Question
Download Solution PDFएक आदर्श गैस के n अणु कोई चक्रीय प्रक्रम ABCA (आरेख देखिए) करते है जिसमें नीचे दिए गए प्रक्रम होते हैं।
A → B : ताप T पर समतापीय प्रसार जिससे आयतन दो गुना, V1 से V2 = 2V1 और दाब P1 से P2 हो जाता है।
B → C : समदाबी संपीड़न-दाब P2 पर आरंभिक आयतन V1 तक होता है।
C → A : समआयतनिक परिवर्तन जिसमें दाब में परिवर्तन P1 से P2 होता है।
इस सम्पूर्ण चक्रीय प्रक्रम ABCA में किया गया कुल कार्य होगा :
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
We know that,
Work Done in a p-v diagram = Area of Region ABCA
Area of Region ABCA = Area of region ABV2V1 - Area of region BV2V1C
Since AB is an Isothermal process so work done in the process AB = Area of region ABV2V1 = nRTln(
Area of region BV2V1C = P2(V2 - V1) = P2V1
As process AB is Isothermal so TA= TB = T=
∴ 2P2V1 = nRT
Hence, Work Done = nRTln(2) - P2V1 = nRTln(2) -
Hence, Option 4) is the correct choice.
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