Charles’s Law Lab Report

Title of the Experiment: Charles’s Law Purpose The purpose of this experiment was to verify Charles’s Law. This was done by establishing volume-to-temperature ratios of air at two different temperatures at constant pressure and if the ratios were the same then Charles’s Law would be verified. The two different temperatures were established by boiling water and cold, icy water in a bucket and then two different volumes were retrieved that were divided by the temperatures in Kelvin of each respective condition and that was the ratio or constant (k) for each condition, respectively. Fundamental Principle or Theory Behind the Experiment Charles’s Law: the relationship between the volume and temperature of a gas at constant pressure. The relationship between the volume of a given mass of a gas and the Kelvin temperature of a gas is directly proportional, therefore when temperature increases, volume increases and when temperature decreases, volume decreases. V = kT We can use Charles’s Law to compare:  Temperature (at 2 volumes)  Volume (at 2 temperatures) ~ If the pressure and the mass of a gas are kept constant, Charles’s law may be applied at two different temperatures. V1 = k1 T1 V2 = k2 T2 If Charles’s Law is valid, then k1 and k2 should be roughly equal. And mathematically, the law will look like: Method I. Set Up 1. Set up the apparatus for the Charles’s Law experiment. This includes attaching an iron clamp to a support stand with a support rod attached to it at a height high enough where a 400-mL beaker on top of a hot plate can fit under the iron clamp. a. The clamp will be holding the 125-mL Erlenmeyer flask and the flask will be dropped into the beaker where the bottom of the flask will drop as far as it could and so that the bottom of the flask is not touching the walls of the beaker. b. The Erlenmeyer flask should have a rubber stopper inserted into it and then glass tubing and then rubber tubing connecting onto the glass tubing. A pinch clamp needs to be added to the rubber tubing but not closed. II. Experiment 1. Fill the 400-mL beaker that should be on your support stand with 125-mL to 200mL of water. 2. Slowly lower the Erlenmeyer flask into the beaker so that the flask is as completely submerged as possible into the beaker. Do not let the flask to touch the bottom or the sides of the beaker. III. 3. Keep the flask submerged in the beaker of boiling water for about 6 minutes. 4. Insert a thermometer into the boiling-water beaker for about 3-4 minutes and measure the temperature of the water to the nearest 0.5 degrees Celsius. a. This will go on your Data Sheet (1). 5. Carefully, close the pinch clamp on the rubber tubing, and use the utility clamp to remove the flask from the boiling-water beaker and invert the flask into the trough of ice water. 6. Keep the flask inverted and in the ice water with the end of the tubing submerged, open the pinch clamp but do not remove the clamp from that position. Keep the flask submerged for about 4 minutes. 7. Keep the rubber tubing and punch camp submerged and raise the flask until the water level inside the flask is at the same height as the water level in the pneumatic trough. 8. Tightly close the pinch clamp and remove the flask from the trough. 9. Measure the volume of the water that was in the flask once the flask was dunked into the trough. This volume measurement will go into Data Sheet (2). 10. Measure the temperature of the ice-water bath/trough once the flask was done being dunked into the trough. This value will go into Data Sheet (3). 11. Completely fill the flask with water to the top of the flask. Open the pinch clamp on the rubber stopper assembly and reinsert the assembly into the flask to the same depth. Tighten the pinch clamp and invert the flask so that any water in the rubber tubing above the clamp drains out. 12. Dry a graduated cylinder and open the pinch clamp and pour all the water into the graduated cylinder. This volume of water will be the volume of the flask which is 4 on the Data Sheet. 13. Record the barometric pressure that should be given by the lab instructor and this will go into (5) on your Data Sheet if the units are mm of Hg and into (6) if the units are torr. Calculations 1. Find the volume of the wet, cold air OR the air from the icy-cold water. Subtract the volume of the water in the flask after being dunked (2) from the volume of the flask (4). a. This is (8) on your Data Sheet. 2. Convert the barometric pressure that was provided to you by the lab instructor and use one of the following equations depending on the units of the barometric pressure given in the problem. a. 1 in. of Hg = 25.4 torr b. 1 mm of Hg = 1 torr c. This is (6) on your Data Sheet. 3. Find the pressure of the dry, cold air OR the air from the boiling water. Subtract the vapor pressure of the water in the ice-water bath from the barometric pressure. a. This is (7) on your Data Sheet. 4. Find the volume of the dry, cold air. Multiply the volume of the wet, cold air OR the air from the icy-cold water (8) by the pressure of the dry, cold air OR the air from the boiling water (7), divided by the barometric pressure (6). a. Volume of air from boiling water = (volume of air from icy, cold water, mL) * (pressure of air from boiling water, torr / barometric pressure, torr) b. This will be (9) on your Data Sheet. 5. Convert the temperatures of the boiling water and the cold, icy water from Celsius to Kelvin. a. K = C + 273.15 b. Record the Kelvin temperature for the boiling water in (10) and the temperature for the icy, cold water in (11). 6. Find the volume-to-temperature ratio for the volume of air from boiling water at the temperature of the water in the boiling-water bath. Divide (4) by (10). a. Record the ratio in (12). 7. Find the volume-to-temperature ratio for the volume of air from the icy, cold water at the temperature of the water in the icy, cold water bath. Divide (9) by (11). a. Record the ratio in (13). Data Analysis (This is a new data sheet I made up with my actual data entered) General Data (5) Barometric pressure, mmHg (given by lab instructor) (6) Barometric pressure, torr (convert (5)) (7) Pressure of air w/ no water vapor in it, torr Tip: (6) - Vapor Pressure Vapor Pressure = 11.2 torr First Determination 757.30 Second Determination 757.30 757.30 757.30 746.1 746.1 Data of Gas (air) from Boiling Water (Hot Air) (1) Temperature of water in the boiling water bath, C (10) Temperature of water in the boiling water bath, K (4) Volume of air held in flask/Volume of water held in flask, mL (12) V/T for gas from boiling water, mL K- 1(k1) Tip: (4) / (10) First Determination 100.2 Second Determination 99.6 373.4 372.8 147 147 0.39 0.394 Data of Gas (air) from Icy, Cold Water (Cold Air) (2) Volume of water in flask once dunked in icy trough, mL (3) Temperature of water in the icy, cold water bath, C (11) Temperature of water in the First Determination 34 Second Determination 32 2.6 4.4 275.6 277.6 icy, cold water bath, K (8) Volume of cold air w/ water vapor, mL Tip: (4) – (2) (9) Volume of cold air w/ no water vapor, mL Tip: ( (8) * (7) ) / (6) (13) V/T for gas from icy, cold water, mL K-1 (k2) Tip: (9) / (11) 113 115 111.3288 113.299221 0.40 0.408 Calculations: = (0.39 / 0.40)-1 * 100 1 = 2.5% = (0.394 / 0.408) -1 * 100 1 = 3.4% After placing the beaker on the support stand and filling it with anywhere from 125-mL to 200-mL of water, the water was boiling and so in the first determination the temperature that was measured of the boiling water bath is 373.4 K for the first determination and 372.8 for the second determination. Then once placed in the bucket with water and ice, the flask pulled in some of the water and the volume of water in the flask once the flask was dunked into the trough was 34-mL for the first determination and 32-mL for the second determination. This value will prove later useful in the calculating of the volume of the air with water vapor in it which can be referred to as the cold, wet air and then this calculation would be used in finding the volume of the cold, dry air and that is the volume of the air without water vapor in it. Filling the Erlenmeyer flask all the way to top, the volume of air that can be held in that flask was about 147-mL. This measurement was used in both trials since the shape and size of the flask did not change whatsoever and so the volume of water that it was able to hold the same amount as it could before. Using the volume of the Erlenmeyer flask and calculating the difference between that and the volume of water that was pulled into the flask when it was dunked into the trough. This difference 147-mL and 34-mL or between 147-mL and 32-mL is what gives you the volume of the wet, cold air, which is the air with water vapor from the icy, cold water and that is 113-mL for the first determination and 115-mL for the second determination. The volume of the wet, cold air, which is air with water vapor in it from the icy, cold water is then used to calculate the volume of the dry, cold air or the air with no water vapor from the icy, cold water available. This was done by multiplying the volume of the air with water vapor in it from the icy, cold water by the pressure of air with no water vapor in it and then that product divided by the barometric pressure. So, ((113-mL)(746.1 torr))/(757.30 torr) for the first determination and that comes out to 111.3288-mL and for the second determination ((115-mL)(746.1 torr))/(757.30 torr) which comes out to 113.299221-mL. Now that the air with no water vapor from the icy, cold water was calculated, the volume-to-temperature ratio was calculated for the gas from the icy, cold water or in other words, cod, dry ice- and for the first determination that was (111.3288-mL)/(275.6 K) = 0.40 and for the second determination (113.299221-mL)/(277.6 K) = 0.408. Since this experiment was looking to verify Charles’s Law, we compared volumes at two different temperatures at a constant pressure. Therefore, the volume of air in the flask which was also how much water was able to fit into the flask all the way to the top was 147-mL and that number was divided by the temperature of the water in the boiling-water bath which for the first determination was 373.4 K and for the second determination was 372.8 K. Dividing this volume by the temperature for each respective determination provided the volume-to-temperature ratio for hot, dry air, or the air with no water vapor, which is why it is called “dry” and that arises from the boiling water which is why it is called “hot.” The whole purpose was to verify Charles’s Law and because the volume-to-temperature ratio at two different temperatures were roughly the same- 0.39 and 0.40 for the first determination and 0.394 and 0.408 for the second determination, Charles’s Law was verified. Some mistakes that could have arisen throughout the experiment was when we had to measure the temperature of the boiling water bath, although we were not supposed to, the thermometer did touch the sides/walls of the beaker and that could possibly lead to a significant change in the temperature reading because the sides/walls of the beaker are definitely cooler than the substance inside the beaker at that moment so the temperature reading could have gone down slightly and so the temperature would have been less than what it should be and therefore the calculated ratio would have then been larger than it was supposed to be. Another possible source of error was in the first determination when the Erlenmeyer flask had to be inverted into the icy trough, the flask was not as raised as it should have been, its bottom was raised for sure, however not as high as it could be, in the second determination we learned from our mistakes and read that we had to raise the flask until the water level inside the flask is at the same height as the water level in the pneumatic trough. Conclusion Charles’s Law was verified through this experiment of comparing volumes at two different temperatures- the cold and the hot. For the first determination, k1 = 0.39 and k2 = 0.40 and for the second determination, k1 = 0.394 and k2 = 0.408, which for both determinations is very close.