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A thermodynamic system undergoes cyclic process `ABCDA` as shown in figure. The work done by the system is A. `P_0V_0`B. `2P_0V_0`C. `(P_0V_0)/(2)`D. zero |
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Answer» Correct Answer - D `DeltaW_(AODA)=(1)/(2)(2P_0-P_0)(2V_0-V_0)=(1)/(2)P_0V_0` `AODA` is clockwise cycle on `P-V` diagram, `DeltaW=+ve` `DeltaW_(COBC)=-(1)/(2)(3P_0-P_0)(2V_0-V_0)=(1)/(2)P_0V_0` `COBC` is anticlockwise cycle on `P-V` diagram, `DeltaW=-ve` `DeltaW_(ABCDA)=DeltaW_(AODA)+DeltaW_(COBC)=0` |
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