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Find the efficiency of the thermodynamics cycle shown in figure for an ideal diatomic gas. |
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Answer» Solution :Let n be the number of moles of the gas and the temperature be `T_(0)` in the state A. Now, work DONE during the CYCLE `W=1/2xx(2V_(0)-V_(0))(2P_(0)-P_(0))=1/2P_(0)V_(0)` For the heat `(DeltaQ_(1))` given during the process `A rarr B`, we have `DeltaQ_(1)=DeltaW_(AB)+DeltaU_(AB)` `DeltaW_(AB)`=area under the STRAIGHT line AB `=1/2(P_(0)+2P_(0))(2V_(0)-V_(0))=(3P_(0)V_(0))/2` APPLYING equation of state for the gas in the state A&B. `(P_(0)V_(0))/T_(0)=((2P_(0))(2V_(0)))/T_(B) rArr T_(B)=4T_(0)` ` :. U_(AB)=nC_(v)DeltaT=n[(5R)/2](4T_(0)-T_(0))=(15nRT_(0))/2=(15P_(0)V_(0))/2` ` :. DeltaQ_(1)=3/2P_(0)V_(0)+15/2P_(0)V_(0)=9P_(0)V_(0)` Obviously, the processes. `BrarrC and CrarrA` involve the abstraction of heat from the gas. `"Efficiency"=("Work done per cycle")/("Total heat supplie per cycle")` i.e., `eta=(1/2P_(0)V_(0))/(9P_(0)V_(0))=1/18` |
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