If 0.75 g of a gas dissolves in 1.0 L of water at 20.0 kPa of pressure, how much will dissolve at 105.4 kPa of pressure?
(0.75g/20) = (x g/105.4)
Solve for x.
To determine how much gas will dissolve at 105.4 kPa of pressure, we need to first understand the relationship between pressure and the solubility of gases in liquids.
According to Henry's Law, the solubility of a gas in a liquid is directly proportional to the partial pressure of that gas above the liquid. Mathematically, it can be expressed as:
C = k * P
Where:
C is the concentration of the gas in the liquid (in mol/L)
k is the Henry's Law constant for the gas and solvent system
P is the partial pressure of the gas (in kPa)
In this case, we can assume that the Henry's Law constant (k) remains constant as the gas and solvent are the same.
Let's begin by calculating the concentration (C1) of the gas in the water at 20.0 kPa.
C1 = k * P1
Given that 0.75 g of the gas dissolves in 1.0 L of water, we can use this information to find the concentration in mol/L.
1.0 L of water is equivalent to 1000 mL.
Therefore, the concentration (C1) in mol/L is calculated as follows:
C1 = (0.75 g / Molar mass of the gas) / (1000 mL / 1000)
After determining C1, we can set up a proportional relationship to find the concentration (C2) of the gas at 105.4 kPa.
C1 / P1 = C2 / P2
Where:
C1 is the initial concentration of the gas (in mol/L) at P1,
P1 is the initial pressure (in kPa),
C2 is the final concentration of the gas (in mol/L) at P2,
and P2 is the final pressure (in kPa).
Rearranging the equation, we can solve for C2:
C2 = (C1 / P1) * P2
Now we can substitute the known values:
C2 = (C1 / 20.0) * 105.4
Using the calculated value of C1, we can now solve for C2.