A student repeated Experiment 2 using methane gas. The table shows the experimental results.

\begin{tabular}{cccc}
Pressure (atm) & Mass [tex]$(g)$[/tex] & Syringe Volume [tex]$(mL)$[/tex] & Temperature [tex]$(^\circ C)$[/tex] \\
\hline
1.05 & 0.127 & 9.10 & 21.5 \\
1.10 & 0.133 & 18.20 & 21.5 \\
1.15 & 0.139 & 27.30 & 21.5 \\
1.35 & 0.151 & 45.50 & 21.5 \\
1.45 & 0.163 & 63.70 & 21.5 \\
\hline
\end{tabular}

Using Excel, plot total volume ([tex]$y$[/tex]-axis) versus number of moles of methane ([tex]$x$[/tex]-axis). What is the slope ([tex]$m$[/tex]) and [tex]$y$[/tex]-intercept ([tex]$b$[/tex]) of the trendline?

Slope:
Enter numeric value

[tex]$y$[/tex]-intercept:
Enter numeric value

Units: [tex]$mL / mol$[/tex]



Answer :

Let's find the slope [tex]\( (m) \)[/tex] and the y-intercept [tex]\( (b) \)[/tex] of the trendline for the given experimental data.

### Step-by-Step Solution:

1. Given Data:

- Pressures (atm): 1.05, 1.10, 1.15, 1.35, 1.45
- Masses (grams): 0.127, 0.133, 0.139, 0.151, 0.163
- Volumes (mL): 9.10, 18.20, 27.30, 45.50, 63.70
- Temperature: 21.5°C

2. Constants and Conversions:

- Ideal gas constant [tex]\( R \)[/tex] = 0.0821 L·atm/(K·mol)
- Molar mass of methane (CH₄) ≈ 16.04 g/mol

3. Convert Mass to Moles:

The number of moles ([tex]\( n \)[/tex]) of methane is calculated using the formula:
[tex]\[ n = \frac{\text{mass}}{\text{molar mass}} \][/tex]
Applying this to each mass in the data:

[tex]\[ \begin{aligned} n_1 & = \frac{0.127}{16.04} \approx 0.00792 \\ n_2 & = \frac{0.133}{16.04} \approx 0.00829 \\ n_3 & = \frac{0.139}{16.04} \approx 0.00867 \\ n_4 & = \frac{0.151}{16.04} \approx 0.00941 \\ n_5 & = \frac{0.163}{16.04} \approx 0.01016 \\ \end{aligned} \][/tex]

4. Plot the Data:

On the x-axis, plot the number of moles ([tex]\( n \)[/tex]) and on the y-axis, plot the corresponding volumes (mL):
[tex]\[ \begin{aligned} & (0.00792, 9.10) \\ & (0.00829, 18.20) \\ & (0.00867, 27.30) \\ & (0.00941, 45.50) \\ & (0.01016, 63.70) \\ \end{aligned} \][/tex]

5. Linear Regression:

Perform a linear regression to find the slope ([tex]\( m \)[/tex]) and y-intercept ([tex]\( b \)[/tex]) of the line that best fits the plotted data.

6. Result:

After performing the regression analysis:

- Slope [tex]\( (m) \)[/tex]: [tex]\( 24327.33 \, \text{mL/mol} \)[/tex]
- Y-intercept [tex]\( (b) \)[/tex]: [tex]\( -183.52 \, \text{mL} \)[/tex]

So, the slope [tex]\( (m) \)[/tex] is approximately [tex]\( 24327.33 \, \text{mL/mol} \)[/tex] and the y-intercept [tex]\( (b) \)[/tex] is approximately [tex]\( -183.52 \, \text{mL} \)[/tex].