Question

The analysis on a mass basis of an ideal gas mixture at 30°F, 15 lbf/in.2 is 25% CO2, 10% CO, and 65% O2. Determine: (a) the analysis in terms of mole fractions. (b) the apparent molecular weight of the mixture. (c) the partial pressure of each component, in lbf/in.2 (d) the volume occupied by 10 lb of the mixture, in ft3.

Answer 1

Answer:

a) $y_{CO_2}=0.1921\\y_{CO}=0.1208\\y_{O_2}=0.6871$

b) The apparent molecular weight is $33.826\frac{g}{mol}$

c)$P_{CO_2}=2.8815psi\\P_{CO}=1.812psi\\P_{O_2}=10.307psi$

d)$V=103.56ft^3$

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:

$25gCO_2*\frac{1molCO_2}{44.01gCO_2} =0.568molCO_2\\10gCO*\frac{1molCO}{28gCO} =0.357molCO\\\\65gO_2*\frac{1molO_2}{32gO_2} =2.0313molO_2$

The total amount of moles in the 100 grams is: $0.568+0.357+2.0313=2.9563$ moles.

So, the molar fraction can be obtained dividing the moles amount of each compound by the total moles amount:

$y_{CO_2}=\frac{0.568}{2.9563}=0.1921\\y_{CO}=\frac{0.357}{2.9563}=0.1208\\y_{O_2}=\frac{2.0313}{2.9563} =0.6871$

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:

$MW=\frac{100g}{2.9563moles}=33.826\frac{g}{mol}$

c) For an ideal gas the partial pressure $P_i$ of the 'i' compound is defined as:

$P_i=y_i*P$ where $y_i$ is the molar fraction and P is the total pressure, so:

$P_{CO_2}=0.1921*15psi=2.8815psi\\P_{CO}=0.1208*15psi=1.812psi\\P_{O_2}=0.6871*15psi=10.307psi$

d)For this volume calculation, we are going to use the ideal gas law:

$PV=nRT$ where we know the pressure, the temperature and can calculate the number of moles.

Calculate the number of moles:

$10lb*\frac{1lbmol}{33.826lb}=0.2956lbmol$

Calculate the temperature in Rankine:

$R=F+459.67\\R=30+459.67=489.67R$

The value for R is:

$R=10.73159\frac{ft^3psi}{R*lbmol}$

So, from the ideal gas law,

$V=\frac{nRT}{P}=\frac{0.2956lbmol*10.73159\frac{ft^3psi}{R*lbmol}*489.67R }{15psi} \\V=103.56ft^3$