Question

. Consider the van der Waals gas for which: a = 3.59 t.atrn/mole b-0.0427 Umole One mole of this gas having a volume of 0.25 C is initially at equilibrium with a temperature of T, and 60 atm pressure. The pressure is suddenly dropped to 37 atm and the gas expands at constant pressure isothermally to 0.54 C at which point equilibrium is achieved. The pressure is then suddenly increased to 60 atm and the gas is compressed isothermally until equilibrium is achieved. a) Calculate ag for this cycle. b) Suppose the same isothermal cycle is carried out reversibly. Find Δq for this path.

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Answer #1

P+ a 7n

Here,R=0.082L,atm.mol-1.K-1.Thus,T1=357.52K.In this cycle of expansion and compression process,

taking equivalent to ideal gas processes,q is zero in 1st n 3rd porcess and q=(P2-P1)(v2-v1)=(60-37)0.54-0.25)=12.17J.

b)In a reverse process,the q value is negative value of this value=-12.17J

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