$${T_{\min }} = \left( {{{{\Delta ^0}H} \over {{\Delta ^0}S}}} \right)$$

$${\Delta ^0}{H_{rxn}} = \left[ {\Delta _f^0H(Fe) + \Delta _f^0H(CO)} \right] - $$

$$ = \left[ {\Delta _f^0H(FeO) + \Delta _f^0H({C_{(graphite)}})} \right]$$

$$ = [0 - 110.5] - [ - 266.3 + 0]$$

= 155.8 kJ/mol

$${\Delta ^0}{S_{rxn}} = \left[ {{\Delta ^0}S(Fe) + {\Delta ^0}S(CO)} \right] - $$

$$\left[ {{\Delta ^0}S(FeO) + {\Delta ^0}S({C_{(graphite)}})} \right]$$

$$ = [27.28 + 197.6] - [57.49 + 5.74]$$

= 161.65 J/mol-K

$${T_{\min }} = {{155.8 \times {{10}^3}J/mol} \over {161.65J/mol - K}} = 963.8$$ K

$$ \simeq 964$$ k (nearest integer)

Applying; (n_I + n_II)_initial = (n_I + n_II)_final

$$\Rightarrow$$ Assuming the system attains a final temperature of T (such that 300 < T < 60)

$$\Rightarrow$$ $$\left( {\matrix{ {Heat\,lost\,by} \cr {{N_2}\,of\,container} \cr I \cr } } \right) = \left( {\matrix{ {Heat\,gained\,by} \cr {{N_2}\,of\,container} \cr {II} \cr } } \right)$$

$$\Rightarrow$$ n₁C_m(300 $$-$$ T) = n_IIC_m(T $$-$$ 60)

$$ \Rightarrow \left( {{{2.8} \over {28}}} \right)(300 - T) = {{0.2} \over {28}}(T - 60)$$

$$\Rightarrow$$ 14(300 $$-$$ T) = T $$-$$ 60

$$\Rightarrow$$ $${{(14 \times 300 + 60)} \over {15}} = T$$

$$\Rightarrow$$ T = 284 K (final temperature)

$$\Rightarrow$$ If the final pressure = P

$$\Rightarrow$$ (n_I + n_II)_final = $$\left( {{{3.0} \over {28}}} \right)$$

$$\Rightarrow$$$${P \over {RT}}({V_I} + {V_{II}}) = {{3.0\,gm} \over {28\,gm/mol}}$$

$$P = \left( {{3 \over {28}}mol} \right) \times 8.31{J \over {mol - K}} \times {{284K} \over {3 \times {{10}^{ - 3}}{m^3}}} \times {10^{ - 5}}{{bar} \over {Pa}}$$

$$\Rightarrow$$ 0.84287 bar

$$\Rightarrow$$ 84.28 $$\times$$ 10^$$-$$2 bar

$$\Rightarrow$$ 84

$$\Rightarrow$$ Millimoles of HCl = 200 $$\times$$ 0.2 = 40

$$\Rightarrow$$ Millimoles of NaOH = 300 $$\times$$ 0.1 = 30

$$\Rightarrow$$ Heat released = $$\left( {{{30} \over {1000}} \times 57.1 \times 1000} \right)$$ = 1713 J

$$\Rightarrow$$ Mass of solution = 500 ml $$\times$$ 1 gm/ml = 500 gm

$$\Rightarrow$$ $$\Delta T = {q \over {m \times c}} = {{1713J} \over {500g \times 4.18{J \over {g - K}}}}$$ = 0.8196 K

= 81.96 $$\times$$ 10^$$-$$2 K

H₂O(l) $$\to$$ H₂O(g) : $$\Delta$$H = 41 $${{kJ} \over {mol}}$$

$$\Rightarrow$$ From the relation : $$\Delta$$H = $$\Delta$$U + $$\Delta$$n_gRT

$$\Rightarrow$$ 41$${{kJ} \over {mol}}$$ = $$\Delta$$U + (1) $$\times$$ $${{8.3} \over {1000}}$$ $$\times$$ 373

$$\Delta$$ DU = 41 $$-$$ 3.0959 = 38 kJ/mol