If one mole of an ideal gas at (P₁, V₁) is allowed to expand reversibly and isothermally (A to B) its pressure is reduced to one-half of the original pressure (see figure). This is followed by a constant volume cooling till its pressure is reduced to one-fourth of the initial value (B → C). Then it is restored to its initial state by a reversible adiabatic compression (C to A). The net workdone by the gas is equal to :

CONCEPT:

• Isothermal process- In the isothermal process the temperature remains constant in the thermodynamic process and work done is written as;

         \(W_{AB}=nRTln(\frac{V_1}{V_2})\)

         Here, V₁, V₂ is the volume of the thermodynamic process, T is the temperature, and R is the ideal gas constant.

• Adiabatic process- In the adiabatic process the heat or mass is not exchanged by the system and surroundings and work done is written as;   

          \(W = \frac{P_2V_2-P_1V_1}{1-\gamma}\)

        Here, V1, V2 is the volume of the thermodynamic process, and  P₁, P₂ is the pressure.

• Isochoric process- In the isochoric process, the volume is constant, and work done is written as;

​           W = PdV

       Volume is constant, W = 0   ​

CALCULATION:

As we go through A to B we can see that the temperature is constant and it is known as the isothermal process and work done is written as;

\(W_{AB}=nRTln(\frac{2V_1}{V_1})\\ W_{AB}=nRTln(2)\)

For n = 1

\(W_{AB}=RTln(2)\)

Now, as we move from B to C we can see that volume is constant and it is isochoric in nature, work done by this process is written as;

\(W_{BC}=0\)

and in C to V, it is an adiabatic process which is written as;

\(W_{CA}=P_1V_1-\frac{\frac{P_1}{4}\times {2V_1}} {1-\gamma}\)

⇒ \(W_{CA}=\frac {P_1V_1}{2{(1-\gamma)}}\)

According to the ideal gas equation,

PV = nRT

P₁V₁ = RT

Now, the total work done is written as;

W_ABCD =W_AB + W_CD

\(W_{ABCD}=RTln2+\frac{RT}{2(1-\gamma)}\)

⇒ \(W_{ABCD}=RT(ln2+\frac{1}{2(1-\gamma)})\)

Hence, option 1) is the correct answer.