Thermodynamics and Gaseous State Previous Year Questions with Solutions

Join ALLEN!

(Session 2026 - 27)

Name

Mobile Number

Class

Choose class

Goal

Choose your goal

Preferred Programs

Preferred Mode

State

Choose State

I agree to

I authorise ALLEN Career Institute Pvt Ltd to send me regular updates via Phone calls, Whatsapp, SMS, Robocalls (Automated Calls), Emails, or on postal address.

Thermodynamics and Gaseous State PYQs with solutions

Thermodynamics and Gaseous State PYQs with Solutions

1.0Thermodynamics

Thermodynamics is the science of energy changes in physical and chemical processes. Chemical thermodynamics focuses on energy transformations in chemical reactions. A system is the part under study, separated from its surroundings by a boundary. Together, they form the universe in thermodynamics.

Key Points

Systems are classified as:

Open (exchanges energy and matter),
Closed (exchanges only energy), and
Isolated (no exchange).

Thermodynamic variables include intensive (independent of mass, e.g., temperature, pressure) and extensive (dependent on mass, e.g., volume, energy). State functions (e.g., enthalpy, entropy) depend only on the current state, while path functions (e.g., heat, work) depend on the path taken.

Processes:

Isothermal: constant temperature
Adiabatic: no heat exchange
Isobaric: constant pressure
Isochoric: constant volume
Cyclic: returns to initial state
Reversible and Irreversible

Thermodynamic potentials include internal energy (U), enthalpy (H), Gibbs free energy (G), and Helmholtz energy (F).

Laws of Thermodynamics:

Zeroth Law: Defines temperature.
First Law: Energy is conserved.
Second Law: Entropy always increases.
Third Law: Entropy approaches zero at absolute zero.

Related Video:

2.0Gaseous State

Key Points

General Properties of Gases:

No definite shape or volume.
Highly compressible and expansible.
Low density due to negligible intermolecular forces.
Exert pressure through elastic collisions.
Diffuse rapidly and form homogeneous mixtures.

Gas Laws:

Boyle’s Law:
$P \propto 1 V \frac{1}{V} (at constant T)$
⇒ P₁V₁=P₂V₂
Charles’ Law:

$V \propto T (at constant P)$
$\Rightarrow V T \frac{V}{T} = co n s t an t$

3. Gay-Lussac’s Law:
P∝T (at constant V)
⇒ PT\frac{P}{T}=constant

Avogadro’s Law:
V∝ n (at constant P and T)

Ideal Gas Equation: PV=nRT R = 0.0821 L·atm·mol⁻¹·K⁻¹ or 8.314 J·mol⁻¹·K⁻¹
1 mol of ideal gas at STP occupies 22.7 L.

Dalton’s Law of Partial Pressures:
Total pressure = sum of partial pressures.

P_{dry gas}=P_total−aqueous tension

Kinetic Molecular Theory:

Gas particles are in constant random motion.
Collisions are perfectly elastic.
Pressure is due to collisions with container walls.
Kinetic energy ∝ temperature.

Real Gases and Deviation:

At high pressure/low temperature, gases deviate from ideal behavior.

$Z = \frac{P V}{n RT}$

Compressibility Factor (Z):
Z = 1 for ideal gases; deviates for real gases.

Liquefaction & Critical Constants:

Critical Temperature (Tc): $\frac{8 a}{27 R b}$
Critical Volume (Vc): 3b
Critical Pressure (Pc): $\frac{a}{27 b ^{2}}$

Specific Heats & γ:

Monoatomic: γ = 1.67
Diatomic: γ = 1.4
Polyatomic: γ = 1.33

JEE Main Previous Year Solved Questions on Thermodynamics and Gaseous State

3.0Thermodynamics and Gaseous State PYQs

Q.1 One mole of an ideal monoatomic gas is subjected to changes as shown in the graph. The magnitude of the work done (by the system or on the system) is _______ J (nearest integer).

One mole of an ideal monoatomic gas is subjected to changes as shown in the graph. The magnitude of the work done (by the system or on the system) is _______ J (nearest integer).

Ans

Solution

Given : log 2 = 0.3, ln 10 = 2.3

1 → 2 Isobaric process

2 → 3 Isochoric process

3 →1Isothermal process

W_{1 → 2}+ W_{2 → 3} + W(3 →1)

$= (- P (V 2 - V 1) + 0) + (- P 1 V 1 l n (V 2 V 1 \frac{V _{2}}{V _{1}}))$

Now solving:

$[- 1 \times (40 - 20) + 0 + (- 1 \times 20 l n (2040 \frac{20}{40}))]$

=−20+20×2.3×0.3

= −6.2 bar L

Thus

∣W∣=6.2 bar L=620 J

Q.2 Which of the following relations are correct?

(A) 𝚫U = q - p𝚫V

(B) 𝚫G = 𝚫H - T𝚫S

(C) 𝚫S= $\frac{q _{re v}}{T}$

(D) 𝚫H = 𝚫U - 𝚫nRT

Choose the most appropriate answer from the options given below :

(1) C and D only (2) B and C only

(3) A and B only (4) B and D only

Ans. (2)

Sol.Only (B) and (C) are correct.

(B)𝚫G = 𝚫H - T𝚫S

At constant T

𝚫G = 𝚫H - T𝚫S

(A) First law is given by

vU = Q + W

If we apply constant P and reversible work.

𝚫U = Q – P𝚫V

(C) By definition of entropy change

𝚫S= $\frac{q _{re v}}{T}$

At constant T

𝚫S= $\frac{q _{re v}}{T}$

(D) H = U + PV

For ideal gas

H = U + nRT

At constant T

𝚫H = 𝚫U + 𝚫nRT

Q.3 An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by tracing the path A ® B ® C ® A as shown in the diagram. The total work done in the process is _____ J.

An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by tracing the path A ® B ® C ® A as shown in the diagram. The total work done in the process is _____ J.

Ans. (200)

Sol. Work done is given by area enclosed in the P vs V cyclic graph or V vs P cyclic graph.Sign of work is positive for clockwise cyclic process for V vs P graph.

$W = 12 \frac{1}{2} \times (30-10) \times (30 - 10) = 200 k P a - d m^{3}$

= 200 × 1000 Pa – L = 2 L-bar = 200 J

Q.4 1 mol of an ideal gas is kept in a cylinder, fitted with a piston, at the position A, at 18°C. If the piston is moved to position B, keeping the temperature unchanged, then 'x' L atm work is done in this reversible process.

Consider the figure provided.

1 mol of an ideal gas is kept in a cylinder, fitted with a piston, at the position A, at 18°C. If the piston is moved to position B, keeping the temperature unchanged, then 'x' L atm work is done in this reversible process.

x = ______ L atm. (nearest integer)

[Given : Absolute temperature = °C + 273.15, R = 0.08206 L atm mol^–1 K^–1]

Ans. (55)

Sol. w = –nRT In $\frac{V _{2}}{V _{1}}$

= –1×0.08206 × 291×15 ln $\frac{100}{10}$

= –55.0128

Work done by system » 55 atm lit.

Q.5 For a certain thermochemical reaction M→N at T = 400 K, 𝚫H^! = 77.2kJ mol^–1, 𝚫S = 122 JK^–1, log equilibrium constant (logK) is –______ × 10^–1.

Ans. (37)

Sol. 𝚫G° = 𝚫H° - T𝚫S°

=77.2×10³− 400×122=28400J

𝚫G° = −2.303RTlogK

⇒28400=−2.303×8.314×400 logK

⇒ log⁡K=−3.708=−37.08×10⁻¹

Q.6 Three bulbs are filled with CH₄, CO₂ and Ne as shown in the picture. The bulbs are connected through pipes of zero volume. When the stopcocks are opened and the temperature is kept constant throughout, the pressure of the system is found to be____atm.(Nearest integer)

Three bulbs are filled with CH4, CO2 and Ne as shown in the picture. The bulbs are connected through pipes of zero volume.

Solution

P_TV_T = n_TRT

For CH₄

2 × 2 = n₁RT

n₁ = 4RT \frac{4}{RT}

For CO₂

⇒ n₂ = $\frac{12}{RT}$

For Ne

⇒ n₃ = $\frac{12}{RT}$

⇒ n_t = $\frac{1}{RT} [4 + 12 + 12] = 28 RT \frac{28}{RT}$

P = $\frac{28}{RT} \frac{RT}{V _{1}}$

P = $\frac{28}{V _{1}} = 3.11$

Q.7 At 600K, the root mean square (rms) speed of gas X (molar mass = 40) is equal to the most probable speed of gas Y at 90K. The molar mass of the gas Y is ________g mol^–1. (Nearest integer)

Ans. (4)

Solution:

$(U_{rms})_{X, 600} = (U_{mp})_{Y, 90} \frac{3 \times R \times 600}{40} = \frac{2 \times R \times 90}{M} M = 4$

Q.8 The number of statements, which are correct with respect to the compression of carbon dioxide from point (A) in the Andrews isotherm from the following is ________.

A. Carbon dioxide remains as a gas upto point (b)

B. Liquid carbon dioxide appears at point (c)

C. Liquid and gaseous carbon dioxide coexist between points (b) and (c)

D. As the volume decreases from (b) to (c), the amount of liquid decreases

Ans. (2)

Solution.

At (a) $C O_{2}$ exist as gas

(b) liquefaction of CO₂starts

(c) liquefaction ends

(d) CO₂ exist as liquid.

Between (b) & (c) liquid and gaseousCO₂ co-exist.

As volume changes from (b) to (c) gas decreases and liquid increases.

(A), (C) correct

Q.9 Arrange the following gases in increasing order of van der Waals constant ‘a’

A. Ar

B. CH₄

C. H₂O

D. C₆H₆

Choose the correct option from the following :-

(1) B, C, D and A (2) C, D, B and A (3) A, B, C and D (4) D, C, B and A

Ans. (3)

Solution. Vanderwaal constant – ‘a’

(i) Ar = 1.34

(ii) CH₄ = 2.25

(iii) H₂O = 5.46

(iv) C₆H₆ = 18.57

‘a’ symbolises force of attraction and directly proportional to surface area

Q.10 Which amongst the given plots is the correct plot for pressure (p) vs density (d) for an ideal gas ?

(1)

(2)

(3)

(4)

Ans. (2)

Sol.

$P = RTM \frac{RT}{M} d$

T₃ > T₂ > T₁

Thermodynamics and Gaseous State PYQs with Solutions

1.0Thermodynamics

Thermodynamics is the science of energy changes in physical and chemical processes. Chemical thermodynamics focuses on energy transformations in chemical reactions. A system is the part under study, separated from its surroundings by a boundary. Together, they form the universe in thermodynamics.

Key Points

Systems are classified as:

Open (exchanges energy and matter),
Closed (exchanges only energy), and
Isolated (no exchange).

Thermodynamic variables include intensive (independent of mass, e.g., temperature, pressure) and extensive (dependent on mass, e.g., volume, energy). State functions (e.g., enthalpy, entropy) depend only on the current state, while path functions (e.g., heat, work) depend on the path taken.

Processes:

Isothermal: constant temperature
Adiabatic: no heat exchange
Isobaric: constant pressure
Isochoric: constant volume
Cyclic: returns to initial state
Reversible and Irreversible

Thermodynamic potentials include internal energy (U), enthalpy (H), Gibbs free energy (G), and Helmholtz energy (F).

Laws of Thermodynamics:

Zeroth Law: Defines temperature.
First Law: Energy is conserved.
Second Law: Entropy always increases.
Third Law: Entropy approaches zero at absolute zero.

Related Video:

2.0Gaseous State

Key Points

General Properties of Gases:

No definite shape or volume.
Highly compressible and expansible.
Low density due to negligible intermolecular forces.
Exert pressure through elastic collisions.
Diffuse rapidly and form homogeneous mixtures.

Gas Laws:

Boyle’s Law:
$P \propto 1 V \frac{1}{V} (at constant T)$
⇒ P₁V₁=P₂V₂
Charles’ Law:

$V \propto T (at constant P)$
$\Rightarrow V T \frac{V}{T} = co n s t an t$

3. Gay-Lussac’s Law:
P∝T (at constant V)
⇒ PT\frac{P}{T}=constant

Avogadro’s Law:
V∝ n (at constant P and T)

Ideal Gas Equation: PV=nRT R = 0.0821 L·atm·mol⁻¹·K⁻¹ or 8.314 J·mol⁻¹·K⁻¹
1 mol of ideal gas at STP occupies 22.7 L.

Dalton’s Law of Partial Pressures:
Total pressure = sum of partial pressures.

P_{dry gas}=P_total−aqueous tension

Kinetic Molecular Theory:

Gas particles are in constant random motion.
Collisions are perfectly elastic.
Pressure is due to collisions with container walls.
Kinetic energy ∝ temperature.

Real Gases and Deviation:

At high pressure/low temperature, gases deviate from ideal behavior.

$Z = \frac{P V}{n RT}$

Compressibility Factor (Z):
Z = 1 for ideal gases; deviates for real gases.

Liquefaction & Critical Constants:

Critical Temperature (Tc): $\frac{8 a}{27 R b}$
Critical Volume (Vc): 3b
Critical Pressure (Pc): $\frac{a}{27 b ^{2}}$

Specific Heats & γ:

Monoatomic: γ = 1.67
Diatomic: γ = 1.4
Polyatomic: γ = 1.33

JEE Main Previous Year Solved Questions on Thermodynamics and Gaseous State

3.0Thermodynamics and Gaseous State PYQs

Q.1 One mole of an ideal monoatomic gas is subjected to changes as shown in the graph. The magnitude of the work done (by the system or on the system) is _______ J (nearest integer).

Ans

Solution

Given : log 2 = 0.3, ln 10 = 2.3

1 → 2 Isobaric process

2 → 3 Isochoric process

3 →1Isothermal process

W_{1 → 2}+ W_{2 → 3} + W(3 →1)

$= (- P (V 2 - V 1) + 0) + (- P 1 V 1 l n (V 2 V 1 \frac{V _{2}}{V _{1}}))$

Now solving:

$[- 1 \times (40 - 20) + 0 + (- 1 \times 20 l n (2040 \frac{20}{40}))]$

=−20+20×2.3×0.3

= −6.2 bar L

Thus

∣W∣=6.2 bar L=620 J

Q.2 Which of the following relations are correct?

(A) 𝚫U = q - p𝚫V

(B) 𝚫G = 𝚫H - T𝚫S

(C) 𝚫S= $\frac{q _{re v}}{T}$

(D) 𝚫H = 𝚫U - 𝚫nRT

Choose the most appropriate answer from the options given below :

(1) C and D only (2) B and C only

(3) A and B only (4) B and D only

Ans. (2)

Sol.Only (B) and (C) are correct.

(B)𝚫G = 𝚫H - T𝚫S

At constant T

𝚫G = 𝚫H - T𝚫S

(A) First law is given by

vU = Q + W

If we apply constant P and reversible work.

𝚫U = Q – P𝚫V

(C) By definition of entropy change

𝚫S= $\frac{q _{re v}}{T}$

At constant T

𝚫S= $\frac{q _{re v}}{T}$

(D) H = U + PV

For ideal gas

H = U + nRT

At constant T

𝚫H = 𝚫U + 𝚫nRT

Q.3 An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by tracing the path A ® B ® C ® A as shown in the diagram. The total work done in the process is _____ J.

Ans. (200)

Sol. Work done is given by area enclosed in the P vs V cyclic graph or V vs P cyclic graph.Sign of work is positive for clockwise cyclic process for V vs P graph.

$W = 12 \frac{1}{2} \times (30-10) \times (30 - 10) = 200 k P a - d m^{3}$

= 200 × 1000 Pa – L = 2 L-bar = 200 J

Q.4 1 mol of an ideal gas is kept in a cylinder, fitted with a piston, at the position A, at 18°C. If the piston is moved to position B, keeping the temperature unchanged, then 'x' L atm work is done in this reversible process.

Consider the figure provided.

x = ______ L atm. (nearest integer)

[Given : Absolute temperature = °C + 273.15, R = 0.08206 L atm mol^–1 K^–1]

Ans. (55)

Sol. w = –nRT In $\frac{V _{2}}{V _{1}}$

= –1×0.08206 × 291×15 ln $\frac{100}{10}$

= –55.0128

Work done by system » 55 atm lit.

Q.5 For a certain thermochemical reaction M→N at T = 400 K, 𝚫H^! = 77.2kJ mol^–1, 𝚫S = 122 JK^–1, log equilibrium constant (logK) is –______ × 10^–1.

Ans. (37)

Sol. 𝚫G° = 𝚫H° - T𝚫S°

=77.2×10³− 400×122=28400J

𝚫G° = −2.303RTlogK

⇒28400=−2.303×8.314×400 logK

⇒ log⁡K=−3.708=−37.08×10⁻¹

Q.6 Three bulbs are filled with CH₄, CO₂ and Ne as shown in the picture. The bulbs are connected through pipes of zero volume. When the stopcocks are opened and the temperature is kept constant throughout, the pressure of the system is found to be____atm.(Nearest integer)

Solution

P_TV_T = n_TRT

For CH₄

2 × 2 = n₁RT

n₁ = 4RT \frac{4}{RT}

For CO₂

⇒ n₂ = $\frac{12}{RT}$

For Ne

⇒ n₃ = $\frac{12}{RT}$

⇒ n_t = $\frac{1}{RT} [4 + 12 + 12] = 28 RT \frac{28}{RT}$

P = $\frac{28}{RT} \frac{RT}{V _{1}}$

P = $\frac{28}{V _{1}} = 3.11$

Q.7 At 600K, the root mean square (rms) speed of gas X (molar mass = 40) is equal to the most probable speed of gas Y at 90K. The molar mass of the gas Y is ________g mol^–1. (Nearest integer)

Ans. (4)

Solution:

$(U_{rms})_{X, 600} = (U_{mp})_{Y, 90} \frac{3 \times R \times 600}{40} = \frac{2 \times R \times 90}{M} M = 4$

Q.8 The number of statements, which are correct with respect to the compression of carbon dioxide from point (A) in the Andrews isotherm from the following is ________.

A. Carbon dioxide remains as a gas upto point (b)

B. Liquid carbon dioxide appears at point (c)

C. Liquid and gaseous carbon dioxide coexist between points (b) and (c)

D. As the volume decreases from (b) to (c), the amount of liquid decreases

Ans. (2)

Solution.

At (a) $C O_{2}$ exist as gas

(b) liquefaction of CO₂starts

(c) liquefaction ends

(d) CO₂ exist as liquid.

Between (b) & (c) liquid and gaseousCO₂ co-exist.

As volume changes from (b) to (c) gas decreases and liquid increases.

(A), (C) correct

Q.9 Arrange the following gases in increasing order of van der Waals constant ‘a’

A. Ar

B. CH₄

C. H₂O

D. C₆H₆

Choose the correct option from the following :-

(1) B, C, D and A (2) C, D, B and A (3) A, B, C and D (4) D, C, B and A

Ans. (3)

Solution. Vanderwaal constant – ‘a’

(i) Ar = 1.34

(ii) CH₄ = 2.25

(iii) H₂O = 5.46

(iv) C₆H₆ = 18.57

‘a’ symbolises force of attraction and directly proportional to surface area

Q.10 Which amongst the given plots is the correct plot for pressure (p) vs density (d) for an ideal gas ?

(1)

(2)

(3)

(4)

Ans. (2)

Sol.

$P = RTM \frac{RT}{M} d$

T₃ > T₂ > T₁

Join ALLEN!

Join ALLEN!

Thermodynamics and Gaseous State PYQs with Solutions

1.0Thermodynamics

2.0Gaseous State

3.0Thermodynamics and Gaseous State PYQs

Table of Contents

Thermodynamics and Gaseous State PYQs with Solutions

1.0Thermodynamics

2.0Gaseous State

3.0Thermodynamics and Gaseous State PYQs

Table of Contents