3 moles of an ideal monoatomic gas performs ABCDA cyclic process as shown in figure below. The gas temperature are TA=400K,TB=800K,TC=2400K and TD=1200K. The work done by the gas is (approximately) (R=8.314J/moleK):
A
10kJ
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B
20kJ
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C
40kJ
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D
100kJ
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Solution
The correct option is A10kJ ForprocessAB:w=0ForprocessBC:w=nRΔT=3×R×[2400−800]=3×R×1600=4800RForprocessCD:w=0ForprocessDA:w=3R[400−1200]=−2400R∴Totalwork=2400R=2400×253J=800×25J=10KJ