The solubility of a gas at 7.0 atm of pressure is 0.52 g/L . How many grams of the gas would be disolved per 1.0 L if the pressure increased 40.0 percent?
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To find the grams of gas dissolved per 1.0 L when the pressure increases by 40.0 percent, we can use the concept of Henry's Law.
Henry's Law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. In equation form, it can be written as:
C = k * P
Where:
C is the concentration or solubility of the gas in the liquid (in g/L),
k is the Henry's Law constant (specific to each gas and solvent), and
P is the partial pressure of the gas (in atm).
Since we are given the initial solubility of the gas at a pressure of 7.0 atm, we can write:
0.52 g/L = k * 7.0 atm
Next, we need to find the new concentration of the gas when the pressure increases by 40.0 percent. To do this, we calculate the new pressure as:
New pressure = 7.0 atm + (40.0% * 7.0 atm)
= 7.0 atm + (0.40 * 7.0 atm)
= 7.0 atm + 2.8 atm
= 9.8 atm
Now, we can apply Henry's Law again to find the new solubility:
C' = k * P'
Where C' is the new concentration of the gas, P' is the new pressure, and k is the same Henry's Law constant.
Substituting the values into the equation:
C' = k * 9.8 atm
We can rearrange the equations to solve for k:
k = C / P = 0.52 g/L / 7.0 atm
Now, we can use this value of k to find the new concentration:
C' = k * P'
= (0.52 g/L / 7.0 atm) * 9.8 atm
= 0.52 g/L * 9.8 atm / 7.0 atm
Simplifying the expression:
C' = 0.74 g/L
Therefore, the new concentration or solubility of the gas with a pressure increase of 40.0 percent would be 0.74 g/L.