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An ideal monoatomic gas is taken round the cycle ABCDA as shown in the P-V diagram. Compute the work done in this process.

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Let the foot of the perpendiculars drawn from points A,<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>,C and D on the volume axis be named as `A^(1), B^(1), C^(1) and D^(1)` respectively. AB,BC,CD and DA will be respectively, given by, the <a href="https://interviewquestions.tuteehub.com/tag/areas-883934" style="font-weight:bold;" target="_blank" title="Click to know more about AREAS">AREAS</a> `(+ABB^(1)A^(1)), (+BC C^(1)B^(1)), (-C C^(1)D^(1) D) and (-D D^(1)A^(1)A)`. Here, the work corresponding to expansion is positive, and that corresponding to compression is <a href="https://interviewquestions.tuteehub.com/tag/negative-570381" style="font-weight:bold;" target="_blank" title="Click to know more about NEGATIVE">NEGATIVE</a>.<br/> :. Total work done=<a href="https://interviewquestions.tuteehub.com/tag/area-13372" style="font-weight:bold;" target="_blank" title="Click to know more about AREA">AREA</a> `(+ABB^(1)A^(1))+"area" (+BC C^(1)B^(1))+ "area" (-C C^(1)D^(1) D) + "area" (-D D^(1)A^(1)A) = "area" (+ABCD)` <br/> `=(6V_(0)-3V_(0)) (5P_(0)-3P_(0))` <br/> `(3V_(0))(2P_(0))=6P_(0)V_(0)` units .</body></html>


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