Question

Benzene is evaporating into a closed, well-mixed 20-L vessel. Assume the surface area of the Benzene is 150 cm2 . Benzene has a molecular weight of 78 g/mol, liquid density of 0.88 g/mL, vapor pressure of 0.125 atm., liquid activity coefficient of 2400 (25 °C and 1 atm), solubility of 1780 mg/L and an octanol-water coefficient of log KBow=2.0 (mass concentration basis).

a) How much chemical (grams) would have to be added to reach a benzene concentration of 200 mg/L in the vessel assuming no liquid remains?

b) If the benzene concentration in the headspace is 200 mg/L after 45 minutes what is the mass transfer coefficient (cm/s)? Hint: Perform a mass balance on benzene in the gas phase

Answer 1

a) Well, to reach benzene concentration = 200 mg/L converting it into g/L = 200 x 10^-3 g/L

Assuming no liquid remains in the vessel means total volume of the vessel is occupied by the benzne.It means volume = 20L

Now, we now molarity = grams / Litre

             200 x 10^-3 g/L = grams / 20 L

           so grams = (200 x 10^-3 g/L) x ( 20 L)

                          = 4 grams

b) Liquid -phase mass transfer coefficient for benzene (cm/s) :

           K_B,liuid = K_rC_B,_dilute (diffusivity benzene)