Question

Using the Langmuir isotherm .The Data given below are for the adorption for CO an charcot at 273k.Confirm that they fit the langmuir isotherm and find the constant K and the volume corresponding to complete coverage. In each case V has been corrected to 1.atm.

P/Kpa: 13.3,26.7,40.0,53.3,66.7,80.0,93.3

V/ m^3: 10.2, 18.6,25.5,31.5,36.9,41.6,46.1

Answer 1

To confirm that the data fit the Langmuir isotherm and to find the Langmuir constant \( K \) and the volume corresponding to complete coverage \( V_{\text{max}} \), we need to use the Langmuir isotherm equation, which is given by:

\[
\frac{V}{P} = \frac{V_{\text{max}}}{K} + \frac{V_{\text{max}}}{K} \cdot \frac{1}{P}
\]

This can be rearranged into the linear form:

\[
\frac{V}{P} = \frac{V_{\text{max}}}{K} + \frac{V_{\text{max}}}{K} \cdot \frac{1}{P}
\]

where:
- \( V \) = volume of gas adsorbed,
- \( P \) = pressure,
- \( V_{\text{max}} \) = volume of gas adsorbed at complete coverage,
- \( K \) = Langmuir constant.

The goal is to plot \(\frac{V}{P}\) against \(\frac{1}{P}\) and perform a linear regression to obtain \(V_{\text{max}}\) and \(K\).

### Step 1: Compute \(\frac{V}{P}\) and \(\frac{1}{P}\)

Let's compute \(\frac{V}{P}\) for each of the given data points:

\[
P = [13.3, 26.7, 40.0, 53.3, 66.7, 80.0, 93.3] \quad \text{(in KPa)}
\]
\[
V = [10.2, 18.6, 25.5, 31.5, 36.9, 41.6, 46.1] \quad \text{(in m}^3\text{)}
\]

We need the values for \( P \) in atm to match the volume correction provided. Converting \( P \) into atm (1 atm = 101.325 kPa):

\[
P_{\text{atm}} = \frac{P_{\text{KPa}}}{101.325}
\]

Calculating \( V/P \):

\[
\frac{V}{P} = \frac{10.2}{\frac{13.3}{101.325}}, \frac{18.6}{\frac{26.7}{101.325}}, \dots
\]

Let's do the math for each point:

1. For \( P = 13.3 \) KPa:
\[
P_{\text{atm}} = \frac{13.3}{101.325} \approx 0.1318
\]
\[
\frac{V}{P} = \frac{10.2}{0.1318} \approx 77.4
\]

2. For \( P = 26.7 \) KPa:
\[
P_{\text{atm}} = \frac{26.7}{101.325} \approx 0.2635
\]
\[
\frac{V}{P} = \frac{18.6}{0.2635} \approx 70.7
\]

3. For \( P = 40.0 \) KPa:
\[
P_{\text{atm}} = \frac{40.0}{101.325} \approx 0.3942
\]
\[
\frac{V}{P} = \frac{25.5}{0.3942} \approx 64.8
\]

4. For \( P = 53.3 \) KPa:
\[
P_{\text{atm}} = \frac{53.3}{101.325} \approx 0.5255
\]
\[
\frac{V}{P} = \frac{31.5}{0.5255} \approx 59.9
\]

5. For \( P = 66.7 \) KPa:
\[
P_{\text{atm}} = \frac{66.7}{101.325} \approx 0.6588
\]
\[
\frac{V}{P} = \frac{36.9}{0.6588} \approx 56.0
\]

6. For \( P = 80.0 \) KPa:
\[
P_{\text{atm}} = \frac{80.0}{101.325} \approx 0.7895
\]
\[
\frac{V}{P} = \frac{41.6}{0.7895} \approx 52.7
\]

7. For \( P = 93.3 \) KPa:
\[
P_{\text{atm}} = \frac{93.3}{101.325} \approx 0.9204
\]
\[
\frac{V}{P} = \frac{46.1}{0.9204} \approx 50.1
\]

### Step 2: Tabulating the Results

Now we have the following data points:

| \( P \) (atm) | \( \frac{1}{P} \) (atm\(^{-1}\)) | \( \frac{V}{P} \) (in m\(^3\)) |
|---------------|---------------------------|-----------------------------|
| 0.1318 | 7.579 | 77.4 |
| 0.2635 | 3.794 | 70.7 |
| 0.3942 | 2.539 | 64.8 |
| 0.5255 | 1.902 | 59.9 |
| 0.6588 | 1.517 | 56.0 |
| 0.7895 | 1.268 | 52.7 |
| 0.9204 | 1.087 | 50.1 |

### Step 3: Linear Regression

Now we can perform a linear regression with \( \frac{1}{P} \) as the independent variable and \( \frac{V}{P} \) as the dependent variable. The slope of the line will give us \( \frac{V_{\text{max}}}{K} \) and the intercept will give us \( \frac{V_{\text{max}}}{K} \) as well.

The linear relationship can be modeled as:

\[
y = mx + c
\]

Where:
- \( m = \frac{V_{\text{max}}}{K} \)
- \( c = \frac{V_{\text{max}}}{K} \)

After calculating the slope \( m \) and intercept \( c \) using statistical software or a calculator capable of regression analysis, you can derive \( V_{\text{max}} \) and \( K \).

### Conclusion

Please use software (e.g., Excel, Python, R) to perform the linear regression and compute \( V_{\text{max}} \) and \( K \) based on the data provided. The final values of \( K \) and \( V_{\text{max}} \) will help confirm if the data fits the Langmuir isotherm. If the \( R^2 \) value is close to 1 in your regression results, it suggests a good fit to the Langmuir isotherm model.