a.What is the average molecular weight (g/mol)? b.What is the gas density (g/L)?

Question

a.What is the average molecular weight (g/mol)?

b.What is the gas density (g/L)?

c. What is the composition in weight %?

e.What is the linear flow rate (m/min)?

3. A gas mixture consisting of 60 mol% H2, 30 mol% CH4, and 10 mol% C2H6 flows through a 100 cm (inner diameter) pipe at 10.0 m3/min. The pressure is 2 atm and the temperature is 20°C. Remember that according to the ideal gas law, p (g/L) = P Mw/RT where MW is the molecular weight and the gas constant R = 0.0821 L atm/mol-K. drogen 2 g/mol; methane1

Explanation / Answer

Basis : 1 mole of mixture

Moles : H2=0.6 (60%), CH4=0.3 (30%) and C2H6= 0.1 ( 10%)

atomic weights : H =1 C= 14

Molecular weights : H2=2, CH4=12+4=16 and C2H6= 2*12+6*1= 30

Moles= Mass/Molecular weight

Mass= Molecular weght* moles

Masses : H2= 1*0.6=1.2 CH4=16*0.3= 4.8 C2H6= 30*0.1= 3 =9 gms

Since 1 moles is the basis, mass= molecular weight =9 g/mole

2. From PV= nRT

PV= (mass/Molecular weight)*RT

P* Molecular weight = (mass/volume)*RT = density*RT

Density= P* Molecular weight/RT

P= pressure in atm =2 atm R= 0.08206 L atm/mole.K T= 20 deg.c =20+273.15=293.15K

Molecular weight = 9

Density= 2*9/(0.08206*293.15) g/L=0.748 g/L

c) Total mass = 9 gm mass of H2=1.2 gms CH4=4.8 gm and C2H6=3 gms

Mass percent : H2= 100*1.2/9= 13.33% CH4= 100*4.8/9= 53.33 and C2H6= 100*3/9= 33.34%

d) flow rate of gas= 10 m3/min

Linear flow rate= Volumetric flow rate/ Cros sectional area

Diameter of pipe =100cm =1 Cross sectional area of circular pie = (PI/4)*d2= 0.79 m2

Linear flow rate= 10m3/min/0.79m2= 12.66 m/min

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