A thermodynamic system is taken from an original state A to an intermediate state B by a linear process as shown in the figure. It's volume is then reduced to the original value from B to C by an isobaric process. The total work done by the gas from A to B and B to C would be :
(A)33800 J
(B)2200 J
(C)600 J
(D)1200 J
Solution
Work done in process AB equals the area under the P-V curve: WAB=21(8000+6000)×4=28000 (in appropriate units). Work done in process BC (isobaric compression): WBC=−4000×4=−16000. Total work = 28000−16000=12000 (in appropriate units). Converting units from Dyne/cm² to Pa and accounting for volume in m³ gives 800 J. The source answer was marked as BONUS, indicating an issue with the question or options.
JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Single correct option
