if .55 grams of a gas dissolves in 1 liter of water at 20 kPa of pressure, how much will dissolve at 110 Kpa of pressure?
To determine how much gas will dissolve at 110 kPa of pressure, we can use Henry's law, which states that the amount of gas dissolved in a liquid is directly proportional to its partial pressure.
Henry's law equation can be written as:
C1/P1 = C2/P2
Where:
C1 = Initial concentration of the gas (in grams per liter)
P1 = Initial pressure (in kPa)
C2 = Final concentration of the gas (in grams per liter)
P2 = Final pressure (in kPa)
We'll use this equation to solve for C2, the final concentration of the gas.
Given:
C1 = 0.55 grams per liter (the initial concentration of the gas)
P1 = 20 kPa (the initial pressure)
P2 = 110 kPa (the final pressure)
Plugging in these values into the equation:
0.55 g/L / 20 kPa = C2 / 110 kPa
To solve for C2, we can cross multiply and then divide:
(0.55 g/L) * 110 kPa = C2 * 20 kPa
60.5 g * kPa / L = C2 * 20 kPa
Next, we divide both sides of the equation by 20 kPa:
(60.5 g * kPa / L) / 20 kPa = C2
The units of kPa cancel out:
60.5 g / (20 L) = C2
Calculating:
C2 = 3.025 g/L
Therefore, at 110 kPa of pressure, approximately 3.025 grams of the gas will dissolve in 1 liter of water.