Question

6.55. A vapor stream that is 65 mole% styrene and 35 mole% toluene is in equilibrium with a liquid mixture of the same two species. The pressure in the system is 150 mm Hg absolute. Use Raoult’s law to estimate the composition of the liquid and the system temperature

Answer 1

Answer:

The composition of vapor is 0.86 and the composition of toluene is 0.14

Explanation:

Given:

Vapor composition $y_{s} = 0.65$

Toluene composition $y_{t} = 0.35$

Pressure of the system $P = 150$ mm Hg

From Raoult's law,

For toluene,

$Py_{t} = x_{t} P _{t}$

$150 \times 0.35 = x_{t} P_{t}$

  $x_{t} = \frac{52.5}{P_{t} }$

For vapor,

 $Py_{s} = x_{s} P _{s}$

$150 \times 0.65 = x_{s} P _{s}$

  $x_{s} = \frac{97.5}{P_{s} }$

Since, $x_{t} + x_{s} = 1$

   $\frac{52.5}{P_{t} } + \frac{97.5}{P_{s} } = 1$

Since $P_{s}, P_{t}$ are dependent on temperature.

At $T = 358.6$ K, value of $P_{s} = 112.876$ mm Hg and $P_{t} = 359.03$ mm Hg

So calculate $x_{t}$

 $x_{t} = \frac{52.5}{359.03} = 0.14$

For $x_{s}$    

 $x_{s} = 1-x_{t}$

 $x_{s} = 1-0.14$

 $x_{s} = 0.86$

Therefore, the composition of vapor is 0.86 and the composition of toluene is 0.14