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Suppose 2.00 mol of an ideal gas of volume V =3.50
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Answer #1

work done =nrTln(v2/v1) = 2 x 300 x 0.0821 ln(7/3.5) =3.6 x 103J

U=Q-W

U=0 as change in temperature=0

change in internal energy =w =3.6 x 103J

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Suppose 2.00 mol of an ideal gas of volume V =3.50m^3 at T = 300K .allowed...
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