(i) For changes to be isothermal walls of container must be perfectly conducting and the changes must be infinitely slow. While for adiabatic changes the walls of container must be perfectly insulating and the changes must be fast.
(ii) For isothermal changes P₂V₂ = P₁ V₁
\(\Rightarrow \quad \frac{P_{2}}{P_{1}}=\frac{V_{1}}{V_{2}}=\frac{V_{1}}{\frac{V_{1}}{2}}=2\) 
(iii) For adiabatic changes \(P_{2} V_{2}^{\gamma}=P_{1} V_{1}^{\gamma}\) 
\(\frac{P_{2}}{P_{1}}=\left(\frac{V_{1}}{V_{2}}\right)^{\gamma}=(2)^{1.4}=2.64\) 
(iv) Here \(V_{2}=\frac{V_{1}}{4}\) 
T₁ = 27 °C = 27 + 273 = 300 K
\(\gamma\) = 1.5, T² - T¹ = ?
As the change is adiabatic, therefore
\(\mathrm{T}_{2} \mathrm{~V}_{2}^{\gamma-1}=\mathrm{T}_{1} \mathrm{~V}_{1}^{\gamma-1}\) 
or \(\mathrm{T}_{2}=\mathrm{T}_{1}\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma-1}\) 
= 300(4)^15-1 = 300 x 2 = 600 K.
(v) 

(vi) Isothermal changes are
(i) Conversion of ice at 0 °C to water at 0 °C
(ii) Conversion of water at 100°C to steam at 100°C.

(i) Temperature, Zeroth law states that when the thermodynamic systems A and B are separately in thermal equilibrium with a third thermodynamic system C, then the system A and B are in thermal equilibrium with each other.
(ii) Thermodynamic process in which state variables of the thermodynamic system are infinitely slow is called quasi-static process. It represents that the system is in thermal and mechanical equilibrium with its surrounding.
(iii) Equation of state represents the connection between the state variables of a system. In adiabatic process. For a fixed amount of gas the State variables temperature, pressure and volume are related as follows.
\(P V^{\gamma}\) = constant, or \(T V^{\gamma-1}\) = constant, or \(P^{1-\gamma} T^{\gamma}\) = constant.
(iv)  During isothermal change \(\Delta T=0,\) Since \(C=\frac{\Delta Q}{M \Delta T}\) , therefore C = \(\infty .\) Specific heat of gas is infinite. and during adiabatic change \(\Delta Q=0\) therefore specific heat of gas C, is zero.
(v) Under isothermal condition
\(K_{i}=\frac{\Delta P}{\frac{\Delta V_{1}}{V}}=P\)          .......(i)
Under adiabatic conditions
\(K_{a}=\frac{\Delta P}{\frac{\Delta V_{2}}{V}}=\gamma P\)        .......(ii)
from eq. we have \(\frac{\Delta V_{1}}{\Delta V_{2}}=\gamma, \text { as } \gamma>1\) 
\(\therefore \quad \Delta V_{1}>\Delta V_{2}\) 
(vi) \(\mathrm{W}=\frac{\mathrm{R}\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)}{(1-\gamma)}, \text { where } \gamma=\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}\) 

In an isothermal process temperature ofthe gas remains constant while pressure and volume changes. Also, internal energy of the gas remains constant if it is an ideal gas.
(ii) In adiabatic process, dQ = 0, i.e., Heat absorbed is zero.
From first law of thermodynamics dQ = dU + dW
\(\Rightarrow\) dU = -dW
= -(-200 J)
= 200 J
i.e., internal energy increases by 200 J.
(iii) From the first law of thermodynamics dU = dQ - dW, since dW = 0 and dQ > O. Hence the change in the internal energy dU> 0 and this increases the internal energy and also temperature.
(iv) Heat and work are two different modes of changing state of a thermodynamic system and hence changing internal energy of the system which is a state variable.
(v) When heat supplied to a system dQ = +ve, while heat is drawn from the system dQ = -ve. When work is done by the system dW =+ve. While work done on the system dW = -ve when temperature increases internal energy dU = +ve and when temperature decreases dU = -ve.
(vi) In an isothermal process dT = 0 so dU = 0
In adiabatic process dQ = 0 therefore dU = -dW i.e., Ans. equal to work done while in cyclic process dU= 0 over a complete cycle.

(i) Molar specific heat of an deal gas at constant pressure (C_p ) and volume (C_v ) relation is given by Mayer's formula as C_p - C_v = R. Where R is universal gas constant.
(ii) First law of thermodynamics is basically the law of conservation of energy.
(iii) Kelvin Planck statement: It is impossible to construe. a heat engine which would absorb heat from a reservoir and convert 100% of the heat absorbed into work.
Clausius statement: It is impossible to design a self-acting machine unaided by any external agency, which would transfer heat from a body at a lower temperature to another body at a higher temperature.
(iv) Thermal efficiency of a heat engine is defined as the ratio of net work done per cycle by the engine to the total amount of heat absorbed per cycle by the working substance from the source.
\(\eta=\frac{\frac{\text { Net workdone }}{\text { Cycle }}(W)}{\frac{\text { Total amount heat absorbed }}{\text { cycle }\left(Q_{1}\right)}}=\frac{Q_{1}-Q_{2}}{Q_{1}}\) 
(v) Cofficient of performance of refrigerator \(\text { (} \beta \text {) }\)is defined as the ratio of quantity of heat removed per cyle from the contents of the refrigerator (Q₂) to the energy spent per cycle (W) to remove this heat. \(\beta=\frac{Q_{2}}{W}=\frac{Q_{2}}{Q_{1}-Q_{2}}\) 
\(\beta=\frac{1-\eta}{\eta}\) 
(vi) Efficiency of heat engine \(\eta=1-\frac{T_{2}}{T_{1}}, \eta\) = 100%, if T₂ = 0 K or T₁ = \(\infty\). Since both these conditions cannot be attained practically, a heat engine cannot have 100% efficiency.

(i) From equation of State at A
\(P_{A} V_{A}=\eta R T_{A} \Rightarrow T_{A}=\frac{P_{A} V_{A}}{\eta R}\) 
or \(T_{A}=\frac{4.15 \times 10^{4} \times 10}{500 \times 8.3}\) = 100 K
(ii) Isochoric process is that which occurs at constant volume.
for isochoric Process A \(\rightarrow\) B.
\(\frac{T_{B}}{T_{A}}=\frac{P_{B}}{P_{A}}=\frac{8.3 \times 10^{4}}{4.15 \times 10^{4}}\) = 2
Thus T_B = 2T_A = 2 x 100K = 200K
(iii) Isobaric process is that which occurs at constant pressure.
For isobaric process B \(\rightarrow\) C
\(\frac{T_{C}}{T_{B}}=\frac{V_{C}}{V_{B}}=\frac{20}{10}\) = 2
Thus T_C = 2T_B = 400 K
For isobaric process A \(\rightarrow\) D
\(\frac{T_{D}}{T_{A}}=\frac{P_{D}}{P_{A}}=\frac{20}{10}\) = 2
Thus T_D = 2T_A = 200 K.
(iv) Workdone in process ABC is
W= W_AB + W_BC = 0 + P_B(V_C- V_B)
= 8.3 x 10⁴ (20 - 10) = 8.3 x 10³ J