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

Using Hess's Law, the enthalpy change of a net reaction can be determined by linearly combining the steps:

Core Logic

The heat of formation of SO₂(g)$\mathrm{SO}_2(g)$ corresponds to the target thermochemical equation:

Let's write out the given equations along with their enthalpy changes (remembering that exothermic reactions release heat, so Δ H = -Q$\Delta H = -Q$): 1. S(g) + (3)/(2)O₂(g) arrow SO₃(g) Δ H₁ = -2x kcal$\mathrm{S}(g) + \frac{3}{2}\mathrm{O}_{2}(g) \rightarrow \mathrm{SO}_{3}(g) \quad \Delta H_{1} = -2x\text{ kcal}$ 2. SO₂(g) + (1)/(2)O₂(g) arrow SO₃(g) Δ H₂ = -y kcal$\mathrm{SO}_{2}(g) + \frac{1}{2}\mathrm{O}_{2}(g) \rightarrow \mathrm{SO}_{3}(g) \quad \Delta H_{2} = -y\text{ kcal}$

To isolate SO₂(g)$\mathrm{SO}_{2}(g)$ on the product side, subtract Equation (2) from Equation (1):

Now apply the same operation to the enthalpy values:

This matches Option (2).

Pattern Recognition

To isolate your target species on the desired side of the equation, use Hess's Law to add or subtract the given elemental equations. Make sure to invert the sign of the enthalpy change if you reverse a reaction.

Chapter Mix

Class 11 Chemistry: Thermodynamics

Related Formula

Core Logic

An endothermic profile specifies that Δ H > 0$\Delta H > 0$. For the system to become spontaneous (Δ G < 0$\Delta G < 0$) specifically when shifting to higher temperatures (T$T$), the temperature-dependent entropic subtraction term (-TΔ S$-T\Delta S$) must outweigh the enthalpic barrier. This transition demands a positive structural entropy step, i.e., Δ S > 0$\Delta S > 0$.

Hence, both Δ H$\Delta H$ and Δ S$\Delta S$ are positive.

Pattern Recognition

Spontaneity driven purely by elevated thermal thresholds mandates matching positive signs for enthalpy and entropy.

Chapter Mix

Class 11 Chemistry: Chemical Thermodynamics

Related Formula

Core Logic

First, calculate the standard entropy change for the reaction system (Δ Sᵣₓₙ⁰$$\Delta S_{\text{rxn}}^{0}$$):

At thermodynamic equilibrium, the change in Gibbs free energy drops to zero (Δ G⁰ = 0$\Delta G^0 = 0$):

Convert the enthalpy value into Joules (Δ H⁰ = -35 × 10³ J/mol$$\Delta H^0 = -35 \times 10^3\text{ J/mol}$$) and substitute the parameters:

Pattern Recognition

Ensure all variables use matching energy units (Joules vs. Kilojoules) before setting up your final division step.

Chapter Mix

Class 11 Chemistry: Chemical Thermodynamics

Related Formula

For any state function like Internal Energy (U$U$), the cyclic integral over a complete closed loop is identically zero:

Core Logic

Internal energy (U$U$) depends only on the initial and final states of the thermodynamic system, not on the path followed.

In all three listed cases, the ideal gas undergoes a complete cyclic path that returns to its original configuration state A$A$.

Step 1: Final Evaluation

Since every transformation begins and ends at point A$A$:

Therefore, Δ U (Case-I) = Δ U (Case-II) = Δ U (Case-III)$\Delta U\text{ (Case-I)} = \Delta U\text{ (Case-II)} = \Delta U\text{ (Case-III)}$.

Pattern Recognition

Do not waste time calculating path areas or values if the question asks for a state function change (Δ U$\Delta U$, Δ H$\Delta H$, Δ S$\Delta S$, Δ G$\Delta G$) over a cyclic loop. The answer is instantly zero for all cases!

Chapter Mix

Class 11 Chemistry: Thermodynamics

Related Formula

Enthalpy of reaction from enthalpy of formation data:

Core Logic

Write out the balanced thermochemical equation for the combustion of benzene (C₆H₆$C_6H_6$):

Given parameters:

• Δ Hc = -3267.0 kJ/mol$\Delta H_c = -3267.0\mathrm{\ kJ/mol}$
• Δ Hf[CO₂] = -393.5 kJ/mol$\Delta H_f[CO_2] = -393.5\mathrm{\ kJ/mol}$
• Δ Hf[H₂O] = -286.0 kJ/mol$\Delta H_f[H_2O] = -286.0\mathrm{\ kJ/mol}$
• Δ Hf[O₂] = 0 kJ/mol$\Delta H_f[O_2] = 0\mathrm{\ kJ/mol}$

Step 1: Applying Hess's Law

Substitute these values into the reaction expression:

Pattern Recognition

Always set up products minus reactants when using heat of formation data. Pay close attention to stoichiometric coefficients (multiply CO₂$CO_2$ by 6$6$ and H₂O$H_2O$ by 3$3$) to ensure accurate bookkeeping.

Chapter Mix

Class 11 Chemistry: Thermodynamics