Question

A device has been designed to measure the flow rate of carbon dioxide evolved from a fermentation reactor. The reactor is sealed except for a tube that allows the generated carbon dioxide to bubble through a soap solution and into a vertical glass tube within internal diameter of 1.2 cm. On leaving the soap solution, the gas forces thin soap films stretched across the tube to traverse the length of the tube. Ambient temperature and pressure are 27°C and 755 mm Mercury. It takes the films 7.4 seconds to traverse the 1.2 m between two calibration marks on the tube

b) what is the rate of the generation of C02 in mol/min?

c) A more refined analysis of the system takes into account that the gas leaving the fermentation reactor contains water with a partial pressure of 26.7mmHg. Calculate the percentage error in part b) using this new information.

Answer 1

diameter of tube= 1.2 cm =1.2/100 m= 0.012m

Cross sectional area of tube = (PI/4)*d²= (22/28)*0.012*0.012 =0.000113 m2,

distance travelled in 7.4 seconds= 1.2m, velocity= 1.2/7.4 m/sec =0.162 m/sec= 0.162*60 m/sec =9.72 m/min

Volumetric flow rate= Cross sectional area* velocity= 0.000113*9.72 m3/min =0.0011 m3/min

V=0.0011 m3/min, T= 27 deg.c =27+273=300K, P= 755 mm Hg= (755/760)atm= 0.993 atm

From gas law equation, PV= nRT, R=0.0821 L.atm/mole.K , n= moles of gas/min=PV/RT=0.993*0.0011/(0.0821*300)= 4.43*10^-5 moles/min

b) For the second case, partial pressure of gas= total pressure-partial pressure of water( saturation vapor pressure of water)= 755-26.7 =728.3 mm Hg= 728.3/760 atm=0.958 atm

moles of gas/min = 0.958*0.0011/(0.0821*300)= 4.28*10^-5 moles/min

difference in moles/min = (4.43-4.28)*10^-5 moles/min =1.6*10^-6 moles/min

percentage error= 100* 1.6*10^-6/4.43*10^-5 =3.6%