The diagram shows a P−V graph of a thermodynamic behaviour of an ideal gas. Find out from this graph (i) work done in the process A→B,B→C,C→D and D→A, (ii) work done in the complete cycle A→B→C→D→A
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Solution
(i) Work done in the process A→B (the process is expansion, hence work done by the gas) =−P×dV=−12×105×5×10−3 =−6000J Work done in the process B→C is zero as volume remains constant. Work done in the process C→D (the process is contraction hence work done on the gas) =P×dV=2×105×6×10−3 =1000J (ii) Work done in the process D→A is zero as volume remains constant. Net work done in the whole cycle =−6000+1000=−5000J i.e., net work is done by the gas.