I need help with this question.
If the solubility of a gas in water is 6.0 g/L when the pressure of the gas above the water is 2.0 atm, what is the pressure of the gas above the water when the solubility of the gas is 3.0 g/L?
Since solubility is proportional to the pressure, wouldn't that be just half the initial P.
To solve this question, we can use Henry's Law, which states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.
Henry's Law can be represented as:
S = k * P
Where:
- S is the solubility of the gas in the liquid (in g/L),
- k is the Henry's Law constant,
- P is the partial pressure of the gas above the liquid (in atm).
We are given that when the solubility is 6.0 g/L, the pressure is 2.0 atm. Let's denote these values as S₁ and P₁ respectively.
S₁ = 6.0 g/L
P₁ = 2.0 atm
Now, we need to find the pressure (P₂) when the solubility is 3.0 g/L, denoted as S₂.
S₂ = 3.0 g/L
By rearranging the equation, we can solve for P₂:
P₂ = (S₂ / S₁) * P₁
Plugging in the values we have:
P₂ = (3.0 g/L / 6.0 g/L) * 2.0 atm
P₂ = 1.0 * 2.0 atm
P₂ = 2.0 atm
Therefore, the pressure of the gas above the water when the solubility of the gas is 3.0 g/L is 2.0 atm.