A perfect gas (0.1mathrm~mol$0.1\mathrm{~mol}$) having barC_v=1.50mathrm~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=0.082mathrm~L~atm~K^-1~mol^-1 = 8.314mathrm~J~K^-1~mol^-1$R=0.082\mathrm{~L~atm~K^{-1}~mol^{-1}} = 8.314\mathrm{~J~K^{-1}~mol^{-1}}$]

### 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). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics

### 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 (Delta_r G^circ$\Delta_r G^{\circ}$) is the difference between standard free energies of pure products (point B) and pure reactants (point A): Delta_r G^circ = G_B^circ - G_A^circ$\Delta_r G^{\circ} = G_B^{\circ} - G_A^{\circ}$. Since B is higher than A, Delta_r G^circ$\Delta_r G^{\circ}$ is positive, not -5.40text kJ mol^-1$-5.40\text{ kJ mol}^{-1}$. Statement A is false. B. Even if Delta_r G^circ$\Delta_r G^{\circ}$ 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 0.84text kJ mol^-1$0.84\text{ kJ mol}^{-1}$. Thus Delta G = -0.84text 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 NO_2$NO_2$) 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) = -6.24text kJ mol^-1$- (5.40 + 0.84) = -6.24\text{ kJ mol}^{-1}$. 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics Class 11 Chemistry: Equilibrium

### Related Formula Delta_r H^circ = sum Delta_f H^circ(textproducts) - sum Delta_f H^circ(textreactants)$$\Delta_r H^{\circ} = \sum \Delta_f H^{\circ}(\text{products}) - \sum \Delta_f H^{\circ}(\text{reactants})$$ Delta_r S^circ = sum S^circ(textproducts) - sum S^circ(textreactants)$$\Delta_r S^{\circ} = \sum S^{\circ}(\text{products}) - \sum S^{\circ}(\text{reactants})$$ Delta_r G^circ = Delta_r H^circ - TDelta_r S^circ$$\Delta_r G^{\circ} = \Delta_r H^{\circ} - T\Delta_r S^{\circ}$$ Delta_r G^circ = -2.303 RT log K$$\Delta_r G^{\circ} = -2.303 RT \log K$$ ### Core Logic For the reaction A_2(g) + B_2(g) rightleftharpoons 2AB(g)$A_2(g) + B_2(g) \rightleftharpoons 2AB(g)$: 1. Enthalpy change (Delta_r H^circ$\Delta_r H^{\circ}$): Delta_r H^circ = [2 times Delta_f H^circ(AB)] - [Delta_f H^circ(A_2) + Delta_f H^circ(B_2)]$$\Delta_r H^{\circ} = [2 \times \Delta_f H^{\circ}(AB)] - [\Delta_f H^{\circ}(A_2) + \Delta_f H^{\circ}(B_2)]$$ Delta_r H^circ = (2 times 32) - (6 + x) = (64 - 6 - x) = (58 - x)text kJ mol^-1$$\Delta_r H^{\circ} = (2 \times 32) - (6 + x) = (64 - 6 - x) = (58 - x)\text{ kJ mol}^{-1}$$ 2. Entropy change (Delta_r S^circ$\Delta_r S^{\circ}$): Delta_r S^circ = [2 times S^circ(AB)] - [S^circ(A_2) + S^circ(B_2)]$$\Delta_r S^{\circ} = [2 \times S^{\circ}(AB)] - [S^{\circ}(A_2) + S^{\circ}(B_2)]$$ Delta_r S^circ = (2 times 222) - (146 + 280) = 444 - 426 = 18text J K^-1text mol^-1$$\Delta_r S^{\circ} = (2 \times 222) - (146 + 280) = 444 - 426 = 18\text{ J K}^{-1}\text{ mol}^{-1}$$ 3. Standard Gibbs Free Energy (Delta_r G^circ$\Delta_r G^{\circ}$) via Equilibrium Constant: Delta_r G^circ = -2.303 times R times T log K$$\Delta_r G^{\circ} = -2.303 \times R \times T \log K$$ Delta_r G^circ = -2.303 times 8.314 times 500 times 2.2 = -21063.8text J mol^-1 = -21.06text kJ mol^-1$$\Delta_r G^{\circ} = -2.303 \times 8.314 \times 500 \times 2.2 = -21063.8\text{ J mol}^{-1} = -21.06\text{ kJ mol}^{-1}$$ 4. Using Gibbs Equation: Delta_r G^circ = Delta_r H^circ - TDelta_r S^circ$$\Delta_r G^{\circ} = \Delta_r H^{\circ} - T\Delta_r S^{\circ}$$ -21.06 = (58 - x) - 500 times left(frac181000right)$$-21.06 = (58 - x) - 500 \times \left(\frac{18}{1000}\right)$$ -21.06 = 58 - x - 9$$-21.06 = 58 - x - 9$$ -21.06 = 49 - x$-21.06 = 49 - x$ x = 49 + 21.06 = 70.06text kJ mol^-1$$x = 49 + 21.06 = 70.06\text{ kJ mol}^{-1}$$ Rounding off to nearest integer, x = 70$x = 70$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics Class 11 Chemistry: Equilibrium

### Related Formula w_textrev = -nRT lnleft(fracV_mathrmfV_mathrmiright)$$w_{\text{rev}} = -nRT \ln\left(\frac{V_{\mathrm{f}}}{V_{\mathrm{i}}}\right)$$ w_textirrev = -P_textext left(V_mathrmf - V_mathrmiright)$$w_{\text{irrev}} = -P_{\text{ext}} \left(V_{\mathrm{f}} - V_{\mathrm{i}}\right)$$ ### Core Logic For isothermal reversible and irreversible steps: 1. **Reversible Path**: Since a reversible compression path retraces the exact coordinates of the reversible expansion path, the magnitudes of work are equal: |w_textrev, expansion| = |w_textrev, compression| implies a = c$$|w_{\text{rev, expansion}}| = |w_{\text{rev, compression}}| \implies a = c$$

2. **Isothermal Expansion**: Reversible work magnitude is the maximum possible work. Hence, for expansion: |w_textrev, expansion| > |w_textirrev, expansion| implies a > b$$|w_{\text{rev, expansion}}| > |w_{\text{irrev, expansion}}| \implies a > b$$

3. **Isothermal Compression**: Irreversible compression requires more work than reversible compression because of sudden pressure adjustments against the surroundings: |w_textirrev, compression| > |w_textrev, compression| implies d > c$$|w_{\text{irrev, compression}}| > |w_{\text{rev, compression}}| \implies d > c$$

### Step 1: Combine the inequalities Combining the results: - We have a = c$a = c$ - We have d > c$d > c$ - We have a > b$a > b$ This leads to the strict inequality sequence: d > c = a > b$d > c = a > b$ ### Pattern Recognition Thermodynamics Principle: Reversible expansion is the most efficient (gives maximum work magnitude), whereas reversible compression is the most efficient (requires minimum work magnitude). Single-stage irreversible compression is always the least efficient, demanding the absolute highest work input. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics

### Related Formula First Law of Thermodynamics expression: mathrmdU = dq + dw$$\mathrm{dU = dq + dw}$$ ### Core Logic The system parameters show an isothermal transformation layout with zero overall heat transfer step: * No change in temperature signifies mathrmdT = 0$\mathrm{dT} = 0$, hence internal energy change for an ideal gas satisfies: mathrmdU = nC_vmathrmdT = 0$$\mathrm{dU} = nC_v\mathrm{dT} = 0$$ * Since it expands freely into an empty chamber (vessel B), external pressure P_textext = 0$P_{\text{ext}} = 0$, meaning work done is: mathrmdw = -P_textextmathrmdV = 0$$\mathrm{dw} = -P_{\text{ext}}\mathrm{dV} = 0$$ * Combining these parameters in the First Law gives mathrmdq = 0$\mathrm{dq} = 0$. * This classic situation of "free expansion" implies vessel B was completely evacuated initially. ### Step 1: Statement Verification Therefore, the pressure inside vessel B before opening the stopcock was precisely zero. ### Pattern Recognition Isothermal + expansion against no opposing force = Free Expansion. For free expansion of an ideal gas, always remember: w = 0$w = 0$, q = 0$q = 0$, and Delta U = 0$\Delta U = 0$ simultaneously. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics