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At a festival, spherical balloons with a radius of 280 cm are to be inflated with hot air and released. The air at the festival will have a temperature of 25° C and must be heated to 100°C to make the balloons float. 3.00 kg of propane (C₃H₈) fuel are available to burn to heat the air. Calculate the number of balloons that can be inflated with hot air.
Density of air at 100°C = [tex]0.946kg/m^3[/tex]
Specific heat capacity of air = 1.009 J/g°C
Answer: The number of balloons inflated with hot air are 7.
Explanation:
The equation used to calculate enthalpy change is of a reaction is:
[tex]\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}][/tex]
The chemical equation for the combustion of propane follows:
[tex]C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H_{rxn}=[(3\times \Delta H_f_{(CO_2(g))})+(4\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(C_3H_8(g))})+(5\times \Delta H_f_{(O_2(g))})][/tex]
We are given:
[tex]\Delta H_f_{(H_2O(g))}=-241.8kJ/mol\\\Delta H_f_{(CO_2(g))}=-393.51kJ/mol\\\Delta H_f_{(C_3H_8(g))}=-103.8kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta H_{rxn}=[(3\times (-393.51))+(4\times (-241.8))]-[(1\times (-103.8))+(5\times (0))]\\\\\Delta H_{rxn}=-2043.93kJ/mol[/tex]
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of propane = 3.00 kg = 3000 g (Conversion factor: 1 kg = 1000 g)
Molar mass of propane = 44 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of propane}=\frac{3000g}{44g/mol}=68.18mol[/tex]
To calculate the heat released, we use the equation:
[tex]\Delta H_{rxn}=\frac{q}{n}[/tex]
where,
q = amount of heat released = ? kJ
n = number of moles = 68.18 moles
[tex]\Delta H_{rxn}[/tex] = enthalpy change of the reaction = -2043.93 kJ/mol
Putting values in above equation, we get:
[tex]-2043.93kJ/mol=\frac{q}{68.18mol}\\\\q=(-2043.93kJ/mol\times 68.18mol)=-1.39\times 10^5kJ[/tex]
To calculate the heat released by the reaction, we use the equation:
[tex]q=mc\Delta T[/tex]
where,
q = heat released = [tex]1.39\times 10^5kJ=1.39\times 10^8J[/tex]
m = mass of air = ?
c = heat capacity of air = 1.009 J/g°C
[tex]\Delta T[/tex] = change in temperature = [tex]T_2-T_1=(100-25)^oC=75^oC[/tex]
Putting values in above equation, we get:
[tex]1.39\times 10^8J=m\times 1.009J/g^oC\times 75^oC\\\\m=\frac{1.39\times 10^8}{1.009\times 75}=1.84\times 10^6g=1.84\times 10^3kg[/tex]
Conversion factor: 1 kg = 1000 g
To calculate the volume of air, we use the equation:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
Density of air = [tex]0.946kg/m^3[/tex]
Mass of air = [tex]1.84\times 10^3kg[/tex]
Putting values in above equation, we get:
[tex]0.946kg/m^3=\frac{1.84\times 10^3kg}{\text{Volume of air}}\\\\\text{Volume of air}=\frac{1.84\times 10^3kg}{0.946kg/m^3}=1945.03m^3[/tex]
To calculate the volume of sphere, we use the equation:
[tex]V=4\pi r^3[/tex]
where,
r = radius of the sphere = 280 cm = 2.8 m (Conversion factor: 1 m = 100 cm)
Putting values in above equation, we get:
[tex]V=4\times 3.14\times (2.8)^3\\\\V=275.7m^3[/tex]
To calculate the number of balloons inflated, we divide the volume of air by the volume of 1 balloon:
[tex]\text{Number of balloons inflated}=\frac{V_{air}}{V_{balloon}}\\\\\text{Number of balloons inflated}=\frac{1945.03m^3}{275.7m^3}=7[/tex]
Hence, the number of balloons inflated with hot air are 7.
Answer:
7 balloons are inflated in the given air of 280 m radius.
Explanation:
The enthalpy change for the combustion of propane gas is -2043.93 kJ/mole.
The moles of propane available in the reaction is:
Moles = [tex]\rm \frac{weight}{molecular weight}[/tex]
moles of propane = [tex]\rm \frac{3000\;grams}{44\;gram/mole}[/tex]
Moles of propane available (n) = 68.18 moles.
We need to calculate the volume of air available and the volume of air for one ballon.
For calculating the volume of air,
First calculate the heat released in the reaction:
heat released (q) = [tex]\Delta[/tex]H [tex]\times[/tex] n
q = -2043.93 [tex]\times[/tex] 68.18
q = -1.39 [tex]\times\;\rm 10^5[/tex] kJ
Change in temperature ([tex]\Delta[/tex]T) = 100 - 25
[tex]\Delta[/tex]T = [tex]\rm 75^\circ[/tex] C
So, mass of air = [tex]\rm \frac{heat\;released}{heat\;capacity\;of\;air\;\times\;change\;in\;temperature}[/tex]
mass of air = [tex]\rm \frac{1.39\;\times\;10^8\;J}{1.009\;\times\;75}[/tex]
mass of air = 1.84 [tex]\rm \times\;10^3[/tex] kg
Volume of air = [tex]\rm \frac{mass}{density}[/tex]
volume of air = [tex]\rm \frac{1.84\;\times\;10^3}{0.946\;kg/m^3}[/tex]
volume of air = 1945.03 [tex]\rm m^3[/tex].
Volume of balloon = [tex]\rm 4\pi r^3[/tex]
Volume of balloon = [tex]\rm 4\;\times\;(2.8\;m)^3\;\times\;\frac{22}{7}\;m^3[/tex]
Volume of balloon = 275.5 [tex]\rm m^3[/tex]
So, total number of ballons inflated by the known volume of air = [tex]\rm \frac{1945.03}{275.5}[/tex]
The total number of ballons inflated = 7.
There are 7 balloons inflated by the 3 kg of propane.
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