### 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

### Related Formula Work done in a polytropic process PV^x = textconstant$PV^x = \text{constant}$: W = fracP_2V_2 - P_1V_11 - x$W = \frac{P_2V_2 - P_1V_1}{1 - x}$ ### Core Logic Here, the polytropic exponent is x = frac32$x = \frac{3}{2}$. Substituting this exponent value into the generic polytropic work integral: W = fracP_2V_2 - P_1V_11 - frac32 = fracP_2V_2 - P_1V_1-frac12$W = \frac{P_2V_2 - P_1V_1}{1 - \frac{3}{2}} = \frac{P_2V_2 - P_1V_1}{-\frac{1}{2}}$ ### Step 1: Simplify Sign Conventions W = -2(P_2V_2 - P_1V_1) = 2(P_1V_1 - P_2V_2)$W = -2(P_2V_2 - P_1V_1) = 2(P_1V_1 - P_2V_2)$ This represents the default standard convention for work executed by the gas medium. ### Pattern Recognition Polytropic model work formula is extremely useful. If x > 1$x > 1$, the denominator 1-x$1-x$ flips the index sequence from (2-1)$(2-1)$ to (1-2)$(1-2)$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics