Question

4. 10 points A monoatomic gas obeys the van der Waals equation: N²a P= NT V - Nb V2 where N is the number of particles and a and b are known constants and t = kbT. The gas has a heat capacity Cy = 3N/2 in the limit V +0. a) Using the thermodynamic identities and the equation of state prove that acv = 0. av т (3 pts) b) Use the result of part a) to determine the entropy of the van-der Waals gas S(V,7) to within an additive constant. (4 pts). c) What is the final temperature when the gas is adiabatically compressed from (V1, 71) to V2? (3 pts)

Answer 1

17 The von-der-Waate equation is н?а p= V-Hb (4) V2 а) We know, cu = T(0) 2 lacu OVN ос, Әу ( since r = K&T las — ә ду ) . ді — ТӘ ду (37) | от ( 39 ) " [18+). | (uning Manwell relation ар ОТ. /у How froom eq (1), we get, NN Р= ӨКТ V-Hb N²a V-НЬ Э (азе ат. - О

= 0 proved). b) a How, we have the result of pant to be, avr=0 How Given > Cv (VN) = f(n) (where f() is only function of KBT Cv = 30 in the limit vt & So, ev (vodin) f (T) 50, f(n) = 3 So, we have Cv= 2 Gold 3N f (0) = 3N 2 SH

ЗМ v = 3N How Cv= - 28 ) v = 34 31. I S(VT) = 34 MT + where grv) is only function of v. How, we have the marewell's solt. melation, & g(v) २ lap of) el grudgu H²a at av NKST Ту-НЫ v2 V 7 daca NKR V-Hb where À is an gu) = MKE m (V-Hb) + A' ambitony additive constant. - SCVT) = 39 T + MKB n (v-Hb) + A' + MKB hn (v-Hb) + A S[v,) = 3H I + MKg In (v-Mb) +A - 30 M KB Islor (KB

2 SCVT) = 34 hr + MKB ln (v - Hb) TA 34 An KB is also an additive constant. where A = A 86 $(v,^) = get him + Hks In (V-Hb) + A 9) For an an adiabatic process, Tds =0 ds=0 as dv + 23 do =0 Əv НКе dv + V- Mb ( 2 3N dr Зн ar da=0 IT S dr HKB / dv V2 dv V-Hb T Vi - 3 m/T2 M ( 72 ) HKB m/₂ - HD VI-Hb ㅋ in (72 2 KB in ( 3 VI- Nb V₂ - Hb 여 2KB M (H 2 Hb 3) exp VI - Hb 3.

> na = y exp a on (vib a Tz = Mi KB exp 2KB hml 3. VI- Mb V2 - No This is the final temperature