A) 15%
B) 50%
C) 20%
D) 25%
Correct Answer: B
Solution :
Maximum efficiency of reversible process is given by \[\eta =1-\frac{T}{T}\] At \[A{{P}_{0}}{{V}_{0}}=nR{{T}_{0}}\] At \[B2{{P}_{0}}{{V}_{0}}=nRT\] \[T=2{{T}_{0}}\] \[\therefore \] \[\eta =1-\frac{{{T}_{0}}}{2{{T}_{0}}}=\frac{1}{2}=50%\]You need to login to perform this action.
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