(i) Total work done in an expansion when the state of an ideal gas is changed reversibly and isothermally from (p _i ,V _i) to ( p _f ,V _f). Reversible work is represented by the combined areas ABC and BCV_iV_f.

(ii) Work against constant pressure, p_f is represented by the area BCV_iV_f .Work(i) > Work (ii) 

25 percent of the gas is wasted due to inefficiency. This means that only 75% of butane gets combusted. Therefore, the energy produced by 75% combustion of butane = \(\frac { 641586\times 75 }{ 100 } \) = 481189.5 kJ
\(\therefore \) the number of days the cylinder will last.

Molecular formula of butane = C₄H₁₀
Molecular mass of butane = 4 x 12 + 10 x 1 = 58
Heat of combustion of butane = 2658 kJ mol^-1
1 mole or 58g of butane on complete combustion gives heat = 2658 kJ
\(\therefore \) 14 x 10³ g of butane on complete combustion will give heat = \(\frac { 2658\times 14\times { 10 }^{ 3 } }{ 58 } \) = 641586
The family needs 20000 kJ of heat per day.
\(\therefore \) 20000 kJ of heat is used for cooking by a family in 1 day
\(\therefore \) 641586 kJ of heat will be used for cooking by a family in =\(\frac { 641586 }{ 20000 } \) = 32days
The cylinder will last for 32days

\(\Delta\) H = q_p and C_p is heat evolved or absorbed per mole for 1^o fall or rise in temperature. Here, fall in temperature 320 - 293 = 27 K
Molar mass of CCl₂F₂ = 12 + 2 \(\times\) 35.5 + 2 \(\times\) 19
= 121 g mol^-1 
\(\therefore\) Heat evolved from 1.25 g of the sample on being cooled from 320 K to 293 K at constant pressu
=\(\frac{80.7}{121}\)\(\times\)1.25 \(\times\)27J = 22.51 J
Further, \(\Delta\)H = \(\Delta\)U + p\(\Delta\)V = -22.51 J
[\(\therefore\) p\(\Delta\)V = 1atm\(\times\) \(\frac{248-27487}{1000}\)L = -0.026L atm 
=  -0.026 \(\times\)101.325J = -2.63 J
-22.51 = \(\Delta\)U -2.63 J
or \(\Delta\)U = -22.51 +2.3 J = -19.88 J

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

100g of glucose is equivalent to 1560 kJ of energy 
Energy utilised in the event = \(\frac { 1560\times 50 }{ 100 } \) = 780
Energy left unutilised = 1560 - 780 = 780 kJ
Enthalpy of vaporisation of water = 44kJ/mol = \(\frac { 44 }{ 18 } \)kJ/g
Water needed to perspire =\(\frac { 44 }{ 18 } \) x 780 = 1906.66g

\(\mathrm{N}_2 \mathrm{H}_4+2 \mathrm{H}_2 \mathrm{O}_2 \longrightarrow \mathrm{N}_2+4 \mathrm{H}_2 \mathrm{O}\)
\( \Delta H_\text {Reachion }=\Delta H f\left(\text { product) }-\Delta H_f\right. \text { (Reactant) } \)
\(=\Delta H_f N_2+4 \Delta H f H_2 O-\left(\Delta H_{N_2 h_9+}\right. \left.2 \Delta \mathrm{H}_4 \mathrm{H}_2 \mathrm{O}_2\right) \)
\(=0+4 x-242.7-(-50.9+2 x-193.2)\)
\(=-970.8+436.8 \)
\(\Delta H_R=-534 \mathrm{~kg} / \mathrm{mL}\)
= -534.0 kJmol^-1

In reversible isothermal compression volume and pressure both are change. Thus system is reached at state 'I' (Change takes place from 3 to 1). Hence, in this change work done  W = - 2.303 nRT log10 V₂/V₁
\(\triangle U=0;\triangle H=0;\triangle S=0\)