LAB REPORT Datasheet_Charles's Law-1

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University of North Florida *

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2046

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Chemistry

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Feb 20, 2024

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CHARLES’S LAW (Volume and Temperature) Name: … Dalia Elkhatib ………………………. Partner’s name: … Ryan Thomas …………………………… Total points: 69 25 pts 125 mL Erlenmeyer Flask 250 mL Erlenmeyer Flask Temperature of boiling water ( o C) 102 o C 102 o C Temperature of ice bath ( o C) 0 o C 0 o C Volume of water inside the flask at the ice water bath temperature (mL 55 mL 81 mL Volume of flask (mL) 144 mL 273 mL Barometric pressure 760 torr 760 torr 1 Temperature of boiling water (K) 375.15 K 375.15 K 2 V/T ratio for dry hot air 0.3838 mL/K 0.7277 mL/K 3 Temperature of ice bath (K) 273.15 K 273.15 K 4 Barometric pressure 760 torr 760 torr 5 Water vapor pressure (torr) at the ice-water bath temperature (from (see Table 4) 4.6 torr 4.6 torr 6 Volume of wet, cold air (mL) {volume flak – volume of backfilled water} 89.00 mL 192.0 mL 7 Volume of dry, cold air (mL) {see slide 18 of ppt) 88.46 mL 190.8 mL 8 V/T ratio for dry cold air 0.3239 mL/K 0.6985 mL/K 9 Average V/T ratios 0.3539 mL/K 1.426 mL/K 10 % difference (V/T of hot vs. V/T of cold) 16.93% 2.048% Work sheet 20 pts 1 102 o C + 375.15 = 375.15 K 2 144.0 mL / 375.15 K 273.0 mL / 375.15 K 3 0 o C + 273.15 = 273.15 K 4 760 torr 5 Use chart - 0 o C -> 4.6 torr 6 144.0 mL – 55.00 mL 273.0 mL – 81.00 mL 7 P dry,cold air = 760 torr – 4.6 torr = 755.4 torr 89.00 mL x (755.4 torr / 760.0 torr) 192.0 mL x (755.4 torr / 760.0 torr) 8 88.46 mL / 273.15 K 190.8 mL /273.15 K

9 (0.3838 mL/K + 0.3239 mL/K)/2 = 0.3539 mL/K (0.7277 mL/K + 0.6985 mL/K)/2 = 1.426 mL/K 10 ((0.3838 mL/K – 0.3239 mL/K)/0.3539 mL/K ) x 100 ((0.7277 mL/K – 0.6985 mL/K)/1.426 mL/K) x 100 Post-lab questions Q. 1. The Volume of a sample is increased three times. How will the following be affected if all other factors are held constant? a. the Kelvin temperature: 2 pts The temperature will triple (increase by a factor of 3) b. the pressure 2 pts The pressure will triple (increase by a factor of 3) Q. 2. If a sample of carbon dioxide gas occupies 20.91 liters at 35.0 o C , determine its volume at 70.0 o C. (assume CO 2 gas shows ideal behavior). 10 pts 35.0 o C + 273.15 = 308.15 K 70.0 o C + 273.15 = 343.15 K 20.91 L / 308.15 K = V2 / 343.15 K V2 = 343.15 K (0.06786 L/K) = 23.28 L Q. 3. A balloon was filled with 5.00 liters of helium at 20.0 degrees o C . The balloon then drifted up into the Earth's atmosphere. Assuming air pressure does not change (even though it does), determine the volume of that helium balloon when it is 15.0 kilometers above the earth where the atmospheric pressure is - 43.0 o C. 20.0 o C + 273.15 = 293.15 K -43.0 o C + 273.15 = 230.15 K (5.00 L / 293.15 K) = (V2 / 230.15 K)

V2 = (1150.75 L/K) / 293.15 K V2 = 3.93 L 10 pts

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