Question

Use Henry's law and the solubilities given below to calculate the total volume of nitrogen and oxygen gas that should bubble out of 1.0 L of water upon warming from 25 ∘C to 50 ∘C. Assume that the water is initially saturated with nitrogen and oxygen gas at 25 ∘C and a total pressure of 1.0 atm. Assume that the gas bubbles out at a temperature of 50 ∘C. The solubility of oxygen gas at 50 ∘C is 27.8 mg/L at an oxygen pressure of 1.00 atm. The solubility of nitrogen gas at 50 ∘C is 14.6 mg/L at a nitrogen pressure of 1.00 atm. Assume that the air above the water contains an oxygen partial pressure of 0.21 atm and a nitrogen partial pressure of 0.78 atm.

Answer 1

Answer:

Explanation:

Given:

• At 25∘C and total pressure of 1.0 atm,

• Partial pressure of Oxygen, PO2 = 0.21 atm and,

• Partial pressure of nitrogen, PN2 = 0.78 atm.

At 50∘C and 1 atm,

• Solubility of oxygen = 27.8 mg/ L and solubility of nitrogen = 14.6 mg/ L

Apply the Henry law, Pi = H x Ci.................equation1

• Where Pi is the partial pressure of species 'i' and H is Henry constant and Ci is the solubility of species 'i'., Henry constant for oxygen is HO2 = 769.23 L atm/mol

From the equation (1) the solubility of oxygen at 25∘C;

• = 0.21/769.23 = 0.000273mol/L

• or C'(O2) = 0.000273 X 32g X gL/mol X mol/L = 8.736mg

Therefore ; the solubility of oxygen at 50oC and pressure @ 1 atm,

• = 27.8 X 1L X mg/L = 27.8mg

• but, the partial pressure of oxygen in the solution is 0.21 atm, hence the solubility will be decreased to 27.8 x 0.21 atm/1 atm = 5.838mg

• from 25∘C - 50∘C ; 8.736 - 5.838 = 2.898 mg of oxygen should be bubble out.

Hence, Volume of oxygen = Mass/density = 2.898mg/1.429mg/L = 2.027mL

• Similarly, For nitrogen, At 50∘C and 1 atm,

• solubility of nitrogen = 14.6 mg/ L

• Hence, Henry constant for Nitrogen is HN2 = 1639.34 L atm/mol,

From the equation (1) the solubility of nitrogen at 25oC ;

• C(N2) = 0.78/1639.34 = 0.0004758mol/L

• or Ci = 0.0004758mol/L X 28g X 1L/mol X mol/L = 13.322mg

Hence, The solubility of Nitrogen at 50oC and pressure @ 1 atm,;

• Ci = 14.6 X 1L X mg/L = 14.6mg

But the partial pressure of Nitrogen in the solution is 0.78 atm, hence the solubility will be reduced to 14.6 x 0.78 atm/1 atm = 11.388 mg

warming from 25∘C - 50∘C; 13.322 - 11.388 = 1.934 mg of Nitrogen

• Hence, volume of nitrogen = mass/density = 1.934mg/1.2506mg/L

• = 1.55mL

Answer 2

Answer:

Explanation:

Henry's law relates the solubility of a gas in a solvent to it's partial pressure above the solution. P=k*solubility, where P is the pressure and k is the Henry's law constant. The Henry's law constants are temperature dependent.

There's basically three steps here (I'll just talk about oxygen here, the procedure for nitrogen is analogous). You first need to calculate how much oxygen is dissolved at 25 C. This is what you need the room temperature Henry's law constant for. You can then plug the constant and partial pressure (.21 atm) into Henry's law to get the concentration, and that can be converted into amount since you know the volume of water (1.0 L).

Alright, next you need to calculate how much oxygen will be dissolved at 50 C. You can do this by first finding the Henry's law constant, which you can do since you know the solubility at 1.00 atm and can plug that into Henry's law (k*27.8 mg/L=1.00 atm), and then use that to figure out the concentration at a pressure of .21 atm. And then translate that to amount of oxygen.

So now you know how much oxygen is dissolved at 25 C, and how much oxygen will be dissolved in 50 C. So, obviously, the difference is how much oxygen is released; translate this into volume using the ideal gas law to figure out what the volume of that amount of oxygen is.

Be careful with units throughout, that may well be the trickiest part.