For a cyclic process the net change in the internal energy is zero because the change in internal energy does not depend on the path.

In adiabatic process, system does not exchange heat with surroundings

$$\eqalign{ & {\text{Given}}\,\,\frac{1}{2}{N_2} + \frac{3}{2}{H_2} \rightleftharpoons N{H_3}; \cr & \Delta {H_f} = - 46.0\,kJ/mol \cr & H + H \rightleftharpoons {H_2};\Delta {H_f} = - 436\,kJ/mol \cr & N + N \rightleftharpoons {N_2};\,\Delta {H_f} = - 712\,kJ/mol \cr & \Delta {H_f}\left( {N{H_3}} \right) = \frac{1}{2}\Delta {H_{N - N}} + \frac{3}{2}\Delta {H_{H - H}} - \Delta {H_{N - F}} \cr & - 46 = \frac{1}{2}\left( { - 712} \right) + \frac{3}{2}\left( { - 436} \right) - \Delta {H_{N - F}} \cr & {\text{On calculation}} \cr & \Delta {H_{N - F}} = - 964\,kJ/mol \cr} $$

$$\eqalign{ & 18\,g\,\,{\text{of water at}}\,\,{100^ \circ }C? \cr & 10\,g\,\,{\text{of }}Cu{\text{ at}}\,\,{\text{2}}{{\text{5}}^ \circ }C\,\,{\text{is added}}{\text{.}} \cr} $$

$$\eqalign{ & {q_p} = {C_{p,m}}\,dT \cr & = 75.32 \times \frac{J}{{K\,mol}} \times \frac{{18g}}{{\frac{{18g}}{{mol}}}}\left( {373 - 298} \right)K \cr & = 75.32\frac{J}{K} \times 75\,K \cr & = 5.649 \times {10^3}J \cr & {\text{If now 10}}g{\text{ of copper is added}} \cr & {C_{p,m}} = \frac{{24.47\,J}}{{mol\,K}} \cr & {\text{Amount of heat gained by }}Cu \cr & = 24.47\frac{J}{{K\,mol}} \times \frac{{10g}}{{\frac{{63g}}{{mol}}}}\left( {373 - 298} \right)\,K \cr & = 291.3\,J \cr & {\text{Heat lost by water = 291}}{\text{.30 J}} = 291.30\,J \cr & - 291.30\,J = 75.32\frac{J}{K} \times \left( {{T_2} - 373\,K} \right) \cr & \Rightarrow - 3.947\,K = {T_2} - 373\,K \cr & \Rightarrow {T_2} = 369.05\,K\, \cr} $$

For isothermal reversible expansion.
$$w = - nRT\,\,{\text{ln}}\,\,\frac{{{V_2}}}{{{V_1}}}$$

$$\Delta G$$  is negative for a spontaneous reaction. Positive sign shows non-spontaneous nature of the reaction.

$$\eqalign{ & {\text{Entropy change is given as,}} \cr & \Delta S = n{C_p}\,\ln \frac{{{T_f}}}{{{T_i}}} + nR\,\ln \frac{{{p_i}}}{{{p_f}}}\,\,\,...{\text{(i)}} \cr & {\text{For isothermal process, }}{T_i} = {T_f} \cr & \therefore \,\,n{C_p}\,\ln \frac{{{T_f}}}{{{T_i}}} = n{C_p}\,\ln \frac{{{T_i}}}{{{T_i}}} = 0\,\,\left[ {\ln \,1 = 0} \right] \cr & {\text{From Eq}}{\text{. (i)}}\,\,\Delta S = nR\,\ln \,\frac{{{p_i}}}{{{p_f}}} \cr} $$

$$\eqalign{ & X + 3Y \rightleftharpoons 2Z \cr & \Delta S = 2 \times 50 - \left( {60 + 3 \times 40} \right) \cr & \,\,\,\,\,\,\,\,\,\, = - 80\,J{K^{ - 1}}\,mo{l^{ - 1}} \cr & \Delta G = \Delta H - T\Delta S\,\,{\text{when}}\,\,\Delta G = 0 \cr & T = \frac{{\Delta H}}{{\Delta S}} \cr & \,\,\,\,\,\, = - 40 \times \frac{{1000}}{{ - 80}} \cr & \,\,\,\,\,\, = 500\,K \cr} $$

$$\eqalign{ & \Delta {{\text{G}}^ \circ } = - RT\,\,{\text{ln}}\,\,{K_p};{K_p} = {\left( {2x} \right)^2}X = 4{x^3} \cr & \Delta {G^ \circ } = - RT\,\,{\text{ln}}\,\,\left( {4{X^3}} \right) \cr & \Delta {G^ \circ } = - RT\,\,{\text{ln}}\,\,4 - 3RT\,\,{\text{ln}}\,\,X \cr} $$

Volume is an extensive property.