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:
\[ C_A = 0 \quad \text{and} \quad C_B = \infty \]
The standard heat of formation, in kcal/mol, of $Ba^{2+}$ is:
Given: Standard heat of formation of SO₄²⁻(aq) = -216 kcal/mol, standard heat of crystallization of BaSO₄(s) = -4.5 kcal/mol, standard heat of formation of BaSO₄(s) = -349 kcal/mol.
Let \( S = \left\{ m \in \mathbb{Z} : A^m + A^m = 3I - A^{-6} \right\} \), where
\[ A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \]Then \( n(S) \) is equal to ______.
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to: