First slide

General discussion of solutions

Question

The ratio of mole fraction of A at point P and point Q in following graph is:

Moderate

A

1.129
B

0.57
C

0.702
D

1.429

Solution

$P_{B}^{o} = 100$
$P_{A}^{o} = 200$
$P_{T o t a l} = P_{A}^{o} x_{A} + P_{B}^{o} x_{B}$
$= P_{B}^{o} + (P_{A}^{o} + P_{B}^{o}) x_{A}$
$140 + 100 + (100) x_{A}$
$x_{A} = 0.4 = \frac{4}{10}$
$y_{A} = \frac{P_{A}^{o} x_{A}}{P_{T o t a l}} = \frac{200 (.4)}{140} = \frac{80}{140} = \frac{4}{7}$
$\frac{y_{A}}{x_{A}} = \frac{\frac{4}{7}}{\frac{4}{10}} = \frac{10}{7} = 1.429$

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