Respuesta :
Explanation:
The chemical reaction is as follows.
[tex]CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O[/tex]
It is given that 2094 [tex]ft^{3}/hr[/tex]. And, it is known that 1 [tex]m^{3}/s[/tex] = 127133 [tex]ft^{3}/hr[/tex]
Hence, convert 2094 [tex]ft^{3}/hr[/tex] into [tex]m^{3}/s[/tex] as follows.
[tex]\frac{2094 ft^{3}/hr}{127133 ft^{3}/hr} \times 1 m^{3}/s[/tex]
= [tex]0.0165 m^{3}/s[/tex]
As ideal gas equation is PV = nRT. So, calculate the number of moles as follows.
n = [tex]\frac{PV}{RT}[/tex]
= [tex]\frac{1 atm \times 0.0165 m^{3}}{0.0821 \times 298 K}[/tex]
= 0.673 mol/sec
According to the stoichiometry of the given reaction, 1 mol of methane reacts with 2 mol of oxygen.
So, 1 mol [tex]CH_{4}[/tex] = 2 mol [tex]O_{2}[\tex]
Hence, [tex]O_{2}[\tex] required theoretically = [tex]2 \times 0.673 mol/s[/tex] = 1.346 mol/s
Hence, air required theoretically = [tex]\frac{1.346}{0.21}[/tex] = 6.4095 mol/s.
Since, 6% of excess air is being supplied. Therefore, total air supplied will be calculated as follows.
Total air supplied = [tex]6.4095 mol/s [1 + \frac{6}{100}][/tex]
= 6.794 mol/s
Now, calculate the volume using ideal gas law equation as follows.
PV = nRT
[tex]1 atm \times V = 6.794 mol/s \times 8.21 \times 10^{-5} Latm/K mol \6 times 298 K[/tex]
V = 0.166229 [tex]m^{3}/s[/tex]
Converting calculated volume into [tex]ft^{3}/hr[/tex] as follows.
1 [tex]m^{3}/s[/tex] = 127133 [tex]ft^{3}/hr[/tex]
So, 0.166229 [tex]m^{3}/s[/tex] = [tex]0.166229 m^{3}/s \times 127133 \frac{ft^{3}/hr}{1 m^{3}/s}[/tex]
= 21133.191 [tex]ft^{3}/hr[/tex]
Thus, we can conclude that 21133.191 [tex]ft^{3}/hr[/tex] of air are drawn from outside per hour by the fan that supplies the air.
