$$\begin{aligned} &\begin{aligned} & \text { I. } \rightarrow(8) \quad \text { J. } \rightarrow(9) \quad \text { K. } \rightarrow(1,12,13) \quad \text { L. } \rightarrow(7,11) \end{aligned}\\ &\text { I. } \rightarrow(8) \quad \text { J. } \rightarrow(9) \quad \text { K. } \rightarrow(1,12,13) \quad \text { L. } \rightarrow(7,11) \end{aligned}$$

Correct matching can be done as

$$\mathrm{A.\to(2)\quad B.\to(3)\quad C.\to(1)}$$

A. When a liquid vaporises, i.e., liquid $\rightarrow$ vapour, entropy increase i.e., $\Delta S=$ positive

B. When $\Delta H=$ positive, i.e., energy factor opposes. The process is non-spontaneous at all temperatures if entropy factor also opposes the process, i.e., $\Delta S=$ negative

C. In the reversible expansion of an ideal gas, the system remains in equilibrium at every stage. Hence, $\Delta S=0$

A. $\rightarrow$ (3)

B. $\rightarrow$ (1)

C. $\rightarrow$ (2)

A. When $\Delta_r G^{\mathrm{s}}$ is positive, reaction is non-spontaneous at all temperatures

B. When $\Delta_r G^{\mathrm{s}}$ is positive at high temperature means the reaction is non-spontaneous at high temperature.

C. When $\Delta_r H^{\mathbb{S}}=$ negative means it favours, $\Delta_r S^{\mathbb{s}}=$ positive means it also favours. $\Delta_r G^{\circ}=$ negative means reaction is spontaneous at all temperatures.

A. $\rightarrow(2,4)$

B. $\rightarrow$ (2)

C. $\rightarrow$ (3)

D. $\rightarrow$ (1)

A. Entropy of vaporisation is always positive. It is equal to $\Delta H_{\text {vap }} / T_b$

B. $\Delta_r G^{\circ}=-R T \ln K$

If $K$ is positive, $\Delta_r G^{\circ}=$ negative and reaction is spontaneous.

C. Crystalline solid state has lowest entropy.

D. During adiabatic expansion of an ideal gas, $q=0$. Hence, $\Delta U=q+W$ gives $\Delta U=W$, i.e., work done is at the cost of internal energy which decreases.