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 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

### Related Formula From the Ideal Gas Law: PV = nRT$PV = nRT$ According to the First Law of Thermodynamics: Delta Q = Delta U + W$$\Delta Q = \Delta U + W$$ ### Core Logic The question states that pressure increases linearly with temperature, which means their ratio is constant : P = kT implies fracPT = textconstant$$P = kT \implies \frac{P}{T} = \text{constant}$$ Since fracPT = fracnRV$\frac{P}{T} = \frac{nR}{V}$, the volume V$V$ must remain constant throughout the process. This identifies it as an isochoric process (Statement E is true). ### Step 1: Evaluate All Statements * Statement A: True. In an isochoric process, dV = 0 implies W = int P dV = 0$dV = 0 \implies W = \int P dV = 0$. * Statement B: False. Since work is zero, the First Law simplifies to Delta Q = Delta U$\Delta Q = \Delta U$, meaning heat added equals the change in internal energy. * Statement C: False. Volume is constant, so it does not increase. * Statement D: True. As pressure increases linearly with temperature, temperature increases, which causes the internal energy of the gas to increase. ### Step 2: Final Selection Gathering the true statements (A, D, and E) points directly to Option (2). ### Pattern Recognition A linear P-T$P-T$ line passing through the origin always indicates a constant volume graph. For constant volume graphs, work done is zero, which simplifies the first law calculation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics

### Related Formula For an ideal gas thermodynamic process: * Total heat added at constant pressure (isobaric process) is : E_1 = n C_P Delta T$$E_{1} = n C_{P} \Delta T$$ * Total change in internal energy is given by : E_2 = n C_V Delta T$$E_{2} = n C_{V} \Delta T$$ The ratio of specific heat capacities is defined as: gamma = fracC_PC_V $$\gamma = \frac{C_{P}}{C_{V}} $$ ### Core Logic Taking the ratio of the two energy expressions[cite: 179, 823]: fracE_1E_2 = fracn C_P Delta Tn C_V Delta T = fracC_PC_V = gamma$$\frac{E_{1}}{E_{2}} = \frac{n C_{P} \Delta T}{n C_{V} \Delta T} = \frac{C_{P}}{C_{V}} = \gamma$$ ### Step 1: Evaluating for a Monoatomic Gas For an ideal monoatomic gas, the degrees of freedom are f = 3$f = 3$. This gives an adiabatic index of : gamma = 1 + frac2f = 1 + frac23 = frac53 $$\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{3} = \frac{5}{3} $$ Equating this value to the given ratio expression [cite: 179, 826]: frac53 = fracx9$$\frac{5}{3} = \frac{x}{9}$$ x = frac5 times 93 = 15$$x = \frac{5 \times 9}{3} = 15$$ ### Pattern Recognition The ratio of heat added to the change in internal energy during an isobaric process is always equal to the adiabatic exponent gamma$\gamma$ of the gas. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics

### Related Formula Delta W = -Delta U = -n C_v Delta T$$\Delta W = -\Delta U = -n C_v \Delta T$$ ### Core Logic In an adiabatic system, no heat exchange occurs (Q=0$Q=0$). By the first law of thermodynamics, Delta W = -Delta U$\Delta W = -\Delta U$. Since internal energy U$U$ depends explicitly on temperature metrics, the total work output shifts uniquely based on temperature variation Delta T$\Delta T$. ### Chapter Mix Class 11 Physics: Thermodynamics