0, q = +9.1 kJ 
 
Q3: The heat of atomisation of methane and ethane are x kJ mol ?¹ and y kJ mol ?¹ 
respectively. 
The longest wavelength (?) of light capable of breaking the C - C bond can be expressed 
in SI unit as: 
A:  
B:  
Page 3


JEE Main Previous Year Questions (2021-2026): 
Thermodynamics  
 
(January 2026) 
 
Q1:  The plot of log 1 0 K vs 1/T gives a straight line. The intercept and slope respectively 
are (where K is the equilibrium constant). 
A:  
B:  
C:  
D:  
Answer: C 
Explanation:
 
 
Q2: 20.0 dm³ of an ideal gas 'X' at 600 K and 0.5 MPa undergoes isothermal reversible 
expansion until the pressure of the gas is 0.2 MPa. 
Which of the following option is correct? 
(Given: log 2 = 0.3010 and log 5 = 0.6989) 
A: w = -9.1 kJ, ?U = 0, ?H = 0, q = 9.1 kJ 
B: w = 9.1 J, ?U = 9.1 J, ?H = 0; q = 0 
C: w = -3.9 kJ, ?U = 0, ?H = 0; q = 3.9 kJ 
D: w = +4.1 kJ, ?U = 0, ?H = 0; q = -4.1 kJ 
Answer: A 
Explanation: 
For an ideal gas undergoing isothermal reversible expansion: 
Temperature is constant ? ?U = 0 (for ideal gas, U depends only on T) 
Hence ?H = 0 (since H also depends only on T for ideal gas) 
Step 1: Given data (convert units) 
V 1 = 20.0 dm³ = 20.0 × 10 ?³ m³ = 0.020 m³ 
P 1 = 0.5 MPa = 0.5 × 10 6 Pa = 5.0 × 10 5 Pa 
P 2 = 0.2 MPa 
T = 600 K 
Step 2: Find number of moles using PV = nRT 
 
Step 3: Work done in reversible isothermal expansion 
For reversible isothermal process: 
 
Convert to base-10 log form (NCERT style): 
 
Now, 
P 1 / P 2 = 0.5 / 0.2 = 2.5 
Given: 
log(2.5) = log(5/2) = log 5 - log 2 = 0.6989 - 0.3010 = 0.3979 
So, 
w = -2.303(2.0)(8.314)(600)(0.3979) 
First calculate nRT: 
nRT = 2.0 × 8.314 × 600 ˜ 9976.8 J 
Then, 
w ˜ -2.303 × 9976.8 × 0.3979 ˜ -9130 J ˜ -9.1 kJ 
Step 4: Heat and energy changes 
For isothermal ideal gas: 
?U = 0 
First law (NCERT sign convention): 
?U = q + w 
So, 
0 = q + (-9.1) ? q = +9.1 kJ 
Also, 
?H = 0 
Correct option A: w = -9.1 kJ, ?U = 0, ?H = 0, q = +9.1 kJ 
 
Q3: The heat of atomisation of methane and ethane are x kJ mol ?¹ and y kJ mol ?¹ 
respectively. 
The longest wavelength (?) of light capable of breaking the C - C bond can be expressed 
in SI unit as: 
A:  
B:  
C:  
D:  
Answer: D 
Explanation: 
For methane: 
?H
atom
 (CH 4) = x = 4D(C–H) ? D(C–H) = x / 4 
For ethane: 
?H
atom
 (C 2H 6) = y = 6D(C–H) + D(C–C) 
? D(C - C) = y - 6(x/4) = y - 6x/4 = (4y - 6x)/4 (kJ mol ?¹) 
Energy per molecule needed to break the C - C bond: 
 
Longest wavelength corresponds to minimum photon energy:
 
Correct Option: D. 
 
Q4: Match the LIST-I with LIST-II 
 
Choose the correct answer from the options given below: 
A: A-IV, B-I, C-III, D-II 
B: A-I, B-III, C-II, D-IV 
C: A-II, B-I, C-III, D-IV 
D: A-IV, B-II, C-III, D-I 
Answer: C 
Explanation: 
Page 4


JEE Main Previous Year Questions (2021-2026): 
Thermodynamics  
 
(January 2026) 
 
Q1:  The plot of log 1 0 K vs 1/T gives a straight line. The intercept and slope respectively 
are (where K is the equilibrium constant). 
A:  
B:  
C:  
D:  
Answer: C 
Explanation:
 
 
Q2: 20.0 dm³ of an ideal gas 'X' at 600 K and 0.5 MPa undergoes isothermal reversible 
expansion until the pressure of the gas is 0.2 MPa. 
Which of the following option is correct? 
(Given: log 2 = 0.3010 and log 5 = 0.6989) 
A: w = -9.1 kJ, ?U = 0, ?H = 0, q = 9.1 kJ 
B: w = 9.1 J, ?U = 9.1 J, ?H = 0; q = 0 
C: w = -3.9 kJ, ?U = 0, ?H = 0; q = 3.9 kJ 
D: w = +4.1 kJ, ?U = 0, ?H = 0; q = -4.1 kJ 
Answer: A 
Explanation: 
For an ideal gas undergoing isothermal reversible expansion: 
Temperature is constant ? ?U = 0 (for ideal gas, U depends only on T) 
Hence ?H = 0 (since H also depends only on T for ideal gas) 
Step 1: Given data (convert units) 
V 1 = 20.0 dm³ = 20.0 × 10 ?³ m³ = 0.020 m³ 
P 1 = 0.5 MPa = 0.5 × 10 6 Pa = 5.0 × 10 5 Pa 
P 2 = 0.2 MPa 
T = 600 K 
Step 2: Find number of moles using PV = nRT 
 
Step 3: Work done in reversible isothermal expansion 
For reversible isothermal process: 
 
Convert to base-10 log form (NCERT style): 
 
Now, 
P 1 / P 2 = 0.5 / 0.2 = 2.5 
Given: 
log(2.5) = log(5/2) = log 5 - log 2 = 0.6989 - 0.3010 = 0.3979 
So, 
w = -2.303(2.0)(8.314)(600)(0.3979) 
First calculate nRT: 
nRT = 2.0 × 8.314 × 600 ˜ 9976.8 J 
Then, 
w ˜ -2.303 × 9976.8 × 0.3979 ˜ -9130 J ˜ -9.1 kJ 
Step 4: Heat and energy changes 
For isothermal ideal gas: 
?U = 0 
First law (NCERT sign convention): 
?U = q + w 
So, 
0 = q + (-9.1) ? q = +9.1 kJ 
Also, 
?H = 0 
Correct option A: w = -9.1 kJ, ?U = 0, ?H = 0, q = +9.1 kJ 
 
Q3: The heat of atomisation of methane and ethane are x kJ mol ?¹ and y kJ mol ?¹ 
respectively. 
The longest wavelength (?) of light capable of breaking the C - C bond can be expressed 
in SI unit as: 
A:  
B:  
C:  
D:  
Answer: D 
Explanation: 
For methane: 
?H
atom
 (CH 4) = x = 4D(C–H) ? D(C–H) = x / 4 
For ethane: 
?H
atom
 (C 2H 6) = y = 6D(C–H) + D(C–C) 
? D(C - C) = y - 6(x/4) = y - 6x/4 = (4y - 6x)/4 (kJ mol ?¹) 
Energy per molecule needed to break the C - C bond: 
 
Longest wavelength corresponds to minimum photon energy:
 
Correct Option: D. 
 
Q4: Match the LIST-I with LIST-II 
 
Choose the correct answer from the options given below: 
A: A-IV, B-I, C-III, D-II 
B: A-I, B-III, C-II, D-IV 
C: A-II, B-I, C-III, D-IV 
D: A-IV, B-II, C-III, D-I 
Answer: C 
Explanation: 
For any process, NCERT uses the expression for work (done on the system): 
 
A. Reversible isothermal expansion (ideal gas) 
In reversible isothermal change, 
 
 
So, A ? II. 
B. Free expansion 
In free expansion,P
ext
 = 0. 
So, B ? I. 
C. Irreversible expansion (against constant external pressure) 
Here P
ext
 is constant. 
W = -P
ext
(V
f
 - V
i
) 
So, C ? III. 
D. Irreversible compression (against constant external pressure) 
Compression means work done on the system is positive. Using the same NCERT form: 
W = -P
ext
(V
f
 - V
i
) = Pext(V
i
 - V
f
) 
This corresponds to the given form IV (written as -P
ext
(V
i 
- V
f
) in the list, which becomes 
positive when  
V
i
 > V
f 
for compression). 
So, D ? IV. 
Final matching 
A-II,B-I,C-III,D-IV 
Correct option: C 
 
Q5: A cup of water at 5
°
C (system) is placed in a microwave oven and the oven is turned on for 
one minute during which the water begins to boil. Which of the following options is true? 
A: q = +ve, w = -ve, ?U = +ve 
B: q = +ve, w = 0, ?U = -ve 
C: q = +ve, w = -ve, ?U = -ve 
D: q = -ve, w = -ve, ?U = -ve 
Answer: A 
Explanation: 
 
When water is heated in the microwave, its temperature increases from 5
°
C to 100
°
C. After 
reaching 100
°
C, it starts changing from liquid to vapour (boiling), so liquid water and water 
vapour can exist together. 
Page 5


JEE Main Previous Year Questions (2021-2026): 
Thermodynamics  
 
(January 2026) 
 
Q1:  The plot of log 1 0 K vs 1/T gives a straight line. The intercept and slope respectively 
are (where K is the equilibrium constant). 
A:  
B:  
C:  
D:  
Answer: C 
Explanation:
 
 
Q2: 20.0 dm³ of an ideal gas 'X' at 600 K and 0.5 MPa undergoes isothermal reversible 
expansion until the pressure of the gas is 0.2 MPa. 
Which of the following option is correct? 
(Given: log 2 = 0.3010 and log 5 = 0.6989) 
A: w = -9.1 kJ, ?U = 0, ?H = 0, q = 9.1 kJ 
B: w = 9.1 J, ?U = 9.1 J, ?H = 0; q = 0 
C: w = -3.9 kJ, ?U = 0, ?H = 0; q = 3.9 kJ 
D: w = +4.1 kJ, ?U = 0, ?H = 0; q = -4.1 kJ 
Answer: A 
Explanation: 
For an ideal gas undergoing isothermal reversible expansion: 
Temperature is constant ? ?U = 0 (for ideal gas, U depends only on T) 
Hence ?H = 0 (since H also depends only on T for ideal gas) 
Step 1: Given data (convert units) 
V 1 = 20.0 dm³ = 20.0 × 10 ?³ m³ = 0.020 m³ 
P 1 = 0.5 MPa = 0.5 × 10 6 Pa = 5.0 × 10 5 Pa 
P 2 = 0.2 MPa 
T = 600 K 
Step 2: Find number of moles using PV = nRT 
 
Step 3: Work done in reversible isothermal expansion 
For reversible isothermal process: 
 
Convert to base-10 log form (NCERT style): 
 
Now, 
P 1 / P 2 = 0.5 / 0.2 = 2.5 
Given: 
log(2.5) = log(5/2) = log 5 - log 2 = 0.6989 - 0.3010 = 0.3979 
So, 
w = -2.303(2.0)(8.314)(600)(0.3979) 
First calculate nRT: 
nRT = 2.0 × 8.314 × 600 ˜ 9976.8 J 
Then, 
w ˜ -2.303 × 9976.8 × 0.3979 ˜ -9130 J ˜ -9.1 kJ 
Step 4: Heat and energy changes 
For isothermal ideal gas: 
?U = 0 
First law (NCERT sign convention): 
?U = q + w 
So, 
0 = q + (-9.1) ? q = +9.1 kJ 
Also, 
?H = 0 
Correct option A: w = -9.1 kJ, ?U = 0, ?H = 0, q = +9.1 kJ 
 
Q3: The heat of atomisation of methane and ethane are x kJ mol ?¹ and y kJ mol ?¹ 
respectively. 
The longest wavelength (?) of light capable of breaking the C - C bond can be expressed 
in SI unit as: 
A:  
B:  
C:  
D:  
Answer: D 
Explanation: 
For methane: 
?H
atom
 (CH 4) = x = 4D(C–H) ? D(C–H) = x / 4 
For ethane: 
?H
atom
 (C 2H 6) = y = 6D(C–H) + D(C–C) 
? D(C - C) = y - 6(x/4) = y - 6x/4 = (4y - 6x)/4 (kJ mol ?¹) 
Energy per molecule needed to break the C - C bond: 
 
Longest wavelength corresponds to minimum photon energy:
 
Correct Option: D. 
 
Q4: Match the LIST-I with LIST-II 
 
Choose the correct answer from the options given below: 
A: A-IV, B-I, C-III, D-II 
B: A-I, B-III, C-II, D-IV 
C: A-II, B-I, C-III, D-IV 
D: A-IV, B-II, C-III, D-I 
Answer: C 
Explanation: 
For any process, NCERT uses the expression for work (done on the system): 
 
A. Reversible isothermal expansion (ideal gas) 
In reversible isothermal change, 
 
 
So, A ? II. 
B. Free expansion 
In free expansion,P
ext
 = 0. 
So, B ? I. 
C. Irreversible expansion (against constant external pressure) 
Here P
ext
 is constant. 
W = -P
ext
(V
f
 - V
i
) 
So, C ? III. 
D. Irreversible compression (against constant external pressure) 
Compression means work done on the system is positive. Using the same NCERT form: 
W = -P
ext
(V
f
 - V
i
) = Pext(V
i
 - V
f
) 
This corresponds to the given form IV (written as -P
ext
(V
i 
- V
f
) in the list, which becomes 
positive when  
V
i
 > V
f 
for compression). 
So, D ? IV. 
Final matching 
A-II,B-I,C-III,D-IV 
Correct option: C 
 
Q5: A cup of water at 5
°
C (system) is placed in a microwave oven and the oven is turned on for 
one minute during which the water begins to boil. Which of the following options is true? 
A: q = +ve, w = -ve, ?U = +ve 
B: q = +ve, w = 0, ?U = -ve 
C: q = +ve, w = -ve, ?U = -ve 
D: q = -ve, w = -ve, ?U = -ve 
Answer: A 
Explanation: 
 
When water is heated in the microwave, its temperature increases from 5
°
C to 100
°
C. After 
reaching 100
°
C, it starts changing from liquid to vapour (boiling), so liquid water and water 
vapour can exist together. 
During boiling, the water changes into steam and the volume increases (expansion). 
Because of expansion, the system does work on the surroundings, so work done by the system 
is negative. 
W = -ve 
Since the microwave supplies energy to the water, heat is absorbed by the system, so q is 
positive. 
Also, water vapour has more internal energy than liquid water at the same temperature, so the 
internal energy of the system increases. Therefore, ?U is positive. 
So,  
q = +ve and ?U = +ve 
 
Q6: Match the LIST -I with LIST -II 
 
A: A-I, B-II, C-III, D-IV 
B: A-III, B-II, C-IV, D-I 
C: A-II, B-I, C-III, D-IV 
D: A-II, B-III, C-I, D-IV 
Answer: D 
Explanation:  
Option (A) 
This is a reversible isothermal expansion of an ideal gas, so we use the NCERT formula for 
isothermal reversible work: