Question

Initially, at a temperature T, and a molar volume vi, a van der Waals gas undergoes a change of state to the final temperature T2 and the molar volume V2. The van der Waals gas is characterized by the two parameters a and b (cf. Eq. (3.3)). a. Show that the change in molar entropy is As = c, In 72 + R In º2 = (3.62) 01 - 6 b. A volume of 1 dm is partitioned by a wall into two equal parts, one con- taining 1 mol xenon (van der Waals parameters a = 4.250 dm bar mol-2 and b = 0.0511 dm mol-?), the other part being evacuated. Calculate the change in the entropy of xenon after the wall is removed and the gas undergoes a free expansion under isothermal conditions (T = 298.15 K). Also, calculate the change in entropy of the surroundings.

Answer 1

) The equation sfate of for Imd wander andls is given by, RT V-b Dusing voJume changes. tamperatire ond the the I'n oider to find te chonge in entrpy, we need to wule the maxweil relition, ds dv Cv OT R.) - 0 as V-b LOT ), - v-b

to calculate The to taud Al ow, have ink grding t conthint entrpy change by tempornatirse and ot constant voleme, At cmstrnt timporatura, Com s(T,K) S(T,*Vs) = s LT,v.) + dv v, s(T, M) + ( E du - sCT, V.) + Rdn [v-b] v, u2-b - S [T,V.) + Rdn At canstont volume, s (to, Vz)= s(T, V;) t dT -s (T,N.) -s T, V,) + Cv en Ie

slTz, Vx) -s(T,vi) = s (T,M) + R dm ( ) - <(P,«) - c h Ta- - Cr AS = R in ) - alo Giren, m= 1 md e xenon 4 = 4 .250 dm 6 bah.md-? 0.4250 m°a mot-2 5.11x 10 5m3 met- b= 6.0511 dm} mot-) %3D v, = 0.5 am3. [: imitial ] Vz =1day} [:Amal] The ontropy change. fos isolte mal fol expan Sim Ewanderundls is gives) by. gos is givess by, Lvano AS gas V, - 7b 5.11x105 - 1x 8.31I4 × 0.5 xio- 5.11x105 As qus = 6. 2215 ķ

the The totul entrp y change in is and Surrcund ngs gos AS= AScar + (AS)gas Surreundi due ti wander walls relings we use Ike daubius egiuation So, im åder to find the enholy A8rur As surreunding's %3D - AU + as it is free the gas No wak is done *jan Sion. -DQ qas F: Came heat is tromsfeced] we know, : a425 x 12 - 425J

(AB)sur (AS)r = %3D =-425 =. - 1.4 25 J K-! 298.15 + A) gas OS) suet AS - = 6.2215 -1.425 =4.7965 J 5