### 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 During phase transition, latent heat of fusion (L_f$L_f$) is absorbed: Q = m L_f$Q = m L_f$ Temperature remains constant because the average kinetic energy of the molecules does not change; instead, potential energy changes during phase transition: textTemperature T propto textAverage Kinetic Energy of molecules$\text{Temperature } T \propto \text{Average Kinetic Energy of molecules}$ ### Core Logic Statement I Analysis: - During a phase transition (such as ice melting at 0^circmathrmC$0^{\circ}\mathrm{C}$), any added heat is utilized as latent heat of fusion. The system remains at a constant temperature of 0^circmathrmC$0^{\circ}\mathrm{C}$ as long as both solid and liquid phases coexist in equilibrium. Thus, Statement I is True. ### Step 1: Analyze Statement II - Statement II explains *why* this happens: The thermal energy is spent entirely to break down the highly ordered crystalline hydrogen-bonded lattice of ice into liquid water. It does not increase the translational kinetic energy of the molecules. Since kinetic energy is constant, temperature remains constant. Thus, Statement II is True. ### Step 2: Conclusion Therefore, both Statement I and Statement II are True, matching Option (3). ### Pattern Recognition For any phase transition (melting, boiling, sublimation): Temperature stays flat. The added energy goes entirely into latent heat (potential energy change to overcome intermolecular forces), meaning average kinetic energy is constant. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics Class 11 Physics: Thermal Properties of Matter

### Related Formula Thermodynamic work done (w$w$) for each step: - Isochoric step (V = textconstant$V = \text{constant}$): w = 0$w = 0$ - Isobaric step (P = textconstant$P = \text{constant}$): w = -P Delta V$w = -P \Delta V$ ### Core Logic The process from point 1 to point 4 consists of three distinct segments: 1. Step 1 rightarrow 2$1 \rightarrow 2$: Isochoric cooling at constant volume V_1 = 1000mathrm~cm^3$V_1 = 1000\mathrm{~cm}^3$. Work done w_1rightarrow 2 = 0$w_{1\rightarrow 2} = 0$. 2. Step 2 rightarrow 3$2 \rightarrow 3$: Isobaric compression at constant pressure P = 3.00mathrm~atm$P = 3.00\mathrm{~atm}$ from volume 2000mathrm~cm^3$2000\mathrm{~cm}^3$ to 1000mathrm~cm^3$1000\mathrm{~cm}^3$. 3. Step 3 rightarrow 4$3 \rightarrow 4$: Isochoric step at constant volume. Work done w_3rightarrow 4 = 0$w_{3\rightarrow 4} = 0$. ### Step 1: Calculate work done in the isobaric step (2 rightarrow 3$2 \rightarrow 3$) The volume changes from V_i = 2000mathrm~cm^3 = 2.0mathrm~L$V_i = 2000\mathrm{~cm}^3 = 2.0\mathrm{~L}$ to V_f = 1000mathrm~cm^3 = 1.0mathrm~L$V_f = 1000\mathrm{~cm}^3 = 1.0\mathrm{~L}$: w_2rightarrow 3 = -P Delta V = -3.00mathrm~atm times (1.0mathrm~L - 2.0mathrm~L) = +3.00mathrm~Lcdot atm$w_{2\rightarrow 3} = -P \Delta V = -3.00\mathrm{~atm} \times (1.0\mathrm{~L} - 2.0\mathrm{~L}) = +3.00\mathrm{~L\cdot atm}$ ### Step 2: Convert work to Joules and analyze direction Convert $ ### Step 2: Convert work to Joules and analyze direction Convert $\mathrm{L\cdot atm} to Joules: $ to Joules: $w_2rightarrow 3 = 3.00 times 101.325mathrm~J = 303.975mathrm~J approx 304mathrm~J$w_{2\rightarrow 3} = 3.00 \times 101.325\mathrm{~J} = 303.975\mathrm{~J} \approx 304\mathrm{~J}$ The question asks for the total work done as ($ The question asks for the total work done as ($-) ________ J, meaning work done *by* the system (expansion) is negative and work done *on* the system (compression) is positive. Since this is compression, work done on the gas is $) ________ J, meaning work done *by* the system (expansion) is negative and work done *on* the system (compression) is positive. Since this is compression, work done on the gas is $+304\mathrm{~J}, which is represented as $, which is represented as $-(-304)\mathrm{~J} in typical IUPAC convention where work of expansion is examined. The absolute magnitude of the work is $ in typical IUPAC convention where work of expansion is examined. The absolute magnitude of the work is $304\mathrm{~J}. ### Pattern Recognition During any cyclic or multi-step path on a P-V graph, work is done *only* when there is a change in volume ($. ### Pattern Recognition During any cyclic or multi-step path on a P-V graph, work is done *only* when there is a change in volume ($W = -\int P dV$). Any vertical line (constant volume) represents an isochoric step where work is exactly zero. The horizontal segment directly represents rectangular area under the path. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics Class 11 Physics: Thermodynamics

### Related Formula Heat released at constant volume (q_v$q_v$) in a bomb calorimeter is: q_v = C_textcal cdot Delta T$q_v = C_{\text{cal}} \cdot \Delta T$ Molar heat of combustion at constant volume (Delta U_textcomb$\Delta U_{\text{comb}}$): Delta U_textcomb = fracq_vn_textfuel$\Delta U_{\text{comb}} = \frac{q_v}{n_{\text{fuel}}}$ ### Core Logic Given parameters: - Mass of n-octane m = 1.14mathrm~g$m = 1.14\mathrm{~g}$ - Heat capacity of calorimeter C_textcal = 5mathrm~kJ/K$C_{\text{cal}} = 5\mathrm{~kJ/K}$ - Temperature rise Delta T = 5mathrm~K$\Delta T = 5\mathrm{~K}$ - Formula of octane: mathrmC_8H_18$\mathrm{C_8H_{18}}$ Rightarrow$\Rightarrow$ Molar mass = 8(12) + 18(1) = 114mathrm~g/mol$= 8(12) + 18(1) = 114\mathrm{~g/mol}$ ### Step 1: Calculate heat absorbed by the calorimeter (q_v$q_v$) q_v = 5mathrm~kJ/K times 5mathrm~K = 25mathrm~kJ$q_v = 5\mathrm{~kJ/K} \times 5\mathrm{~K} = 25\mathrm{~kJ}$ ### Step 2: Calculate moles of octane n = frac1.14mathrm~g114mathrm~g/mol = 0.01mathrm~mol$n = \frac{1.14\mathrm{~g}}{114\mathrm{~g/mol}} = 0.01\mathrm{~mol}$ ### Step 3: Calculate molar heat of combustion $ ### Step 3: Calculate molar heat of combustion $Delta U_textcomb = frac25mathrm~kJ0.01mathrm~mol = 2500mathrm~kJ/mol$\Delta U_{\text{comb}} = \frac{25\mathrm{~kJ}}{0.01\mathrm{~mol}} = 2500\mathrm{~kJ/mol}$ The magnitude of the heat of combustion is $ The magnitude of the heat of combustion is $2500\mathrm{~kJ~mol^{-1}}. ### Pattern Recognition A bomb calorimeter operates at rigid constant volume, meaning boundary work $. ### Pattern Recognition A bomb calorimeter operates at rigid constant volume, meaning boundary work $w = 0. Thus, by the first law of thermodynamics, the measured heat flow represents the internal energy change ($. Thus, by the first law of thermodynamics, the measured heat flow represents the internal energy change ($\Delta U), not the enthalpy change ($), not the enthalpy change ($\Delta H$, which occurs at constant pressure). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics