Question

3. A solution consists of chloroform and acetone. At 300 K, the partial vapor pressure for chloroform is 20.0 and 220.0 mmHg when the mole fraction in the solution is 0.12 and 0.80, respectively. Calculate the change in the chemical potential of chloroform in the solution, if, (a)chloroform - acetone solution is a real solution, and (b)chloroform - acetone solution is an ideal solution

Answer 1

Change in chemical potential is less in ideal solution than real solution .

In thermodynancics, Chemical that can be absorbed number of particles of a potential due to given of a species change of species, is energy the Given Temprature when Temprature = 3ook P. - 20 mm-ig P2 = 220 mm-rg x = 0.12 (mole fraction) x2 = 0.80 (") Coefficient Case I): Chloroform - acetone solution is real solution: Since for real solution the chemical potential is given as Mi = 4 + RT ls (2.) ] - where hi = Chemical potential of pure species - ai - activity of species for real solution ait air where (r= activity ] A r= pi. now from eg total Partial up of Mi - Mi +RT solution which is constant Since we need to find diffrence at 2 pressure hence Ma Mi = -( ut + RT in his hie - w - RT don Mis his M.-H, = – RT / In Ris Pis ) =-RT in (ris lis) live Xi, Pia xiz Piz on letting values Difference in chemical Potential (M2-M.) = -8.314 J x 300k In (0,12x80) K-mol 0.80x220)

(M2-M.) = 8.314 J x 300k +2.303 log (0680x220 1042 x 20 x-mol mol hol = 10.67 KI hid (1₂-14) = 5.74 x 103 dog (73.33) (12-14) = 5174x 1086*103 I = 10.67 kJ (12--) =10.67 Kate Case II.- when chloroform -acetone Solution is ideal! when solution is ideal than [r - 1] Hence Chemical potential is as: - Hi = 48 + RT ln Bc.)] hence diffrence En Chemical concentration is asar H₂ H = xo + RT dn (X) - 4 - RT ln (4) (M-14) = RT ln (x) = 2.303 x 8.314 x 300 log ( 0.80) (42-41) = 5.74 kJ) x log (6.66) = 5.74x0.82 kJ [(H2-H1) = 4.70 kJ/ma] Hence it is clear that Chemical potential es law in case of ideal solution, nood

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