107 774
Assignments Done
100%
Successfully Done
In June 2024

# Answer to Question #225827 in Chemical Engineering for Lokika

Question #225827

The solubility of a gaseous substance A(molecular weight 26) in water is given by Henry’s law: pA=105xA ., pA in mm Hg. Convert the equilibrium relation to the following forms, a) yA=m.xA ., if the total pressure is 10 bar; (b) pA=m’CA ,CA in gmol/litre. Also write down the equilibrium relation using the mole ratio unit.

1
2021-08-16T02:19:15-0400

Part a

We know that "y_A= \\frac{p_A}{P} \\implies p_A=y_AP"

Putting in the equation

"y_A*7500=105 x_A\\\\\ny_A=0.014 x_A"

Part b

"p_A=105x_A\\\\\nx_A=\\frac{n_A}{n_A+n_B}\\\\\np_A=105 \\space\\frac{n_A}{n_A+n_B}\\\\\np_A= 105 \\space\\frac{c_A}{c_A+c_B}"

Part C

Given equilibrium relation

"y_A=0.014x_A"

Convert this to mole ratio unit

Y and X - mole ratio in vapour

"X= \\frac{x_A}{1-x_A}\\\\\nY= \\frac{y_A}{1-y_A}\\\\\nX= \\frac{x_A}{1+x_A}\\\\\nY= \\frac{y_A}{1+y_A}\\\\\n\\frac{Y_A}{1+Y_A}=0.014 \\frac{X_A}{1+X_A}\\\\"

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!