ઉપર દર્શાવેલ ત્રણ કિસ્સાઓમાં બતાવ્યા પ્રમાણે, એક આદર્શ વાયુ બિંદુ \(A\) થી શરૂ થઈને અને પાથ \(\mathrm{A} \rightarrow \mathrm{B} \rightarrow \mathrm{C} \rightarrow \mathrm{D} \rightarrow \mathrm{A}\) ને અનુસરીને તે જ બિંદુ પર પાછા ફરીને ચક્રીય રૂપાંતરણમાંથી પસાર થાય છે.
\(\Delta \mathrm{U}\) સંબંધિત સાચો વિકલ્પ પસંદ કરો :

A \(\Delta \mathrm{U}(\) Case-I \()=\Delta \mathrm{U}(\) Case-II \()=\Delta \mathrm{U}(\) Case-III \()\)
B \(\Delta \mathrm{U}(\) Case-I \()\gt\Delta \mathrm{U}(\) Case-III \()\gt\Delta \mathrm{U}(\) Case-II \()\)
C \(\Delta \mathrm{U}(\) Case-III \()\gt\Delta \mathrm{U}(\) Case-II \()\gt\Delta \mathrm{U}(\) Case-I \()\)
D \(\Delta \mathrm{U}(\) Case-I \()\gt\Delta \mathrm{U}(\) Case-II \()\gt\Delta \mathrm{U}(\) Case-III \()\)

Verified Solution

Answer & Solution

Correct Answer

(A) \(\Delta \mathrm{U}(\) Case-I \()=\Delta \mathrm{U}(\) Case-II \()=\Delta \mathrm{U}(\) Case-III \()\)

Step-by-step Solution

Detailed explanation

જેમ કે આંતરિક ઊર્જા ' \(U\) ' એ અવસ્થા વિધેય છે, ચક્રીય પ્રક્રિયામાં તેનું ચક્રીય સંકલન શૂન્ય હોવું જોઈએ. \(\therefore \Delta U \text { case }(I)=\Delta U \text { case }(\text { II })=\Delta U \text { case }(\text { III })\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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