### Related Formula P_i V_i^gamma = P_f V_f^gamma$P_i V_i^{\gamma} = P_f V_f^{\gamma}$ where, P_i, P_f$P_i, P_f$ = initial and final pressures V_i, V_f$V_i, V_f$ = initial and final volumes gamma$\gamma$ = adiabatic exponent ### Core Logic Since the gas is stored in a "thermally insulated container" and is compressed "suddenly", the process is **adiabatic**. From the adiabatic relation: fracP_fP_i = left(fracV_iV_fright)^gamma$\frac{P_f}{P_i} = \left(\frac{V_i}{V_f}\right)^{\gamma}$ Given: - V_f = frac18 V_i implies fracV_iV_f = 8$V_f = \frac{1}{8} V_i \implies \frac{V_i}{V_f} = 8$ - gamma = frac53$\gamma = \frac{5}{3}$ ### Step 1: Computation Substitute the values to find the pressure ratio: fracP_fP_i = (8)^5/3 = left(2^3right)^5/3$\frac{P_f}{P_i} = (8)^{5/3} = \left(2^3\right)^{5/3}$ fracP_fP_i = 2^5 = 32$\frac{P_f}{P_i} = 2^5 = 32$ ### Pattern Recognition Sees: "suddenly compressed" or "thermally insulated container" → Adiabatic process. Shortcut: P V^gamma = textconstant$P V^\gamma = \text{constant}$. Since the volume goes down by 8$8$ times, the pressure increases by 8^gamma = 8^5/3 = 32$8^{\gamma} = 8^{5/3} = 32$ times. ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics

### Related Formula left(fracdPdVright)_textisothermal = -fracPV$\left(\frac{dP}{dV}\right)_{\text{isothermal}} = -\frac{P}{V}$ left(fracdPdVright)_textadiabatic = -gamma fracPV$\left(\frac{dP}{dV}\right)_{\text{adiabatic}} = -\gamma \frac{P}{V}$ ### Core Logic The slope of an adiabatic process on a P-V$P-V$ diagram is gamma$\gamma$ times steeper than that of an isothermal process: left|left(fracdPdVright)_textadiabaticright| > left|left(fracdPdVright)_textisothermalright|$\left|\left(\frac{dP}{dV}\right)_{\text{adiabatic}}\right| > \left|\left(\frac{dP}{dV}\right)_{\text{isothermal}}\right|$

When pressure increases (compression), the volume drops. Because the adiabatic curve is steeper, the pressure rises more rapidly for a given drop in volume, or conversely, for a specified increase in pressure, the volume falls off more rapidly in the isothermal process than the adiabatic process. Hence, Assertion (A) is true. Reason (R) states the governing equations PV = C$PV = C$ and PV^gamma = C$PV^{\gamma} = C$, which directly lead to these slope expressions via differentiation. Thus, Reason (R) is true and correctly explains Assertion (A). ### Pattern Recognition Adiabatic curves are steeper than isothermal curves on a P-V$P-V$ diagram because gamma > 1$\gamma > 1$. For any expansion or compression process, remember that slope magnitude satisfies textSlope_textadi = gamma cdot textSlope_textiso$\text{Slope}_{\text{adi}} = \gamma \cdot \text{Slope}_{\text{iso}}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics

### 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

### 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