### Related Formula 1. First Law of Thermodynamics: dQ = dU + dW$dQ = dU + dW$ In an adiabatic process: dQ = 0 implies dW = -dU$dQ = 0 \implies dW = -dU$ 2. Change in Internal Energy: dU = n C_v dT = n C_v (T_2 - T_1)$dU = n C_v dT = n C_v (T_2 - T_1)$ ### Core Logic Let's analyze each statement: - **(A) Internal energy is constant:** Incorrect. Since temperature changes during an adiabatic expansion/compression, internal energy (U propto T$U \propto T$) must change. - **(B) Work done is equal to the change in internal energy:** Correct in magnitude (|dW| = |dU|$|dW| = |dU|$). By definition, dW = -dU$dW = -dU$, which correlates the magnitude of work to the change in internal energy. - **(C) Product of temperature and volume is constant:** Incorrect. The adiabatic equation of state is T V^gamma-1 = textconstant$T V^{\gamma-1} = \text{constant}$. - **(D) Product of pressure and volume is constant:** Incorrect. The relation is P V^gamma = textconstant$P V^\gamma = \text{constant}$. - **(E) Work done is proportional to (T_2 - T_1)$(T_2 - T_1)$:** Correct. Since dW = -dU = -n C_v (T_2 - T_1)$dW = -dU = -n C_v (T_2 - T_1)$, work done is directly proportional to the temperature change (T_2 - T_1)$(T_2 - T_1)$. ### Step 1: Determine the correct option Since only statements (B) and (E) are correct, the correct option is (3). ### Pattern Recognition Sees: Characteristics of adiabatic thermodynamic process. Trap: Confusing adiabatic state relations (PV^gamma = C$PV^\gamma = C$, TV^gamma-1 = C$TV^{\gamma-1} = C$) with isothermal state relations (PV = C$PV = C$, T = C$T = C$). Shortcut: First law of thermodynamics under dQ=0$dQ=0$ strictly enforces |dW| = |dU|$|dW| = |dU|$, which validates statement B and E immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics

### Related Formula dQ = n C dT$dQ = n C dT$ dQ = 0 quad text(for adiabatic process)$dQ = 0 \quad \text{(for adiabatic process)}$ ### Core Logic An adiabatic process involves no heat exchange between the system and its surroundings (dQ = 0$dQ = 0$). The molar heat capacity C$C$ is defined as: C = frac1nfracdQdT$C = \frac{1}{n}\frac{dQ}{dT}$ Since dQ = 0$dQ = 0$ while the temperature changes (dT neq 0$dT \neq 0$): C = 0$C = 0$ Thus, the molar heat capacity for any adiabatic process is always zero. ### Step 1: Check other options - Option (1): Isothermal processes have infinite molar heat capacity (dT = 0$dT = 0$). - Option (2): From the First Law (dQ = dU + dW implies dW = -dU$dQ = dU + dW \implies dW = -dU$), the work done equals the *decrease* in internal energy. - Option (4): The internal energy of an ideal gas (dU = n C_mathrmv dT$dU = n C_{\mathrm{v}} dT$) increases directly as temperature increases. ### Step 2: Final Conclusion The statement 'The molar heat capacity is zero' is true. ### Pattern Recognition Adiabatic = no heat flow (dQ=0$dQ=0$). Since molar heat capacity tracks the ratio of heat input to temperature change, C$C$ must be 0. Conversely, isothermal has infinite capacity because heat is absorbed without any temperature change (dT=0$dT=0$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics

### Related Formula For a cyclic thermodynamic process, the net change in internal energy is zero: Delta U_textcyclic = 0$\Delta U_{\text{cyclic}} = 0$ By the First Law of Thermodynamics, the total heat exchanged Q_T$Q_T$ equals the net work done W_textnet$W_{\text{net}}$: Q_T = W_textnet = W_1 + W_2 + W_3$Q_T = W_{\text{net}} = W_1 + W_2 + W_3$ ### Core Logic The cycle consists of three steps: 1. Isothermal expansion from (P_0, V_0)$(P_0, V_0)$ to volume 4V_0$4V_0$. 2. Isobaric compression to the original volume V_0$V_0$. 3. Isochoric heating back to the initial state.

### Step 1: Work in Isothermal Expansion (W_1$W_1$) Initial state: (P_0, V_0)$(P_0, V_0)$. Final state volume: 4V_0$4V_0$. W_1 = P_0 V_0 lnleft(frac4V_0V_0right) = P_0 V_0 ln(4) = 2 P_0 V_0 ln(2)$W_1 = P_0 V_0 \ln\left(\frac{4V_0}{V_0}\right) = P_0 V_0 \ln(4) = 2 P_0 V_0 \ln(2)$ Also, the pressure at the end of this process is: P_1 = fracP_0 V_04V_0 = fracP_04$P_1 = \frac{P_0 V_0}{4V_0} = \frac{P_0}{4}$ ### Step 2: Work in Isobaric Compression (W_2$W_2$) The process occurs at constant pressure P = P_1 = P_0/4$P = P_1 = P_0/4$. The volume goes from 4V_0$4V_0$ back to V_0$V_0$: W_2 = P Delta V = fracP_04 (V_0 - 4V_0) = fracP_04 (-3V_0) = -0.75 P_0 V_0$W_2 = P \Delta V = \frac{P_0}{4} (V_0 - 4V_0) = \frac{P_0}{4} (-3V_0) = -0.75 P_0 V_0$ ### Step 3: Work in Isochoric Heating (W_3$W_3$) Since the volume is held constant at V_0$V_0$, no boundary work is done: W_3 = 0$W_3 = 0$ ### Step 4: Total Heat Exchanged (Q_T$Q_T$) Q_T = W_textnet = W_1 + W_2 + W_3$Q_T = W_{\text{net}} = W_1 + W_2 + W_3$ Q_T = 2 P_0 V_0 ln(2) - 0.75 P_0 V_0 = P_0 V_0 (2ln(2) - 0.75)$Q_T = 2 P_0 V_0 \ln(2) - 0.75 P_0 V_0 = P_0 V_0 (2\ln(2) - 0.75)$ ### Pattern Recognition In any cyclic system returning to its initial state, finding total heat is mathematically equivalent to calculating the enclosed area on a P-V diagram. Here, the isobaric step occurs at the lowest expanded pressure, resulting in a simple negative rectangular area correction subtracted from the logarithmic isothermal expansion curve. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics

### Related Formula The work done W$W$ in a cyclic thermodynamic process is equal to the area enclosed by the loop on a Pressure-Volume (P-V$P-V$) diagram: W = textArea of Closed Loop$W = \text{Area of Closed Loop}$ For an ellipse with semi-major axis a$a$ and semi-minor axis b$b$: textArea = pi a b$\text{Area} = \pi a b$ ### Core Logic Determine the semi-axes of the elliptical cycle on the P-V$P-V$ plane: - On the Pressure axis (x-axis): a = fracP_textmax - P_textmin2 = frac500 - 3002 mathrm~kPa = 100 mathrm~kPa = 10^5 mathrm~Pa$a = \frac{P_{\text{max}} - P_{\text{min}}}{2} = \frac{500 - 300}{2} \mathrm{~kPa} = 100 \mathrm{~kPa} = 10^5 \mathrm{~Pa}$ - On the Volume axis (y-axis): b = fracV_textmax - V_textmin2 = frac350 - 1502 mathrm~cm^3 = 100 mathrm~cm^3 = 100 times 10^-6 mathrm~m^3 = 10^-4 mathrm~m^3$b = \frac{V_{\text{max}} - V_{\text{min}}}{2} = \frac{350 - 150}{2} \mathrm{~cm}^3 = 100 \mathrm{~cm}^3 = 100 \times 10^{-6} \mathrm{~m}^3 = 10^{-4} \mathrm{~m}^3$ ### Step 1: Calculate Area Substitute a$a$ and b$b$ in standard SI units into the area equation: W = pi a b = 3.14 times (10^5 mathrm~Pa) times (10^-4 mathrm~m^3)$W = \pi a b = 3.14 \times (10^5 \mathrm{~Pa}) \times (10^{-4} \mathrm{~m}^3)$ W = 3.14 times 10 = 31.4 mathrm~J$W = 3.14 \times 10 = 31.4 \mathrm{~J}$ Express in terms of times 10^-1 mathrm~J$\times 10^{-1} \mathrm{~J}$: W = 314 times 10^-1 mathrm~J$W = 314 \times 10^{-1} \mathrm{~J}$ Thus, the multiplier is 314$314$. ### Pattern Recognition Sees: Circular/elliptical thermodynamic cycle. Shortcut: Work done is always pi Delta P Delta V / 4$\pi \Delta P \Delta V / 4$. Simply compute the semi-axes difference 100 mathrm~kPa$100 \mathrm{~kPa}$ and 100 mathrm~cm^3$100 \mathrm{~cm^3}$ and multiply by pi$\pi$ directly, keeping tracking of metric prefixes (10^3 times 10^-6 = 10^-3$10^3 \times 10^{-6} = 10^{-3}$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics

### Related Formula Delta U = mgh quad textand quad Q = msDelta T$\Delta U = mgh \quad \text{and} \quad Q = ms\Delta T$ By conservation of energy (assuming all potential energy goes into heating the water): mgh = msDelta T implies Delta T = fracghs$mgh = ms\Delta T \implies \Delta T = \frac{gh}{s}$ where, g$g$ = acceleration due to gravity h$h$ = height of the fall s$s$ = specific heat of water Delta T$\Delta T$ = rise in temperature ### Core Logic Given parameters: - h = 200mathrm~m$h = 200\mathrm{~m}$ - g = 10mathrm~m/s^2$g = 10\mathrm{~m/s}^{2}$ - s = 4200mathrm~J/(kgcdot K)$s = 4200\mathrm{~J/(kg\cdot K)}$ Substitute the values to find Delta T$\Delta T$: Delta T = frac10 times 2004200 = frac20004200$\Delta T = \frac{10 \times 200}{4200} = \frac{2000}{4200}$ Delta T = frac1021 approx 0.476mathrm~K approx 0.48mathrm~K$\Delta T = \frac{10}{21} \approx 0.476\mathrm{~K} \approx 0.48\mathrm{~K}$ ### Pattern Recognition Sees: "Water falling from height h$h$ raises temperature" → Mass cancels out. Delta T = fracghs$\Delta T = \frac{gh}{s}$. Shortcut: Always use SI units (s_textwater = 4200mathrm~J/kgcdot K$s_{\text{water}} = 4200\mathrm{~J/kg\cdot K}$ is given; if given in mathrmcal/gcdot^circ C$\mathrm{cal/g\cdot^\circ C}$, convert using 1mathrm~cal = 4.184mathrm~J$1\mathrm{~cal} = 4.184\mathrm{~J}$). ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics Class 11 Physics: Work, Energy and Power