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If 3.40 g of O2(g) are heated from 25.0 °C to 75.0 °C at constant volume...

If 3.40 g of O2(g) are heated from 25.0 °C to 75.0 °C at constant volume (19.5 J/mol°C) and then
expanded from 0.75 L to 1.25 L at constant temperature, what are the changes in heat (q), work (w),
and internal energy (U)?
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Answer #1

Number of moles of O2 = 3.4g/32g/mol = 0.10625mol

qv = ∆U1 = n.Cv.∆T = 0.10625mol × 19.5J/mol.℃ × (75.0 - 25.0)℃ = 103.59375 J

Work done = - 0.10625mol×8.314J/K.mol × 348.15K × ln (1.25/0.75) = -157.100668 J

∆U2 = 0 = q2 + work done

q2 = 157.100668 J

Change in heat = 103.59375 + 157.100668 = 260.7 J (answer)

Work done = - 157.1 J (answer )

∆U = 103.6 J ( Answer)

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