A steel cylinder containing compressed air is stored in a fermentation laborator
ID: 888149 • Letter: A
Question
A steel cylinder containing compressed air is stored in a fermentation laboratory ready to provide aeration gas to a small-scale bioreactor. The capacity of the cylinder is 48 litres, the absolute pressure is 0.35 MPa and the temperature is 22 degree C. One day in mid-summer when the air conditioning breaks down, the temperature in the laboratory rises to 33 degree C and the valve at the top of the cylinder is accidentally left open. Estimate the proportion of air that will be lost. Note: this problem requires you to get an expression for the number of moles before (n1) and after the leak (n2). You have to calculate the ratio n2/n1. To get an expression for n1 simply write the ideal gas equation for the conditions before the leak, and to get an expression for n2 simply write the ideal gas equation again for the conditions after the leak. To get the proportion, we need to consider that n1 (the original number of moles contained in the tank) was 100%. The ratio n2/n1 reflects how much is let in the tank. You need to calculate how much of that initial 100% was lost due to the leak. Your result will be a % value.Explanation / Answer
before leakage
From gas law , PV= n1RT
V= 48 liters
P =0.35 Mpa
1 atm =1.013*105Pa
0.35X106 Pa= 3.5X105 Pa equals =3.5/1.013=3.455 atm
R =0.08205lit.atm/k.nol and T=21+273.15 K =294.15 K
n1 =moles before leakage
n= PV/RT = 3.455*48/ (0.08205*294.15)=6.87 moles
when there is a leak, the temperarue rises to 33 deg.c =33+273.15= 306.15K
P =1 atma ( since there is a leakager, the pressure equals atmospheric pressure
V=48 liters
n2= 1*48/(0.08205*306.15)=1.9108
inital moles/ final moles in the tank = n1/n2= 6.87/1.9108=3.59
n1-n2 =moles lost =6.87-1.9108 =4.95902
proportion of air that is lost = (4.95902/ 6.87)*100 =72.18%
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