Question

The small bubbles that form on the bottom of a water pot that is being heated (before boiling) are due to dissolved air coming out of solution.

Part A

Use Henry's law and the solubilities given below to calculate the total volume of nitrogen and oxygen gas that should bubble out of 1.1 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

At 25 C and a total pressure of 1.0 atm,

The partial pressure of Oxygen, P O₂ = 0.21 atm

The partial pressure of nitrogen, P N₂ = 0.78 atm

.

Given solubility at 50 C and 1 atm

Solubility of oxygen = 27.8 mg/ L

For nitrogen = 14.6 mg/ L

.

According to Henry's law, Henry's law constant, k = solubility/ pressure

.

So k for oxygen = 27.8 mg/L / 1 atm

                        = 27.8 mg/ (L.atm)

For nitrogen = 14.6 mg/ L/ 1 atm

                   =14.6 mg / (L.atm)

At 25 C, solubility = k. P

solubility of oxygen = 27.8 mg/ L.atm * (0.21 atm)

                            = 5.84 mg / L

Similarlly solubility of nitrogen = 14.6 mg/ (L.atm) * (0.78atm)

                                             =11.39 mg/ L

We are given 1.1 L of water

At 25 C, solubility = 5.84 mg / L

mass of oxygen present in 1.1 L of water = (5.84 mg / L) *(1.1 L)

                                                                    =6.424 mg

At 50 C, solubility = 27.8 mg/ L

Mass of oxygen = (27.8 mg/ L) * (1.1 L)

                         =30.58 mg

.

Hence the difference in mass of oxgyen at the two temperatures= 30.58 mg - 6.424 mg = 24.156 mg

This is the amount of oxygen which would be given out.

.

molar mass = 32 g/mol

Pressure = 1 atm

Temperature = 25 C = 298 K

mass = 24.156 mg = 0.024156 g

R = 0.0821 L-atm/ (mol.K)

Volume, V = ( mass/ molar mass) * R * T / P

V= [ ( 0.024156 g/ 32 g/mol) * (0.0821 L-atm/ (mol.K)) *(298 K) ]/ (1 atm)= 0.01847 L

           

the volume of oxygen is 0.01847 L

Similarly for nitrogen;

At 25 C, solubility = 11.39 mg/ L

Mass of nitrogen in 1.9 L = 11.39 mg / L * 1.1 L = 12.53mg

.

At 50 C solubility = 14.6 mg/ L

Mass of nitrogen = 14.6 mg / L * 1.1 L =16.06 mg

.

Mass of nitrogen released = 16.06 - 12.53 = 3.53 mg

.

molar mass = 28 g/mol

Pressure = 1 atm

Temperature = 25 C = 298 K

mass = 3.53 mg = 0.00353 g

R = 0.0821 L-atm/ (mol.K)

Volume, V = ( mass/ molar mass) * R * T / P

V =  [ ( 0.00353 g/ 28g/mol) * (0.0821 L-atm/ (mol.K)) *(298 K) ]/ (1 atm)= 0.00308 L

           

the total volume of nitrogen and oxygen in Liters is 0.01847 L + 0.00308 L = 0.02155 L = 21.55 mL