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mash [69]
3 years ago
9

a balloon filled with 750.0 ml of helium at a temperature of 25 degrees celsius and an atmosphere pressure of 745 mm Hg. What vo

lume will the balloon have when it reaches an altitude where the temperature is -35 degrees celsius and the pressure is 0.60 atm?
Chemistry
1 answer:
Softa [21]3 years ago
4 0

Answer:

V₂ = 0.98 L

Explanation:

Given data:

Initial volume = 750 mL  =0.75 L

Initial pressure = 745 mmHg (745/760 =0.98 atm)

Initial temperature = 25 °C (25 +273 = 298 K)

Final temperature = -35°C (-35+273 = 238 K)

Final volume = ?

Final pressure = 0.60 atm

Formula:

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

P₁V₁/T₁ = P₂V₂/T₂  

V₂ = P₁V₁T₂/T₁ P₂

V₂ = 0.98 atm × 0.75 L × 238 K / 298 K × 0.60 atm

V₂ = 174.93 atm .L. K / 178.8  K.atm

V₂ = 0.98 L

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Answer:

1.75272\ \text{m}^3

Explanation:

The breakdown reaction of ozone is as follows

CF_3Cl + UV \rightarrow CF_3 + Cl

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So for 10 cycles, 20 moles of ozone is required

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Number of moles is given by

n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{15.5}{104.46}\\\Rightarrow n=0.1484\ \text{moles}

20\ \text{moles} = 20\times 0.1484 = 2.968\ \text{moles}

From ideal gas law we have

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

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