InterviewSolution
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InterviewSolution
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100 moles of an ideal monatomic gas undergoes the thermodynamic process as shown in the figure `ArarrB:` isothermal expansion `BrarrC:` adiabatic expansion `CrarrD:` isobaric compression `DrarrA:` isochoric process The heat transfer along the process AB is `9xx10^4J`. The net work done by the gas during the cycle is [Take `R=8JK^-1mol^-1`]A. (a) `-0.5xx10^4J`B. (b) `+0.5xx10^4J`C. (c) `-5xx10^4J`D. (d) `+5xx10^4J` |
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Answer» Correct Answer - D `Q_(BC)=0` `Q_(CD)=nC_pDeltaT` `=n(5/2R)(T_D-T_C)` `=5/2(p_DV_D-p_CV_C)` `=5/2(10^5-2xx10^5)` `=-25xx10^4J` `Q_(DA)=nC_VDeltaT` `=n(3/2R)(T_A-T_D)` `=3/2(p_AV_A-p_DV_D)` `=3/2(2.4xx10^5-10^5)` `=21xx10^4J` Now, `W_(n et)=Q_(n et)` `=(9-25+21)xx10^4J=5xx10^4J` |
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