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Sagot :
The solubility of the gas if the pressure is changed to 100 kPa is 0.742 g/L
Effect of Pressure on Solubility
As the pressure of a gas increases, the solubility increases, and as the pressure of a gas decreases, the solubility decreases.
Thus, Solubility varies directly with Pressure
If S represents Solubility and P represents Pressure,
Then we can write that
S ∝ P
Introducing proportionality constant, k
S = kP
S/P = k
∴ We can write that
[tex]\frac{S_{1} }{P_{1} } = \frac{S_{2} }{P_{2} }[/tex]
Where [tex]S_{1}[/tex] is the initial solubility
[tex]P_{1}[/tex] is the initial pressure
[tex]S_{2}[/tex] is the final solubility
[tex]P_{2}[/tex] is the final pressure
From the given information
[tex]S_{1} = 0.890 \ g/L[/tex]
[tex]P_{1} = 120 \ kPa[/tex]
[tex]P_{2} = 100 \ kPa[/tex]
Putting the parameters into the formula, we get
[tex]\frac{0.890}{120}=\frac{S_{2}}{100}[/tex]
[tex]S_{2}= \frac{0.890 \times 100}{120}[/tex]
[tex]S_{2}= 0.742 \ g/L[/tex]
Hence, the solubility of the gas if the pressure is changed to 100 kPa is 0.742 g/L
Learn more on Solubility here: https://brainly.com/question/4529762
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