Question

Assume that silver and gold form ideal, random mixtures. Calculate the mass of pure Ag needed to cause an entropy increase of 20 J/K when mixed with 100g of pure Au

Answer 1

Answer:

$m_{Ag}=2,265.9g$

Explanation:

Hello!

In this case, since the definition of entropy in a random mixture is:

$\Delta S=-n_TR\Sigma[x_i*ln(x_i)]$

For this silver-gold mixture we write:

$\Delta S=-(n_{Au}+n_{Ag})R\Sigma[\frac{n_{Au}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Au}}{n_{Au}+n_{Ag}} )+\frac{n_{Ag}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Ag}}{n_{Au}+n_{Ag}} )]$

By knowing the moles of gold:

$n_{Au}=100g*\frac{1mol}{197g} =0.508mol$

It is possible to write the aforementioned formula in terms of the variable $x$ representing the moles of silver:

$20\frac{J}{mol}=-(0.508+x)8.314\frac{J}{mol*K} \Sigma[\frac{0.508}{0.508+x} *ln(\frac{0.508}{0.508+x} )+\frac{x}{0.508+x} *ln(\frac{x}{0.508+x} )]$

Which can be solved via Newton-Raphson or a solver software, in this case, I will provide you the answer:

$x=n_{Ag}=21.0molAg$

So the mass is:

$m_{Ag}=21.0mol*\frac{107.9g}{1mol}\\ \\m_{Ag}=2,265.9g$

Best regards!