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…

Q27 jee_main_2025_04_april_morning Spontaneity and Gibbs Energy

Let us consider a reversible reaction at temperature,

T

. In this reaction, both

Δ H

and

Δ S

were observed to have positive values. If the equilibrium temperature is

Tₑ

, then the reaction becomes spontaneous at:

A. $T = Tₑ$ $T = T_e$
B. $Tₑ > T$ $T_e > T$
C. $T > Tₑ$ $T > T_e$
D. $Tₑ = 5T$ $T_e = 5T$

Solution

Related Formula

Δ G = Δ H - TΔ S

Core Logic

For a reaction to be spontaneous, the change in Gibbs free energy must be negative:

Δ G < 0 Δ H - TΔ S < 0

Given that both $Δ H > 0$ $\Delta H > 0$ and $Δ S > 0$ $\Delta S > 0$:

Δ H < TΔ S T > (Δ H)/(Δ S)

At the equilibrium temperature $Tₑ$ $T_e$, $Δ G = 0$ $\Delta G = 0$, which gives:

Tₑ = (Δ H)/(Δ S)

Substituting this back into the inequality reveals that the reaction is spontaneous when:

$T > Tₑ$ $T > T_e$

Pattern Recognition

When both $Δ H$ $\Delta H$ and $Δ S$ $\Delta S$ are positive, the reaction is entropy-driven and becomes spontaneous only at higher temperatures ( $T > Tₑ$ $T > T_e$).

Chapter Mix

Class 11 Chemistry: Chemical Thermodynamics

Q30 jee_main_2025_04_april_morning Isothermal and Reversible Expansion

One mole of an ideal gas expands isothermally and reversibly from

10~dm³

to

20~dm³

at

300~K

.

Δ U

,

q

and work done in the process respectively are: Given:

R = 8.3 ~J~K⁻¹~mol⁻¹

,

ln 10 = 2.3

,

2 = 0.30

,

3 = 0.48

A. $0, 21.84~kJ, -1.26~kJ$
B. $0, -17.18~kJ, 1.718~J$
C. $0, 21.84~kJ, 21.84~kJ$
D. $0, 1.718~kJ, -1.718~kJ$

Solution

Related Formula

Δ U = n Cv Δ T

w = -n R T ln((V₂)/(V₁))

Δ U = q + w

Core Logic

Since the expansion step is strictly isothermal ( $Δ T = 0$ $\Delta T = 0$):

$Δ U = 0$ $\Delta U = 0$

Now compute the work command parameter $w$ $w$:

w = -n R T ln((V₂)/(V₁)) = -1 \cdot 8.3 \cdot 300 \cdot ln((20)/(10))

w = -2490 \cdot ln(2) = -2490 \cdot (2.3 \cdot 2)

w = -2490 \cdot (2.3 \cdot 0.30) = -2490 \cdot 0.69 = -1718.1~J = -1.718~kJ

Applying the first law equation constraint:

q = -w = +1.718~kJ

Hence, $Δ U = 0$ $\Delta U = 0$, $q = 1.718~kJ$ $q = 1.718\mathrm{~kJ}$, $w = -1.718~kJ$ $w = -1.718\mathrm{~kJ}$.

Pattern Recognition

Isothermal expansion of an ideal gas ALWAYS yields $Δ U = 0$ $\Delta U = 0$. Work is negative (done by system) and heat exchange $q$ $q$ matches work magnitude inversely.

Chapter Mix

Class 11 Chemistry: Chemical Thermodynamics

Q28 jee_main_2025_07_april_evening Lattice Enthalpy and Born-Haber Cycle

The hydration energies of

K^+

and

Cl^-

are

-x

and

-y kJ/mol

respectively. If lattice energy of

KCl

is

-z kJ/mol

, then the heat of solution of

KCl

is:

A. $+x - y - z$ $+\text{x} - \text{y} - \text{z}$
B. $x + y + z$ $\text{x} + \text{y} + \text{z}$
C. $z - (x + y)$ $\text{z} - (\text{x} + \text{y})$
D. $-z - (x + y)$ $-\text{z} - (\text{x} + \text{y})$

Solution

Related Formula

Δ Hsol = Lattice Energy (L.E.) + Δ Hhyd(Cation) + Δ Hhyd(Anion)

Core Logic

According to Hess's Law, the dissolution process can be mapped as follows:

Lattice Enthalpy and Born-Haber Cycle diagram for Q28 - JEE Main 2025 Evening

Given parameters:

Lattice Energy of $KCl$ $\text{KCl}$ breaking into gaseous ions $= -(-z) = z kJ/mol$ (since lattice energy released on formation is given as $-z$ $-\text{z}$).
Hydration energy of $K^+ = -x kJ/mol$ $\text{K}^+ = -\text{x}\text{ kJ/mol}$
Hydration energy of $Cl^- = -y kJ/mol$ $\text{Cl}^- = -\text{y}\text{ kJ/mol}$

Step 1: Computation

Substituting the values into the governing formulation:

Δ Hsol = z + (-x) + (-y)

Δ Hsol = z - x - y = z - (x + y)

Pattern Recognition

To dissolve an ionic crystal, energy equal to the lattice energy must be supplied (endothermic step, $+z$ $+\text{z}$), and hydration releases energy (exothermic steps, $-x$ $-\text{x}$ and $-y$ $-\text{y}$). Net heat of solution is simply the sum of these parts: $z - x - y$ $\text{z} - \text{x} - \text{y}$.

Chapter Mix

Class 11 Chemistry: Chemical Thermodynamics Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q33 jee_main_2025_07_april_evening Standard Enthalpy of Formation

The correct statement amongst the following is:

A. $The term 'standard state' implies that the temperature is 0°C$
B. $The standard state of pure gas is the pure gas at a pressure of 1 and temperature 273 K$
C. $ΔfH298θ is zero for O(g)$
D. $ΔfH500θ is zero for O2(g)$

Solution

Related Formula

ΔfHθ = 0 for an element in its reference/most stable standard state

Core Logic

Standard state conditions prescribe a pressure of $1$ $1\text{ \bar}$. Temperature is not fixed by definition but is explicitly specified (often reference tables use $298.15 K$ $298.15\text{ K}$).
Oxygen naturally and stably exists as diatomic gas molecules ( $O₂(g)$ $\text{O}_2(g)$) at standard thresholds.
The enthalpy of formation of an element in its reference elemental state is identically zero at any reference temperature:

ΔfH₅₀₀θ[O2(g)] = 0

Conversely, atomic oxygen gas ( $O(g)$ $\text{O}(g)$) is not the reference phase, so its formation enthalpy is non-zero.

Step 1: Verification of Options

Statement (4) accurately aligns with thermodynamic core definitions, while statement (1) and (2) mistakenly conflate standard ambient reference states with STP conditions ( $273.15 K, 1 atm$ $273.15\text{ K}, 1\text{ atm}$).

Pattern Recognition

Standard state definitions checklist: Pressure $= 1$ $= 1\text{ \bar}$. Temperature is variable/assigned independently. Elements in their most stable natural form take $ΔfHθ = 0$ $\Delta{\text{f}}H^{\theta} = 0$ at all thermal profiles.

Chapter Mix

Class 11 Chemistry: Chemical Thermodynamics

Q31 jee_main_2025_24_jan_evening Enthalpy of Neutralization

Which of the following mixing of 1M base and 1M acid leads to the largest increase in temperature?

A. \text{30 mL HCl and 30 mL NaOH}
B. \text{30 mL } \mathrm{CH_{3}COOH} \text{ and 30 mL NaOH}
C. \text{50 mL HCl and 20 mL NaOH}
D. \text{45 mL } \mathrm{CH_{3}COOH} \text{ and 25 mL NaOH}

Solution

Related Formula

Q = nreacted \cdot Δ Hneutralization

Δ T = (Q)/(m \cdot c)

Core Logic

The temperature rise depends directly on the total heat released ( $Q$ $Q$) normalized by the total heat capacity of the resulting mixed volume ( $m \cdot c$ $m \cdot c$). Let's evaluate the millimoles of $H^+$ $\mathrm{H}^+$ and $OH^-$ $\mathrm{OH}^-$ that react in each mixture:

Option 1: $30 mL of 1M HCl + 30 mL of 1M NaOH$

$Reactive millimoles = 30 mmol$ . Both are strong electrolytes, releasing full neutralization energy ( $-57.3 kJ/mol$ $\sim -57.3\text{ kJ/mol}$). Total volume = $60 mL$ $60\text{ mL}$.

Option 2: $30 mL of 1M CH₃COOH + 30 mL of 1M NaOH$

$Reactive millimoles = 30 mmol$ . However, since acetic acid is a weak acid, part of the heat is consumed in its ionization. Thus, less total heat is evolved compared to Option 1.

Option 3: $50 mL of 1M HCl + 20 mL of 1M NaOH$

$Limiting reagent = NaOH = 20 mmol$ . Only $20 mmol$ $20\text{ mmol}$ reacts. Total volume = $70 mL$ $70\text{ mL}$.

Option 4: $45 mL of 1M CH₃COOH + 25 mL of 1M NaOH$

$Limiting reagent = 25 mmol$ weak neutralization profile.

Comparing Option 1 and Option 3, Option 1 releases significantly more heat ( $30 mmol$ $30\text{ mmol}$ vs $20 mmol$ $20\text{ mmol}$) into a smaller volume ( $60 mL$ $60\text{ mL}$ vs $70 mL$ $70\text{ mL}$), yielding the largest increase in temperature $Δ T$ $\Delta T$.

Pattern Recognition

To maximize $Δ T$ $\Delta T$, look for the option that maximizes the amount of reacting strong acid and strong base equivalents while keeping the total solution volume as small as possible.

Chapter Mix

Class 11 Chemistry: Thermodynamics Class 11 Chemistry: Equilibrium

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

Core Logic

Pattern Recognition

Chapter Mix

More Chemical Thermodynamics Previous-Year Questions — Page 6

Solution

Related Formula

Core Logic

Pattern Recognition

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Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Computation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Verification of Options

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

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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: