Answer:
a.6531.53 mole/hr
b. 32.76 degC
c. 3.78 deg.c
Explanation:A gas mixture containing 85.0 mole% N2 and the balance n-hexane flows through a pipe at a rate of 100.0 m3/h. The pressure is 2.00 atm absolute and the temperature is 140.0°C.
a. What is the molar flow rate of the gas? kmol/h
b. To what temperature would the gas have to be cooled at constant pressure in order to begin condensing hexane? °C
c. To what temperature would the gas have to be cooled at constant pressure in order to condense 75.0% of the hexane? °C
Given V= Volume of gas mixture= 100m3/hr=10^5 Lt/hr P= 2.0 atm and T= 100 deg.c =100+273.15= 373.15K
n= moles of mixture= PV/RT , where R=0.08206 L.atm/mole.K
n= 2.0*10^5/0.08206*373.15)=6531.53 mole/hr
b. To what temperature would the gas have to be cooled at constant pressure in order to begin condensing hexane? A gas mixture containing 85 mole% N2 and the°C
Condensation begins at a point at which the partial pressure of vapor =vapor pressure of liquid at the given temperature
Partial pressure of hexane in the mixture= 0.15*2.0= 0.3 atm
so for saturation to begin, the vapor pressure shoudl correspond to 0.3 atm=30.39 Kpa
Antoine constant for Hexane
lnP (Kpa)= 13.82- 2696/(T-48.833) ( T is in K
ln(30.39 )= 13.8193- 2696/ (T-48.833)
, 3.414= 13,82-2696/(T-48.333)
2696/(T-48.833)= 13.82-3.414=10.405
T-46.833= 2696/10.414=259.08
259.08+46.833 K=305.91k
305.91k-273.15K
32.76C
c. To what temperature would the gas have to be cooled at constant pressure in order to condense 75% of the hexane? °C?
Vapor present in the gas mixture= 25%, Its partial pressure=0.15* 0.25*2.0 =0.075 atm= 7.59Kpa
from ln(7.59)= 13.82- 2696/ (T-48.333)
2.02 = 13.82- 2696/(T-48.333)
2696/(T-48.333)= 11.179
T-48.333= 2696/11.79=228.60
T= 228.6+48.333= 276.93 K= 3.78 deg.c
