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Boyle's Law: Experimental Data and Analysis, Lab Reports of Chemistry

Experimental data and analysis of boyle's law, which relates the pressure and volume of a gas at constant temperature. The data includes multiple runs with varying volumes and pressures, and the text explains the constant relationship between pressure and volume, as well as the reason why runs with different starting volumes produce different constants.

Typology: Lab Reports

Pre 2010

Uploaded on 08/09/2009

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koofers-user-zs0 🇺🇸

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Boyle’s Law page 5
Data
Run #1 Run #2 Run #3
Vol., mL P, mmHg P x V P, mmHg P x V P, mmHg P x V
10.0 752.7 7.53 x 103744.3 7.44 x 103742.3 7.42 x 103
5.0 1415.9 7.1 x 1031343.1 6.7 x 1031361.9 6.8 x 103
7.5 1000.1 7.5 x 103964.7 7.2 x 103975.1 7.3 x 103
12.5 615.4 7.69 x 103590.5 7.38 x 103590.5 7.38 x 103
15.0 511.5 7.67 x 103494.8 7.42 x 103501.1 7.52 x 103
mean, 0-------- 7.5 x 103-------- 7.2 x 103-------- 7.3 x 103
1. run #1: 0 P x V = _______7.5 x 103_mmHg mL________________
run #2: 0 P x V = _______7.2 x 103_mmHg mL________________
run #3: 0 P x V = _______7.3 x 103_mmHg mL________________
0 for the three runs = ____7.3333 x 103 mmHg mL_______________
Run #4 Run #5
Vol., mL P, mmHg P x V Vol., mL P, mmHg P x V
15.0 746.4 1.12 x 1045.0 758.9 3.8 x 103
5.0 2031.3 1.0 x 1047.5 511.5 3.8 x 103
7.5 1461.7 1.1 x 10410.0 386.7 3.87 x 103
10.0 1122.8 1.12 x 10412.5 305.6 3.82 x 103
12.5 900.3 1.13 x 10415.0 253.7 3.81 x 103
mean, 0-------- 1.1 x 104-------- 3.8 x 103
2. run #4: 0 P x V = _____1.1 x 104_mmHg mL__________________
run #5: 0 P x V = _____3.8 x 103_mmHg mL__________________
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Boyle’s Law page 5

Data

Run #1 Run #2 Run #

Vol., mL P, mmHg P x V P, mmHg P x V P, mmHg P x V

10.0 752.7 7.53 x 10 3 744.3 7.44 x 10^3 742.3 7.42 x 10^3

5.0 1415.9 7.1 x 10 3 1343.1 6.7 x 10^3 1361.9 6.8 x 10^3

7.5 1000.1 7.5 x 10 3 964.7 7.2 x 10^3 975.1 7.3 x 10^3

12.5 615.4 7.69 x 10 3 590.5 7.38 x 10^3 590.5 7.38 x 10^3

15.0 511.5 7.67 x 10 3 494.8 7.42 x 10^3 501.1 7.52 x 10^3

mean, 0 -------- 7.5 x 10 3 -------- 7.2 x 10 3 -------- 7.3 x 10 3

1. run #1: 0 P x V = _______7.5 x 10 3 _mmHg mL________________

run #2: 0 P x V = _______7.2 x 10 3 _mmHg mL________________

run #3: 0 P x V = _______7.3 x 10 3 _mmHg mL________________

0 for the three runs = ____7.3 333 x 10^3 mmHg mL_______________

Run #4 Run #

Vol., mL P, mmHg P x V Vol., mL P, mmHg P x V

15.0 746.4 1.12 x 10^4 5.0 758.9 3.8 x 10^3

5.0 2031.3 1.0 x 10^4 7.5 511.5 3.8 x 10^3

7.5 1461.7 1.1 x 10^4 10.0 386.7 3.87 x 10^3

10.0 1122.8 1.12 x 10^4 12.5 305.6 3.82 x 10^3

12.5 900.3 1.13 x 10^4 15.0 253.7 3.81 x 10^3

mean, 0 -------- 1.1 x 10 4 -------- 3.8 x 10 3

2. run #4: 0 P x V = _____1.1 x 10 4 _mmHg mL__________________

run #5: 0 P x V = _____3.8 x 10 3 _mmHg mL__________________

Boyle’s Law page 6

  1. Do runs 4 and 5 produce a constant value for P x V? Are they the same values as runs 1, 2 and 3. Why or why not?

Run 4 yields a constant value for PV as does run 5. But they are not the same values. Runs 4 and 5 produce different values than Runs 1, 2 and 3. Runs 1, 2 and 3 produce the same value.

Runs 1, 2 and 3 produce the same value because our starting point, 10.0 mL, is the same for each of these runs. Thus we are starting with the same amount of air in the syringe. Runs 4 and 5 are starting at different points: 15.0 mL for run 4 and 5.0 mL for Run 5. Thus we are starting with different amounts of air in each of those runs yielding constants that are not the same.

Remember that when we introduced Boyl’s Law we stated that both the amount of gas and the temperature must remain constant. The temperature is constant but the amount of gas in Runs 4 and 5 is not constant from run to run.

  1. For run #1 if the pressure is 250.0 mmHg, what is the volume?

P x V = 0 for run

250.0 mmHg x V = 7.5 x 10 3 mmHg mL; V = 3.0 x 10^1 mL

OR V 1 = 10.0 mL V 2 =? P 1 = 752.7 mmHg P 2 = 250.0 mmHg

10.0 mL x 752.7 mmHg = 250.0 mmHg x V 2 ; V 2 = 30.1 mL

Answer _______________________

  1. For run #5 if the volume is 25.0 mL, what is the pressure?

P x V = 0 for run

P x 25.0 mL = 3.8 x 10^3 mmHg mL; P = 1.5 x 10^2 mmHg

OR V 1 = 5.0 mL V 2 = 25.0 mL P 1 = 758.9 mmHg P 2 =?

5.0 mL x 758.9 mmHg = 25.0 mL x P 2 ; P 2 = 1.5 x 10^2 mmHg

Answer _______________________