Ice and water are placed in a closed container at a pressure of 1 atm and temperature 273.15K . If pressure of the system is increased 2 times, keeping temperature constant, then identify correct observation from following:

Water has a unique property where the density of the liquid phase is greater than the density of the solid phase (ice). Consequently, the molar volume of ice is larger than that of liquid water: According to Le Chatelier's Principle, increasing the pressure favors the phase that occupies a smaller volume to alleviate the applied stress. Thus, shifting the system forward converts ice into liquid water: If the pressure is increased considerably (such as doubling it to 2 atm) at 273.15K, the melting point decreases, causing the entire solid phase (ice) to disappear completely. Sees: Ice-water system under pressure change. Trap: Assuming that an increase in pressure always favors the solid phase. Water has an anomalous phase curve with a negative slope. Class 11 Chemistry: Chemical Thermodynamics

Ice and water are placed in a closed container at a pressure of 1 atm and temperature 273.15K . If pressure of the system is increased 2 times, keeping temperature constant, then identify correct obse

Water has a unique property where the density of the liquid phase is greater than the density of the solid phase (ice). Consequently, the molar volume of ice is larger than that of liquid water: According to Le Chatelier's Principle, increasing the pressure favors the phase that occupies a smaller volume to alleviate the applied stress. Thus, shifting the system forward converts ice into liquid water: If the pressure is increased considerably (such as doubling it to 2 atm) at 273.15K, the meltin

Ice and water are placed in a closed container at a pressure of 1 atm and temperature 273.15K . If pressure of the system is increased 2 times, keeping temperature constant, then identify correct observation from following:

Water has a unique property where the density of the liquid phase is greater than the density of the solid phase (ice). Consequently, the molar volume of ice is larger than that of liquid water: According to Le Chatelier's Principle, increasing the pressure favors the phase that occupies a smaller volume to alleviate the applied stress. Thus, shifting the system forward converts ice into liquid water: If the pressure is increased considerably (such as doubling it to 2 atm) at 273.15K, the melting point decreases, causing the entire solid phase (ice) to disappear completely. Sees: Ice-water system under pressure change. Trap: Assuming that an increase in pressure always favors the solid phase. Water has an anomalous phase curve with a negative slope. Class 11 Chemistry: Chemical Thermodynamics

Ice and water are placed in a closed container at a pressure of 1 atm and temperature 273.15K . If pressure of the system is increased 2 times, keeping temperature constant, then identify correct obse

Water has a unique property where the density of the liquid phase is greater than the density of the solid phase (ice). Consequently, the molar volume of ice is larger than that of liquid water: According to Le Chatelier's Principle, increasing the pressure favors the phase that occupies a smaller volume to alleviate the applied stress. Thus, shifting the system forward converts ice into liquid water: If the pressure is increased considerably (such as doubling it to 2 atm) at 273.15K, the meltin

Ice and water are placed in a closed container at a pressure of 1 atm and temperature 273.15K . If pressure of the…

Q68 jee_main_2026_21_jan_morning Work Done in PV Graph

Which of the following graphs between pressure ‘P’ versus volume ‘V’ represent the maximum work done?

A. $Option 1$ $\text{Option 1}$
B. $Option 2$ $\text{Option 2}$
C. $Option 3$ $\text{Option 3}$
D. $Option 4$ $\text{Option 4}$

Solution

Core Logic

The magnitude of work done by or on a gas is given by the area under the P-V curve projected onto the volume axis.

Graph 1: Cyclic process forming a triangle. Area is bounded, represents net work. Graph 2: Isochoric drop (vertical line at $V=22.4L$ $V=22.4L$). Area = 0, so work done is zero. Graph 3: Expansion process forming a cycle. Area is enclosed in a convex shape. Graph 4: Direct expansion from $V=22.4$ $V=22.4$ to $V=44.8$ $V=44.8$ at pressure $P=1$ $P=1$ up to $P=2$ $P=2$ (a rectangle combined with a triangle). The total area under the upper curve from $V=22.4$ $V=22.4$ to $V=44.8$ $V=44.8$ covers the entire shaded region under the path down to the V-axis.

Option (4) provides the largest total area under the curve extending down to the horizontal axis (maximum magnitude of work done).

Chapter Mix

Class 11 Chemistry: Thermodynamics

Q69 jee_main_2026_21_jan_morning Gibbs Free Energy and Equilibrium

For the reaction,

N₂O₄ leftharpoons 2NO₂

, graph is plotted as shown below. Identify correct statements. A. Standard free energy change for the reaction is

-5.40 kJ mol⁻¹

. B. As

Δ G

in graph is positive,

N₂O₄

will not dissociate into

NO₂

at all. C. Reverse reaction will go to completion. D. When 1 mole of

N₂O₄

changes into equilibrium mixture, value of

Δ G = -0.84 kJ mol⁻¹

. E. When 2 mole of

NO₂

changes into equilibrium mixture,

Δ G

for equilibrium mixture is

-6.24 kJ mol⁻¹

.

Gibbs Free Energy and Equilibrium diagram for Q69 - JEE Main 2026 Morning — A plot of Gibbs free energy against the fraction of N2O4 dissociated.

Choose the correct answer from the options given below :

A. $D and E only$ $\text{D and E only}$
B. $C and E only$ $\text{C and E only}$
C. $A and D only$ $\text{A and D only}$
D. $B and C only$ $\text{B and C only}$

Solution

Core Logic

Let's analyze the statements based on the given Gibbs free energy (G) vs. extent of reaction plot.

A. Standard free energy change ( $Δᵣ G°$ $\Delta_r G^{\circ}$) is the difference between standard free energies of pure products (point B) and pure reactants (point A): $Δᵣ G° = GB° - GA°$ . Since B is higher than A, $Δᵣ G°$ $\Delta_r G^{\circ}$ is positive, not $-5.40 kJ mol⁻¹$ $-5.40\text{ kJ mol}^{-1}$. Statement A is false.

B. Even if $Δᵣ G°$ $\Delta_r G^{\circ}$ is positive, the minimum of the curve (equilibrium state E) lies between the pure reactant and product states. Therefore, partial dissociation occurs to reach equilibrium. It is false to say it will not dissociate at all. Statement B is false.

C. The minimum E is not at fraction = 0 or 1, meaning an equilibrium mixture exists. Reverse reaction does not go to completion. Statement C is false.

D. From 1 mole of pure reactant (point A) to the equilibrium mixture (point E), the drop in Gibbs energy is $0.84 kJ mol⁻¹$ $0.84\text{ kJ mol}^{-1}$. Thus $Δ G = -0.84 kJ mol⁻¹$ $\Delta G = -0.84\text{ kJ mol}^{-1}$ is correct. Statement D is true.

E. The difference from pure products (point B, equivalent to 2 moles $NO₂$ $NO_2$) to equilibrium (E) is the entire vertical distance. The drop from B to A is $5.40$ $5.40$, and A to E is $0.84$ $0.84$. So drop from B to E is $- (5.40 + 0.84) = -6.24 kJ mol⁻¹$ . Statement E is true.

Step 1: Final Conclusion

Only statements D and E are correct.

Pattern Recognition

The free energy of mixing creates a minimum 'dip' (Equilibrium) below both pure reactants and pure products, preventing either forward or reverse reactions from going absolutely to 100% completion.

Chapter Mix

Class 11 Chemistry: Thermodynamics Class 11 Chemistry: Equilibrium

Q73 jee_main_2026_21_jan_morning Gibbs Free Energy and Equilibrium Constant

Use the following data :

Substance	$Δf H(500K)kJ mol⁻¹$	$S(500K)J K⁻¹ mol⁻¹$
AB(g)	32	222
$A₂(g)$ $A_2(g)$	6	146
$B₂(g)$ $B_2(g)$	x	280

One mole each of

A₂(g)

and

B₂(g)

are taken in a 1L closed flask and allowed to establish the equilibrium at 500K.

A₂(g) + B₂(g) leftharpoons 2AB(g)

The value of x (in

kJ mol⁻¹

) is .... (Nearest integer) (Given:

K=2.2, R=8.314 J K⁻¹ mol⁻¹

)

Numerical Answer. Answer: 70 to 70

Solution

Related Formula

Δᵣ H° = Σ Δf H°(products) - Σ Δf H°(reactants)

Δᵣ S° = Σ S°(products) - Σ S°(reactants)

Δᵣ G° = Δᵣ H° - TΔᵣ S°

Δᵣ G° = -2.303 RT K

Core Logic

For the reaction $A₂(g) + B₂(g) leftharpoons 2AB(g)$ :

Enthalpy change ( $Δᵣ H°$ $\Delta_r H^{\circ}$):

Δᵣ H° = [2 \times Δf H°(AB)] - [Δf H°(A₂) + Δf H°(B₂)]

Δᵣ H° = (2 \times 32) - (6 + x) = (64 - 6 - x) = (58 - x) kJ mol⁻¹

Entropy change ( $Δᵣ S°$ $\Delta_r S^{\circ}$):

Δᵣ S° = [2 \times S°(AB)] - [S°(A₂) + S°(B₂)]

Δᵣ S° = (2 \times 222) - (146 + 280) = 444 - 426 = 18 J K⁻¹ mol⁻¹

Standard Gibbs Free Energy ( $Δᵣ G°$ $\Delta_r G^{\circ}$) via Equilibrium Constant:

Δᵣ G° = -2.303 \times R \times T K

Δᵣ G° = -2.303 \times 8.314 \times 500 \times 2.2 = -21063.8 J mol⁻¹ = -21.06 kJ mol⁻¹

Using Gibbs Equation:

Δᵣ G° = Δᵣ H° - TΔᵣ S°

-21.06 = (58 - x) - 500 \times ((18)/(1000))

-21.06 = 58 - x - 9

$-21.06 = 49 - x$ $-21.06 = 49 - x$

x = 49 + 21.06 = 70.06 kJ mol⁻¹

Rounding off to nearest integer, $x = 70$ $x = 70$.

Chapter Mix

Class 11 Chemistry: Thermodynamics Class 11 Chemistry: Equilibrium

Q56 jee_main_2026_21_jan_evening Bond Enthalpy and Enthalpy of Atomization

Consider the following data:

Δf H(methane, g) = -X kJ mol⁻¹

Enthalpy of sublimation of graphite

= Y kJ mol⁻¹

Dissociation enthalpy of

H₂ = Z kJ mol⁻¹

The bond enthalpy of

C-H

bond is given by:

A. $(1) (X + Y + 2Z)/(4)$ $(1) \ \frac{X + Y + 2Z}{4}$
B. $(2) (X + Y + 4Z)/(2)$ $(2) \ \frac{X + Y + 4Z}{2}$
C. $(3) X + Y + Z$ $(3) \ X + Y + Z$
D. $(4) (-X + Y + Z)/(4)$ $(4) \ \frac{-X + Y + Z}{4}$

Solution

Core Logic

Reaction for formation of methane:

C(s) + 2H₂(g) arrow CH₄(g)

$-X = (Δ Hsub of carbon) + 2 \times (B.E. of H-H) - 4 \times (B.E. of C-H)$

$-X = Y + 2Z - 4(B.E. of C-H)$ $-X = Y + 2Z - 4(\text{B.E. of C-H})$

Step 1: Rearranging for Bond Enthalpy

Rearranging the equation for C-H bond enthalpy:

B.E. of C-H = (X + Y + 2Z)/(4)

Pattern Recognition

Sees: bond enthalpy derivation from heat of formation, sublimation, and dissociation. Trap: Sign convention errors when substituting formation enthalpies.

Chapter Mix

Class 11 Chemistry: Chemical Thermodynamics

Q60 jee_main_2026_22_january_morning Work and Internal Energy

Match the LIST-I with LIST-II

List-I (Thermodynamic Process)	List-II (Magnitude in kJ)
A. Work done in reversible, isothermal expansion of 2 mol of ideal gas from $2 dm³$ $2\text{ dm}^{3}$ to $20 dm³$ $20\text{ dm}^{3}$ at 300 K.	I. 4
B. Work done in irreversible isothermal expansion of 1 mol ideal gas from $1 m³$ $1\text{ m}^{3}$ to $3 m³$ $3\text{ m}^{3}$ at 300 K against a constant pressure of 3kPa.	II. 11.5
C. Change in internal energy for adiabatic expansion of a 1 mol ideal gas with change of temperature = 320 K and $Cv = (3)/(2)R$ $\overline{C}_{v} = \frac{3}{2}R$	III. 6
D. Change in enthalpy at constant pressure of 1 mole ideal gas with change of temperature = 337 K and $Cₚ = (5)/(2)R$ $\overline{C}_{p} = \frac{5}{2}R$	IV. 7

Choose the correct answer from the option given below:

A. $A-III, B-II, C-IV, D-I$ $\text{A-III, B-II, C-IV, D-I}$
B. $A-II, B-III, C-I, D-IV$ $\text{A-II, B-III, C-I, D-IV}$
C. $A-I, B-II, C-III, D-IV$ $\text{A-I, B-II, C-III, D-IV}$
D. $A-II, B-I, C-III, D-IV$ $\text{A-II, B-I, C-III, D-IV}$

Solution

Core Logic

Evaluate each option one by one:

Option (A): Work done in reversible isothermal expansion $W = -nRT ln((V₂)/(V₁))$ Magnitude $= (2 \times 8.314 \times 300)/(1000) \times ln(10) kJ$ $= 4.9884 \times 2.303 \approx 11.5 kJ$ So, A $arrow$ $\rightarrow$ II.

Option (B): Work done in irreversible isothermal expansion against constant external pressure $W = -Pₑₓₜ (V₂ - V₁)$ $W = -P_{\text{ext}} (V_2 - V_1)$ Magnitude $= 3 kPa \times (3 - 1) m³ = 3 \times 10³ Pa \times 2 m³ = 6000 J = 6 kJ$ So, B $arrow$ $\rightarrow$ III.

Option (C): Change in internal energy $Δ U = n Cv Δ T$ $\Delta U = n C_v \Delta T$ Magnitude $= 1 \times ((3)/(2) \times 8.314) \times 320 = (12.471 \times 320)/(1000) kJ \approx 3.99 kJ \approx 4 kJ$ So, C $arrow$ $\rightarrow$ I.

Option (D): Change in enthalpy $Δ H = n Cₚ Δ T$ $\Delta H = n C_p \Delta T$ Magnitude $= 1 \times ((5)/(2) \times 8.314) \times 337 = (20.785 \times 337)/(1000) kJ \approx 7 kJ$ So, D $arrow$ $\rightarrow$ IV.

Step 1: Final Matching

Matches are A-II, B-III, C-I, D-IV.

Pattern Recognition

Standard match-list question. Calculate A and D first, as they quickly eliminate wrong options.

Chapter Mix

Class 11 Chemistry: Thermodynamics

Ice and water are placed in a closed container at a pressure of 1 atm and temperature $273.15K$ $273.15\mathrm{K}$ . If pressure of the system is increased 2 times, keeping temperature constant, then identify correct observation from following:

Solution & Explanation

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Ice and water are placed in a closed container at a pressure of 1 atm and temperature 273.15K$273.15\mathrm{K}$ . If pressure of the system is increased 2 times, keeping temperature constant, then identify correct observation from following:

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Ice and water are placed in a closed container at a pressure of 1 atm and temperature $273.15K$ $273.15\mathrm{K}$ . If pressure of the system is increased 2 times, keeping temperature constant, then identify correct observation from following: