\(20\) g of gold (Au) and \(20\) g of silver (Ag) are mixed to form a single phase solid solution (assume ideal mixing). The atomic weight of Au is \(197\) g/mol and the atomic weight of Ag is \(108\) g/mol. The universal gas constant \(R = 8.314\ \text{J/mol-K}\). Find the total entropy of mixing (rounded off to two decimal places), in J/K.
Show Hint
Convert masses to mole fractions, then apply \(\Delta S_{mix}=-Rn_{total}\sum x_i\ln x_i\).
Step 1: Set out the mixing entropy formula first.
For an ideal solid solution of two elements, the total entropy of mixing is
\[ \Delta S_{mix}=-R\,n_{total}\left(x_{Au}\ln x_{Au}+x_{Ag}\ln x_{Ag}\right) \]
writing it this way keeps the total mole count $n_{total}$ inside the same expression instead of scaling it up at the end.
Step 2: Convert the given masses to moles.
$n_{Au}=\dfrac{20}{197}=0.10152$ mol and $n_{Ag}=\dfrac{20}{108}=0.18519$ mol, so $n_{total}=0.28671$ mol.
Step 3: Get the mole fractions.
\[ x_{Au}=\frac{0.10152}{0.28671}=0.3541,\qquad x_{Ag}=\frac{0.18519}{0.28671}=0.6459 \]
Step 4: Work out $\ln x_{Au}$ and $\ln x_{Ag}$ together.
$\ln(0.3541)=-1.0382$ and $\ln(0.6459)=-0.4371$. Multiplying each by its mole fraction gives $-0.3676$ and $-0.2823$, which sum to $-0.6499$.
Step 5: Plug everything into the one shot formula.
\[ \Delta S_{mix}=-8.314\times0.28671\times(-0.6499) \]
First multiply $8.314\times0.28671=2.3835$, then $2.3835\times0.6499=1.5493$.
\[ \Delta S_{mix}=1.5493\ \text{J/K} \]
Step 6: Round off.
This rounds to $1.55$ J/K, which lies inside the accepted range of $1.50$ to $1.60$ J/K.
\[ \boxed{\Delta S_{mix}\approx1.55\ \text{J/K}} \]
