### Related Formula Q = n_textreacted cdot Delta H_textneutralization$Q = n_{\text{reacted}} \cdot \Delta H_{\text{neutralization}}$ Delta T = fracQm cdot c$\Delta T = \frac{Q}{m \cdot c}$ ### 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 mathrmH^+$\mathrm{H}^+$ and mathrmOH^-$\mathrm{OH}^-$ that react in each mixture: 1. **Option 1:** 30text mL of 1mathrmM mathrmHCl + 30text mL of 1mathrmM mathrmNaOH$30\text{ mL of } 1\mathrm{M } \mathrm{HCl} + 30\text{ mL of } 1\mathrm{M } \mathrm{NaOH}$ textReactive millimoles = 30text mmol$\text{Reactive millimoles} = 30\text{ mmol}$. Both are strong electrolytes, releasing full neutralization energy (sim -57.3text kJ/mol$\sim -57.3\text{ kJ/mol}$). Total volume = 60text mL$60\text{ mL}$. 2. **Option 2:** 30text mL of 1mathrmM mathrmCH_3COOH + 30text mL of 1mathrmM mathrmNaOH$30\text{ mL of } 1\mathrm{M } \mathrm{CH_3COOH} + 30\text{ mL of } 1\mathrm{M } \mathrm{NaOH}$ textReactive millimoles = 30text 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. 3. **Option 3:** 50text mL of 1mathrmM mathrmHCl + 20text mL of 1mathrmM mathrmNaOH$50\text{ mL of } 1\mathrm{M } \mathrm{HCl} + 20\text{ mL of } 1\mathrm{M } \mathrm{NaOH}$ textLimiting reagent = mathrmNaOH = 20text mmol$\text{Limiting reagent} = \mathrm{NaOH} = 20\text{ mmol}$. Only 20text mmol$20\text{ mmol}$ reacts. Total volume = 70text mL$70\text{ mL}$. 4. **Option 4:** 45text mL of 1mathrmM mathrmCH_3COOH + 25text mL of 1mathrmM mathrmNaOH$45\text{ mL of } 1\mathrm{M } \mathrm{CH_3COOH} + 25\text{ mL of } 1\mathrm{M } \mathrm{NaOH}$ textLimiting reagent = 25text mmol$\text{Limiting reagent} = 25\text{ mmol}$ weak neutralization profile. Comparing Option 1 and Option 3, Option 1 releases significantly more heat (30text mmol$30\text{ mmol}$ vs 20text mmol$20\text{ mmol}$) into a smaller volume (60text mL$60\text{ mL}$ vs 70text mL$70\text{ mL}$), yielding the largest increase in temperature Delta T$\Delta T$. ### Pattern Recognition To maximize Delta 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics Class 11 Chemistry: Equilibrium

### Related Formula Using Hess's Law, the enthalpy change of a net reaction can be determined by linearly combining the steps: Delta Htextnet = sum Delta Htextproducts - sum Delta Htextreactants$\Delta H{\text{net}} = \sum \Delta H{\text{products}} - \sum \Delta H{\text{reactants}}$ ### Core Logic The heat of formation of mathrmSO_2(g)$\mathrm{SO}_2(g)$ corresponds to the target thermochemical equation: textTarget: mathrmS(g) + mathrmO2(g) ightarrow mathrmSO2(g) quad Delta H_f = ?$\text{Target: } \mathrm{S}(g) + \mathrm{O}{2}(g) ightarrow \mathrm{SO}{2}(g) \quad \Delta H_f = ?$ Let's write out the given equations along with their enthalpy changes (remembering that exothermic reactions release heat, so Delta H = -Q$\Delta H = -Q$): 1. mathrmS(g) + frac32mathrmO_2(g) ightarrow mathrmSO_3(g) quad Delta H_1 = -2xtext kcal$\mathrm{S}(g) + \frac{3}{2}\mathrm{O}_{2}(g) ightarrow \mathrm{SO}_{3}(g) \quad \Delta H_{1} = -2x\text{ kcal}$ 2. mathrmSO_2(g) + frac12mathrmO_2(g) ightarrow mathrmSO_3(g) quad Delta H_2 = -ytext kcal$\mathrm{SO}_{2}(g) + \frac{1}{2}\mathrm{O}_{2}(g) ightarrow \mathrm{SO}_{3}(g) \quad \Delta H_{2} = -y\text{ kcal}$ To isolate mathrmSO_2(g)$\mathrm{SO}_{2}(g)$ on the product side, subtract Equation (2) from Equation (1): left[mathrmS(g) + frac32mathrmO2(g) ight] - left[mathrmSO2(g) + frac12mathrmO2(g) ight] ightarrow mathrmSO3(g) - mathrmSO3(g)$\left[\mathrm{S}(g) + \frac{3}{2}\mathrm{O}{2}(g) ight] - \left[\mathrm{SO}{2}(g) + \frac{1}{2}\mathrm{O}{2}(g) ight] ightarrow \mathrm{SO}{3}(g) - \mathrm{SO}{3}(g)$ mathrmS(g) + mathrmO2(g) ightarrow mathrmSO2(g)$\mathrm{S}(g) + \mathrm{O}{2}(g) ightarrow \mathrm{SO}{2}(g)$ Now apply the same operation to the enthalpy values: Delta H_f = Delta H1 - Delta H_2 = -2x - (-y) = y - 2xtext kcal$\Delta H_f = \Delta H{1} - \Delta H_{2} = -2x - (-y) = y - 2x\text{ kcal}$ 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics

### Related Formula From the Ideal Gas Law: PV = nRT$PV = nRT$ According to the First Law of Thermodynamics: Delta Q = Delta U + W$\Delta Q = \Delta U + W$ ### Core Logic The question states that pressure increases linearly with temperature, which means their ratio is constant : P = kT implies fracPT = textconstant$P = kT \implies \frac{P}{T} = \text{constant}$ Since fracPT = fracnRV$\frac{P}{T} = \frac{nR}{V}$, the volume V$V$ must remain constant throughout the process. This identifies it as an isochoric process (Statement E is true). ### Step 1: Evaluate All Statements * Statement A: True. In an isochoric process, dV = 0 implies W = int P dV = 0$dV = 0 \implies W = \int P dV = 0$. * Statement B: False. Since work is zero, the First Law simplifies to Delta Q = Delta U$\Delta Q = \Delta U$, meaning heat added equals the change in internal energy. * Statement C: False. Volume is constant, so it does not increase. * Statement D: True. As pressure increases linearly with temperature, temperature increases, which causes the internal energy of the gas to increase. ### Step 2: Final Selection Gathering the true statements (A, D, and E) points directly to Option (2). ### Pattern Recognition A linear P-T$P-T$ line passing through the origin always indicates a constant volume graph. For constant volume graphs, work done is zero, which simplifies the first law calculation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics

### Related Formula For an ideal gas thermodynamic process: * Total heat added at constant pressure (isobaric process) is : E_1 = n C_P Delta T$E_{1} = n C_{P} \Delta T$ * Total change in internal energy is given by : E_2 = n C_V Delta T$E_{2} = n C_{V} \Delta T$ The ratio of specific heat capacities is defined as: gamma = fracC_PC_V $\gamma = \frac{C_{P}}{C_{V}} $ ### Core Logic Taking the ratio of the two energy expressions[cite: 179, 823]: fracE_1E_2 = fracn C_P Delta Tn C_V Delta T = fracC_PC_V = gamma$\frac{E_{1}}{E_{2}} = \frac{n C_{P} \Delta T}{n C_{V} \Delta T} = \frac{C_{P}}{C_{V}} = \gamma$ ### Step 1: Evaluating for a Monoatomic Gas For an ideal monoatomic gas, the degrees of freedom are f = 3$f = 3$. This gives an adiabatic index of : gamma = 1 + frac2f = 1 + frac23 = frac53 $\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{3} = \frac{5}{3} $ Equating this value to the given ratio expression [cite: 179, 826]: frac53 = fracx9$\frac{5}{3} = \frac{x}{9}$ x = frac5 times 93 = 15$x = \frac{5 \times 9}{3} = 15$ ### Pattern Recognition The ratio of heat added to the change in internal energy during an isobaric process is always equal to the adiabatic exponent gamma$\gamma$ of the gas. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics

### Related Formula For any state function like Internal Energy (U$U$), the cyclic integral over a complete closed loop is identically zero: oint dU = 0 implies Delta U_textcyclic = 0$\oint dU = 0 \implies \Delta U_{\text{cyclic}} = 0$ ### 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$A$: Delta U_textCase-I = 0$\Delta U_{\text{Case-I}} = 0$ Delta U_textCase-II = 0$\Delta U_{\text{Case-II}} = 0$ Delta U_textCase-III = 0$\Delta U_{\text{Case-III}} = 0$ Therefore, Delta Utext (Case-I) = Delta Utext (Case-II) = Delta Utext (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 (Delta U$\Delta U$, Delta H$\Delta H$, Delta S$\Delta S$, Delta G$\Delta G$) over a cyclic loop. The answer is instantly zero for all cases! ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics