One mole of an ideal diatomic gas is taken through a cyclic process ABCDA as shown in the figure. The thermodynamic efficiency of cycle is (nearly)
Options:
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10% |
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20% |
|
|
50% |
|
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25% |
Asked by dipanshusingla029
31st May 2018,
11:07 AM
Answered by Expert
Answer:
Process along AB and CD are constant volume process. Hence no work is done along these path.
Workdone along the path BC = 2P0×(2V0-V0) = 2P0×V0
Workdone along the path DA = P0×(2V0-V0) = P0×V0
efficiency = P0×V0 / (2P0×V0) = 1/2 = 50 %
Answered by Expert
31st May 2018,
2:40 PM
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