The solubility of a gas is 0.34 g/L at STP. What is its solubility at a pressure of 0.80 atm and the same temperature?
Henry's law requires that solubility of a gas in a liquid depends directly on pressure.
ANS: .34g/L * .80
To determine the solubility of a gas at a given pressure and temperature, we can use Henry's law. Henry's law states that the solubility of a gas is directly proportional to the partial pressure of the gas above the liquid solvent.
The equation for Henry's law is:
C = k * P
where:
C is the solubility of the gas in moles per liter (mol/L)
k is the Henry's law constant (mol/(L * atm))
P is the partial pressure of the gas (atm)
To solve this problem, we need to find the value of k.
At standard temperature and pressure (STP), the solubility of the gas is given as 0.34 g/L. The molar mass of the gas is needed to convert grams to moles.
Let's assume the molar mass of the gas is M grams/mol.
To convert the solubility from grams per liter to moles per liter, we use the following conversion:
C (in moles/L) = (solubility in grams / molar mass) (in moles/grams)
Using the given solubility at STP:
C (in moles/L) = 0.34 g/L / M g/mol
Now, let's rearrange the Henry's law equation and solve for k:
k = C / P
Substituting the known values:
k = (0.34 g/L / M g/mol) / 1 atm
So, the value of k is (0.34 g/L / M g/mol) / 1 atm.
Next, we can determine the solubility at a pressure of 0.80 atm using Henry's law.
Let's call the solubility at this pressure C2. We can rearrange the Henry's law equation to solve for C2:
C2 = k * P2
Substituting the known values:
C2 = [0.34 g/L / M g/mol) / 1 atm] * 0.80 atm
Simplifying the expression:
C2 = (0.34 g/L * 0.80 atm) / (M g/mol * 1 atm)
C2 = (0.272 g-atm / L) / (M g/mol)
Therefore, the solubility of the gas at a pressure of 0.80 atm and the same temperature is (0.272 g-atm / L) / (M g/mol).
To determine the solubility of a gas at a different pressure, we can use Henry's law. According to Henry's law, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.
Mathematically, Henry's law is expressed as:
C = k * P
Where C is the solubility of the gas, P is the partial pressure of the gas, and k is the proportionality constant.
In this case, we want to find the solubility of the gas at a pressure of 0.80 atm but have the solubility value at STP (standard temperature and pressure), which is 0.34 g/L.
Since the temperature is constant, we assume that the proportionality constant remains the same.
To solve for the solubility (C) at the new pressure (P), we can set up a proportion:
C1 / C2 = P1 / P2
Where C1 is the solubility at the first pressure (STP), C2 is the solubility at the second pressure, P1 is the first pressure (STP - 1 atm), and P2 is the second pressure (0.80 atm).
Plugging in the given values:
0.34 g/L / C2 = 1 atm / 0.80 atm
Now, we can cross-multiply and solve for C2:
0.34 g/L * 0.80 atm = C2 * 1 atm
C2 = (0.34 g/L * 0.80 atm) / 1 atm
C2 = 0.272 g/L
Therefore, the solubility of the gas at a pressure of 0.80 atm and the same temperature is 0.272 g/L.