In an isobaric process, the change in temperature for 0.1 mole of N2 is 300K. The work done by the gas is (given CP – CV = 8.3 J/mole – K)

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2490 J

249 J

24.90 J

2.490 J

answer is B.

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Detailed Solution

W=PΔV=nRΔT=0.1×8.3×300=249J

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