Given below are two statements: Statement I: When a system containing ice in equilibrium with water (liquid) is heated, heat is absorbed by the system and there is no change in the temperature of the system until whole ice gets melted. Statement II: At melting point of ice, there is absorption of heat in order to overcome intermolecular forces of attraction within the molecules of water in ice and kinetic energy of molecules is not increased at melting point. In the light of the above statements, choose the correct answer from the options given below:

During phase transition, latent heat of fusion (Lf) is absorbed: Q = m Lf Temperature remains constant because the average kinetic energy of the molecules does not change; instead, potential energy changes during phase transition: Statement I Analysis: - During a phase transition (such as ice melting at 0), any added heat is utilized as latent heat of fusion. The system remains at a constant temperature of 0 as long as both solid and liquid phases coexist in equilibrium. Thus, Statement I is True. - Statement II explains *why* this happens: The thermal energy is spent entirely to break down the highly ordered crystalline hydrogen-bonded lattice of ice into liquid water. It does not increase the translational kinetic energy of the molecules. Since kinetic energy is constant, temperature remains constant. Thus, Statement II is True. Therefore, both Statement I and Statement II are True, matching Option (3). For any phase transition (melting, boiling, sublimation): Temperature stays flat. The added energy goes entirely into latent heat (potential energy change to overcome intermolecular forces), meaning average kinetic energy is constant. Class 11 Chemistry: Thermodynamics Class 11 Physics: Thermal Properties of Matter

Given below are two statements: Statement I: When a system containing ice in equilibrium with water (liquid) is heated, heat is absorbed by the system and there is no change in the temperature of the

During phase transition, latent heat of fusion (Lf) is absorbed: Q = m Lf Temperature remains constant because the average kinetic energy of the molecules does not change; instead, potential energy changes during phase transition: Statement I Analysis: - During a phase transition (such as ice melting at 0), any added heat is utilized as latent heat of fusion. The system remains at a constant temperature of 0 as long as both solid and liquid phases coexist in equilibrium. Thus, Statement I is Tru

Given below are two statements: Statement I: When a system containing ice in equilibrium with water (liquid) is heated, heat is absorbed by the system and there is no change in the temperature of the system until whole ice gets melted. Statement II: At melting point of ice, there is absorption of heat in order to overcome intermolecular forces of attraction within the molecules of water in ice and kinetic energy of molecules is not increased at melting point. In the light of the above statements, choose the correct answer from the options given below:

During phase transition, latent heat of fusion (Lf) is absorbed: Q = m Lf Temperature remains constant because the average kinetic energy of the molecules does not change; instead, potential energy changes during phase transition: Statement I Analysis: - During a phase transition (such as ice melting at 0), any added heat is utilized as latent heat of fusion. The system remains at a constant temperature of 0 as long as both solid and liquid phases coexist in equilibrium. Thus, Statement I is True. - Statement II explains *why* this happens: The thermal energy is spent entirely to break down the highly ordered crystalline hydrogen-bonded lattice of ice into liquid water. It does not increase the translational kinetic energy of the molecules. Since kinetic energy is constant, temperature remains constant. Thus, Statement II is True. Therefore, both Statement I and Statement II are True, matching Option (3). For any phase transition (melting, boiling, sublimation): Temperature stays flat. The added energy goes entirely into latent heat (potential energy change to overcome intermolecular forces), meaning average kinetic energy is constant. Class 11 Chemistry: Thermodynamics Class 11 Physics: Thermal Properties of Matter

Given below are two statements: Statement I: When a system containing ice in equilibrium with water (liquid) is heated, heat is absorbed by the system and there is no change in the temperature of the

During phase transition, latent heat of fusion (Lf) is absorbed: Q = m Lf Temperature remains constant because the average kinetic energy of the molecules does not change; instead, potential energy changes during phase transition: Statement I Analysis: - During a phase transition (such as ice melting at 0), any added heat is utilized as latent heat of fusion. The system remains at a constant temperature of 0 as long as both solid and liquid phases coexist in equilibrium. Thus, Statement I is Tru

Given below are two statements: Statement I: When a system containing ice in equilibrium with water (liquid) is heated…

Year	2026	2025	2024	Total
Questions	12	24	8	44

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

During phase transition, latent heat of fusion ( $Lf$ $L_f$) is absorbed:

$Q = m Lf$ $Q = m L_f$

Temperature remains constant because the average kinetic energy of the molecules does not change; instead, potential energy changes during phase transition:

Temperature T \propto Average Kinetic Energy of molecules

Core Logic

Statement I Analysis:

During a phase transition (such as ice melting at $0°C$ $0^{\circ}\mathrm{C}$), any added heat is utilized as latent heat of fusion. The system remains at a constant temperature of $0°C$ $0^{\circ}\mathrm{C}$ as long as both solid and liquid phases coexist in equilibrium. Thus, Statement I is True.

Step 1: Analyze Statement II

Statement II explains why this happens: The thermal energy is spent entirely to break down the highly ordered crystalline hydrogen-bonded lattice of ice into liquid water. It does not increase the translational kinetic energy of the molecules. Since kinetic energy is constant, temperature remains constant. Thus, Statement II is True.

Step 2: Conclusion

Therefore, both Statement I and Statement II are True, matching Option (3).

Pattern Recognition

For any phase transition (melting, boiling, sublimation): Temperature stays flat. The added energy goes entirely into latent heat (potential energy change to overcome intermolecular forces), meaning average kinetic energy is constant.

Chapter Mix

Class 11 Chemistry: Thermodynamics Class 11 Physics: Thermal Properties of Matter

More Thermodynamics Previous-Year Questions — Page 9

Q83 jee_main_2024_30_january_evening Hess's Law of Constant Heat Summation

Two reactions are given below:

2Fe(s) + (3)/(2)O2(g) arrow Fe₂O3(s), Δ H° = -822 kJ/mol

C(s) + (1)/(2)O2(g) arrow CO(g), Δ H° = -110 kJ/mol

Then enthalpy change for following reaction

3C(s) + Fe₂O3(s) arrow 2Fe(s) + 3CO(g)

Numerical Answer. Answer: 492 to 492

Solution

Related Formula

According to Hess's Law, the net enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps into which it can be divided.

Core Logic

Let the given reactions be: (1) $2Fe(s) + (3)/(2)O2(g) arrow Fe₂O3(s), Δ H₁ = -822 kJ/mol$ (2) $C(s) + (1)/(2)O2(g) arrow CO(g), Δ H₂ = -110 kJ/mol$

Target Reaction (3):

3C(s) + Fe₂O3(s) arrow 2Fe(s) + 3CO(g), Δ H₃ = ?

To construct the target reaction:

We need $3 CO(g)$ $3 \mathrm{CO}_{(g)}$ on the product side, so we multiply reaction (2) by 3.
We need $Fe₂O3(s)$ $\mathrm{Fe}_2\mathrm{O}_{3(s)}$ on the reactant side and $2 Fe(s)$ $2 \mathrm{Fe}_{(s)}$ on the product side, so we reverse reaction (1).

Step 1: Calculate Net Enthalpy

Target Reaction (3) = $3 \times (2) - (1)$ $3 \times (2) - (1)$

Δ H₃ = 3 \times Δ H₂ - Δ H₁

Δ H₃ = 3(-110) - (-822)

Δ H₃ = -330 + 822 = 492 kJ/mol

Chapter Mix

Class 11 Chemistry: Thermodynamics

Q jee_main_2024_30_jan_morning Work Done in Cyclic Process

An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by tracing the path

Aarrow Barrow Carrow A

as shown in the diagram. The total work done in the process is ________ J.

Work Done in Cyclic Process diagram for Q83 - JEE Main 2024 Morning — The image is a graph of Volume (dm3) vs Pressure (kPa) showing a triangular cyclic process starting from A(10,10) to B(10,30) to C(30,10) and back to A.

Numerical Answer. Answer: 200 to 200

Solution

Related Formula

Wcyclic = Area enclosed in P-V graph

Core Logic

The work done in a cyclic process is equal to the magnitude of the area enclosed by the cycle on a Pressure-Volume graph. Note that the provided graph is Volume ( $V$ $V$) on the y-axis versus Pressure ( $P$ $P$) on the x-axis. The path $A arrow B arrow C arrow A$ is traced in a clockwise direction on the $V$ $V$- $P$ $P$ graph. Clockwise on a $V$ $V$- $P$ $P$ graph corresponds to anti-clockwise on a standard $P$ $P$- $V$ $V$ graph, meaning net expansion work is done by the gas, making it positive conventionally (or negative depending on chemistry sign convention, but magnitude is asked for).

Step 1: Calculating Area

The enclosed region is a right-angled triangle. Base of triangle on P-axis = $30 - 10 = 20 kPa$ $30 - 10 = 20 \text{ kPa}$ Height of triangle on V-axis = $30 - 10 = 20 dm³$ $30 - 10 = 20 \text{ dm}^3$

Area = (1)/(2) \times base \times height

Area = (1)/(2) \times 20 \times 20 = 200 kPa ³

Step 2: Unit conversion

$1 kPa = 10³ Pa$ $1 \text{ kPa} = 10^3 \text{ Pa}$ $1 dm³ = 1 Litre = 10⁻³ m³$

W = 200 \times 10³ Pa \times 10⁻³ m³

W = 200 J

Pattern Recognition

$1 kPa \cdot 1 L = 1 Joule$ . This direct conversion saves time without converting explicitly to standard SI units ( $Pa$ $Pa$ and $m³$ $m^3$).

Chapter Mix

Class 11 Chemistry: Thermodynamics

Q90 jee_main_2024_31_jan_evening Work Done in Isothermal Reversible Expansion

5 moles

of an ideal gas expands from

10 L

to a volume of

100 L

300 K

under isothermal and reversible condition then work

w

, is

-x J

. The value of

x

is ________ (Given

R = 8.314 J K⁻¹mol⁻¹

)

Numerical Answer. Answer: 28720 to 28721

Solution

Related Formula

W = -2.303 nRT ( (V₂)/(V₁) )

Core Logic

For an isothermal and reversible expansion of an ideal gas, work is done by the system on the surroundings, hence it is negative by IUPAC convention. Given: $n = 5 moles$ $n = 5\text{ moles}$ $R = 8.314 J K⁻¹mol⁻¹$ $T = 300 K$ $T = 300\text{ K}$ $V₁ = 10 L$ $V_1 = 10\text{ L}$ $V₂ = 100 L$ $V_2 = 100\text{ L}$

Step 1: Calculating Work Done

W = -2.303 \times 5 \times 8.314 \times 300 \times ( (100)/(10) )

W = -2.303 \times 5 \times 8.314 \times 300 \times (10)

W = -2.303 \times 12471 \times 1

W = -28720.713 J

Step 2: Final Formatting

The question asks for work $w = -x J$ $w = -x\text{ J}$. So $x = 28720.713$ $x = 28720.713$, which rounds to $28721$ $28721$.

Chapter Mix

Class 11 Chemistry: Thermodynamics

Q88 jee_main_2024_31_jan_morning Gibbs Free Energy and Equilibrium

Consider the following reaction at 298 K.

(3)/(2)O2(g) leftharpoons O3(g). Kₚ = 2.47 \times 10⁻²⁹

ΔᵣG

for the reaction is ________ kJ. (Given

R = 8.314 J K⁻¹ mol⁻¹

)

Numerical Answer. Answer: 163 to 164

Solution

Related Formula

ΔᵣG = -RT ln Kₚ

Step 1: Calculation

ΔᵣG = -8.314 \times 10⁻³ kJ K⁻¹ mol⁻¹ \times 298 K \times ln(2.47 \times 10⁻²⁹)

= -8.314 \times 10⁻³ \times 298 \times (-65.87)

= 163.19 kJ

Step 2: Nearest Integer

Rounding $163.19$ $163.19$ to the nearest integer gives $163$ $163$.

Chapter Mix

Class 11 Chemistry: Thermodynamics

Solution & Explanation

Related Formula

Core Logic

Step 1: Analyze Statement II

Step 2: Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More Thermodynamics Previous-Year Questions — Page 9

Solution

Related Formula

Core Logic

Step 1: Calculate Net Enthalpy

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculating Area

Step 2: Unit conversion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculating Work Done

Step 2: Final Formatting

Chapter Mix

Solution

Related Formula

Step 1: Calculation

Step 2: Nearest Integer

Chapter Mix

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