### Related Formula Work done (W$W$) in a thermodynamic process corresponds to the area under the curve on a P-V$P-V$ diagram: W = int P \, dV$W = \int P \, dV$ ### Core Logic Let's compute the work done along the distinct steps AB$AB$ and BC$BC$ matching the values shown in the graph: From graph coordinates: P_A = 8000 mathrm~Dyne/cm^2, quad V_A = 3 mathrm~m^3$P_A = 8000 \mathrm{~Dyne/cm^2}, \quad V_A = 3 \mathrm{~m^3}$ P_B = 4000 mathrm~Dyne/cm^2, quad V_B = 7 mathrm~m^3$P_B = 4000 \mathrm{~Dyne/cm^2}, \quad V_B = 7 \mathrm{~m^3}$ P_C = 4000 mathrm~Dyne/cm^2, quad V_C = 3 mathrm~m^3$P_C = 4000 \mathrm{~Dyne/cm^2}, \quad V_C = 3 \mathrm{~m^3}$

### Step 1: Calculate Work Done in Process AB Process AB$AB$ is a linear expansion. Fulfilling the area of the trapezoid underneath path AB$AB$: W_AB = frac12 (P_A + P_B) Delta V_AB$W_{AB} = \frac{1}{2} (P_A + P_B) \Delta V_{AB}$ W_AB = frac12 (8000 + 4000) mathrm~Dyne/cm^2 times (7 - 3) mathrm~m^3$W_{AB} = \frac{1}{2} (8000 + 4000) \mathrm{~Dyne/cm^2} \times (7 - 3) \mathrm{~m^3}$ W_AB = 6000 times 4 = 24000 mathrm~Dyne/cm^2 cdot mathrmm^3$W_{AB} = 6000 \times 4 = 24000 \mathrm{~Dyne/cm^2} \cdot \mathrm{m^3}$ ### Step 2: Calculate Work Done in Process BC Process BC$BC$ is an isobaric compression at P = 4000 mathrm~Dyne/cm^2$P = 4000 \mathrm{~Dyne/cm^2}$: W_BC = P_B cdot (V_C - V_B) = 4000 times (3 - 7) = -16000 mathrm~Dyne/cm^2 cdot mathrmm^3$W_{BC} = P_B \cdot (V_C - V_B) = 4000 \times (3 - 7) = -16000 \mathrm{~Dyne/cm^2} \cdot \mathrm{m^3}$ ### Step 3: Total Work and Unit Conversion Net work done is: W_texttotal = W_AB + W_BC = 24000 - 16000 = 8000 mathrm~Dyne/cm^2 cdot mathrmm^3$W_{\text{total}} = W_{AB} + W_{BC} = 24000 - 16000 = 8000 \mathrm{~Dyne/cm^2} \cdot \mathrm{m^3}$ Converting units to Joules (1 mathrm~N/m^2 = 10 mathrm~Dyne/cm^2 implies 1 mathrm~Dyne/cm^2 = 0.1 mathrm~N/m^2$1 \mathrm{~N/m^2} = 10 \mathrm{~Dyne/cm^2} \implies 1 \mathrm{~Dyne/cm^2} = 0.1 \mathrm{~N/m^2}$): W_texttotal = 8000 times (0.1 mathrm~N/m^2) cdot mathrmm^3 = 800 mathrm~J$W_{\text{total}} = 8000 \times (0.1 \mathrm{~N/m^2}) \cdot \mathrm{m^3} = 800 \mathrm{~J}$ Since 800 mathrm~J$800 \mathrm{~J}$ is not present in the options, this question was declared a **BONUS**. ### Pattern Recognition Always be highly vigilant of mixed unit systems in thermodynamic graphics. Here, pressure is in CGS (textDyne/cm^2$\text{Dyne/cm}^2$) while volume is in MKS (textm^3$\text{m}^3$). Converting early keeps you clear of algebraic traps. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics

### Related Formula textIsothermal: PV = textconstant$\text{Isothermal: } PV = \text{constant}$ textAdiabatic: PV^gamma = textconstant$\text{Adiabatic: } PV^{\gamma} = \text{constant}$ textAdiabatic T-P relation: T^gamma P^1-gamma = textconstant implies fracT^gammaP^gamma - 1 = textconstant$\text{Adiabatic T-P relation: } T^{\gamma} P^{1-\gamma} = \text{constant} \implies \frac{T^{\gamma}}{P^{\gamma - 1}} = \text{constant}$ ### Core Logic In a P-V$P-V$ diagram starting from the same initial state and undergoing expansion, the adiabatic curve is steeper than the isothermal curve. Looking at the graph, Curve (B) is steeper, so it represents an adiabatic process. Curve (A) is less steep, so it represents an isothermal process. ### Step 1: Check Options Process A (Isothermal): PV = k$PV = k$ or T = k$T = k$. Process B (Adiabatic): PV^gamma = k$PV^{\gamma} = k$. Option (1) states: PV^gamma = k$PV^{\gamma} = k$ for B and PV = k$PV = k$ for A. This is entirely correct. Note: For an adiabatic process, T^gamma P^1-gamma = k$T^{\gamma} P^{1-\gamma} = k$, which can be written as fracT^gammaP^gamma-1 = k$\frac{T^{\gamma}}{P^{\gamma-1}} = k$. Option (3) proposes fracP^gamma-1T^gamma = k$\frac{P^{\gamma-1}}{T^{\gamma}} = k$ for B, which is mathematically equivalent, but Option (1) is the standard and most direct NTA choice. ### Pattern Recognition Slope of isothermal = -P/V$= -P/V$. Slope of adiabatic = -gamma P/V$= -\gamma P/V$. Since gamma > 1$\gamma > 1$, adiabatic is always steeper. In expansions, adiabatic lies below isothermal. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics

### Related Formula C = C_v + fracR1 - n quad (textfor PV^n = textconst)$C = C_v + \frac{R}{1 - n} \quad (\text{for } PV^n = \text{const})$ textIf P = k V^x, text then n = -x$\text{If } P = k V^x, \text{ then } n = -x$ ### Core Logic For a generic straight line on a log P$\log P$ vs log V$\log V$ plot, the slope defines the exponent. log P = x log V + c Rightarrow P = V^x cdot e^c Rightarrow PV^-x = textConstant$\log P = x \log V + c \Rightarrow P = V^x \cdot e^c \Rightarrow PV^{-x} = \text{Constant}$. The molar heat capacity is computed using n = -x$n = -x$. ### Step 1: Evaluate Process A For process A: slope = gamma$= \gamma$ (given tan theta > 1$\tan \theta > 1$). log P = gamma log V Rightarrow PV^-gamma = textConstant quad (texthere n = -gamma)$\log P = \gamma \log V \Rightarrow PV^{-\gamma} = \text{Constant} \quad (\text{here } n = -\gamma)$ C_A = C_V + fracR1 - (-gamma) = C_V + fracR1 + gamma quad dots (i)$C_A = C_V + \frac{R}{1 - (-\gamma)} = C_V + \frac{R}{1 + \gamma} \quad \dots (i)$ ### Step 2: Evaluate Process B For process B: slope = 1$= 1$ (given tan 45^circ = 1$\tan 45^\circ = 1$). log P = 1 cdot log V Rightarrow PV^-1 = textConstant quad (texthere n = -1)$\log P = 1 \cdot \log V \Rightarrow PV^{-1} = \text{Constant} \quad (\text{here } n = -1)$ C_B = C_v + fracR1 - (-1) = C_v + fracR2 quad dots (ii)$C_B = C_v + \frac{R}{1 - (-1)} = C_v + \frac{R}{2} \quad \dots (ii)$ ### Step 3: Analyze Heat Capacities We also know: C_P = C_v + R quad dots (iii)$C_P = C_v + R \quad \dots (iii)$ Comparing (i), (ii), and (iii): Since gamma > 1$\gamma > 1$ for process A, fracR1+gamma < fracR2$\frac{R}{1+\gamma} < \frac{R}{2}$. Thus, C_P > C_B > C_A > C_v$C_P > C_B > C_A > C_v$. Note: Ans. By NTA (1 or 2). Ans. By our answer (Bonus). Since no option fully matches the mathematical derivation C_P > C_B > C_A > C_v$C_P > C_B > C_A > C_v$, this was treated as a bonus or flawed option set. We assign index 0 based on official NTA preliminary keys marking 1 or 2. ### Pattern Recognition Slopes on a log-log PV plot represent the polytropic exponent modifier. Positive slopes imply -n > 0$-n > 0$, pulling heat capacity structurally between C_v$C_v$ and C_P$C_P$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics

### Related Formula PV = nRT$PV = nRT$ ### Core Logic From the Ideal Gas Law: V = left(fracnRPright) T$V = \left(\frac{nR}{P}\right) T$ In a V-T$V-T$ graph, the equation of the line represents y = mx$y = mx$, where the slope m$m$ is: textSlope = fracnRP$\text{Slope} = \frac{nR}{P}$ textSlope propto frac1P$\text{Slope} \propto \frac{1}{P}$ Thus, a higher slope corresponds to a lower pressure. ### Step 2: Compare Slopes From the given figure, the slope of line 2 is greater than the slope of line 1: (textSlope)_2 > (textSlope)_1$(\text{Slope})_2 > (\text{Slope})_1$ Therefore, inversely: P_2 < P_1$P_2 < P_1$ or P_1 > P_2$P_1 > P_2$. ### Pattern Recognition In V-T graphs, steeper lines mean lower Pressure. In P-T graphs, steeper lines mean lower Volume. It's an inverse inverse slope relationship. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics