The solubility of a gas is 0.54 g/L at 1.5 atm. what is its solubility at pressure of 0.80 atm and the same temperature?
0.54 g/L x (0.80/1.50) = ?
show work pls!
To determine the solubility of the gas at a different pressure, we can use Henry's Law, which states that the solubility of a gas is directly proportional to the partial pressure of that gas.
The equation that represents Henry's Law is:
C1/P1 = C2/P2
Where:
C1 is the initial solubility of the gas (0.54 g/L),
P1 is the initial pressure (1.5 atm),
C2 is the solubility at the target pressure (unknown),
P2 is the target pressure (0.80 atm).
Now we can plug the values into the equation and solve for C2:
C1/P1 = C2/P2
(0.54 g/L) / (1.5 atm) = C2 / (0.80 atm)
Cross-multiply and solve for C2:
C2 = (0.54 g/L) * (0.80 atm) / (1.5 atm)
C2 ≈ 0.288 g/L
Therefore, the solubility of the gas at a pressure of 0.80 atm and the same temperature is approximately 0.288 g/L.
To find the solubility of a gas at a different pressure but the same temperature, we can use the combined gas law equation. The combined gas law states that the ratio of the initial pressure to the final pressure is equal to the ratio of the initial solubility to the final solubility.
Let's assume that the initial solubility at 1.5 atm is S1, and the final solubility at 0.80 atm is S2.
According to the combined gas law:
(P1 / P2) = (S1 / S2)
Substituting the given values, we have:
(1.5 atm / 0.80 atm) = (0.54 g/L / S2)
To find S2, we can cross-multiply and solve for S2:
(1.5 atm * S2) = (0.80 atm * 0.54 g/L)
1.5S2 = 0.432
Divide both sides by 1.5:
S2 = 0.432 / 1.5
S2 ≈ 0.288 g/L
Therefore, the solubility of the gas at a pressure of 0.80 atm and the same temperature is approximately 0.288 g/L.