Solution Exercise 1
The solution to Exercise 1 has two parts: a) and b).
a) Equation (4) is duplicated here.
[latex]C_{i,eq} = \frac{\left( P - e(T) \right)x_{i}}{K_{w,i}(T,S)}[/latex]
Common values needed for all gases are P and e(T). We calculate P as:
[latex]P = P_{0}\exp\left( - \frac{Z}{Z_{s}} \right) = 1\ atm\ exp\left( - \frac{450\ m}{8300\ m} \right)\ = \ 0.947\ atm[/latex]
We calculate e at 10 °C and convert to atmospheres as shown in the next equation.
[latex]e = 10^{A - \frac{B}{C + T}} = 10^{\left( 8.07131 - \ \frac{1730.63}{233.426 + 10} \right)} = 9.16\ mmHg\ \frac{1atm}{760\ mmHg} = 0.0121\ atm[/latex]
Using Kh,i at 10 °C from Table 1 and [latex]{x\ }_{i}[/latex]from Table 2, we have the values shown in this table.
| i | Kh,i
(atm g/ccSTP) |
xi | Ci,eq
(ccSTP/g) |
|---|---|---|---|
| He | 111 | 5.24[latex]\mathbf{\times}[/latex]10–6 | [latex]\frac{\left( \mathbf{0.947 - 0.0121} \right)\mathbf{5.24}\mathbf{\times}\mathbf{10}^{\mathbf{- 6}}}{\mathbf{111}}\mathbf{= 4.41}\mathbf{\times}\mathbf{10}^{\mathbf{- 8}}[/latex] |
| Ne | 89.0 | 1.82[latex]\mathbf{\times}[/latex]10–5 | [latex]\frac{\left( \mathbf{0.947 - 0.0121} \right)\mathbf{1.82}\mathbf{\times}\mathbf{10}^{\mathbf{- 5}}}{\mathbf{89.0}}\mathbf{= 1.91}\mathbf{\times}\mathbf{10}^{\mathbf{- 7}}[/latex] |
| N2 | 53.2 | 0.781 | [latex]\frac{\left( \mathbf{0.947 - 0.0121} \right)\mathbf{0.781}}{\mathbf{53.2}}\mathbf{= 1.37}\mathbf{\times}\mathbf{10}^{\mathbf{- 2}}[/latex] |
| Ar | 23.9 | 9.34[latex]\mathbf{\times}[/latex]10–3 | [latex]\frac{\left( \mathbf{0.947 - 0.0121} \right)\mathbf{9.34}\mathbf{\times}\mathbf{10}^{\mathbf{- 3}}}{\mathbf{23.9}}\mathbf{= 3.65}\mathbf{\times}\mathbf{10}^{\mathbf{- 4}}[/latex] |
| Kr | 12.4 | 1.14[latex]\mathbf{\times}[/latex]10–6 | [latex]\frac{\left( \mathbf{0.947 - 0.0121} \right)\mathbf{1.14}\mathbf{\times}\mathbf{10}^{\mathbf{- 6}}}{\mathbf{12.4}}\mathbf{= 8.60}\mathbf{\times}\mathbf{10}^{\mathbf{- 8}}[/latex] |
| Xe | 6.52 | 8.7[latex]\mathbf{\times}[/latex]10–8 | [latex]\frac{\left( \mathbf{0.947 - 0.0121} \right)\mathbf{8.7}\mathbf{\times}\mathbf{10}^{\mathbf{- 8}}}{\mathbf{6.52}}\mathbf{= 1.25}\mathbf{\times}\mathbf{10}^{\mathbf{- 8}}[/latex] |
b) Equation (9) is shown here.
[latex]C_{i} = \frac{\left( P - e(T) \right)x_{i}}{K_{w,i}(T,S)} + A'x_{i}[/latex]
Values for [latex]x_{i}[/latex] are given in Table 2, and are shown in the table in Solution Exercise 1a. The first term in Equation (9) was computed in Solution Exercise 1a. The second term (with A’ = 0.003 ccSTP/g), concentration (Ci), and percent change are shown in the next table.
| i | A’xi
(ccSTP/g) |
Ci
(ccSTP/g) |
Percent Change
[latex]\frac{\mathbf{C}_{\mathbf{i}}\mathbf{-}\mathbf{C}_{\mathbf{i,eq}}}{\mathbf{C}_{\mathbf{i,eq}}}\mathbf{\ X\ 100\%}[/latex] |
|
|---|---|---|---|---|
| He | 1.57 X 10–8 | 5.99 X 10–8 | 36.0% | |
| Ne | 5.45 X 10–8 | 2.46 X 10–7 | 29.0% | |
| N2 | 2.34 X 10–3 | 1.61 X 10–2 | 17.0% | |
| Ar | 2.80 X 10–5 | 3.93 X 10–4 | 7.7% | |
| Kr | 3.42 X 10–9 | 8.94 X 10–8 | 4.0% | |
| Xe | 2.61 X 10–10 | 1.27 X 10–8 | 2.1% |
As seen in this table, the addition of excess air has a much larger effect on the low solubility gases such as [latex]He[/latex] and [latex]Ne[/latex] than the higher solubility gases such as [latex]Kr[/latex] and [latex]Xe[/latex].