InterviewSolution
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InterviewSolution
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A reversible heat engine carries `1 mol` of an ideal monatomic gas around the cycle `ABCA`, as shown in the diagram. The process `BC` is adiabatic. Call the processes `AB, BC` and `CA` as `1,2` and `3` and the heat `( DeltaQ)_(r)`, change in internal energy `(DeltaU)`, and work done `( DeltaW)_(r), r=1,2,3` respectively. The temperature at `A,B,C` are `T_(1)=300K,T_(2)=600K` and `T_(3)=455K`. Indicate the pressure and volume at `A,B` and` C` by `P_(r)` and `V_(r), r=1,2,3,` respectively. Assume that intially pressure `P_(1)=1.00atm.` Which of the following represents the correct values of the quantities indicated ? A. `(DeltaQ)=450RJ, (DeltaU)_(1)=450RJ,(DeltaW)_(1)=300RJ`B. `(Delta Q)_(2)=0, (DeltaU)_(2)=450RJ, (DeltaW)_(2)=-217.5 RJ`C. `(DeltaQ)_(3)=0, (DeltaU)_(3)=-232.5RJ,(DeltaW)_(3)=0`D. `(DeltaQ)_(1)=450RJ, (DeltaU)_(3)=-232.5RJ, (DeltaW)_(2)=217.5RJ` |
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Answer» Correct Answer - D Process 1 is isochoric. Now work is done `Dw=0` Hence, `(DeltaQ)_(1)=dU=C_(V)(T_(2)-T_(1))` `=(3R)/(2)[600-300]=450RJ` Process 2 is adaibatic expansion `(Delta Q)_(2)=0,(DeltaW)_(2)=-(DeltaU)_(2)` `(DeltaW)_(2)=-[C_(v)(T_(3)-T_(2))]` `=-[(3R)/(2)(455-600)]` `=(3xx145)/(2)R=217.5RJ` `(DeltaV)_(3)=C_(v)-217.5RJ` Process 3 is isobaric compression. Value of pressure `= 1 atm = 1.013 xx 10^(5)N//m^(2)` `(DeltaW)_(3)=1.013xx10^(5)(V_(c)-V_(A))J=0` `(DeltaW)_(3)=C_(v)(300-455)=(-3R)/(2) xx 155=-232.5R` |
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