Problem on weight fraction
A gas contains 350 ppm of H2S in CO2 at 72°F and 1.53 atm pressure. If the gas is liquified, what is the weight fraction H2S?
Expert
Now we consider 1liter of CO2 at 72deg F {[K] ≡ (72+ 459.67) × 5/9}=295.3722 K 1.53 atm
So No moles n= PV/RT =1.53atm*1liter/(0.08205liter-atm/K-mole*295.3722) =0.063131mole
So wt of CO = 0.063131mole*28g/mole =1.76767g
Now 350 ppm H2S would have a volume 350 microlitre ( µL ) per litre, = 350 ppm =350E-6 litre
So No of moles H2S = n= PV/RT =1.53atm*(350E-6)liter/(0.08205liter-atm/K-mole*295.3722K) = 2.20959E-05 mole
The MW H2S = 34.116g/mole
So wt of H2S = 0.000753823 g
So total weight =1.768423823g
So %wt H2S = (0.0007538236/1.768423823g)*100%=0.04262%
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A stream of propane is throttled from 4.298 bar and 450 K to 0.22 bar. This process is irreversible due to frictional losses at the nozzle. a. In words or equations, why are frictional losses irreversible? Discover Q & A Leading Solution Library Avail More Than 1450995 Solved problems, classrooms assignments, textbook's solutions, for quick Downloads No hassle, Instant Access Start Discovering 18,76,764 1951048 Asked 3,689 Active Tutors 1450995 Questions Answered Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!! Submit Assignment
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