Question

I need help on how to find the atmospheric pressure from the scatter plot and trend line I made with the results from my Boyle's Law lab. I don't know if my scatter plot is correct. I need the atmospheric pressure in order to fill in Table 5 and answer some questions.

24. Use the data in Data Table 1 and Data Table 3 to create a scatter-plot graph of 1/Average Volume vs. Pressure; with “1/Average Volume (mL-1)” on the x-axis and “Pressure (Pa)” on the y-axis.

25. Add a linear trend-line to the scatter-plot, and display the equation of the line on the graph (this is a typical function of graphing programs).

26. Resize and insert an image of the graph into Data Table 4. 27. Use the equation of the line to determine atmospheric pressure; As the trend-line equation is displayed as y=mx + b, b is the value where y is 0, thus the absolute value of b is roughly representative of atmospheric pressure. Record the “atmospheric pressure” into Data Table 5 of your Lab Report Assistant.

27. Use the equation of the line to determine atmospheric pressure; As the trend-line equation is displayed as y=mx + b, b is the value where y is 0, thus the absolute value of b is roughly representative of atmospheric pressure. Record the “atmospheric pressure” into Data Table 5 of your Lab Report Assistant.

(This graph is created with the results from my chem lab and I'm not sure if I should make the trend line starting at 0)

W UCalibri CHM150 Lab 3-Boyles Law Lab Assistant Microsoft Word Home Insert Page Layout References Rricur Fil Mailings View rind XCut 21 Aa A A A Calibri 12 AaBbCcDd AaBbCcDd AaBbCcDd Aa BbCcDc AaBbCcDc AaB AaBb ae Replace Copy Change Styles BIUabu x. x Paste AA 1 List Paragr.. List Paragr... List Paragr. 1Normal No Spacing Heading Heading 2 1 Solect Format Painter Clipboard Styles Font Paraaph Cditing 1/Average Volume vs. Pressure Pressure(Pa) 140000 120000 100000 80000 60000 Pressure(Pa) 40000 20000 0 0.01 0.02 0.03 0.04 0,05 0.06 0.07 1/Average Volume (1/mL) nglish (U.S.) 120 % - Page: 3 ul 6 Words: 635 Pressure (pa

28. For each number of books calculate the total pressure, by adding the atmospheric pressure to the pressure values in Data Table 1. Record the total pressure for each number of books in Data Table 5.

29. Calculate the product of “Total Pressure” multiplied by the “Average Volume” for each number of books and record in Data Table 5. Relating this back to the information presented in the Background section, Boyle’s law is written as PV = k, therefore the values calculated in this step represent the constant (k) in Boyle’s law.

Questions:

(A)

***(Atmospheric pressure in NJ is 101,014.17 pa) --> I need this to answer the question (A) above

(B) As stated by Boyle’s Law, the products PV should equal a constant, k. Did your results of P_t × V_avg come out approximately the same for k?

Answer 1

When the book is placed on the object, the pressure it experiences is due to the weight of the object, plus the atmospheric weight. Hence the change in volume is due to the net pressure.

So, we can write, the net pressure P_t = P_o + P where P_o is the atmospheric pressure and P is pressure due to the object.

Now, Boyle's Law states that PV is a constant.

Let the constant be k. Then,

P_t x V = k

(P_o + P) x V =k

or P = (k/V) - P_o

This is of the form y=mx+b

where y =P , x= (1/V), m =k and b = -P_o  

Hence the y intercept (the point where x =0) gives -P_o .

For your data, the scatter plot looks like,

Boyle's Law P vs 1/V 140000 120000 100000 80000 60000 40000 20000 0.01 0,02 0.03 0.04 0.05 0.06 0.07 1/V (mL1) P (Pa)

The equation is of the form

3 100 P = -35016

Hence P_o = -b = 35016 Pa

Final Pressure Pt in Pa	Pt x Vavg in Pa mL
56872.76	2359866
78994.78	2762055
107098.2	2999950
128437.8	2899272
151196.4	2615854