Answer:
V₂ = 120 dm³
Explanation:
Given data:
Volume of oxygen = 160 dm³
Initial temperature = 91 °C
Final volume = ?
Final temperature = 0 °C
Solution:
First of all we will convert the temperature into kelvin.
91 + 273 = 364 k
0 + 273 = 273 K
According to Charle's law,
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 160 dm³ × 273 k / 364 k
V₂ = 43680 dm³.K / 364 k
V₂ = 120 dm³
The volume of the oxygen would be 120 dm³
From the question, we are determine the final volume of the oxygen
From Charles's law equation, we have
[tex]\frac{V_{1} }{T_{1}} = \frac{V_{2} }{T_{2}}[/tex]
Where V₁ is the initial volume
T₁ is the initial temperature
V₂ is the final volume
T₂ is the final temperature
From the given information
V₁ = 160 dm³ = 160 L
T₁ = 91 °C = 91 + 273.15 K = 364.15 K
V₂ = ?
T₂ = 0.00 °C = 0.00 + 273.15 K = 273.15 K
Putting the parameters into the formula,
[tex]\frac{160}{364.15} =\frac{V_{2} }{273.15}[/tex]
[tex]V_{2}= \frac{160 \times 273.15}{364.15}[/tex]
[tex]V_{2} = 120.01 \ L[/tex]
V₂ ≅ 120 L = 120 dm³
Hence, the volume of the oxygen would be 120 dm³
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