InterviewSolution
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InterviewSolution
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The figure shows the P-V plot of an ideal gas taken through a cycle ABCDA. The part ABC is a semi-circle and CDA is half of an ellipse. Then, A. the process during the path `AtoB` is isothermalB. work done during the path `BtoCtoD`C. heat flows out of the gas during the path `BtoCtoD`D. the process during the path `AtoB` is isothermal |
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Answer» Correct Answer - B::D (b,d) In case of an isothermal process we get a rectangular hyperbola in a P-V diagram. Therefore option (a) is wrong. `T_DltT_B.` Therefore in process `BtoCtoD, DeltaU` is negative. PV decreases and volume also decrease, therefore W is negative. From first law os thermodynamic, Q is negative i.e. There is a heat loss option (b) is correct. `W_(AB)gtW_(BC).` Therefore work done during path `AtoBtoC` is positive, option (c) is wrong Work done is clockwise cycle in a PV diagram is positive. Option (d) is correct. |
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