what is the solubility of N2 at a pressure of 10 atm?
partial pressure at 1 atm=0.78 atm
solubility at 1 atm= 5.3 x 10^-3 mol/L
To find the solubility of N2 at a pressure of 10 atm, you can utilize 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 this case, you are given the solubility of N2 at a partial pressure of 1 atm, which is 5.3 x 10^-3 mol/L. To find the solubility at 10 atm, you can use the following formula:
solubility at 10 atm = (partial pressure at 10 atm / partial pressure at 1 atm) x solubility at 1 atm
Given that the partial pressure at 1 atm is 0.78 atm, you can substitute the values into the formula:
solubility at 10 atm = (10 atm / 0.78 atm) x (5.3 x 10^-3 mol/L)
After performing the calculation, you will find the solubility of N2 at a pressure of 10 atm.