$$\eqalign{ & {\text{At constant volume}},\,\frac{{{P_1}}}{{{T_1}}} = \frac{{{P_2}}}{{{T_2}}} \cr & \Rightarrow {P_2} = 1 \times \frac{{300}}{{200}} = \frac{3}{2} \cr & {\text{and}}\,\,{V_1} = 24.63\,L \cr & {\text{for single phase}} \cr & \because \,\,dG = Vdp - SdT \cr & \Delta G = V\Delta P - \int {\left( {2 + {{10}^{ - 2}}T} \right)dT} \cr & = 1231.5 - 200 - \left( {\frac{{{{10}^{ - 2}} \times 50,000}}{2}} \right) \cr & = 781.5\,J \cr} $$

Work is equal to area under $$P - V$$   graph ( when $$P$$  is plotted along $$y$$ - axis ). As area under graph 3 is maximum and area under graph 1 is minimum, so $${w_3}$$  is minimum and $${w_1}$$  is minimum.
$$\therefore \,\,{w_1} < {w_2} < {w_3}$$

$$\eqalign{ & {q_p} = \Delta H = {C_p}\,dT \cr & \Rightarrow {q_p} = 75.32\frac{J}{{K\,mol}} \times \left( {299 - 298} \right)K \cr & \Rightarrow {q_p} = 75.32\frac{J}{{K\,mol}} \cr & {\text{For }}180{\text{ }}kg{\text{ of water, no}}{\text{. of moles of water}} \cr & {\text{ = }}\frac{{180 \times {{10}^3}g}}{{18g/mol}} \cr & = {10^4}g\,moles \cr & {q_p} = 75.32\frac{J}{{mol}} \times {10^4}\,moles \cr & = 753.2 \times {10^3}\,J \cr & = 753.2\,kJ \cr & \Delta H\,{\text{for}}\,ATP = 7\,kcal/mol \cr & = 7 \times 4.184\,kJ/mol \cr & = 29.2\,kJ/mol \cr & 6.022 \times {10^{23}}\,{\text{molecules of }}ATP{\text{ produce}} \cr & = 29.2\,kJ \cr & 29.2\,kJ\,{\text{produced from}} \cr & 6.022 \times {10^{23}}{\text{molecules}} \cr & 753.2\,kJ\,{\text{produced from}} \cr & 6.022 \times {10^{23}} \times \frac{{75.8}}{{29.2}} \cr & = 1.5 \times {10^{25}}\,{\text{molecules}} \cr} $$

In question $$0.01\,gev$$   of $$NaOH$$  is being neutralised by $$0.01\,gev$$   of $$HCl$$  and value is $$x\,kJ,$$  for $$1\,gev$$  the value is $$ - 100x$$  ( exothermic process ).

When one mole of a substance is directly formed from its constituent elements, then the enthalpy change is called heat of formation.
For the reaction,
$$\eqalign{ & Xe\left( g \right) + 2{F_2}\left( g \right) \to \mathop {Xe{F_4}\left( g \right)}\limits_{1\,\,mol} \cr & \Delta {H^ \circ }_{{\text{react}}} = \Delta H_f^ \circ \cr} $$

NOTE : In a reversible process the work done is greater than in irreversible process. Hence the heat absorbed in reversible process would be greater than in the latter case. So
$${T_f}\left( {rev.} \right) < {T_f}\left( {irr.} \right)$$

Condition of equilibrium, hence $$\Delta G = 0$$

$${q_2}$$ is heat of neutralization of strong acid with strong base and $${q_1}$$ is that of weak acid $$HA$$  with strong base. The difference gives the heat of dissociation of $$HA$$  i.e.,
$$\eqalign{ & \Delta H = \Delta {H_2} - \Delta {H_1} \cr & \,\,\,\,\,\,\,\,\,\,\, = \left( { - {q_2}} \right) - \left( { - {q_1}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\, = {q_1} - {q_2} \cr} $$

$$\eqalign{ & q = m \times c \times \Delta T,m = \frac{q}{{\left( {c \times \Delta T} \right)}} \cr & = \frac{{\left( {24 \times {{10}^6} \times 0.7} \right)}}{{\left( {4.18 \times 50} \right)}} \cr & = 80383\,g\,\,{\text{or}}\,\,80.383\,kg \cr} $$

For monoatomic gas $${C_v} = \frac{3}{2}\,R\,{C_p} = \frac{5}{2}\,R$$
At constant volume, $$\Delta U = {q_v} = n{C_{v,m}}\,\Delta T$$
At constant pressure, $$\Delta H = {q_p} = n{C_{p,m}}\,\Delta T$$