(A) $$\to$$ (R and T)

Freezing of water,

H₂O_(l) $$\mathrel{\mathop{\kern0pt\rightleftharpoons} \limits_{273\,K}^{1\,atm}} $$ H₂O_(s)

The system is cooled i.e.; heat is released during the process

so, q < 0

$$\mathop {Water}\limits_{(Less\,volume)} $$ $$\rightleftharpoons$$ $$\mathop {Ice}\limits_{(More\,volume)} $$ + heat

Volume is increased i.e.; $$\Delta$$V = +ve.

w = $$-$$P$$\Delta$$V = $$-$$ve

i.e.; w < 0 (expansion)

Entropy of system is decreased, $$\Delta$$S_sys < 0.

$$\Delta$$U = q + w

As q < 0, w < 0 so, $$\Delta$$U < 0.

At equilibrium, $$\Delta$$G = 0.

(B) $$\to$$ (P, Q and S)

Expansion of 1 mol of an ideal gas into a vacuum under isolated conditions,

w = 0, q = 0 so, $$\Delta$$U = 0

For expansion, $$\Delta$$S_sys > 0 as entropy increases.

$$\Delta G = - nRT\ln {{{V_2}} \over {{V_1}}}$$

For expansion, V₂ > V₁

$$\Delta$$G = $$-$$ve i.e.; $$\Delta$$G < 0.

(C) $$\to$$ (P, Q and S)

Mixing of equal volumes of two ideal gases at constant temperature and pressure in an isolated container.

q = 0 (isolated)

w = $$-$$P$$\Delta$$V

w = 0 ($$\because$$ $$\Delta$$V = 0)

$$\Delta$$S_sys > 0 (mixing of gases)

$$\Delta$$U = q + w = 0

$$\Delta$$G = $$\Delta$$H $$-$$ T$$\Delta$$S

$$\Delta$$G = q_p $$-$$ T$$\Delta$$S (at constant P, T)

$$\Delta$$G = 0 $$-$$ T$$\Delta$$S = $$-$$T$$\Delta$$S

$$\Delta$$G < 0 ($$\because$$ $$\Delta$$S_sys > 0)

(D) $$\to$$ (P, Q, S and T)

$$\mathop {{H_{2(g)}}}\limits_{(300\,K)} \mathrel{\mathop{\kern0pt\rightleftharpoons} \limits_{Cool}^{Heat,\,1\,atm}} \mathop {{H_{2(g)}}}\limits_{(600\,K)} $$

Internal energy (U), entropy (S) and free energy (G) are state functions which depend only upon the state of the system and do not depend upon the path by which the state is attained.

Thus, $$\Delta$$U = 0, $$\Delta$$S = 0 and $$\Delta$$G = 0

Work and heat are path functions but the same path is retraced so, q = 0 and w = 0.