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.

T₁ 100 °C = 373 K and T₂ = 0 °C = 273 K
\(η={T_1-T_2\over T_1}={373-273\over 373}={100\over 373}=0.268\)
As we want to increase its efficiency by 20%, hence new efficiency is
η' = 26.8% + 20% = 46.8%
(a) If keeping temperature of colder reservoir fixed the temperature of hot reservoir is changed to T_I', then
\(46.8={T_1'-273\over T_1'}\times100\)
⇒ 46.8 T₁' = 100   T₁' - 27300
⇒ 53.2 T₁' = 27300 or \(T_1'={27300\over 53.2}=513.2K\)
∴ T₁' - T₁ = 513.2 - 373 = 140.2 K
It means that temperature of hot reservoir be raised by 140.2 K.
(b) If keeping the temperature of hot reservoir fixed, the temperature of colder reservoir is changed to T₂', then
\(46.8={T_1-T_2'\over T_1}\times100={373-T_2'\over 373}\times100\)
∴ 373 x 46.8 = 373 x 100 - 100 T₂'
⇒ 100 T₂ = 373 x (100 - 46.8) = 373 x 53.2
⇒ T₂' = \({373\times53.2\over 100}=198.4K\)
∴ T₂ - T²' 273 - ]98.4 = 74.6 K = 74.6 ⁰C.
It means that temperature of colder reservoir be lowered by 74.6 ⁰C.

The processes are shown in Fig.
The process B to C is at constant volume, hence

\({P_3\over P_2}={T_3\over T_2}\)
or \(T_2=T_3\times{P_2\over P_3}\)
\(=366\times{7\times10^4\over 1.1\times10^5}\)
= 233K
The process A to B is adiabatic, hence
\({T_1\over T_2}=\left(P_1\over P_2\right)^{γ-1/γ}\)
or \({366\over 233}=\left(3.5\times10^5\over 7\times10^4\right)^{(γ-1)/γ}\)
Solving we get ⋎ = 1.39
Now γ = 1.39
Now \({C_P\over C_V}=1.39\ \ (∵\ ⋎=C_P/C_V)\)
Again  C_P - C_V R or C_P - C_V= 8.3
From eq. (1) C_P = 1.39 C_V
Substituting the value of C_P in eq. (2), we have
1.39 C_v - C_p 8.3 or 0.39 C_v = 8
∴  \(C_V={8.3\over 0.39}=21.28\ J mol^{-1}K^{-1}\)
Now  C_P = 8.3 + C_V = 8.3 + 21.28 = 29.58 J mol^-1.
Hence C_P = 29.58 J mol^-1 K^-1 and C_V = 21.28 J mol^-1 K^-1

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., \(2250={80\times4.2\times10^3\over t}\)
\(∴\ \ t={80\times4.2\times10^3\over 2250}=149.33s\) 

(a) Let capacity of each cylinder be V and atmospheric pressure be p.
p₁ = p
Initial volume of gas = Volume of cylinder A
V₁ = V
When stopcock is opened, then volume available for gas becomes 2V
V₂ = 2V
Final pressure (p₂) = ?
As system is thermally insulated, therefore there is no change in temperature during the process and hence it is an isothermal process.
For an isothermal process (according to Boyle's law),
p₁V₁ = p₂ V₂
or    \({ p }_{ 2 }={ p }_{ 1 }\frac { { V }_{ 1 } }{ { V }_{ 2 } } =pX\frac { V }{ 2V }\)
\( \\ =\frac { p }{ 2 } =\frac { 1 }{ 2 } atm=0.5\ atm\)
(b) Change in internal energy, \(\triangle \)U = 0, as work is done on or by the gas.
(c) Change in temperature of the gas is zero as gas does no work in expansion.
(d) No, because free expansion of gas is rapid and cannot be controlled. The intermediate states are non-equilibrium states and so not satisfy the gas equation. Therefore, the intermediate state of the gas does not be on the p - V - T surface.

Given, Q₂ = 263J/s,T₂ = -10⁰ C = -10 + 273 = 263K
and T₁ = 25⁰C = 25 + 273 = 298K,
\(\beta =\frac { { Q }_{ 2 } }{ W } =\frac { { T }_{ 2 } }{ { T }_{ 1 }-{ T }_{ 2 } } \) 
\(\\ \Rightarrow Average\ power,\ W=\frac { { Q }_{ 2 }({ T }_{ 1 }-{ T }_{ 2 }) }{ { T }_{ 2 } } =\frac { 263(298-263) }{ 263 }\)
\( \\ =35J/s=35W\)

Here, T₁ = 500K, T₂ = 300K, W = 1KJ = 1000J
As, efficiency, \(\eta =\frac { W }{ Q } =\frac { { T }_{ 1 }-{ T }_{ 2 } }{ { T }_{ 1 } } \)
So, heat transferred to engine by the reservoir in cycle
\(Q_{ 1 }=\frac { W{ T }_{ 1 } }{ { T }_{ 1 }-{ T }_{ 2 } } =\frac { 1000\times 500 }{ 500-300 } =2500J\quad or\quad 2.5kJ\)
and heat transferred by the engine to the cold reservoir in one cycle
Q₂ = Q₁ - W  = 2.5kJ-  1kJ = 1.5kJ

Given, Mass of water, m = 1g = \(1\times { 10 }^{ -3 }kg\)
\(Hence,\ pressure\ is\ p=1.013\times { 10 }^{ 5 }N/{ m }^{ 2 }\)
\(\\ Volume \ of \ steam,{ V }_{ s }=1551{ cm }^{ 3 }=1551\times { 10 }^{ -6 }{ m }^{ 3 }\)
\(\\ Volume \ of \ water, \ V_{ w }=\frac { mass }{ density } =\frac { 1\times { 10 }^{ -3 } }{ { 10 }^{ 3 } } =2\times { 10 }^{ -6 }{ m }^{ 3 }\)
\(\\ First \ law \ of \ thermodynamics \ gives\)
\(\\ \triangle Q=\triangle U+p\triangle V\)
\(\\ \Rightarrow mL=\triangle U+p({ V }_{ s }-{ V }_{ w })\)
\(\\ \therefore \ change \ in \ internal \ energy \ is\)
\(\\ \triangle U=mL-p({ V }_{ s }-{ V }_{ w })\)
\(=1\times { 10 }^{ -3 }\times 2256\times { 10 }^{ 3 }-1.013\times { 10 }^{ 5 }\times (1551\times { 10 }^{ -6 }-{ 10 }^{ -6 })\)
\(=2256-0.1013\times 1550\cong 2099J\)

In an isobaric process, p= constant
\(\therefore C={ C }_{ p }\)
\(\\ and\ \frac { \triangle Q }{ \triangle U } =\frac { n{ C }_{ p }\triangle T }{ n{ C }_{ v }\triangle T } =\frac { { C }_{ p } }{ { C }_{ v } } =\gamma \)
\(\\ also\ \ \frac { \triangle Q }{ \triangle U } =\frac { \triangle Q }{ \triangle Q-\triangle U } \)
\(\\ =\frac { n{ C }_{ p }\triangle T }{ n{ C }_{ p }\triangle T-n{ C }_{ V }\triangle T } =\frac { { C }_{ p } }{ { C }_{ p }-{ C }_{ v } } =\frac { \gamma }{ \gamma -1 } \)