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Determination of the Ideal Gas Constant "R" using Carbon Dioxide. (Report 3 digits for this exercise. No points will be deducted for significant digit errors.) (20 points)

Mass of beaker (optional)

Mass of plastic bag (optional)   :

Temperature (in the bag)  : 10 oC = 283.15 K

Volume of plastic bag : 4.000 L = 0.004 m3

Barometric pressure                  : 753.6  mm/Hg = 100471.75 Pa

Using the ideal gas equation : PV = nRT

Here, is the number of moles of CO2 .

Calculated value of R for each trial is shown below-

Trial

1

2

3

4 (wet)

5 (wet)

Mf

288.045

282.169

276.341

263.162

255.967

Mi

294.635

288.045

282.169

276.346

263.162

R

9.4767

10.6282

10.7157

6.6747

12.2306

Rave

10.2735

9.4527

% Error

22.16%

13.70%

Room temperature – TR     :     28 oC   or  301.15 K

Water temperature - Tw     :     24 oC   or   297.15 K 

Barometric Pressure – Patm   :   753.6 mm/Hg = 100471.75 Pa

Vapor Pressure of Water – Pwater   :   2986.421 Pa

Pressure of Butane – Pbutane (Patm- P­water)   :   95485.329 Pa


Trial

1

2

3

Total

Mf

12.6293 g

12.5400 g

12.4405 g

MI

12.7084 g

12.6293 g

12.5400 g

Mbutane

0.0791 g

0.0893 g

0.0995 g

0.0893 g

Vbutane

35.10 mL

40.50 mL

45.00 mL

40.2 mL

Molar Mass

58.138 g/mol

Pressure/Volume Measurements. (Report 3 digits for this exercise. No points will be deducted for significant figure errors.) (20 points)

V

20 cm3

16 cm3

9 cm3

6 cm3

4 cm3

P

363

447

773

1137

1662

PV

7260

7152

6957

6822

6648

V

5 cm3

10 cm3

20 cm3

P

1113

635

320

PV

5565

6350

6400

V

20 cm3

10 cm3

5 cm3

P

414

788

1444

PV

8280

7880

7220

How do you explain the large percent error in the gas constant determination? There are both experimental and scientific factors involved. Include both.

While opening and closing the bag, a lot of dry ice leaves into the surrounding thereby changing the value of the initial mass we took. Also, the temperature of the bag changes since there is no proper insulation. Lastly, we assumed the gas to be ideal but in practice, no gas in the universe is ideal. All these factors lead to large percentage error in the calculation of Gas Constant.

If some water remained on the lighter at the time of the second weighing (after collecting the gas), how would this affect your determination of molecular weight? Be specific. Show a sample calculation to clarify your answer.

Mater remaining on the lighter will increase the measured final mass of butane. Let us assume the water droplets weigh 2 gm, then,

Initial mass = 12.6293 g

Final Mass = 14.7084 g

Mass of butane gas = 2.0791 g

Volume of gas = 35.10 mL

Using ideal gas equation : PV = nRT

(100471.75)(0.0000351) = ( )(8.314)(297.15)

M = 1456.5 g/mol

As can be observed above, just because of the presence of water droplets on the bag surface, the molecular mass of butane which is expected to be 52 g/mol is coming out as 1456.5 g/mol. Hence, before measurement of the mass, water droplets must be cleaned and dried.

Since, PV = constant in the measurements, therefore, Boyle’s Law is confirmed by the results of part-3.

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