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A gas (gamma=1.5) undergoes a cycle of adiabatic, isobaric and isochoric processes in an order. If the volume of the gas is doubled in the adiabatic process then the efficiency of the cycle is approximately, |
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Answer» `18%` In isochoric process, `(p_(A))/T_(A)=p_(B)/T_(B)` As work done in ADIABATIC process, `W=(nR(T_(1)-T_(2)))/(y-1)=(R(T-T/sqrt(2)))/((1.5-1)) ""("Taking, " n=1)` `V_(2)=2V_(1)` `rArr ""T_(2)=T_(1)/sqrt(2) " or "T_(2)=T/sqrt(2)` `W_(t)=RT 2(1-1/sqrt(2))=0.58 ""(because r=i " Given ")` Work done in isobaric process, `""W_(2)=pDeltaV=nRDeltaT` `rArr W=nR(T/(2sqrt(2))-T/sqrt(2))=-RT(1/sqrt(2)-1/(2sqrt(2)))=-0.35` Now, heat given to system `Q_(1)=nC_(p)DeltaT=7/2R(1/sqrt(2)-1/(2sqrt(2)))=-1.23 J` Negative, `Q_(1)` shows that heat is given to the system. Efficiency of the CYCLE, `""eta="net work done"/"heat input"=(W_(1)+W_(2))/Q_(1)` `""=(0.58-0.35)/1.23` `rArr ""eta=0.23/1.23 TIMES 100 =18.6%=18%` Hence, the efficiency of cycle is 18%. |
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