### 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

### Related Formula Delta H_texttotal = n C_p(ell) Delta T_1 - nDelta H_textfusion + n C_p(s) Delta T_2$\Delta H_{\text{total}} = n C_{p(\ell)} \Delta T_1 - n\Delta H_{\text{fusion}} + n C_{p(s)} \Delta T_2$ ### Core Logic We need to compute the enthalpy change for the pathway: mathrmH_2O(ell, 10^circC) rightarrow mathrmH_2O(s, -10^circC)$\mathrm{H_2O(\ell, 10^{\circ}C)} \rightarrow \mathrm{H_2O(s, -10^{\circ}C)}$ This can be divided into three consecutive steps: 1. Cool liquid water from 10^circC$10^{\circ}C$ to 0^circC$0^{\circ}C$: Delta H_1 = n cdot C_p[mathrmH_2O(ell)] cdot (0 - 10) = 1 cdot y cdot (-10) = -10y text J$\Delta H_1 = n \cdot C_p[\mathrm{H_2O(\ell)}] \cdot (0 - 10) = 1 \cdot y \cdot (-10) = -10y \text{ J}$ 2. Freeze water to ice at 0^circC$0^{\circ}C$: Delta H_2 = -n cdot Delta_textfusH = -1 cdot x text kJ = -1000x text J$\Delta H_2 = -n \cdot \Delta_{\text{fus}}H = -1 \cdot x \text{ kJ} = -1000x \text{ J}$ 3. Cool ice from 0^circC$0^{\circ}C$ to -10^circC$-10^{\circ}C$: Delta H_3 = n cdot C_p[mathrmH_2O(s)] cdot (-10 - 0) = 1 cdot z cdot (-10) = -10z text J$\Delta H_3 = n \cdot C_p[\mathrm{H_2O(s)}] \cdot (-10 - 0) = 1 \cdot z \cdot (-10) = -10z \text{ J}$

Adding these three steps yields: Delta H_texttotal = -10y - 1000x - 10z = -10(100x + y + z) text Joule$\Delta H_{\text{total}} = -10y - 1000x - 10z = -10(100x + y + z) \text{ Joule}$ ### Pattern Recognition Freezing is an exothermic process, so all three steps (cooling water, freezing, and cooling ice) must carry a negative sign. Factoring out -10$-10$ cleanly yields the expression -10(100x+y+z)$-10(100x+y+z)$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics

### Related Formula Resonance energy equation related to experimental and theoretical formation enthalpies: Delta H_textR.E. = Delta H_f(textexp) - Delta H_f(textTheo)$\Delta H_{\text{R.E.}} = \Delta H_{f(\text{exp})} - \Delta H_{f(\text{Theo})}$ Theoretical enthalpy calculation using bond energies: Delta H_f(textTheo) = sum textB.E._textreactants - sum textB.E._textproducts$\Delta H_{f(\text{Theo})} = \sum \text{B.E.}_{\text{reactants}} - \sum \text{B.E.}_{\text{products}}$ ### Execution Step 1: Write the chemical equation to calculate the theoretical enthalpy of formation based on the localized structure X=X=Y$X=X=Y$: X equiv X(g) + frac12Y=Y(g) longrightarrow X=X=Y(g)$X \equiv X(g) + \frac{1}{2}Y=Y(g) \longrightarrow X=X=Y(g)$ Step 2: Substitute the localized bond energies into the reactant-minus-product relation: Delta H_f(textTheo) = left[ textB.E._X equiv X + frac12textB.E._Y=Y right] - left[ textB.E._X=X + textB.E._X=Y right]$\Delta H_{f(\text{Theo})} = \left[ \text{B.E.}_{X \equiv X} + \frac{1}{2}\text{B.E.}_{Y=Y} \right] - \left[ \text{B.E.}_{X=X} + \text{B.E.}_{X=Y} \right]$ Delta H_f(textTheo) = left[ 940 + frac12(500) right] - [410 + 602]$\Delta H_{f(\text{Theo})} = \left[ 940 + \frac{1}{2}(500) \right] - [410 + 602]$ Delta H_f(textTheo) = [940 + 250] - 1012 = 1190 - 1012 = 178 text kJ mol^-1$\Delta H_{f(\text{Theo})} = [940 + 250] - 1012 = 1190 - 1012 = 178 \text{ kJ mol}^{-1}$ Step 3: Calculate the resonance energy: Delta H_textR.E. = Delta H_f(textexp) - Delta H_f(textTheo) = 80 - 178 = -98 text kJ mol^-1$\Delta H_{\text{R.E.}} = \Delta H_{f(\text{exp})} - \Delta H_{f(\text{Theo})} = 80 - 178 = -98 \text{ kJ mol}^{-1}$ Taking the magnitude as requested: |Delta H_textR.E.| = 98$|\Delta H_{\text{R.E.}}| = 98$. ### Pattern Recognition Resonance energy is always a stabilizing factor, meaning Delta H_f(textexp)$\Delta H_{f(\text{exp})}$ is more exothermic (or less endothermic) than the theoretical localized state. The magnitude is simply the absolute value of this difference (|80 - 178| = 98$|80 - 178| = 98$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics Class 11 Chemistry: Chemical Bonding and Molecular Structure

### Core Logic 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: V_m(textice) > V_m(textwater)$V_m(\text{ice}) > V_m(\text{water})$ 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 273.15mathrmK$273.15\mathrm{K}$, the melting point decreases, causing the entire solid phase (ice) to disappear completely. ### Pattern Recognition 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics