### Related Formula Delta Q_textnet = Delta Q_AB + Delta Q_BC$\Delta Q_{\text{net}} = \Delta Q_{AB} + \Delta Q_{BC}$ Delta Q_textisothermal = nRT lnleft(fracV_fV_i ight) = P_i V_i lnleft(fracV_fV_i ight)$\Delta Q_{\text{isothermal}} = nRT \ln\left(\frac{V_f}{V_i} ight) = P_i V_i \ln\left(\frac{V_f}{V_i} ight)$ ### Core Logic For path A rightarrow B$A \rightarrow B$: Since it is given as an adiabatic path: Delta Q_AB = 0$\Delta Q_{AB} = 0$ For path B rightarrow C$B \rightarrow C$: Since it is given as an isothermal path, the change in internal energy Delta U_BC = 0$\Delta U_{BC} = 0$. From the first law of thermodynamics, heat absorbed equals work done: Delta Q_BC = W_BC = nRT_B lnleft(fracV_CV_B ight)$\Delta Q_{BC} = W_{BC} = nRT_B \ln\left(\frac{V_C}{V_B} ight)$ Using the ideal gas state at point B$B$, nRT_B = P_B V_B$nRT_B = P_B V_B$: P_B V_B = (5 times 10^4 mathrm~Pa) times (6 times 10^-6 mathrm~m^3) = 0.3 mathrm~J$P_B V_B = (5 \times 10^4 \mathrm{~Pa}) \times (6 \times 10^{-6} \mathrm{~m}^3) = 0.3 \mathrm{~J}$ Wait, let's express it in terms of the gas constant R$R$ for a single mole (n=1$n=1$) using temperature data directly provided in the original figure labels (T_B = 450mathrm~K$T_B = 450\mathrm{~K}$): Delta Q_BC = (1) cdot R cdot (450) cdot lnleft(frac8 times 10^-66 times 10^-6right)$\Delta Q_{BC} = (1) \cdot R \cdot (450) \cdot \ln\left(\frac{8 \times 10^{-6}}{6 \times 10^{-6}}\right)$ Delta Q_BC = 450 R lnleft(frac43 ight) = 450 R (ln 4 - ln 3)$\Delta Q_{BC} = 450 R \ln\left(\frac{4}{3} ight) = 450 R (\ln 4 - \ln 3)$ Thus, the total net heat absorbed per unit mole is: Delta Q = 0 + 450 R (ln 4 - ln 3) = 450 R (ln 4 - ln 3)$\Delta Q = 0 + 450 R (\ln 4 - \ln 3) = 450 R (\ln 4 - \ln 3)$ ### Pattern Recognition Adiabatic paths have zero heat exchange by baseline definition. The calculation boils down directly to the work done during the isothermal stage B rightarrow C$B \rightarrow C$ matching RT ln(V_f/V_i)$RT \ln(V_f/V_i)$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics

### Core Logic Let's treat the given parameters as terms for exothermic heats evolved on the product side: 1) C(textdiamond) ightarrow C(textgraphite)$C(\text{diamond}) ightarrow C(\text{graphite})$, Delta H = -X$Delta H = -X$ 2) C(textdiamond) + O_2(g) ightarrow CO_2(g)$C(\text{diamond}) + O_{2}(g) ightarrow CO_{2}(g)$, Delta H = -Y$Delta H = -Y$ 3) C(textgraphite) + O_2(g) ightarrow CO_2(g)$C(\text{graphite}) + O_{2}(g) ightarrow CO_{2}(g)$, Delta H = -Z$Delta H = -Z$ By subtracting Equation (3) from Equation (2): C(textdiamond) - C(textgraphite) ightarrow 0 implies C(textdiamond) ightarrow C(textgraphite)$C(\text{diamond}) - C(\text{graphite}) ightarrow 0 implies C(\text{diamond}) ightarrow C(\text{graphite})$ Delta H = (-Y) - (-Z) = Z - Y$Delta H = (-Y) - (-Z) = Z - Y$ Matching this to Equation (1): -X = Z - Y implies X = Y - Z$-X = Z - Y implies X = Y - Z$ ### Pattern Recognition Hess's Law states that the enthalpy change of an overall reaction is equal to the sum of the enthalpy changes of its individual steps. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics

### Related Formula eta = 1 - fracmathrmT_LmathrmT_H$\eta = 1 - \frac{\mathrm{T}_L}{\mathrm{T}_H}$ ### Core Logic Let's compute the efficiency parameters explicitly: eta_12 = 1 - frac273473 = frac200473 approx 0.423$\eta_{12} = 1 - \frac{273}{473} = \frac{200}{473} \approx 0.423$ eta_1 = 1 - frac373473 = frac100473 approx 0.211$\eta_1 = 1 - \frac{373}{473} = \frac{100}{473} \approx 0.211$ eta_2 = 1 - frac273373 = frac100373 approx 0.268$\eta_2 = 1 - \frac{273}{373} = \frac{100}{373} \approx 0.268$ Evaluating the linear sum of fractional bounds: eta_1 + eta_2 = 0.211 + 0.268 = 0.479$\eta_1 + \eta_2 = 0.211 + 0.268 = 0.479$ Comparing the outputs clearly demonstrates: eta_12 < eta_1 + eta_2$\eta_{12} < \eta_1 + \eta_2$ ### Step 1: Final Conclusion Thus, the inequality satisfies option (1). ### Pattern Recognition The joint efficiency of cascading perfect thermodynamic steps is bounded multiplicatively as (1-eta_12) = (1-eta_1)(1-eta_2)$(1-\eta_{12}) = (1-\eta_1)(1-\eta_2)$, which algebraically forces eta_12 = eta_1 + eta_2 - eta_1eta_2 < eta_1 + eta_2$\eta_{12} = \eta_1 + \eta_2 - \eta_1\eta_2 < \eta_1 + \eta_2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics

### Related Formula Delta U = Delta Q + Delta W_texton system$\Delta U = \Delta Q + \Delta W_{\text{on system}}$ where, Delta U$\Delta U$ = change in internal energy, Delta Q$\Delta Q$ = heat exchange, Delta W_texton system$\Delta W_{\text{on system}}$ = work done on the system. ### Core Logic During the melting of ice at 273mathrm~K$273\mathrm{~K}$, the density of water is higher than the density of ice. This means the volume of the ice-water system decreases during melting: V_f < V_i implies Delta V < 0$V_f < V_i \implies \Delta V < 0$ Since the system contracts, the atmosphere performs positive work on it: W_texton system = -P Delta V > 0$W_{\text{on system}} = -P \Delta V > 0$ Additionally, heat is absorbed by the system to melt the ice, so Delta Q > 0$\Delta Q > 0$. By the first law of thermodynamics, since both Delta Q$\Delta Q$ and Delta W_texton system$\Delta W_{\text{on system}}$ are positive, the internal energy of the system increases: Delta U = Delta Q + W_texton system > 0$\Delta U = \Delta Q + W_{\text{on system}} > 0$ ### Step 1: Evaluation of Options Let's check the given options: 1. Internal energy remains unchanged rightarrow$\rightarrow$ False (it increases). 2. Positive work is done by the system rightarrow$\rightarrow$ False (work done by the system is negative since it contracts). 3. Internal energy decreases rightarrow$\rightarrow$ False. 4. Positive work is done on the ice-water system by the atmosphere rightarrow$\rightarrow$ True (since volume decreases under atmospheric pressure). ### Pattern Recognition Remember: Ice contracts upon melting (unlike most solids). Shrinking volume (V downarrow$V \downarrow$) under positive pressure (P > 0$P > 0$) means the surroundings (atmosphere) compress it, performing positive work on the system. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermal Properties of Matter Class 11 Physics: Thermodynamics

### Related Formula First Law of Thermodynamics: Delta Q = Delta U + W_textby gas$\Delta Q = \Delta U + W_{\text{by gas}}$ Work done by an ideal gas during isothermal expansion: W_textgas = nRTlnleft(fracV_1V_0right) = nRTlnleft(fracL_1L_0 ight)$W_{\text{gas}} = nRT\ln\left(\frac{V_1}{V_0}\right) = nRT\ln\left(\frac{L_1}{L_0} ight)$ Conservation of Energy (Work-Energy Theorem): Total energy delivered by the heating filament (W_textfilament$W_{\text{filament}}$) must equal the total work needed to lift the piston against gravity and compress the non-linear spring. ### Core Logic Since the process is isothermal, the change in internal energy of the ideal gas is zero (Delta U = 0$\Delta U = 0$). Hence: Q = W_textgas$Q = W_{\text{gas}}$ By the Work-Energy Theorem for the piston: W_textgas + W_textfilament = Delta U_textgravity + Delta U_textspring$W_{\text{gas}} + W_{\text{filament}} = \Delta U_{\text{gravity}} + \Delta U_{\text{spring}}$ Let's evaluate each term: - Increase in gravitational potential energy: Delta U_textgravity = Mg(L_1 - L_0)$\Delta U_{\text{gravity}} = Mg(L_1 - L_0)$ - Increase in spring potential energy: U_textspring = -int_L_0^L_1 F_textrestoring dx = int_L_0^L_1 kx^3 dx = frack4(L_1^4 - L_0^4)$U_{\text{spring}} = -\int_{L_0}^{L_1} F_{\text{restoring}} dx = \int_{L_0}^{L_1} kx^3 dx = \frac{k}{4}(L_1^4 - L_0^4)$ ### Step 1: Finding Total Energy Delivered Isolating W_textfilament$W_{\text{filament}}$ (the net external energy delivered to the gas system): W_textfilament = W_textgas + Mg(L_1 - L_0) + frack4(L_1^4 - L_0^4)$W_{\text{filament}} = W_{\text{gas}} + Mg(L_1 - L_0) + \frac{k}{4}(L_1^4 - L_0^4)$ Since W_textgas = nRTlnleft(fracL_1L_0 ight)$W_{\text{gas}} = nRT\ln\left(\frac{L_1}{L_0} ight)$: W_textfilament = nRTlnleft(fracL_1L_0 ight) + Mg(L_1 - L_0) + frack4(L_1^4 - L_0^4)$W_{\text{filament}} = nRT\ln\left(\frac{L_1}{L_0} ight) + Mg(L_1 - L_0) + \frac{k}{4}(L_1^4 - L_0^4)$ ### Pattern Recognition Notice how energy conservation instantly frames this complex thermodynamics question. The heating filament's energy simply goes into three distinct stores: the isothermal work of gas expansion, raising the mass against gravity (Mgh$Mgh$), and the potential energy of the spring (integrated from kx^3$kx^3$). Keeping this total energy ledger in mind prevents tedious mathematical tangents. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics: First Law Class 11 Physics: Work, Energy and Power: Variable Force Integration