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Question

Figure shows the variation in the internal energy U with the volume V of 2.0 mol of an ideal gas in a cyclic process abcda. The temperatures of the gas at b and c are 500 K and 300 K respectively. Calculate the heat absorbed by the gas during the process.

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Solution

Given, n = 2 moles.

ΔV=0

In ad and bc

Hence ΔW=ΔWAB+ΔWCD

=nRT1 ln(2V0V0)+nRT2 ln(V02V0)

=nR×2.303×log 2×(500300)

=2×8.314×2.303×0.301×200

=2305.31 J


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