\(Process \ A \ \longrightarrow \ B(expansion),\)
\( p=12\times { 10 }^{ 5 }{ Nm }^{ -2 },\triangle V=8-2=6L\)
\( =6\times { 10 }^{ -3 }{ m }^{ 3 }\)
\( \therefore \ Work \ done \ = \ -p\triangle V\)
\( = -\left( 12\times { 10 }^{ 5 } \right) \times \left( 6\times { 10 }^{ -3 } \right) J=-7200 \ J\)
\(Process \ B \ \longrightarrow \ C.No \ change \ in \ volume,i.e \ \triangle V=0\)
\(\therefore Work \ done \ = \ 0\)
\( Process \ C\longrightarrow \ D(contraction)\)
\( \triangle V=8-2=6L=6\times { 10 }^{ -3 }m^{ 3 },p=4\times { 10 }^{ 5 }{ Nm }^{ -2 }\)
\( Work \ done\)
\( p\triangle V=\left( 4\times { 10 }^{ 5 } \right) \left( 6\times { 10 }^{ -3 } \right) =2400 \ J\)
\(Process \ D\longrightarrow A.\ No \ change \ in \ volume,\ i.e.\ \triangle V=0\)
\( \therefore Work \ done=0\)
\(\therefore Net \ work \ done \ in \ the \ complete \ cyclic \ process\)
\(=-7200+2400 \ J=-4800 \ J\)
\(Minus \ sign \ shows \ that \ net \ work \ has \ been \ done \ by \ the \ gas.\)

In general, if enthalpy of an overall reaction A\(\rightarrow \) B along one route is  \(\Delta _{ r }H \ and \ \Delta _{ r }H_{ 1 },\Delta _{ r }H_{ 2 },\Delta _{ r }H_{ 3 }....\) representing enthalpies of reactions leading to same product B along another route, then we have
\(\Delta _{ r }H=\Delta _{ r }H_{ 1 }+\Delta _{ r }H_{ 2 }+\Delta _{ r }H_{ 3 }+....\)
Note
For a general reaction Hess's law of constant heat summation can be represented as

2 NaHCO₃(s) \(\longrightarrow \) Na₂CO₃ (s) + CO₂(g) + H₂O(g)
\({ \triangle }_{ r }{ H }^{ o }={ \triangle }_{ f }{ H }^{ o }({ Na }_{ 2 }{ CO }_{ 3 })+{ \triangle }_{ f }{ H }^{ o }({ CO }_{ 2 })+{ \triangle }_{ r }{ H }^{ o }({ H }_{ 2 }O)-2{ \triangle }_{ f }{ H }^{ o }({ S }_{ m }^{ o }(NaHCO_{ 3 })\)
\(=-1130+(-393.51)+(-241.82)-2X(-947.7)\)
\( =-1766.23+1895.4=129.17KJ{ mol }^{ -1 }\)
\( { \triangle }_{ r }{ S }^{ o }={ \triangle }_{ r }{ S }_{ m }^{ o }({ Na }_{ 2 }{ CO }_{ 3 })+{ S }_{ m }^{ o }({ CO }_{ 2 })+{ S }_{ m }^{ o }({ H }_{ 2 }O)-2{ S }_{ m }^{ o }(NaHCO_{ 3 })\)
\(=136.0+188.83+231.74-2\times102.1\)
\( =538.57-204.2=334.37J{ K }^{ -1 }{ mol }^{ -1 }\)
From second law of thermodynamics \({ \triangle }_{ r }{ S }^{ o }=\frac { { \triangle }_{ r }{ H }^{ o } }{ T } \)
 \(\therefore T=\frac { { \triangle }_{ r }{ H }^{ o } }{ { \triangle }_{ r }{ S }^{ o } } =\frac { 129.17 }{ 334.37\times{ 10 }^{ -3 } } =386.3 \ K\)
Reaction will be spontaneous above 386.3 K.

\(N_{ 2 }{ O }_{ 4 }(g)\leftrightharpoons 2NO_{ 2 }(g)\)
If N₂O₄ is 50% dissociated,d, the mole fraction of both the substances is given by
\(x_{ N_{ 2 }O_{ 4 } }=\frac { 1-0.5 }{ 1+0.5 } \Rightarrow x_{ NO_{ 2 } }=\frac { 2\times 0.5 }{ 1+0.5 }\)
\( p_{ N_{ 2 }O_{ 4 } }=\frac { 0.5 }{ 1.5 } \times 1atm,\ p_{ NO_{ 2 } }=\frac { 1 }{ 1.5 } \times 1atm\)
The equilibrium constant K_p is given by
\(K_{ p }=\frac { (p_{ NO_{ 2 } })^{ 2 } }{ p_{ { N }_{ 2 }{ O }_{ 4 } } } =\frac { 1.5 }{ (1.5)^{ 2 }(0.5) } =1.33atm\)
Since,
\(\Delta _{ r }G^{ \circ }=-RT \ ln \ K_{ p }\)
\( \Delta _{ r }G^{ \circ }=(-8.314 \ JK^{ - } \ mol^{ - })\times (333K)\times (2.303)\times (0.1239)\)
\( =-763.8 \ kJmol^{ -1 }\)

A heat capacity of water = heat capacity of calorimeter, the heat gained by water = heat lost by calorimeter
\(=125\times (296.5-286.4)\times 4.184 \ J=5282J=5.282kJ\)

(i) Yes in an electrical fan, the electrical energy is converted into mechanical work that moves the blades.In a heater, electrical is converted into heat energy.
(ii) Running water of river or lake has kinetic energy. By stopping the flow in the dam, the kinetic energy is converted into potential energy.When the water is allowed to fall on the turbine,it is converted into kinetics energy which is then converted into mechanical work that runs the turbine and ultimately converted into electrical energy.
(iii) The stone at a height has potential energy. When it falls its potential energy is converted into kinetic energy and hence, is associated with force that hits the object or breaks our head.
(iv) sonu is intelligent, curious, practical and scientific.

Suppose q is the quantity of heat from the reaction mixture and C_V is the heat capacity of the calorimeter, then the quantity of heat absorbed by the calorimeter.
q = C_V × ∆T
Quantity of heat from the reaction will have the same magnitude but opposite sign because the heat lost by the system (reaction mixture) is equal to the heat gained by the calorimeter.
q = –C_V × ∆T = – 20.7 kJ/K × (299 – 298) K = – 20.7 kJ
(Here, negative sign indicates the exothermic nature of the reaction) Thus, ∆U for the combustion of the 1g of graphite = – 20.7 kJK^–1 For combustion of 1 mol of graphite,
\(=\frac { 12.0gmol^{ -1 }\times (-20.7kJ) }{ 1g }\)
= – 2.48 ×102 kJ mol^–1 , Since ∆ n_g = 0, ∆ H = ∆ U = – 2.48 ×102 kJ mol^–1

The heat of combustion of C₂H₆(g)  per mole (\(\triangle H\)) = \(\frac { -3129 }{ 2 } =-1564.5KJ\)
The combustion equation may be written as :
\({ C }_{ 2 }H_{ 6 }(g)+\frac { 7 }{ 2 } O_{ 2 }(g)\longrightarrow 2CO_{ 2 }(g)+3{ H }_{ 2 }O(l);\triangle H=-1564.5KJ\)
\(\triangle H=\sum { { \triangle }_{ f } } { H }^{ \ominus }(products)-\sum { { \triangle }_{ f } } { H }^{ \ominus }(reactants)\)
\(=[2{ \triangle }_{ f }{ H }^{ \ominus } \ CO_{ 2 }(g)+3{ \triangle }_{ f }{ H }^{ \ominus } \ { H }_{ 2 }O(l)]\)
\(-\left[ { \triangle }_{ f }{ H }^{ \ominus }C_{ 2 }H_{ 6 }(g)+\frac { 7 }{ 2 } { \triangle }_{ f }{ H }^{ \ominus }O_{ 2 }(g) \right] \)
On substituting the values in the above equation
1564.5 = [2 x (- 393.5) + 3 (- 286)] - [\({ \triangle }_{ f }{ H }^{ \ominus }C_{ 2 }H_{ 6 }\)(g) + zero]
- 1564.5 = - 787 - 858 - \({ \triangle }_{ f }{ H }^{ \ominus }C_{ 2 }H_{ 6 }\)(g)
or \({ \triangle }_{ f }{ H }^{ \ominus }C_{ 2 }H_{ 6 }\)(g) - 787 - 858 + 1564.5 = - 80.5 k].

N₂O₄(g) + 3co(g) \(\longrightarrow \) N₂O(g) + 3CO₂(g)
Given that ; \(\triangle _{ f }\) HCO (g) = -110 KJ mol^-1; \(\triangle _{ f }\)HCO₂(g) = -393 KJ  mol^-1
\(\triangle _{ f }\)HN₂O(g) = 81 KJ mol^-1 \(\triangle _{ f }\) HN₂O₄(g) = 9.4 KJ  mol^-1
Enthalpy of reaction (\(\triangle _{ f }\)H) = [81 + 3(-393)] - [9.7 + 3(-110)] = [81 - 1179] - [9.7-330] = 778  -KJ  mol^-1

The equation are :
(i) \( \mathrm{CH}_3 \mathrm{COOH}(l)+2 \mathrm{O}_2(g) \rightarrow 2 \mathrm{CO}_2(g)+2 \mathrm{H}_2 \mathrm{O}(l) \)
\( \Delta \mathrm{H}=-867 k J \)
\( \text { (ii) } \mathrm{C}(\mathrm{s})+\mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g) \Delta \mathrm{H}=-393.5 k J\)
(ii) \(\mathrm{C}(\mathrm{s})+\mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g) \Delta H=-393.5 k J\)
(iii) \(H_2(g)+\frac{1}{2} O_2(g) \rightarrow H_2 O(l) \Delta H=-285.9 k J\)
The required equation is
\( 2 \mathrm{C}(s)+2 \mathrm{H}_2(g)+\mathrm{O}_2(g) \rightarrow \mathrm{CH}_3 \mathrm{COOH}(l) \)
\( \Delta H=\text { ? }\)
Multiply eq.(ii) by 2 and (iii) by 2 and add
\( 2 \times(i i) 2 \mathrm{C}(s)+2 \mathrm{O}_2(g) \rightarrow 2 \mathrm{CO}_2(g) \)
\(\Delta H=-787.0 k J \)
\(2 \times(i i i) 2 \mathrm{H}_2(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(\mathrm{l})\)
\(\Delta H=-571.8 k J\)
Adding (iv) \(2 \mathrm{C}(\mathrm{s})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l})\)
\(\Delta H=-1358.8 k J\)
Subtract eq.(i) from eq.(iv)
(i) \(\mathrm{CH}_3 \mathrm{COOH}(l)+2 \mathrm{O}_2(g) \rightarrow 2 \mathrm{CO}_2(g)+2 \mathrm{H}_2 \mathrm{O}(l)\)
\(\Delta H=-867.0 k J\)
Subtracting \(2 \mathrm{C}(s)+2 \mathrm{H}_2(g) \mathrm{O}_2(g) \rightarrow \mathrm{CH}_3 \mathrm{COOH}(l)\)
\(\Delta H=-491.8 k J\)

\(3 C (graphite) + 4H2(g) -> C3H8 (g)\)
\(\triangle _{ r }S=\sum { { s }_{ m }^{ o } } (products)-{ S }_{ m }^{ o }(reactants){ 3 }{ H }_{ 8 }(g)-{ 3S }_{ m }^{ o }[C(graphite)]+{ 4S }_{ m }^{ o }[{ H }_{ 2 }]\)
\(\triangle _{ r }S={ S }_{ m }^{ o }[{ C }]\)
= (270.2 - 3 x 5.70 - 4 x 130.7)JK^-1mol^-1
= (270.2 - 17.10 - 522.80) JK^-1 mol^-1
= (270.2 - 539.90)
= -269.7 JK^-1mol-1
\(\triangle _{ r }{ H }^{ o }={ \triangle }_{ f }{ H }^{ o }product-{ \triangle }_{ f }{ H }^{ o }reactants\\ \triangle _{ r }{ H }^{ o }={ \triangle }_{ f }{ H }^{ o }({ C }_{ 3 }{ H }_{ 8 })-4{ \triangle }_{ f }{ H }^{ o }[{ H }_{ 2 }(g)]\)
\( -3{ \triangle }_{ f }{ H }^{ o }[C(graphite)]\)
\(\triangle _{ r }{ H }^{ o }=-103.8KJ{ mol }^{ -1 }\)
\(\triangle _{ r }{ G }^{ o }=\triangle _{ r }{ H }^{ o }-T\triangle _{ r }{ S }^{ o }\)
\( =\left( -103.8-\frac { 298X(-269.7) }{ 1000 } \right) KJ \ { mol }^{ -1 }\)
\(=(-103.80+80.370)KJ{ mol }^{ -1 }=-23.43KJ{ mol }^{ -1 } \)

For vaporization of diethyl ether
 \(\therefore \ \triangle _{ vap }{ S }^{ o }=\frac { \triangle _{ vap }{ H }^{ o } }{ T }\)
\( \triangle _{ vap }{ H }^{ o }=26.0kJ \ mo{ l }^{ -1 }, \ T=273+35=308K\)
\(\triangle _{ vap }{ S }^{ o }=\frac { 26.0\times{ 10 }^{ 3 }J{ mol }^{ -1 } }{ 308K } =84.4J{ K }^{ -1 }{ mol }^{ -1 }\)

(i) If work done is zero, the first law reduced to \(\Delta U=-q.\) For such a system, the walls must be thermally conducting.
(ii) If W≠0 and q≠0, then from first law of thermodynamics
ΔU=q−W (as work is done by the system)
Thus, it is a closed system.

\(\triangle _{ r }{ G }^{ \ominus }\) = -RT InK, thus if \(\triangle { G }^{ \ominus }\) is less than zero i.e., it is negative, then InK will be positive and hence K will be greater than one.

On increasing temperature, entropy of a substance increases.

The net enthalpy change, ΔH  for a cyclic process, is zero as enthalpy change is a state function.

\(G=H-TS,\quad G_{ 1 }=H_{ 1 }-TS_{ 1 },{ G }_{ 2 }=H_{ 2 }-TS_{ 2 }\)
\( G_{ 2 }-G_{ 1 }=H_{ 2 }-H_{ 1 }-T({ S }_{ 2 }-S_{ 1 })\)
\( \Delta G=\Delta H-T\Delta S\)
\( \Delta S_{ total }=\Delta S_{ sys }+\Delta S_{ sure }\)
\( \Rightarrow \Delta S_{ total }=\Delta S_{ sys }-\frac { q }{ T } \)
\( \Rightarrow T\Delta S_{ total }=T\Delta S_{ sys }-q=T\Delta S_{ sys }-\Delta H\)
\(T\Delta S_{ total }=-\Delta G\Rightarrow \Delta G=-T\Delta S_{ total }\)

\({ \Delta }_{ f }{ H }^{ \Theta } \ of \ { O }_{ 2 } \ (g)=0\) by convention
\({ \Delta }_{ f }{ H }^{ - }=\Sigma { \Delta }_{ f }{ H }^{ \Theta } \ (products) \ -\Sigma { \Delta }_{ f }{ H }^{ \Theta } \ (reactants)\)
Substituting the given values
\(-2646=[4(-393.5)+4\times (-249.0)]-[{ \Delta }_{ f }{ H }^{ \Theta } \ { C }_{ 4 }{ H }_{ 8 })+0]\)
\( = -1574 \ -996] \ -{ \Delta }_{ f }{ H }^{ \Theta } \ ({ C }_{ 2 }{ H }_{ 4 })\)
\( = -2570 = \ { \Delta }_{ f }{ H }^{ \Theta } \ ({ C }_{ 2 }{ H }_{ 4 })\)
\(or \ { \Delta }_{ f }{ H }^{ \Theta } \ ({ C }_{ 2 }{ H }_{ 4 } \ = \ 2646 \ - \ 2570 \ = \ 76 \ kJ \ { mol }^{ -1 }\)

Δ_rG^∘ = −2.303 RT logK_p
Here K_p= 0.98, i.e. K_p < 1 therefore, Δ_rG^∘ is positive. Hence, the reaction is non-spontaneous.

Given, \({ N }_{ 2 }(g)+{ 3H }_{ 2 }(g)\longrightarrow 2{ NH }_{ 3 }(g); \ { \Delta }_{ r }{ H }^{ o }\)
= -92.4 kJ mol^-1 
Chemical reaction for the enthalpy of formation of NH₃ (g) is as follows.
\(\frac { 1 }{ 2 } { N }_{ 2 }(g)+\frac { 3 }{ 2 } { H }_{ 2 }(g)\longrightarrow { NH }_{ 3 }(g)\)
Therefore, \({ \Delta }_{ f }{ H }^{ o }=\frac { -92.4 }{ 2 } =-46.2 \ kJ\ { mol }^{ -1 }\)

Yes, the temperature of system and surroundings be the same when they are in thermal equilibrium.