(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) \[{{W}_{b\to c}}+{{W}_{d\to a}}=0\] |
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 \[{{W}_{a-b}}+{{W}_{b-c}}+{{W}_{c-d}}+{{W}_{d-a}}>0\] \[\because \] \[{{W}_{b-c}}={{W}_{d-a}}\] \[\therefore \] \[{{W}_{b-c}}+(-){{W}_{d-a}}=0\]You need to login to perform this action.
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