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Home/JEE MAINS/physics/thermodynamics
Curated PYQ Collection

Top 50 Most Repeated THERMODYNAMICS PYQs | JEE MAINS

A curated collection of the most important questions from THERMODYNAMICS, fully solved with step-by-step concepts to prepare for JEE MAINS.

Question #1

Practice Question

Concept Applied

Given $C_p = \frac{7}{2}R$, then $C_v = C_p - R = \frac{7}{2}R - R = \frac{5}{2}R$. So $\gamma = \frac{C_p}{C_v} = \frac{7/2}{5/2} = \frac{7}{5} = 1.4...

Read Full Step-by-Step Solution →

Question #2

Practice Question

Concept Applied

For any ideal gas, $C_p - C_v = R = 8.31\,\text{J/molĀ·K}$....

Read Full Step-by-Step Solution →

Question #3

Practice Question

A.
.40$
B.
.67$
C.
.33$
D.
.00$

Concept Applied

Diatomic gas has $C_v = \frac{5}{2}R$, $C_p = \frac{7}{2}R$, so $\gamma = 7/5 = 1.40$....

Read Full Step-by-Step Solution →

Question #4

Practice Question

A.50 J
B.350 J
C.-50 J
D.-350 J

Concept Applied

ΔU = Q - W = 200 - 150 = 50 J....

Read Full Step-by-Step Solution →

Question #5

Practice Question

Concept Applied

Carnot efficiency $\eta = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{600} = 0.50$....

Read Full Step-by-Step Solution →

Question #6

Practice Question

A.Monatomic gas (e.g., Helium)
B.Diatomic gas at room temperature (e.g., Nitrogen)
C.Polyatomic gas (e.g., Carbon dioxide)
D.Ionized plasma

Concept Applied

For a diatomic gas at moderate temperatures, $C_p/C_v = \frac{7}{5}=1.4$. Monatomic gases give $\gamma=5/3\approx1.67$, while polyatomic gases have lo...

Read Full Step-by-Step Solution →

Question #7

Practice Question

A.$\gamma = \frac{5}{3}$
B.$\gamma = \frac{7}{5}$
C.$\gamma = \frac{4}{3}$
D.$\gamma = \frac{3}{2}$

Concept Applied

A diatomic molecule has 3 translational and 2 rotational degrees of freedom, giving $f=5$. Hence $C_v = \frac{f}{2}R = \frac{5}{2}R$ and $C_p = C_v + ...

Read Full Step-by-Step Solution →

Question #8

Practice Question

A.0.33
B.0.50
C.0.67
D.0.75

Concept Applied

Efficiency $\eta = 1 - T_c/T_h = 1 - 300/600 = 0.50$....

Read Full Step-by-Step Solution →

Question #9

Practice Question

A.8.314 J·K⁻¹·mol⁻¹
B.0 J·K⁻¹·mol⁻¹
C.4.157 J·K⁻¹·mol⁻¹
D.16.628 J·K⁻¹·mol⁻¹

Concept Applied

Mayer's relation directly gives $C_p - C_v = R$. Substituting $R = 8.314$ yields the same value....

Read Full Step-by-Step Solution →

Question #10

Practice Question

A.0 J (no net work)
B.ā‰ˆā€Æ150 J done by the gas
C.ā‰ˆā€Æ620 J done on the gas
D.ā‰ˆā€Æ620 J done by the gas

Concept Applied

Work for each segment: Isothermal AB: $W_{AB}=P_{A}V_{A}\ln(V_{B}/V_{A})=2\times1\times\ln2=1.386\,\text{atmĀ·L}$. Adiabatic BC: $W_{BC}=\frac{P_{B}V_{...

Read Full Step-by-Step Solution →

Question #11

Practice Question

A.1726 J
B.2490 J
C.0 J
D.830 J

Concept Applied

$W = nRT \ln(V_f/V_i) = 1 \cdot 8.3 \cdot 300 \cdot \ln(2) \approx 1726$ J....

Read Full Step-by-Step Solution →

Question #12

Practice Question

A.20 J
B.30 J
C.50 J
D.80 J

Concept Applied

Using $\Delta U = Q - W$: $\Delta U = 50 - 30 = 20$ J....

Read Full Step-by-Step Solution →

Question #13

Practice Question

A.A fundamental principle of Physics.
B.A complex derivation in JEEMains syllabus.
C.An experimental observation.
D.A theoretical assumption.

Concept Applied

This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Thermodynamics....

Read Full Step-by-Step Solution →

Question #14

Practice Question

A.A fundamental principle of Physics.
B.A complex derivation in JEEMains syllabus.
C.An experimental observation.
D.A theoretical assumption.

Concept Applied

This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Thermodynamics....

Read Full Step-by-Step Solution →

Question #15

Practice Question

Concept Applied

Given $C_p = \frac{7}{2}R$, then $C_v = C_p - R = \frac{7}{2}R - R = \frac{5}{2}R$. So $\gamma = \frac{C_p}{C_v} = \frac{7/2}{5/2} = \frac{7}{5} = 1.4...

Read Full Step-by-Step Solution →

Question #16

Practice Question

Concept Applied

For any ideal gas, $C_p - C_v = R = 8.31\,\text{J/molĀ·K}$....

Read Full Step-by-Step Solution →

Question #17

Practice Question

A.
.40$
B.
.67$
C.
.33$
D.
.00$

Concept Applied

Diatomic gas has $C_v = \frac{5}{2}R$, $C_p = \frac{7}{2}R$, so $\gamma = 7/5 = 1.40$....

Read Full Step-by-Step Solution →

Question #18

Practice Question

A.50 J
B.350 J
C.-50 J
D.-350 J

Concept Applied

ΔU = Q - W = 200 - 150 = 50 J....

Read Full Step-by-Step Solution →

Question #19

Practice Question

Concept Applied

Carnot efficiency $\eta = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{600} = 0.50$....

Read Full Step-by-Step Solution →

Question #20

Practice Question

A.Monatomic gas (e.g., Helium)
B.Diatomic gas at room temperature (e.g., Nitrogen)
C.Polyatomic gas (e.g., Carbon dioxide)
D.Ionized plasma

Concept Applied

For a diatomic gas at moderate temperatures, $C_p/C_v = \frac{7}{5}=1.4$. Monatomic gases give $\gamma=5/3\approx1.67$, while polyatomic gases have lo...

Read Full Step-by-Step Solution →

Question #21

Practice Question

A.$\gamma = \frac{5}{3}$
B.$\gamma = \frac{7}{5}$
C.$\gamma = \frac{4}{3}$
D.$\gamma = \frac{3}{2}$

Concept Applied

A diatomic molecule has 3 translational and 2 rotational degrees of freedom, giving $f=5$. Hence $C_v = \frac{f}{2}R = \frac{5}{2}R$ and $C_p = C_v + ...

Read Full Step-by-Step Solution →

Question #22

Practice Question

A.0.33
B.0.50
C.0.67
D.0.75

Concept Applied

Efficiency $\eta = 1 - T_c/T_h = 1 - 300/600 = 0.50$....

Read Full Step-by-Step Solution →

Question #23

Practice Question

A.8.314 J·K⁻¹·mol⁻¹
B.0 J·K⁻¹·mol⁻¹
C.4.157 J·K⁻¹·mol⁻¹
D.16.628 J·K⁻¹·mol⁻¹

Concept Applied

Mayer's relation directly gives $C_p - C_v = R$. Substituting $R = 8.314$ yields the same value....

Read Full Step-by-Step Solution →

Question #24

Practice Question

A.0 J (no net work)
B.ā‰ˆā€Æ150 J done by the gas
C.ā‰ˆā€Æ620 J done on the gas
D.ā‰ˆā€Æ620 J done by the gas

Concept Applied

Work for each segment: Isothermal AB: $W_{AB}=P_{A}V_{A}\ln(V_{B}/V_{A})=2\times1\times\ln2=1.386\,\text{atmĀ·L}$. Adiabatic BC: $W_{BC}=\frac{P_{B}V_{...

Read Full Step-by-Step Solution →

Question #25

Practice Question

A.1726 J
B.2490 J
C.0 J
D.830 J

Concept Applied

$W = nRT \ln(V_f/V_i) = 1 \cdot 8.3 \cdot 300 \cdot \ln(2) \approx 1726$ J....

Read Full Step-by-Step Solution →

Question #26

Practice Question

A.20 J
B.30 J
C.50 J
D.80 J

Concept Applied

Using $\Delta U = Q - W$: $\Delta U = 50 - 30 = 20$ J....

Read Full Step-by-Step Solution →

Question #27

Practice Question

A.A fundamental principle of Physics.
B.A complex derivation in JEEMains syllabus.
C.An experimental observation.
D.A theoretical assumption.

Concept Applied

This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Thermodynamics....

Read Full Step-by-Step Solution →

Question #28

Practice Question

A.A fundamental principle of Physics.
B.A complex derivation in JEEMains syllabus.
C.An experimental observation.
D.A theoretical assumption.

Concept Applied

This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Thermodynamics....

Read Full Step-by-Step Solution →

Question #29

Practice Question

Concept Applied

Given $C_p = \frac{7}{2}R$, then $C_v = C_p - R = \frac{7}{2}R - R = \frac{5}{2}R$. So $\gamma = \frac{C_p}{C_v} = \frac{7/2}{5/2} = \frac{7}{5} = 1.4...

Read Full Step-by-Step Solution →

Question #30

Practice Question

Concept Applied

For any ideal gas, $C_p - C_v = R = 8.31\,\text{J/molĀ·K}$....

Read Full Step-by-Step Solution →

Question #31

Practice Question

A.
.40$
B.
.67$
C.
.33$
D.
.00$

Concept Applied

Diatomic gas has $C_v = \frac{5}{2}R$, $C_p = \frac{7}{2}R$, so $\gamma = 7/5 = 1.40$....

Read Full Step-by-Step Solution →

Question #32

Practice Question

A.50 J
B.350 J
C.-50 J
D.-350 J

Concept Applied

ΔU = Q - W = 200 - 150 = 50 J....

Read Full Step-by-Step Solution →

Question #33

Practice Question

Concept Applied

Carnot efficiency $\eta = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{600} = 0.50$....

Read Full Step-by-Step Solution →

Question #34

Practice Question

A.Monatomic gas (e.g., Helium)
B.Diatomic gas at room temperature (e.g., Nitrogen)
C.Polyatomic gas (e.g., Carbon dioxide)
D.Ionized plasma

Concept Applied

For a diatomic gas at moderate temperatures, $C_p/C_v = \frac{7}{5}=1.4$. Monatomic gases give $\gamma=5/3\approx1.67$, while polyatomic gases have lo...

Read Full Step-by-Step Solution →

Question #35

Practice Question

A.$\gamma = \frac{5}{3}$
B.$\gamma = \frac{7}{5}$
C.$\gamma = \frac{4}{3}$
D.$\gamma = \frac{3}{2}$

Concept Applied

A diatomic molecule has 3 translational and 2 rotational degrees of freedom, giving $f=5$. Hence $C_v = \frac{f}{2}R = \frac{5}{2}R$ and $C_p = C_v + ...

Read Full Step-by-Step Solution →

Question #36

Practice Question

A.0.33
B.0.50
C.0.67
D.0.75

Concept Applied

Efficiency $\eta = 1 - T_c/T_h = 1 - 300/600 = 0.50$....

Read Full Step-by-Step Solution →

Question #37

Practice Question

A.8.314 J·K⁻¹·mol⁻¹
B.0 J·K⁻¹·mol⁻¹
C.4.157 J·K⁻¹·mol⁻¹
D.16.628 J·K⁻¹·mol⁻¹

Concept Applied

Mayer's relation directly gives $C_p - C_v = R$. Substituting $R = 8.314$ yields the same value....

Read Full Step-by-Step Solution →

Question #38

Practice Question

A.0 J (no net work)
B.ā‰ˆā€Æ150 J done by the gas
C.ā‰ˆā€Æ620 J done on the gas
D.ā‰ˆā€Æ620 J done by the gas

Concept Applied

Work for each segment: Isothermal AB: $W_{AB}=P_{A}V_{A}\ln(V_{B}/V_{A})=2\times1\times\ln2=1.386\,\text{atmĀ·L}$. Adiabatic BC: $W_{BC}=\frac{P_{B}V_{...

Read Full Step-by-Step Solution →

Question #39

Practice Question

A.1726 J
B.2490 J
C.0 J
D.830 J

Concept Applied

$W = nRT \ln(V_f/V_i) = 1 \cdot 8.3 \cdot 300 \cdot \ln(2) \approx 1726$ J....

Read Full Step-by-Step Solution →

Question #40

Practice Question

A.20 J
B.30 J
C.50 J
D.80 J

Concept Applied

Using $\Delta U = Q - W$: $\Delta U = 50 - 30 = 20$ J....

Read Full Step-by-Step Solution →

Question #41

Practice Question

A.A fundamental principle of Physics.
B.A complex derivation in JEEMains syllabus.
C.An experimental observation.
D.A theoretical assumption.

Concept Applied

This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Thermodynamics....

Read Full Step-by-Step Solution →

Question #42

Practice Question

A.A fundamental principle of Physics.
B.A complex derivation in JEEMains syllabus.
C.An experimental observation.
D.A theoretical assumption.

Concept Applied

This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Thermodynamics....

Read Full Step-by-Step Solution →

Question #43

Practice Question

Concept Applied

Given $C_p = \frac{7}{2}R$, then $C_v = C_p - R = \frac{7}{2}R - R = \frac{5}{2}R$. So $\gamma = \frac{C_p}{C_v} = \frac{7/2}{5/2} = \frac{7}{5} = 1.4...

Read Full Step-by-Step Solution →

Question #44

Practice Question

Concept Applied

For any ideal gas, $C_p - C_v = R = 8.31\,\text{J/molĀ·K}$....

Read Full Step-by-Step Solution →

Question #45

Practice Question

A.
.40$
B.
.67$
C.
.33$
D.
.00$

Concept Applied

Diatomic gas has $C_v = \frac{5}{2}R$, $C_p = \frac{7}{2}R$, so $\gamma = 7/5 = 1.40$....

Read Full Step-by-Step Solution →

Question #46

Practice Question

A.50 J
B.350 J
C.-50 J
D.-350 J

Concept Applied

ΔU = Q - W = 200 - 150 = 50 J....

Read Full Step-by-Step Solution →

Question #47

Practice Question

Concept Applied

Carnot efficiency $\eta = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{600} = 0.50$....

Read Full Step-by-Step Solution →

Question #48

Practice Question

A.Monatomic gas (e.g., Helium)
B.Diatomic gas at room temperature (e.g., Nitrogen)
C.Polyatomic gas (e.g., Carbon dioxide)
D.Ionized plasma

Concept Applied

For a diatomic gas at moderate temperatures, $C_p/C_v = \frac{7}{5}=1.4$. Monatomic gases give $\gamma=5/3\approx1.67$, while polyatomic gases have lo...

Read Full Step-by-Step Solution →

Question #49

Practice Question

A.$\gamma = \frac{5}{3}$
B.$\gamma = \frac{7}{5}$
C.$\gamma = \frac{4}{3}$
D.$\gamma = \frac{3}{2}$

Concept Applied

A diatomic molecule has 3 translational and 2 rotational degrees of freedom, giving $f=5$. Hence $C_v = \frac{f}{2}R = \frac{5}{2}R$ and $C_p = C_v + ...

Read Full Step-by-Step Solution →

Question #50

Practice Question

A.0.33
B.0.50
C.0.67
D.0.75

Concept Applied

Efficiency $\eta = 1 - T_c/T_h = 1 - 300/600 = 0.50$....

Read Full Step-by-Step Solution →
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