Question

The theoretical minimum work required to separate one mole of a liquid mixture at 1 atm, containing 50 mole % each of n-heptane and n-octane into pure compounds, each at 1 atm, is:

• A. $-2RT \ln 0.5$
• B. $2RT$
• C. $-RT \ln 0.5$
• D. $0.5RT$

Hint:

Use the formula $W = -RT \sum x_i \ln x_i$ to calculate the minimum work for separating ideal mixtures. For equimolar binary mixtures, this simplifies to $-RT \ln 0.5$.

Answer 1

The Correct Option is C

The minimum work required to separate an ideal binary mixture into its pure components is derived from the concept of entropy of mixing.
The reversible work of separation per mole of mixture is given by the formula:
\[ W = -RT \sum x_i \ln x_i \]
Where:
$R$ = universal gas constant
$T$ = absolute temperature in Kelvin
$x_i$ = mole fraction of component $i$ in the mixture
For a binary mixture containing 50 mol% of each component:
$x_1 = x_2 = 0.5$
\[ W = -RT [0.5 \ln 0.5 + 0.5 \ln 0.5] = -RT [2 \cdot 0.5 \ln 0.5] = -RT \ln 0.5 \]
Thus, the theoretical minimum work required to separate one mole of this mixture into pure components is:
\[ \boxed{-RT \ln 0.5} \]