A) \[{{Q}_{1}}>{{Q}_{2}}>{{Q}_{3}}\] and\[\Delta {{U}_{1}}=\Delta {{U}_{2}}=\Delta {{U}_{3}}\]
B) \[{{Q}_{3}}>{{Q}_{2}}>{{Q}_{1}}\] and \[\Delta {{U}_{1}}=\Delta {{U}_{2}}=\Delta {{U}_{3}}\]
C) \[{{Q}_{1}}={{Q}_{2}}={{Q}_{3}}\] and \[\Delta {{U}_{1}}>\Delta {{U}_{2}}>\Delta {{U}_{3}}\]
D) \[{{Q}_{3}}>{{Q}_{2}}>{{Q}_{1}}\] and \[\Delta {{U}_{1}}>\Delta {{U}_{2}}>\Delta {{U}_{3}}\]
Correct Answer: A
Solution :
[a] AU is path independent and depends only on initial & find state as the initial & final states in 3 processes are same \[-\text{ }A{{U}_{1}}=\Delta {{U}_{2}}=\Delta {{U}_{3}}\] While \[W\text{ }=\text{ }P\Delta V=\]area under P ? V graph \[Area=curv{{e}_{1}}>curv{{e}_{2}}>curv{{e}_{3}}\] \[{{W}_{1}}>{{W}_{2}}>{{W}_{3}},\,and\,\Delta {{U}_{1}}A{{U}_{2}}\,\Delta {{U}_{3}}\] Hence \[\Delta {{Q}_{1}}\,>\Delta {{Q}_{2}}>\Delta {{Q}_{3}}\]You need to login to perform this action.
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