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(a) Conversions: 1 atm=14.696 psig, 1 ft3=28.317 L, 1 lb mol=453.59 mol Convert the units of each term P( atm )=(P^(')(p sig)+14.696)/(14.696) V(L)=V^(')(ft^(3))**28.317 ft^(3)//L n(mol)=n^(')(lbmol)**453.59mol//lbmol T(K)=(T^(')(^(@)F)-32)/(1.8)+273.15 The ideal gas equation is PV=0.08206 nT =>(P^(')+14.696)/(14.696)**V^(')**28.317=0.08206**n^(')**453.59**((T^(')-32)/(1.8)+273.15) where P' is the Pressure in psig V' is the Volume in ft3 n' is the Number of moles in lb mol T' is the Temperature in 0F =>(P^(')+14.696)**V^(')=(14.696**0.08206**453.59)/(28.317**1.8)**n^(')**(T^(')-32+491.67) (P^(')+14.696)V^(')=10.7318n^(')(T^(')+459.67) (b) Given: Mole fraction of CO, xCO=30 mole%=0.30 lb mole CO/lb mole Mole fraction of N2, xN2=70 mole%=0.70 lb mole N2/lb mole Volume of the cylinder, V=3.5 ft3 Temperature, T=850F Bourdon ... See the full answer