Gases and gas laws/Boyle's law lab

An excellent worksheet for this experiment can be found at Mr. Smith's Page:

http://www.phs.d211.org/science/smithcw/Chemistry%20332/Quarter%204%20Unit%201/2%20Boyle's%20Law.pdf

Mr. Smith's discussion is far superior to this one, and should be used by instructors wishing to perform this lab (see also Appendix:Comment on Mr. Smith's page).



Materials
The materials used for this lab were a medical syringe, two blocks of wood to act as bases, and multiple weights as seen in the photo to the right. The lab kit from Eisco included directions for set up and use. The entire kit including instructions may or may not be available at: http://www.eiscolabs.com/Store/Product/PH0145A

Experimental Methods and results
Boyle's Law Apparatus Simple Form

1. The apparatus was set up according to instructions, set to an initial volume of 40mL (without added weight). This constitutes one data point (with zero mass) in the calculation of the next section.

2. One kg weight was added to compress the air in the syringe and the new volume was recorded.

3. Four more 1kg weights were added for a total of 5kg and the volume was recorded after addition of each one.

4. The process was then repeated in reverse order. We started by adding 5kg and took one weight off at a time recording the resulting volumes. This was done to see if any air leaked out of the syringe, or if a heating or cooling effect was influencing the pressure.

5. Volumes and weights were recorded in an Excel spreadsheet. Excel functions were also used to calculate the values Pressure in Pascals and Area in cm^2.

Finding atmospheric pressure with one data point


All you need to calculate atmospheric pressure is two measurements of the volume, one with zero weight added, and the other with a known mass. You also need to know the area and height of the cylinder.

By the ideal gas law, using the fact that both the number of atoms and the temperature is the same:

(1)   $$P_0V_0 = P_1V_1$$

where $$P_0$$ is the atmospheric pressure and $$V_0$$ is the initial volume of the cylinder. Simple algebra yields:

(2)   $$\frac{P_1}{P_0}=\frac{V_0}{V_1} = \frac{h_0}{h_1} $$

where $$h_0$$ and $$h_1$$ are the two cylinder heights. When a weight is added, the total pressure exerted by the plunger is the sum of the atmospheric pressure, plus the added pressure caused by the added mass. Since pressure is F/A and the force of gravity is mg, we have:

(3)   $$P_1 = P_0 + \frac{m_1g}{A}=P_0 + \Delta P$$

where

(4)    $$\Delta P = \frac{m_1g}{A}$$

Substituting Eq.(3) into Eq.(2), we have:

(5)   $$\frac{P_0 + \Delta P}{P_0}= \frac{h_0}{h_1} $$

can be solved for $$P_0$$.

Assignment: Collect a number of data points abd make a graph. Then select one value of mass and volume, and calculate atmospheric pressure using equation (5).

Calculation
Here, we show that the experiment appears to have been successful.

From the first of the two graphs shown above, we have two data points: The volume was approximately 40 ml at zero weight and approximately 30 ml at 2kg weight. Since height is proportional to volume, we have


 * $$\frac{h_0}{h_1}=\frac{V_0}{V_1}\approx\frac{40}{30}=\frac{4}{3}$$

(We truncate 4/3 at four significant digits because we are fairly certain that this experiments lack such accuracy.)

To find the area, calipers were used to obtain an inner diameter of 28.6&plusmn;0.4 mm. Care must be taken not to scratch the inside of the plastic tube with the metal calipers. Also, there is a rim that can be detected with the finger, making the measurement using calipers to be a lower bound. An upper bound can be obtained by measuring the rubber stopper, which is compressed in diameter while inside the cylinder. This uncertainty represents an 4/286 = 1.4% uncertainty in diameter, or equivalently an error of approximately 2.8% in area.

With a weight of 2 kg, we have a pressure of:

$$\Delta P = \frac{mg}{\pi R^2}= \frac{4mg}{\pi D^2}$$ $$= \frac{4}{\pi}\times 2\,kg \frac{9.8 m}{s^2}\frac{1}{(28.6 mm)^2}\left(\frac{10^3 mm}{m}\right)^2$$$$\approx 3.011\times 10^5\frac{kg\cdot m}{s^2}\frac{1}{m^2}\approx 3.051\times 10^4\frac{N}{m^2}=3.05\times 10^4 Pa$$,

where the Pascal (Pa) is used as the SI unit of pressure.

Solving Eq. (5) for $$P_0$$, we have:


 * $$P_0 = \frac{\Delta P}{\frac{V_0}{V_1}-1}=\frac{3.051\times 10^4}{4/3-3/3} = 9.15\times 10^4 Pa$$

One atmosphere is defined as 1.01325x105 Pascals, but the actual atmospheric pressure will differ (depending on altitude and current weather conditions).

Linearization
It's always nice if the theory predicts a straight line. For that reason we seek an equation the form y=mx+b, where either m or b is related to pressure, and (x,y) are either volume or mass.

Strategy: We can easily calculate &Delta;P, so make that one of our variables (x or y). Another easy variable to calculate is &alpha; = h1/h0. If this variable doesn't give us a straight line, perhaps &beta;=1/&alpha; will.


 * $$\frac{\Delta P}{P_0} = \frac{V_0}{V_1}-1$$


 * $$\Delta P = P_0V_0 \left(\frac{1}{V_1}\right) -P_0$$

Appendix: Comment on Mr. Smith's page
Mr. Smith's effort is far superior to this one, and it could probably be copied with impunity by any teacher for classroom use. The biggest problem with Mr. Smith's page is that it is not permanent. It is my experience that such pages are often removed (e.g. soon after the instructor retires). Also, Mr. Smith's page cannot be edited. At some point, every useful resource on the web should be replaced by a corresponding page on a wiki that uses a Creative Commons license.