Question

At a certain temperature, the solubility of N2 gas in water at 4.07 atm is 95.7 mg of N2 gas/100 g water . Calculate the solubility of N2 gas in water, at the same temperature, if the partial pressure of N2 gas over the solution is increased from 4.07 atm to 10.0 atm .

Answer 1

By applying Henry's Law, which states that solubility of a gas in a liquid is proportional to its partial pressure, the solubility of N₂ gas in water increases from 95.7 mg/100 g water at 4.07 atm to approximately 234.89 mg/100 g water at 10.0 atm.

Step-by-step explanation:

The question relates to the concept of gas solubility. Specifically, it applies Henry's Law, which states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the liquid. The solubility of N₂ gas at a certain temperature and a pressure of 4.07 atm is given as 95.7 mg/100 g water.

Using Henry's Law, we can set up a direct proportion to calculate the new solubility under a pressure of 10.0 atm:

• S1/P1 = S2/P2
  where S1 is the initial solubility (95.7 mg/100 g water), P1 is the initial pressure (4.07 atm), S2 is the final solubility, and P2 is the final pressure (10.0 atm).

• To find S2, we rearrange the equation: S2 = S1 * (P2/P1).

• Plugging in the known values and solving for S2 gives:
  S2 = (95.7 mg/100 g water) * (10.0 atm / 4.07 atm) = 234.89 mg/100 g water.

Therefore, the solubility of N₂ gas in water at a pressure of 10.0 atm is approximately 234.89 mg/100 g water.

Answer 2

Answer: Thus the solubility of
`N_2` gas in water, at the same temperature, if the partial pressure of gas is 10.0 atm is 235mg/100g.

Explanation:-

The Solubility of
`N_(2)` in water can be calculated by Henry’s Law. Henry’s law gives the relation between gas pressure and the concentration of dissolved gas.

Formula of Henry’s law,
`C=k_(H)P`.

`k_(H)`= Henry’s law constant = ?

The partial pressure (P) of
`N_(2)` in water = 4.07 atm

\
`C= k_(H)* P\\95.7mg=k_(H)* 4.07`

`k_(H)=23.5`

At pressure of 10.0 atm

`C= k_(H)* P\\C=23.5* 10.0=235mg/100mg`

Thus the solubility of
`N_2` gas in water, at the same temperature, is 235mg/100g