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rewona [7]
2 years ago
12

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 ,

Engineering
1 answer:
aliya0001 [1]2 years ago
8 0

The correct responses are;

(i) Weight percent of nitrogen: <u>58.6%</u>

Weight percent of oxygen: <u>14.65%</u>

Weight percent of carbon dioxide: <u>29.59%</u>

(ii) Volume percent of nitrogen: <u>64.38%</u>

Weight percent of oxygen: <u>14.1%</u>

Weight percent of carbon dioxide: <u>21.53%</u>

(iii) Partial pressure of nitrogen: <u>19.314 psia</u>

Partial pressure of oxygen: <u>4.23 psia</u>

Partial pressure of carbon dioxide: <u>6.459 psia</u>

(iv) Partial pressure of nitrogen: <u>19.314 psia</u>

Partial pressure of oxygen: <u>4.23 psia</u>

Partial pressure of carbon dioxide: <u>6.459 psia</u>

(v) The average molecular weight is approximately <u>32.02 g/mole</u>

(vi) At 20 psia, 60 °F, the density is<u> 0.11275 lb/ft.³</u>

At 14.7 psia, 60 °F, the density is <u>0.0844 lb/ft.³</u>

At 14.7 psia, 32 °F, the density is <u>0.0892 lb/ft.³</u>

(vii) The specific gravity of the mixture <u>0.169</u>

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 <em>x</em> 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 ≈ <u>19.314 psia</u>

Partial pressure of oxygen, P_{O_2} = 0.141 × 30 psia ≈ <u>4.23 psia</u>

Partial pressure of carbon dioxide, P_{CO_2} = 0.2153 × 30 psia ≈ <u>6.459 psia</u>

(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:

brainly.com/question/19813237

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Summary Table

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Unit                     mA     mA          μA            V           V          V

Value                  4.65    4.6039   46.039    10.7      10     4.6039

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q_{in} = 2745.8\,\frac{kJ}{kg} - 2217.2\,\frac{kJ}{kg}

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