Answer:
a) [tex]y_{CO_2}=0.1921\\y_{CO}=0.1208\\y_{O_2}=0.6871[/tex]
b) The apparent molecular weight is [tex]33.826\frac{g}{mol}[/tex]
c)[tex]P_{CO_2}=2.8815psi\\P_{CO}=1.812psi\\P_{O_2}=10.307psi[/tex]
d)[tex]V=103.56ft^3[/tex]
Explanation:
a)
Let's take a calculation basis of 100g. This can be seen just as analyze a sample of the gas of 100g. A sample of 100g will have the composition: 25g CO2, 10g CO and 65 g O2 because the given analysis on mass basis. We are going to pass that mass to its respective moles amount:
[tex]25gCO_2*\frac{1molCO_2}{44.01gCO_2} =0.568molCO_2\\10gCO*\frac{1molCO}{28gCO} =0.357molCO\\\\65gO_2*\frac{1molO_2}{32gO_2} =2.0313molO_2\\[/tex]
The total amount of moles in the 100 grams is: [tex]0.568+0.357+2.0313=2.9563[/tex] moles.
So, the molar fraction can be obtained dividing the moles amount of each compound by the total moles amount:
[tex]y_{CO_2}=\frac{0.568}{2.9563}=0.1921\\y_{CO}=\frac{0.357}{2.9563}=0.1208\\y_{O_2}=\frac{2.0313}{2.9563} =0.6871[/tex]
b) The apparent molecular weight can be obtained with the data we have for 100 g sample, because now we know that 2.9563 gas moles are in 100 g, so:
[tex]MW=\frac{100g}{2.9563moles}=33.826\frac{g}{mol}[/tex]
c) For an ideal gas the partial pressure [tex]P_i[/tex] of the 'i' compound is defined as:
[tex]P_i=y_i*P[/tex] where [tex]y_i[/tex] is the molar fraction and P is the total pressure, so:
[tex]P_{CO_2}=0.1921*15psi=2.8815psi\\P_{CO}=0.1208*15psi=1.812psi\\P_{O_2}=0.6871*15psi=10.307psi[/tex]
d)For this volume calculation, we are going to use the ideal gas law:
[tex]PV=nRT[/tex] where we know the pressure, the temperature and can calculate the number of moles.
Calculate the number of moles:
[tex]10lb*\frac{1lbmol}{33.826lb}=0.2956lbmol[/tex]
Calculate the temperature in Rankine:
[tex]R=F+459.67\\R=30+459.67=489.67R[/tex]
The value for R is:
[tex]R=10.73159\frac{ft^3psi}{R*lbmol}[/tex]
So, from the ideal gas law,
[tex]V=\frac{nRT}{P}=\frac{0.2956lbmol*10.73159\frac{ft^3psi}{R*lbmol}*489.67R }{15psi} \\V=103.56ft^3[/tex]
