Answer to Question #225827 in Chemical Engineering for Lokika

Question

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.

Answer 1

Part a

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

Putting in the equation

$y_A*7500=105 x_A\\ y_A=0.014 x_A$

Part b

$p_A=105x_A\\ x_A=\frac{n_A}{n_A+n_B}\\ p_A=105 \space\frac{n_A}{n_A+n_B}\\ p_A= 105 \space\frac{c_A}{c_A+c_B}$

Part C

Given equilibrium relation

$y_A=0.014x_A$

Convert this to mole ratio unit

Y and X - mole ratio in vapour

$X= \frac{x_A}{1-x_A}\\ Y= \frac{y_A}{1-y_A}\\ X= \frac{x_A}{1+x_A}\\ Y= \frac{y_A}{1+y_A}\\ \frac{Y_A}{1+Y_A}=0.014 \frac{X_A}{1+X_A}\\$