According to first law of thermodynamics
dQ = dU + dW
or Q = U_f - U_i + W
U_f = internal energy
in final state and U, internal energy in initial state For path i a f.
Q = +50 cal, and W = 20 cal
∵  U_f - U_i = Q - W = 50 - 20 = 30 cal
Here it should be remembered that the change in internal energy between i and f state remains the same i.e., 20 cal. whatever path is followed.
(i) For path ibf,
Q = 36 cal. and dU = U_f - U_i = 30 cal
W = Q - (U_f - U_i) = 36 - 30 = 6 cal.
(ii) For path f_i,
W = -13 ca!. dU = 30 cal
∴ Q = W + (U_f - U_i) = .13 - 30 = -43 cal.
(iii) U_i = 10cal
dU = U_f - U_i = 30
∴ U_f = 30 + U_i = 30 + 10 = 40 cal.
(iv) For process bf, volume is constant i.e., workdone is zero
∴ Q dU = U_f - U_b = 40 - 22 = 18 cal.
For path ib,
∴ Q = Q_ibf - Q_bf = 36 - 18 = 18 cal.

We know that
\(β={Q\over W}={Q_2\over Q_1-Q_2}={T_2\over T_1-T_2}\)
Here, 
T₁ = 300 K, T₂ = 270K
W = 250 W = 250 Js-1
Q ?, t = ?
\(∴\ Q_2=Wβ=W\left(T_2\over T_1-T_2\right)\)
\(=250\left(270\over 300-270\right)\)
\(=250\times{270\over 30}\)
= 2250 J S^-1
(i) Let Q be the heat extracted from the freezing compartment in 10 minutes
∴ Q = Q₂ x 10 min = 2250 x 10 x 60
= 1350000 J
\(={135\times10^4\over 4.2\times10^3}kcal = 321.4 kcal\)
(ii) Heat required to convert 1 kg of water at 273 K into ice,
Q' = m x L = 1 x 80 kcal
= 80 x 4.2 x 10³ J
Let Q' be extracted in a time t.
∴ Rate of extraction of heat from freezing compartment
\(={80\times4.2\times10^3\over t}J S^{-1}\)
This rate must be equal to Q₂
i.e., \(2250={80\times4.2\times10^3\over t}\)
\(∴\ \ t={80\times4.2\times10^3\over 2250}=149.33s\) 

Efficiency , \(\eta =1-\frac { { T }_{ 2 } }{ { T }_{ 1 } } \)
where, T₂  = sink temperature
T₁ = source temperature.
\(1-\frac { { T }_{ 2 } }{ { T }_{ 1 } } =\frac { 1 }{ 2 } \quad ...(i)\)
\(\\ 1-\left( \frac { { T }_{ 2 }-100 }{ { T }_{ 1 } } \right) =\frac { 2 }{ 3 } \ ...(ii)\)
From Eq. (i), \(\frac { { T }_{ 2 } }{ { T }_{ 1 } } =\frac { 1 }{ 2 } \) and Eq. (ii)
\(\frac { { T }_{ 2 }-100 }{ { T }_{ 1 } } =\frac { 1 }{ 3 } \)
On dividing, we get 
 \(\frac { { T }_{ 2 } }{ { T }_{ 2 }-100 } =\frac { 3 }{ 2 } \Rightarrow { T }_{ 2 }=300K\) 
(ii) Substituting in Eq.(i) T₁ = 600K
(iii) As efficiency, \({ \eta }_{ 2 }\Rightarrow 1-\frac { { T }_{ 2 } }{ { T }_{ 1 } } \)
It equals to 1 only when \(\frac { { T }_{ 2 } }{ { T }_{ 1 } } =0\quad or\quad { T }_{ 2 }=0K\)
But absolute zero is not possible.

(a). As there is no work done by or on the system, also there is no heat interaction, therefore internal energy of the system remains the same.
Change in internal energy, \(\triangle \)U = 0, as work is done on or by the gas.
(b). The expansion of the gas is not due to any external work done on the system and hence, temperature of the system will not change.
(c). Free expansion is a very fast process and it cannot be controlled. As the intermediate states are non-equilibrium states, hence they can’t be on the P−V−T surface of the system.

неге, \(\frac{V_1}{V_2}=\frac{1}{6}, \gamma=1.4\),
\(T_1=327^{\circ} \mathrm{C}=327+273=600 \mathrm{~K}\)
(a) For an adiabatic change,
\( T_2 V_2^{\gamma-1}=T_1 V_1^{\gamma-1} \)
\(T_2=T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}=600\left(\frac{1}{6}\right)^{1.4-1} \)
\( =600 \times 0.4884=293 \mathrm{~K}=293-273  =20^{\circ} \mathrm{C}\)
This is lowest temperature.
(b) Net work done during each cycle by 1 mole of air.
\( W=R T_1 \log _e \frac{V_2}{V_1}-R T_2 \log _e \frac{V_3}{V_4} \)
\(=R\left(T_1-T_2\right) \log _e \frac{V_2}{V_1}\left(\therefore \frac{V_3}{V_4}=\frac{V_2}{V_1}\right) \)
\( =2.303 R\left(T_1-T_2\right) \log _{10} \frac{V_2}{V_1} \)
\(=2.303 \times 8.31(600-293) \log _{10} 6 \)
\( W=2.303 \times 8.31 \times 307 \times 0.7782  =4572.2 J\)
(C) \(\eta=1-\frac{T_2}{T_1}=1-\frac{293}{600}=\frac{307}{600}\)
\(=0.512=51.2 \%\)