InterviewSolution
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InterviewSolution
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Figure shows a cycle `ABCDA` undergone by `2` moles of an ideal diatomic gas. The curve `AB` is a rectangular hyperbola and `T_(1)=300K` and `T_(2)=500K`. Determine the work done by the gas in the process `ArarrB`. A. `-3.320kJ`B. `4.326kJ`C. `2.326kJ`D. `3.326kJ` |
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Answer» Correct Answer - A Evidently, for the process `ArarrB` `Vprop(1)/(T) ` or `VT`=constant (say,K) or `TdV+VdT=0`or `dV=-(VdT)/(T)` (a) Now, work done by the gas in a process is given by `W=SigmaPdV` or `W_(AB)=Sigma(nRTdV)/(V)=Sigma(-nRdT)` `W_(AB)=overset(500)underset(300)(int)-nRdT=-(2 mol) (8.314 J//mol-K)` `[500-300]K=-3.326kJ` |
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