Answer : The solubility of gas at pressure 250 kPa is, 1.39 g/L
Explanation:
According top the Henry's Law, the solubility of a gas in a liquid is directly proportional to the pressure of the gas.
[tex]S\propto P[/tex]
or,
[tex]\frac{S_1}{S_2}=\frac{P_1}{P_2}[/tex]
where,
[tex]S_1[/tex] = initial solubility of gas = 0.58 g/L
[tex]S_2[/tex] = final solubility of gas = ?
[tex]P_1[/tex] = initial pressure of gas = 104 kPa
[tex]P_2[/tex] = final pressure of gas = 250 kPa
Now put all the given values in the above formula, we get the final solubility of the gas.
[tex]\frac{0.58g/L}{S_2}=\frac{104kPa}{250kPa}[/tex]
[tex]S_2=1.39g/L[/tex]
Therefore, the solubility of gas at pressure 250 kPa is, 1.39 g/L
