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Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: Donaldson Tan on May 22, 2006, 07:53:22 PM

Title: ?S mixing for Perfect Gas Mixture
Post by: Donaldson Tan on May 22, 2006, 07:53:22 PM

I am stuck at trying to derive the expression for ?S_mixing

For a perfect gas mixture, the chemical potential of component i is given as:
?_i,mixture = ?_i,REF + RT.ln(P_i/P_REF)

from G = H - TS, we arrive at dG = dH - TdS - SdT
since dH = TdS + V.dP, then dG = VdP - SdT

At constant pressure, dP = 0 => dG = -SdT => S = -(dG/dT)_P

Hence, s_i,mixture = -(d?_i/dT)_P

s_i,mixture = -(d?/dT)_P = -d(?_i,REF)/dT - (d/dT)(RT.ln(P_i/P_REF)) = -d(?_REF)/dT - (d/dT)(RT.ln(n_iRT/P_REFV))

s_i,mixture = s_i,REF - R.ln(n_iRT/P_REFV) - RT/T = s_i,REF - R.ln(n_iRT/P_REFV) - R

s_i,mixture = s_i,REF - R( ln(n_iRT/P_REFV) + 1 ) =s_i,REF - R( ln(P_i/P_REF) + 1 )

Then I got stuck here...

Title: Re: ?S mixing for Perfect Gas Mixture
Post by: Donaldson Tan on May 22, 2006, 10:23:34 PM

Think I figure it out already...

Prior to mixture, the chemical potential of each compound is given as

?_i,initial = ?_i,REF + RT.ln(P/P_REF)

s_i,initial = -(d?_i,initial/dT)_P = s_i,REF - R( ln(P/P_REF) + 1 )

?S_mixing = sum{ n_is_i,mixture } - sum{n_is_i,initial} = sum{n_i.(s_i,mixture - s_i,initial)} = sum{ -n_i.R.ln(P_i/P) }


?S_mixing =  sum{ -n_i.R.ln(P_i/P) } = sum{ -n_iR.ln(x_i) }

since 0 < x_i < 1, then ln (x_i) < 0
?S_mixing > 0 for all possible values of x_i