Question

2. At 298 K. 0.30-mol Ar at 1 atm mixes with 0.20-mol He at 1 atm. The final mixture remains at 1 atm and 290 the standard state, the chemical potential of Ar and He are given as "(Ar) and i'(He), respectively. At the standard state, pº = 1 atm for any gases. (a) (10pt) Write the expression of Gibbs free energy before mixing. (b) (10pt) Write the expression of Gibbs free energy after mixing. (c) (15pt) Determine the mixing Gibbs free energy in Joule. (d) (15pt) Determine the mixing entropy in Joule.

Answer 1

2. (a) now, before mixing the expression of Gibbs free energy will be the Gibbs free energy of individual Species as and the From additivity rule we know Go- Tuini where Mi is the chemical potential of theith Species and ni is the no. of the moles of ith species. So here Go- Marnas - MAs x 0.30 The Mirente. Mite to.ro row, if the As and the behave idcally then we know Mu+kili (1 bar) Mo is the chemical potential at standard state Now, as P = 1 bar, So Mu (1) So, before mixing GAR= Mar) x 0:30 Gite= M(te) x 0.20

Nows We know. Armix - MRT I where xi is the mole Praction of ith species Now for the system (binary) Ris universal gan Coast whole value of AGmix = nRT ( x, lor, tezlenz) is 8.314 5. malik-1 | ART (xar la sant the like). So, Gmix = RRT (Per Insert the latte) will be the expression of Gibbs free energy after xmixing Darix = niet (xArln sertate duane) = 0.50x9-314x298 (0.30 80030 0.20 ) = 9:56x8 314x238 (495){ osoucosi) terusnya = 8.314x298 * 4-0.3357 =-.829.38 Joute. Here negative sign indicate at Const. temperature and pressure Gasses are mixed Spontanewusly.. 0.50 losol 0.50 wow, Entropy of mixing Smix =R xilnei where si su s Saix = one lean Inxant @ xite In wenzsofiresteeris is the mulcfrection = (+) 2. 785 Joule Entrophy of mixing is always busive.