Step 1. Understanding the Slopes in the log P vs. log V Diagram:
Process A has a slope of \( \tan^{-1} \gamma \), where \( \gamma = \frac{C_P}{C_V} \), indicating an adiabatic process (since \( PV^\gamma = \text{constant} \)). Process B has a slope of \( 45^\circ \) or \( \tan^{-1} 1 \), suggesting that it is an isothermal process (since \( PV = \text{constant} \)).
Step 2. Using Heat Capacities for Adiabatic and Isothermal Processes:
For an adiabatic process (\( PV^\gamma = \text{constant} \)), the heat capacity \( C_A \) is effectively zero because no heat exchange occurs (\( dQ = 0 \) for adiabatic). For an isothermal process (\( PV = \text{constant} \)), the heat capacity \( C_B \) tends to infinity because any heat added is used to perform work without changing temperature.
Conclusion:
Therefore, the correct statement is:
\[ C_A = 0 \quad \text{and} \quad C_B = \infty \]
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32