Question

A tank with a volume of 40 cuft is filled with a carbon dioxide and air mixture. The pressure within the tank is 30 psia at 70oF. It is known that 2 lb of carbon dioxide was placed in the tank. Assume that air is 80% nitrogen and 20% oxygen and use the ideal gas laws. Calculate ,

Answer 1

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 = $\displaystyle \mathbf{ \frac{907.1847 \, g}{44.01 \, g/mol}} \approx 20.613 \, moles$

Assuming the gas is an ideal gas, we have;

$\displaystyle n = \mathbf{\frac{206843\times 1.132674}{8.314 \times 294.2611} }\approx 95.7588$

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;

$\displaystyle \mathbf{ \frac{0.8 \cdot x}{28.014} + \frac{0.2 \cdot x}{32}}= 75.1458$

Solving gives;

x ≈ 2158.92 grams

Mass of the mixture = 2158.92 g + 907.1847 g ≈ 3066.1047 g

$\displaystyle Weight \ percent \ of \ nitrogen= \frac{0.8 \times 2158.92}{3066.105} \times 100 \approx \underline{58.6 \%}$

$\displaystyle Weight \ percent \ of \ oxygen = \frac{0.8 \times 2158.92}{3066.105} \times 100 \approx \underline{14.65\%}$

$\displaystyle Weight \ percent \ of \ carbon \ dioxide = \frac{907.184}{3066.105} \times 100 \approx \underline{29.59\%}$

ii.

$\displaystyle Number \ of \ moles \ nitrogen= \frac{0.8 \times 2158.92 \, g}{28.014 \, g/mol} \approx 61.65\, moles$

$\displaystyle Number \ of \ moles \ oxygen= \frac{0.2 \times 2158.92 \, g}{32 \, g/mol} \approx 13.5\, moles$

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

$\displaystyle Volume \ percent \ of \ nitrogen= \frac{61.65}{95.763} \times 100 \approx \underline{64.38 \%}$

$\displaystyle Volume \ percent \ of \ oxygen = \frac{13.5}{95.763} \times 100 \approx \underline{14.1\%}$

$\displaystyle Volume \ percent \ of \ carbon \ dioxide = \frac{20.613 }{95.763} \times 100 \approx \underline{21.53\%}$

(iv) The partial pressure of a gas in a mixture, $P_A = \mathbf{X_A \cdot P_{total}}$

Partial pressure of nitrogen, $P_{N_2}$ = 0.6438 × 30 psia ≈ 19.314 psia

Partial pressure of oxygen, $P_{O_2}$ = 0.141 × 30 psia ≈ 4.23 psia

Partial pressure of carbon dioxide, $P_{CO_2}$ = 0.2153 × 30 psia ≈ 6.459 psia

(v) The average molecular weight is given as follows;

$\displaystyle Average \ molecular \ weight = \frac{3066.105\, g}{95.7588 \, moles} = 32.02 \, g/mole$

(vi) At 20 psia 70 °F, we have;

Converting to SI units, we have;

$\displaystyle n = \frac{137895.1456\times 1.132674}{8.314 \times 294.2611} \approx 63.84$

The number of moles, n ≈ 63.84 moles

The mass = 63.84 moles × 32.02 g/mol  ≈ 2044.16 grams ≈ 4.51 lb

$\displaystyle Density = \frac{4.51\ lb}{40 \ ft.^3} \approx \underline{0.11275\, lb/ft.^3}$

When the pressure is 14.7 psia = 101352.93 Pa, and 60 °F

$\displaystyle n = \frac{101352.9\times 1.132674}{8.314 \times 288.7056} \approx 42.8247$

The mass = 47.8247 moles × 32.02 g/mol  ≈ 1531.35 grams = 3.376 lb.

$\displaystyle Density = \frac{3.376 \ lb}{40 \ ft.^3} \approx \underline{ 0.0844\, lb/ft.^3}$

When the pressure is 14.7 psia = 101352.93 and 32 °F

$\displaystyle n = \frac{101352.9\times 1.132674}{8.314 \times 273.15} \approx 50.5482$

The mass = 50.5482 moles × 32.02 g/mol  ≈ 1618.55 grams = 3.568 lb.

$\displaystyle Density = \frac{3.568 \ lb}{40 \ ft.^3} \approx \underline{ 0.0892 \, lb/ft.^3}$

(vii) $\displaystyle The \ specific \ gravity \ of \ the \ mixture \ = \frac{\frac{3066.1047\, g}{40 \, ft.^3} }{1.00} = \frac{\frac{6.7596 \, lbs}{40 \, ft.^3} }{1.00} \approx \underline{0.169}$

Learn more about partial pressure here:

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