To solve this problem, let's go step by step:

1. Graphing the Data:

You are provided with data showing the relationship between Temperature (K) and Pressure (atm). To graph the dataTo solve this problem, let's go step by step:

1. Graphing the Data:

You are provided with data showing the relationship between Temperature (K) and Pressure (atm). To graph the

• X-axis: Represent Temperature (K). Since the temperature values range from 50 to 500 K, you can space the x-axis accordingly, ensuring the values are evenly distributed.

• Y-axis: Represent Pressure (atm). The pressure values range from 0.1314 to 1.3161 atm. You can distribute the y-axis to accommodate this range, ensuring enough space between each value.

After plotting the points, draw a best-fit line through them.

2. Determine the Slope:

To find the slope ($m$) of the line, use the formula:

$m = \frac{\Delta y}{\Delta x} = \frac{\text{Change in Pressure}}{\text{Change in Temperature}}$

Pick two points from the table, for example:

• $(50, 0.1314)$
• $(500, 1.3161)$

Now calculate the slope: $m = \frac{1.3161 - 0.1314}{500 - 50} = \frac{1.1847}{450} \approx 0.00263 \ \text{atm/K}$

3. Determine the Intercept:

The intercept is the value of pressure when the temperature is 0 K. You can estimate this by extending the best-fit line back to the y-axis or using the slope-intercept form of the equation. The line equation is:

$P = mT + b$

Using one of the points, say $(50, 0.1314)$, substitute into the equation:

$0.1314 = 0.00263(50) + b$

Solve for $b$:

$b = 0.1314 - 0.1315 = -0.0001 \ \text{atm}$

Thus, the equation of the line is approximately: $P = 0.00263T - 0.0001$

4. Determine the Pressure at 350 K:

Using the equation of the line $P = 0.00263T - 0.0001$, substitute $T = 350 \ \text{K}$:

$P = 0.00263(350) - 0.0001 = 0.9205 - 0.0001 = 0.9204 \ \text{atm}$

Thus, the pressure at 350 K is approximately 0.9204 atm.

Conclusion:

• Slope: $0.00263 \ \text{atm/K}$
• Intercept: $-0.0001 \ \text{atm}$
• Equation of the line: $P = 0.00263T - 0.0001$
• Pressure at 350 K: $0.9204 \ \text{atm}$

Would you like me to help you graph this, provide further details, or clarify anything?

5 Additional Questions:

1. How would a change in volume affect the pressure in this scenario?
2. What assumptions are we making about the behavior of the gas?
3. How can you interpret the slope in terms of physical significance?
4. What would happen to the graph if we doubled the temperature range?
5. Could the data still follow the same pattern at extremely high temperatures?

Tip: When graphing manually, use a ruler to ensure your best-fit line is as accurate as possible!