Answer:
1. Use Henry’s law to find the initial concentrations of oxygen and nitrogen gas in water at 25°C. Henry’s law states that the concentration of a gas in a liquid is proportional to the partial pressure of the gas above the liquid. The proportionality constant is called the Henry’s law constant and it depends on the temperature and the nature of the gas and the liquid. The formula for Henry’s law is:
C = kP
where C is the concentration of the gas in the liquid, k is the Henry’s law constant, and P is the partial pressure of the gas.
The solubilities given in the problem are actually the values of k for oxygen and nitrogen gas at 50°C. To find the values of k at 25°C, we need to use a table or a graph that shows how k changes with temperature.
Using this table, we can estimate that k for oxygen at 25°C is about 40 mg/L/atm and k for nitrogen at 25°C is about 20 mg/L/atm.
Now we can plug in the values of k and P to find C for oxygen and nitrogen at 25°C:
C_O2 = k_O2 * P_O2 = 40 mg/L/atm * 0.21 atm = 8.4 mg/L C_N2 = k_N2 * P_N2 = 20 mg/L/atm * 0.78 atm = 15.6 mg/L
2. Use the ideal gas law to find the initial moles of oxygen and nitrogen gas in water at 25°C. The ideal gas law states that the pressure, volume, temperature, and moles of a gas are related by the formula:
PV = nRT
where P is the pressure, V is the volume, n is the moles, R is the universal gas constant, and T is the temperature.
We can rearrange this formula to solve for n:
n = PV/RT
We know that P is 1 atm, V is 1.2 L, R is 0.0821 Latm/molK, and T is 298 K (25°C + 273). We can plug in these values to find n for oxygen and nitrogen:
n_O2 = PV/RT = (1 atm * 1.2 L) / (0.0821 Latm/molK * 298 K) = 0.049 mol n_N2 = PV/RT = (1 atm * 1.2 L) / (0.0821 Latm/molK * 298 K) = 0.049 mol
However, these are not the actual moles of oxygen and nitrogen gas in water, because some of them are dissolved in water. To find the moles of dissolved gas, we need to use the concentration and the volume of water:
n_dissolved_O2 = C_O2 * V = (8.4 mg/L) * (1.2 L) / (1000 mg/g) / (32 g/mol) = 0.00032 mol n_dissolved_N2 = C_N2 * V = (15.6 mg/L) * (1.2 L) / (1000 mg/g) / (28 g/mol) = 0.00067 mol
To find the moles of gas that are not dissolved, we need to subtract the moles of dissolved gas from the total moles:
n_undissolved_O2 = n_O2 - n_dissolved_O2 = 0.049 mol - 0.00032 mol = 0.049 mol n_undissolved_N2 = n_N2 - n_dissolved_N2 = 0.049 mol - 0.00067 mol = 0.048 mol
3. Use Henry’s law again to find the final concentrations of oxygen and nitrogen gas in water at 50°C. We can use the same formula as before, but with different values of k and P:
C_O2 = k_O2 * P_O2 = 27.8 mg/L/atm * 0.21 atm = 5.8 mg/L C_N2 = k_N2 * P_N2 = 14.6 mg/L/atm * 0.78 atm = 11.4 mg/L
4. Use the ideal gas law again to find the final moles of oxygen and nitrogen gas in water at 50°C. We can use the same formula as before, but with different values of V and T:
n_O2 = PV/RT = (1 atm * 1.2 L) / (0.0821 Latm/molK * 323 K) = 0.045 mol n_N2 = PV/RT = (1 atm * 1.2 L) / (0.0821 Latm/molK * 323 K) = 0.045 mol
However, these are not the actual moles of oxygen and nitrogen gas in water, because some of them are dissolved in water. To find the moles of dissolved gas, we need to use the concentration and the volume of water:
n_dissolved_O2 = C_O2 * V = (5.8 mg/L) * (1.2 L) / (1000 mg/g) / (32 g/mol) = 0.00022 mol n_dissolved_N2 = C_N2 * V = (11.4 mg/L) * (1.2 L) / (1000 mg/g) / (28 g/mol) = 0.00049 mol
To find the moles of gas that are not dissolved, we need to subtract the moles of dissolved gas from the total moles:
n_undissolved_O2 = n_O2 - n_dissolved_O2 = 0.045 mol - 0.00022 mol = 0.045 mol n_undissolved_N2 = n_N2 - n_dissolved_N2 = 0.045 mol - 0.00049 mol = 0.044 mol
5. Use the ideal gas law one more time to find the final volume of oxygen and nitrogen gas that bubbles out of water at 50°C. We can use the same formula as before, but with different values of n and P:
V_O2 = nRT/P = (0.045 mol * 0.0821 Latm/molK * 323 K) / (1 atm) = 1.20 L V_N2 = nRT/P = (0.044 mol * 0.0821 Latm/molK * 323 K) / (1 atm) = 1.17 L
6. Add the volumes of oxygen and nitrogen gas to get the total volume of gas that bubbles out of water:
V_total = V_O2 + V_N2 = 1.20 L + 1.17 L = 2.37 L
Therefore, the total volume of nitrogen and oxygen gas that should bubble out of 1.2 L of water upon warming from 25°C to 50°C is 2.4 L using two significant figures.