AMU Medical AMU Solved Paper-2012

  • question_answer
    Refer to the Carnot cycle of an ideal gas shown in the figure. Let \[{{W}_{a\to b}},{{W}_{b\to c}},{{W}_{c\to d}}\] and \[{{W}_{d\to a}}\]represent the work done by the system during the processes \[a\to b,\,b\to c,\,c\to d\] and \[d\to a\]respectively. Consider the following relations
    (i) \[{{W}_{a\to b}}+{{W}_{b\to c}}+{{W}_{c\to d}}+{{W}_{d\to a}}>0\]
    (ii) \[{{W}_{a\to b}}+{{W}_{b\to c}}+{{W}_{c\to d}}+{{W}_{d\to a}}<0\]
    (iii) \[{{W}_{a\to b}}+{{W}_{c\to d}}>0\]
    (iv) \[{{W}_{b\to c}}+{{W}_{d\to a}}=0\]
                    Which of the above relation(s) is/are true?

    A)  (i) and (iii)          

    B)  (iii) and (iv)

    C)  (i) (iii) and (iv)                  

    D)  (ii),(iii) and (iv)

    Correct Answer: C

    Solution :

                    \[{{W}_{a-b}}\] is isothermal \[{{W}_{b-c}}\] is adiabatic \[{{W}_{c-d}}\] is isothermal (Negative) \[{{W}_{d-a}}\] is adiabatic (Negative) Net work done during the complete cycle                 \[W={{W}_{a-b}}+{{W}_{b-c}}+(-{{W}_{c-d}})+(-{{w}_{d-a}})\]                 \[W={{W}_{a-b}}-{{W}_{c-d}}\]        \[[As\,{{W}_{b-c}}={{W}_{d-a}}]\] i.e.,        \[{{W}_{a-b}}+{{W}_{c-d}}>0\] and        \[{{W}_{a-b}}+{{W}_{b-c}}+{{W}_{c-d}}+{{W}_{d-a}}>0\] \[\because \]     \[{{W}_{b-c}}={{W}_{d-a}}\] \[\therefore \]  \[{{W}_{b-c}}+(-){{W}_{d-a}}=0\]


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