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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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