Problem: Calculate the total volume of N2 gas that comes out of 2.5 L of water warming from 20°C to 50°C. The water is initially saturated with air at 1.00 atm. The solubility of O2(g) at 50°C is 27.8mg/L at 1.00 atm. The solubility of N2(g) at 50°C is 14.6mg/L at 1.00 atm. Henry’s Law constant for O2 at 20°C is 1.4 x 10-3 M/atm and for N2 is 6.3 x 10-4 M/atm. Air is 80% nitrogen and the rest oxygen.A. 7.29 mL B. 5.83 mL C. 2.75 mL D. 11.4 mL E. 15.6 mL

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FREE Expert Solution

We’re being asked to calculate for the total volume of N2 gas that comes out of water warming from 20°C to 50°C. The water is initially saturated with air at 1.00 atm.

Recall that saturated means that the maximum amount of air is dissolved in water (or solubility).

The solubility of a gas is given by Henry’s law:

Sgas=kH·Pgas

where:
Sgas = solubility of the gas (in mol/L or M)
KH = Henry’s law constant for the gas
Pgas = partial pressure of the gas


We’re going to calculate the volume of N2 gas using the following steps:

Step 1: Calculate the partial pressure of N2 gas at 20°C.
Step 2: Calculate the solubility of N gas at 20°C.
Step 3: Calculate the solubility of N2 gas at 50°C.
Step 4: Calculate the moles of N2 gas that comes out of water.
Step 5: Calculate the volume of N2 gas that comes out of the water using the ideal gas equation.

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Problem Details

Calculate the total volume of N2 gas that comes out of 2.5 L of water warming from 20°C to 50°C. The water is initially saturated with air at 1.00 atm. The solubility of O2(g) at 50°C is 27.8mg/L at 1.00 atm. The solubility of N2(g) at 50°C is 14.6mg/L at 1.00 atm. Henry’s Law constant for O2 at 20°C is 1.4 x 10-3 M/atm and for N2 is 6.3 x 10-4 M/atm. Air is 80% nitrogen and the rest oxygen.

A. 7.29 mL 

B. 5.83 mL 

C. 2.75 mL 

D. 11.4 mL 

E. 15.6 mL