A perfect gas (0.1~mol$0.1\mathrm{~mol}$) having Cv=1.50~R$\bar{C}_{v}=1.50\mathrm{~R}$ (independent of temperature) undergoes the transformation from point 1 to point 4 as shown in the pressure-volume diagram below. If each step is reversible, the total work done (w) while going from point 1 to point 4 is (-$-$) ________ J. (nearest integer)

[Given: R=0.082 ~L~atm~K⁻¹~mol⁻¹ = 8.314 ~J~K⁻¹~mol⁻¹$R=0.082\mathrm{~L~atm~K^{-1}~mol^{-1}} = 8.314\mathrm{~J~K^{-1}~mol^{-1}}$]

Related Formula

Core Logic

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

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

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

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

Related Formula

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$w$:

Applying the first law equation constraint:

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

Related Formula

Core Logic

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

Given parameters:

• Lattice Energy of KCl$\text{KCl}$ breaking into gaseous ions = -(-z) = z kJ/mol$= -(-\text{z}) = \text{z}\text{ kJ/mol}$ (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:

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

Related Formula

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:

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

Related Formula

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 · 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$30\text{ mL of } 1\mathrm{M } \mathrm{HCl} + 30\text{ mL of } 1\mathrm{M } \mathrm{NaOH}$

Reactive millimoles = 30 mmol$\text{Reactive millimoles} = 30\text{ mmol}$. 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$30\text{ mL of } 1\mathrm{M } \mathrm{CH_3COOH} + 30\text{ mL of } 1\mathrm{M } \mathrm{NaOH}$

Reactive millimoles = 30 mmol$\text{Reactive millimoles} = 30\text{ mmol}$. 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$50\text{ mL of } 1\mathrm{M } \mathrm{HCl} + 20\text{ mL of } 1\mathrm{M } \mathrm{NaOH}$

Limiting reagent = NaOH = 20 mmol$\text{Limiting reagent} = \mathrm{NaOH} = 20\text{ mmol}$. 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$45\text{ mL of } 1\mathrm{M } \mathrm{CH_3COOH} + 25\text{ mL of } 1\mathrm{M } \mathrm{NaOH}$

Limiting reagent = 25 mmol$\text{Limiting reagent} = 25\text{ mmol}$ 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