Given below are two statements :

Statement I : For an ideal gas, heat capacity at constant volume is always greater than the heat capacity at constant pressure.

Statement II : In a constant volume process, no work is produced and all the heat withdrawn goes into the chaotic motion and is reflected by a temperature increase of the ideal gas.

In the light of the above statements, choose the correct answer from the options given below

Gas 'A' undergoes change from state 'X' to state 'Y'. In this process, the heat absorbed and work done by the gas is 10 J and 18 J respectively. Now gas is brought back to state 'X' by another process during which 6 J of heat is evolved. In the reverse process of 'Y' to 'X',

Consider the following data.

(i) $2\text{Al(s)} + 6\text{HCl(aq)} \rightarrow \text{Al}_2\text{Cl}_6\text{(aq)} + 3\text{H}_2\text{(g)} + 1200~\text{kJ/mol}$

(ii) $\text{H}_2\text{(g)} + \text{Cl}_2\text{(g)} \rightarrow 2\text{HCl(g)} + 164~\text{kJ/mol}$

(iii) $\text{HCl(g)} + \text{aq} \rightarrow \text{HCl(aq)} + 83~\text{kJ/mol}$

(iv) $\text{Al}_2\text{Cl}_6\text{(s)} + \text{aq} \rightarrow \text{Al}_2\text{Cl}_6\text{(aq)} + 663~\text{kJ/mol}$

The enthalpy of formation of anhydrous solid $\text{Al}_2\text{Cl}_6$ is :

The plot of $\log_{10} K$ vs $\frac{1}{T}$ gives a straight line. The intercept and slope respectively are (where K is equilibrium constant).