the solubility of a gas is 0.34g/L at STP. What is it's solubility at a pressure of 0.80 atm and the same temperature?
Use Henry's Law.
To determine the solubility of a gas at a different pressure, you can use Henry's law, which states that the solubility of a gas is directly proportional to its partial pressure. Henry's law equation can be written as:
C1/P1 = C2/P2
Where:
C1 = initial solubility of the gas (in g/L)
P1 = initial pressure of the gas (in atm)
C2 = final solubility of the gas (in g/L)
P2 = final pressure of the gas (in atm)
Now let's plug in the values given in the problem:
C1 = 0.34 g/L (initial solubility)
P1 = 1 atm (STP conditions)
P2 = 0.80 atm (final pressure)
We can rearrange Henry's law equation to solve for C2:
C2 = (C1 * P2) / P1
Now let's substitute the values into the equation:
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.