Match List-I with List-II.

List-I	List-II
(A) Isothermal	(I) Delta W$\Delta W$ (work done) =0$=0$
(B) Adiabatic	(II) Delta Q$\Delta Q$ (supplied heat) =0$=0$
(C) Isobaric	(III) Delta U$\Delta U$ (change in internal energy) ne0$\ne0$
(D) Isochoric	(IV) Delta U=0$\Delta U=0$

Choose the correct answer from the options given below: [cite: 151, 152]

### 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

### Related Formula For an adiabatic process: T V^gamma - 1 = textconstant$$T V^{\gamma - 1} = \text{constant}$$ where, T$T$ = absolute temperature in Kelvin, V$V$ = volume of the gas, gamma$\gamma$ = adiabatic exponent. ### Core Logic Given values: - Initial volume, V_1 = 800mathrm~cm^3$V_1 = 800\mathrm{~cm}^3$ - Final volume, V_2 = 200mathrm~cm^3$V_2 = 200\mathrm{~cm}^3$ - Initial temperature, T_1 = 27^circmathrmC = 27 + 273 = 300mathrm~K$T_1 = 27^{\circ}\mathrm{C} = 27 + 273 = 300\mathrm{~K}$ - Adiabatic exponent, gamma = 1.5 implies gamma - 1 = 0.5$\gamma = 1.5 \implies \gamma - 1 = 0.5$ ### Step 1: Calculating Final Temperature Apply the adiabatic relation: T_1 V_1^gamma - 1 = T_2 V_2^gamma - 1$$T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$$ T_2 = T_1 left(fracV_1V_2right)^gamma - 1$$T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma - 1}$$ Substitute the values: T_2 = 300 left(frac800200right)^0.5 = 300 times (4)^0.5$$T_2 = 300 \left(\frac{800}{200}\right)^{0.5} = 300 \times (4)^{0.5}$$ T_2 = 300 times 2 = 600mathrm~K$$T_2 = 300 \times 2 = 600\mathrm{~K}$$ ### Step 2: Calculating Change in Temperature Now compute the change in temperature (Delta T$\Delta T$): Delta T = T_2 - T_1 = 600mathrm~K - 300mathrm~K = 300mathrm~K$$\Delta T = T_2 - T_1 = 600\mathrm{~K} - 300\mathrm{~K} = 300\mathrm{~K}$$ ### Pattern Recognition Always read carefully to see if the question asks for the **final temperature** or the **change in temperature**. Many students lose marks by choosing 600mathrm~K$600\mathrm{~K}$ (the final temperature) instead of the difference 300mathrm~K$300\mathrm{~K}$! Stay sharp. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics

### Related Formula First Law of Thermodynamics: Delta Q = Delta U + Delta W$$\Delta Q = \Delta U + \Delta W$$ ### Core Logic - **Isobaric**: Pressure is constant, work done Delta W = PDelta V$\Delta W = P\Delta V$. Thus, Delta Q = Delta U + PDelta V$\Delta Q = \Delta U + P\Delta V$ (Matches IV). - **Isochoric**: Volume is constant, Delta V = 0 implies Delta W = 0$\Delta V = 0 \implies \Delta W = 0$. Thus, Delta Q = Delta U$\Delta Q = \Delta U$ (Matches II). - **Adiabatic**: No heat transfer, Delta Q = 0$\Delta Q = 0$ (Matches III). - **Isothermal**: Temperature is constant, internal energy change Delta U = 0$\Delta U = 0$ for an ideal gas. Thus, Delta Q = Delta W$\Delta Q = \Delta W$ (Matches I). ### Step 1: Match Compilation Combining everything gives: (A)-(IV), (B)-(II), (C)-(III), (D)-(I). ### Pattern Recognition Adiabatic definition is always heat-isolated (Q=0$Q=0$), isochoric implies rigid boundaries (W=0$W=0$). Identifying these two immediately isolates the correct match permutation in seconds. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics