Question

in some cases, the relationship between two variables is not linear, but the data can be manipulated to make a linear relationship. using the data above, make a graph of the volume vs. pressure of a container, with volume being the independent variable. is the relationship linear? if it is not, make a graph of volume vs. 1/pressure. what would the pressure of the container be if the volume were 2.4 l? show your calculation. (table data: volume (l) and pressure (atm) with rows: 0.86, 22.9; 1.94, 10.3; 2.96, 6.7; 4.03, 5.0; 5.02, 3.9; 5.59, 3.5)

Explanation:

Step1: Check linearity of Volume vs Pressure

Plotting Volume (x) vs Pressure (y) shows a curve, not linear. So we check Volume vs \( \frac{1}{\text{Pressure}} \) (let \( y' = \frac{1}{P} \)).

Step2: Calculate \( y' \) for each data point

For \( V = 0.86 \), \( y' = \frac{1}{22.9} \approx 0.0437 \); \( V = 1.94 \), \( y' = \frac{1}{10.3} \approx 0.0971 \); \( V = 2.96 \), \( y' = \frac{1}{6.7} \approx 0.1493 \); \( V = 4.03 \), \( y' = \frac{1}{5.0} = 0.2 \); \( V = 5.02 \), \( y' = \frac{1}{3.9} \approx 0.2564 \); \( V = 5.59 \), \( y' = \frac{1}{3.5} \approx 0.2857 \).

Step3: Find linear regression for \( V \) vs \( y' \)

Using linear regression (or trendline), the equation is approximately \( y' = 0.049V + 0.002 \) (approximate from data).

Step4: Predict \( y' \) at \( V = 2.4 \)

Substitute \( V = 2.4 \) into \( y' = 0.049V + 0.002 \): \( y' = 0.049(2.4) + 0.002 = 0.1176 + 0.002 = 0.1196 \).

Step5: Find Pressure from \( y' \)

Since \( y' = \frac{1}{P} \), \( P = \frac{1}{y'} = \frac{1}{0.1196} \approx 8.36 \) atm. (Alternative: Use Boyle's Law \( P_1V_1 = P_2V_2 \), take average \( P_1V_1 \): \( 0.86×22.9≈19.694 \), \( 1.94×10.3≈20.0 \), \( 2.96×6.7≈19.832 \), average \( k≈20 \). Then \( P = \frac{k}{V} = \frac{20}{2.4}≈8.33 \) atm, consistent.)

Answer:

The pressure when volume is 2.4 L is approximately \(\boldsymbol{8.3}\) to \(\boldsymbol{8.4}\) atm (more precisely ~8.33 - 8.36 atm).