Answer:
[tex]1.75272\ \text{m}^3[/tex]
Explanation:
The breakdown reaction of ozone is as follows
[tex]CF_3Cl + UV \rightarrow CF_3 + Cl[/tex]
[tex]Cl + O_3 \rightarrow ClO + O_2[/tex]
[tex]O_3 + UV \rightarrow O_2 + O[/tex]
[tex]ClO + O \rightarrow Cl + O_2[/tex]
It can be seen that 2 moles of ozone is required in the complete cycle
So for 10 cycles, 20 moles of ozone is required
m = Mass of [tex]CF_3Cl[/tex] = 15.5 g
M = Molar mass of [tex]CF_3Cl[/tex] = 104.46 g/mol
P = Pressure = 24.5 mmHg
T = Temperature = 232 K
R = Gas constant = [tex]62.363\ \text{L mmHg/K mol}[/tex]
Number of moles is given by
[tex]n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{15.5}{104.46}\\\Rightarrow n=0.1484\ \text{moles}[/tex]
[tex]20\ \text{moles} = 20\times 0.1484 = 2.968\ \text{moles}[/tex]
From ideal gas law we have
[tex]PV=nRT\\\Rightarrow V=\dfrac{nRT}{P}\\\Rightarrow V=\dfrac{2.968\times 62.363\times 232}{24.5}\\\Rightarrow V=1752.72\ \text{L}=1.75272\ \text{m}^3[/tex]
For 20 cycles of the reaction the volume of the ozone is [tex]1.75272\ \text{m}^3[/tex].
The Total volume of Ozone measured is = 1.7527 m³
Given data :
Mass of CF₃Cl = 15.5 g
Molar mass of CF₃Cl = 104.46 g/mol
Pressure ( P ) = 24.5 mmHg
T = 232 K
Gas constant ( R ) = 62.363 L mmHg/K mol
Note : For a complete ozone cycle 2 moles of ozone is required therefore for 10 cycles 20 moles of ozone will be required
First step : Determine number of moles in each cycle reaction
n = m / M ----- ( 1 )
m = mass = 15.5 g
M = molar mass = 104.46 g/mol
∴ n = 15.5 / 104.46 = 0.1484
Hence the number of moles in 10 cycle = 20 * 0.1484 = 2.968 moles
Next step : Apply Ideal gas law to determine the volume of Ozone
PV = nRT ----- ( 2 )
n = 2.968
V = ?
R = 62.363 L mmHg/K mol
T = 232 K
P = 24.5 mmHg
∴ V = nRT / P --- ( 3 )
= ( 2.968 * 62.363 * 232 ) / 24.5
= 1752.72 L ≈ 1.7527 m³
Hence we can conclude that the Total volume of Ozone measured is = 1.7527 m³
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