(P₁, V₁) पर किसी आदर्श गैस के 1 मोल को उत्क्रमणीय और समतापीय (A से B तक) स्थिति में फैलने दिए जाने पर इसका दाब अपने मूल दाब का आधा हो जाता है (आरेख देखिए)। इसके पश्चात् नियत आयतन पर, इसको अपना दाब आरम्भिक दाब का एक चौथाई होने तक (B →C), ठंडा किया जाता है। इसके पश्चात् उत्क्रमणीय रुद्धोष्म संपीडन द्वारा इसको आरम्भिक अवस्था (C से A) पर लाया जाता है। गैस द्वारा किया गया कुल कार्य है :

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.