Thermodynamics

JEE Main level, numerical answer, medium

63 of 190

Question 63 of 190

A thermodynamic system is taken through the cyclic process ABCA B C as shown in the figure. The total work done by the system during the cycle ABCA B C is ____\_\_\_\_ J.

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Answer: 300

From the P-V diagram: A(2,100), B(5,300), C(5,100) in Pa and m³ units.
Process A→B is linear. Equation: P=100+2003(V−2)P = 100 + \frac{200}{3}(V-2)
WAB=∫25200V−1003dV=13[100V2−100V]25=18003=600W_{AB} = \int_2^5 \frac{200V - 100}{3} dV = \frac{1}{3}[100V^2 - 100V]_2^5 = \frac{1800}{3} = 600 J
Process B→C is isobaric: WBC=300(5−5)=0W_{BC} = 300(5-5) = 0 J
Process C→A is isobaric: WCA=100(2−5)=−300W_{CA} = 100(2-5) = -300 J
Total work done by system: W=600+0−300=300W = 600 + 0 - 300 = 300 J

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