### 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

### Core Logic Let's evaluate each process condition based on the first law of thermodynamics: * **(A) Isothermal:** Continuous constant temperature (Delta T = 0$\Delta T = 0$) implies that the internal energy change of an ideal gas is zero, so Delta U = 0 implies text(IV)$\Delta U = 0 \implies \text{(IV)}$ [cite: 730]. * **(B) Adiabatic:** No thermal energy transfer occurs between the system and surroundings, meaning Delta Q = 0 implies text(II)$\Delta Q = 0 \implies \text{(II)}$ [cite: 731]. * **(C) Isobaric:** Constant pressure process where both volume and temperature typically vary, so internal energy changes continuously, Delta U neq 0 implies text(III)$\Delta U \neq 0 \implies \text{(III)}$ [cite: 732]. * **(D) Isochoric:** Rigid boundary condition at constant volume (Delta V = 0$\Delta V = 0$) ensures work done Delta W = PDelta V = 0 implies text(I)$\Delta W = P\Delta V = 0 \implies \text{(I)}$ [cite: 733]. Putting these together yields: (A)-(IV), (B)-(II), (C)-(III), (D)-(I)[cite: 155, 729]. ### Pattern Recognition Matching 'Isochoric' with zero work done (Delta W=0$\Delta W=0$) or 'Adiabatic' with zero heat exchange (Delta Q=0$\Delta Q=0$) are fundamental definitions that let you rapidly break down multi-choice grids[cite: 731, 733]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics

### Related Formula T_1 V_1^gamma-1 = T_2 V_2^gamma-1$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$ ### Core Logic Assertion (A) review: An insulated container means the process is adiabatic (Q=0$Q=0$). When the gas is shrunk (compressed) to half its volume (V_2 = V_1 / 2$V_2 = V_1 / 2$), T_2 = T_1 left(fracV_1V_2 ight)^gamma-1 = T_1 (2)^gamma-1$T_2 = T_1 \left(\frac{V_1}{V_2} ight)^{\gamma-1} = T_1 (2)^{\gamma-1}$ Since gamma > 1$\gamma > 1$, T_2 > T_1$T_2 > T_1$. Therefore, temperature increases during adiabatic compression, which makes Assertion (A) false. Reason (R) review: Free expansion occurs when a gas expands into a vacuum inside an insulated container. No work is done (W=0$W=0$) and no heat is exchanged (Q=0$Q=0$), hence it is adiabatic. It cannot spontaneously reverse, so it is irreversible. Thus, Reason (R) is true. ### Pattern Recognition Adiabatic compression always raises temperature due to work being done on the gas, while free expansion keeps the temperature of an ideal gas constant (dI=0$dI=0$ as W=0, Q=0$W=0, Q=0$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics

### Related Formula From the first law of thermodynamics for a complete cycle: Delta U = 0 implies Q = W = textArea of the loop$\Delta U = 0 \implies Q = W = \text{Area of the loop}$ ### Core Logic The graph shows a semicircle in a Ptext-V$P\text{-}V$ indicator diagram.

- Pressure dimension diameter: Delta P = 400 - 200 = 200\ mathrmkPa = 200 times 10^3\ mathrmPa$\Delta P = 400 - 200 = 200\ \mathrm{kPa} = 200 \times 10^{3}\ \mathrm{Pa}$ - Volume dimension diameter: Delta V = 400 - 200 = 200\ mathrmcc = 200 times 10^-6\ mathrmm^3$\Delta V = 400 - 200 = 200\ \mathrm{cc} = 200 \times 10^{-6}\ \mathrm{m^3}$ Radius along Pressure axis: R_P = 100 times 10^3\ mathrmPa$R_P = 100 \times 10^3\ \mathrm{Pa}$ Radius along Volume axis: R_V = 100 times 10^-6\ mathrmm^3$R_V = 100 \times 10^{-6}\ \mathrm{m^3}$ Area of the closed semicircular path: W = frac12 pi R_P R_V$W = \frac{1}{2} \pi R_P R_V$ W = frac12 times pi times (100 times 10^3) times (100 times 10^-6)$W = \frac{1}{2} \times \pi \times (100 \times 10^3) \times (100 \times 10^{-6})$ W = frac10pi2 = 5pi\ mathrmJ$W = \frac{10\pi}{2} = 5\pi\ \mathrm{J}$ Since Q = W$Q = W$, the heat exchanged has a magnitude of 5pi\ mathrmJ$5\pi\ \mathrm{J}$. ### Pattern Recognition For a cycle on an indicator chart with mismatched scales, use the elliptic area template pi a b$\pi a b$ (or frac12pi a b$\frac{1}{2}\pi a b$ for a half-ellipse/semicircle). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics