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Vsevolod [243]
3 years ago
15

Calculate the number of pounds of CO2CO2 released into the atmosphere when a 22.0 gallon22.0 gallon tank of gasoline is burned i

n an automobile engine. Assume that gasoline is primarily octane, C8H18,C8H18, and that the density of gasoline is 0.692 g⋅mL−1.0.692 g⋅mL−1. This assumption ignores additives. Also, assume complete combustion. CO2CO2 released:
Chemistry
1 answer:
Gnesinka [82]3 years ago
3 0

Answer:

391.28771 pounds of carbon-dioxide was released into the atmosphere when 22.0 gallon tank of gasoline is burned in an automobile engine.

Explanation:

Density of the gasoline ,d= 0.692 g/mL

Volume of gasoline in an tanks,V = 22.0 gallons = 83,279.02 mL

Let mass of the gasoline be M

Density= \frac{Mass}{Volume}

M = V × d = 83,279.02 mL × 0.692 g/mL=57,629.081 g

Given that gasoline is primarily octane.

2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O

Mass of octane burnt in the tank = M = 57,629.081 g

Moles of octane =\frac{57,629.081 g}{114.08g/mol}=505.1637 mol

According to reaction, 2 moles of octane gives 16 moles of carbon-dioxide.

Then 505.1637 mol of octane will give:

\frac{16}{2}\times 505.1637 mol=4,041.3100 mol of carbon-dioxide

Mass of 4,041.3100 mol of carbon-dioxide:

4,041.3100 mol × 44.01 g/mol = 177,858.05 g

Mass of carbon-dioxide produced in pounds = 391.28771 pounds

391.28771 pounds of carbon-dioxide was released into the atmosphere when 22.0 gallon tank of gasoline is burned in an automobile engine.

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3 years ago
10-kg of R-134a at 300 kPa fills a rigid container whose volume is 14 L. Determine the temperature and total enthalpy in the con
Mariulka [41]

Answer:

Temperature = 0.605°C

Total enthalpy at 300kpa = 545.2 kJ

Total enthalpy at 600kpa = 846.45 kJ.

Explanation:

Checking the table for 134a pressure table. It is given that the specific volume of saturated liquid and the specific volume of the saturated vapor of 280kpa is 0.0007699 m^3/kg and 0.072352 m^3/kg respectively.

Also, the specific volume of saturated liquid and the specific volume of the saturated vapor of 320kpa is 0.0007772 m^3/kg and 0.063604 m^3/kg respectively.

The first thing to do is to determine the value for the specific volume of saturated liquid.

At 300 kpa, the specific volume of saturated liquid,n is given below as;

300 - 280/ 320 - 280 = (n - 0.0007699)/ 0.0007772 - 0.0007699.

Therefore, n = specific volume of saturated liquid = 0.0007735 m^3/kg.

300 - 280/ 320 - 280 = n - 0.072352/ (0.063604 - 0.072352).

n = 0.0679 m^3/kg.

The second thing to do is to determine the value of the specific volume.

Specific volume = 14 × 10^-3/ 10 = 0.0014 m^3/kg.

Determine the enthalpy of the mixture,b(I). This is given below as;

300 - 280/ 320 - 280 = b(I) - 199.54/ (196.7 - 199.54).

b(I) = 198.125 kJ/Kg.

Hence, b = [ 300 - 280/ 320 - 280 = j - 50.18 / 55.16 - 50.18] + [ ( 0.0014 - 0.00077735) / 0.067978 - 0.00077735] × 198.125.

b = 54.517 KJ/Kg.

Total enthalpy = 10 × 54.517 = 545.17 kJ.

Temperature can be Determine as below;

300 - 280/ 320 - 280 = T + 1.25 / 2.46 - 1.25.

Temperature = 0.605°C.

Hence, at 600kpa, the total enthalpy = [81.51 + ( 0.0014 - 0.0008199/ 0.034295 - 0.0008199) × 180.90] × 10

Total enthalpy at 600kpa = 846.45 kJ.

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2 years ago
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