### 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 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