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First law of thermodynamics, heat capacity

Problem 2.58

One mole of oxygen is expanded from a volume V1V_{1} =1.00l=1.00 \mathrm{l} to V2=5.01V_{2}=5.01 at a constant temperature T=280 K.T=280 \mathrm{~K} . Calculate: (a) the increment of the internal energy of the gas: (b) the amount of the absorbed heat. The gas is assumed to be a Van der Waals gas.

Reveal Answer
2.58. (a) ΔU=a/V1−a/V2=0.11 kJ;\Delta U=a / V_{1}-a / V_{2}=0.11 \mathrm{~kJ} ; (b) Q=RTln⁡V2−bV1−b=Q=R T \ln \frac{V_{2}-b}{V_{1}-b}= =3.8 kJ=3.8 \mathrm{~kJ}
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