Answer :
The correct responses are;
(i) Weight percent of nitrogen: 58.6%
Weight percent of oxygen: 14.65%
Weight percent of carbon dioxide: 29.59%
(ii) Volume percent of nitrogen: 64.38%
Weight percent of oxygen: 14.1%
Weight percent of carbon dioxide: 21.53%
(iii) Partial pressure of nitrogen: 19.314 psia
Partial pressure of oxygen: 4.23 psia
Partial pressure of carbon dioxide: 6.459 psia
(iv) Partial pressure of nitrogen: 19.314 psia
Partial pressure of oxygen: 4.23 psia
Partial pressure of carbon dioxide: 6.459 psia
(v) The average molecular weight is approximately 32.02 g/mole
(vi) At 20 psia, 60 °F, the density is 0.11275 lb/ft.³
At 14.7 psia, 60 °F, the density is 0.0844 lb/ft.³
At 14.7 psia, 32 °F, the density is 0.0892 lb/ft.³
(vii) The specific gravity of the mixture 0.169
Reasons:
The volume of the tank = 40 ft.³ = 1.132675 m³
Content of the tank = Carbon dioxide and air
The pressure inside the tank = 30 psia = 206843 Pa
The temperature of the tank = 70 °F ≈ 294.2611 K
Mass of carbon dioxide placed in the tank = 2 lb.
Percent of nitrogen in the tank = 80%
Percent of oxygen in the tank = 20%
(i) 2 lb ≈ 907.1847 g
Molar mass of carbon dioxide = 44.01 g/mol
Number of moles of carbon dioxide = [tex]\displaystyle \mathbf{ \frac{907.1847 \, g}{44.01 \, g/mol}} \approx 20.613 \, moles[/tex]
Assuming the gas is an ideal gas, we have;
[tex]\displaystyle n = \mathbf{\frac{206843\times 1.132674}{8.314 \times 294.2611} }\approx 95.7588[/tex]
The number of moles of nitrogen and oxygen = 95.7588 - 20.613 = 75.1458
Let x represent the mass of air in the mixture, we have;
[tex]\displaystyle \mathbf{ \frac{0.8 \cdot x}{28.014} + \frac{0.2 \cdot x}{32}}= 75.1458[/tex]
Solving gives;
x ≈ 2158.92 grams
Mass of the mixture = 2158.92 g + 907.1847 g ≈ 3066.1047 g
[tex]\displaystyle Weight \ percent \ of \ nitrogen= \frac{0.8 \times 2158.92}{3066.105} \times 100 \approx \underline{58.6 \%}[/tex]
[tex]\displaystyle Weight \ percent \ of \ oxygen = \frac{0.8 \times 2158.92}{3066.105} \times 100 \approx \underline{14.65\%}[/tex]
[tex]\displaystyle Weight \ percent \ of \ carbon \ dioxide = \frac{907.184}{3066.105} \times 100 \approx \underline{29.59\%}[/tex]
ii.
[tex]\displaystyle Number \ of \ moles \ nitrogen= \frac{0.8 \times 2158.92 \, g}{28.014 \, g/mol} \approx 61.65\, moles[/tex]
[tex]\displaystyle Number \ of \ moles \ oxygen= \frac{0.2 \times 2158.92 \, g}{32 \, g/mol} \approx 13.5\, moles[/tex]
Number of moles of carbon dioxide = 20.613 moles
Sum = 61.65 moles + 13.5 moles + 20.613 moles ≈ 95.763 moles
In an ideal gas, the volume is equal to the mole fraction
[tex]\displaystyle Volume \ percent \ of \ nitrogen= \frac{61.65}{95.763} \times 100 \approx \underline{64.38 \%}[/tex]
[tex]\displaystyle Volume \ percent \ of \ oxygen = \frac{13.5}{95.763} \times 100 \approx \underline{14.1\%}[/tex]
[tex]\displaystyle Volume \ percent \ of \ carbon \ dioxide = \frac{20.613 }{95.763} \times 100 \approx \underline{21.53\%}[/tex]
(iv) The partial pressure of a gas in a mixture, [tex]P_A = \mathbf{X_A \cdot P_{total}}[/tex]
Partial pressure of nitrogen, [tex]P_{N_2}[/tex] = 0.6438 × 30 psia ≈ 19.314 psia
Partial pressure of oxygen, [tex]P_{O_2}[/tex] = 0.141 × 30 psia ≈ 4.23 psia
Partial pressure of carbon dioxide, [tex]P_{CO_2}[/tex] = 0.2153 × 30 psia ≈ 6.459 psia
(v) The average molecular weight is given as follows;
[tex]\displaystyle Average \ molecular \ weight = \frac{3066.105\, g}{95.7588 \, moles} = 32.02 \, g/mole[/tex]
(vi) At 20 psia 70 °F, we have;
Converting to SI units, we have;
[tex]\displaystyle n = \frac{137895.1456\times 1.132674}{8.314 \times 294.2611} \approx 63.84[/tex]
The number of moles, n ≈ 63.84 moles
The mass = 63.84 moles × 32.02 g/mol ≈ 2044.16 grams ≈ 4.51 lb
[tex]\displaystyle Density = \frac{4.51\ lb}{40 \ ft.^3} \approx \underline{0.11275\, lb/ft.^3}[/tex]
When the pressure is 14.7 psia = 101352.93 Pa, and 60 °F
[tex]\displaystyle n = \frac{101352.9\times 1.132674}{8.314 \times 288.7056} \approx 42.8247[/tex]
The mass = 47.8247 moles × 32.02 g/mol ≈ 1531.35 grams = 3.376 lb.
[tex]\displaystyle Density = \frac{3.376 \ lb}{40 \ ft.^3} \approx \underline{ 0.0844\, lb/ft.^3}[/tex]
When the pressure is 14.7 psia = 101352.93 and 32 °F
[tex]\displaystyle n = \frac{101352.9\times 1.132674}{8.314 \times 273.15} \approx 50.5482[/tex]
The mass = 50.5482 moles × 32.02 g/mol ≈ 1618.55 grams = 3.568 lb.
[tex]\displaystyle Density = \frac{3.568 \ lb}{40 \ ft.^3} \approx \underline{ 0.0892 \, lb/ft.^3}[/tex]
(vii) [tex]\displaystyle The \ specific \ gravity \ of \ the \ mixture \ = \frac{\frac{3066.1047\, g}{40 \, ft.^3} }{1.00} = \frac{\frac{6.7596 \, lbs}{40 \, ft.^3} }{1.00} \approx \underline{0.169}[/tex]
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