Match List-I with List-II:

List-I	List-II
(A) Isobaric	(I) Delta Q=Delta W$\Delta Q=\Delta W$
(B) Isochoric	(II) Delta Q=Delta U$\Delta Q=\Delta U$
(C) Adiabatic	(III) Delta Q=textzero$\Delta Q=\text{zero}$
(D) Isothermal	(IV) Delta Q=Delta U+PDelta V$\Delta Q=\Delta U+P\Delta V$

Choose the correct answer from the options given below:

### Related Formula Delta U = n C_v Delta T$$\Delta U = n C_v \Delta T$$ C_v = C_p - R$C_v = C_p - R$ ### Core Logic Given values: n = 10text moles$n = 10\text{ moles}$ Delta T = 40^circtextC - 30^circtextC = 10^circtextC$\Delta T = 40^{\circ}\text{C} - 30^{\circ}\text{C} = 10^{\circ}\text{C}$ C_p = 7text cal/molcdot^circtextC$C_p = 7\text{ cal/mol}\cdot^{\circ}\text{C}$ R = 2text cal/molcdot^circtextC$R = 2\text{ cal/mol}\cdot^{\circ}\text{C}$ First, find C_v$C_v$ using Mayer's relation: C_v = C_p - R = 7 - 2 = 5text cal/molcdot^circtextC$$C_v = C_p - R = 7 - 2 = 5\text{ cal/mol}\cdot^{\circ}\text{C}$$ ### Step 1: Calculate Internal Energy Change Delta U = n C_v Delta T$$\Delta U = n C_v \Delta T$$ Delta U = 10 times (7 - 2) times (40 - 30)$$\Delta U = 10 \times (7 - 2) \times (40 - 30)$$ Delta U = 10 times 5 times 10 = 500text cal$$\Delta U = 10 \times 5 \times 10 = 500\text{ cal}$$ ### Pattern Recognition Change in internal energy of an ideal gas is ALWAYS Delta U = n C_v Delta T$\Delta U = n C_v \Delta T$, regardless of the process (constant volume or not). Use C_v = C_p - R$C_v = C_p - R$ when C_p$C_p$ is given. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics Class 11 Physics: Kinetic Theory of Gases

### 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