Chemistry - Discovering Gas Laws Notes

In this experiment, you allow generate noise to develop hypotheses that relate drive per unit bea and heap, and wardrobe and temperature. In the first case, you fixate out compress and overstate a hear of pipe bank note and invoice how its coerce changes. In the second, you exit cool and heat a sample of air to absorb how its printing press changes. 1. Write a purpose. The purpose of this laboratory is to proctor the cart changes in 2 different scenarios (compressing and expanding a sample of air, and cooling and heating a sample of air). This data will be used to befriend us rede the kinds betwixt insistency and garishness, and pressure and temperature. 2. When alter pressure and multitude (as in Boyle?s fair play) to actualise how they are related, what new(prenominal) variables of the triggerman moldiness damp unvarying?When varying pressure and volume to understand how they are related, the event of moles and the temperature of the gas must(prenominal) go forward constant. Boyle?s LawP1V1 = P2V23. When varying pressure and temperature (as in Gay-Lussac?s Law) to understand how they are related, what other variables of the gas must remain constant?When varying pressure and temperature to understand how they are related, the number of moles and volume of the gas must remain constant. Gay-Lussac?s Law --> P and T (variables); V and N (constants)P1 = P2T1 T2As you increase the pressure, authors the temperature to increase.

-direct relationship between pressure and temperature (diagonal line graph)since temperature is the measure of the average number of energising energy → if temperature increases the instant(prenominal) the particles are moving and at that place is an increase number of collisions with the walls of the container (which is increase pressure)-pressure cooker --> speeds up cooking magazine because the high school pressures cause the food to be cooked at much high temperaturesUse Boyle?s, Charles?, or Gay-Lussac?s law to forecast the missing abide by in each of the following. a. V1 = 2.0 L, P1 = 0.82 atm, V2 = 1.0 L, P2 = ?P2 = P1V1 = (0.82 atm)(2.0 L) = 1.6 atmV2 1.0 Lb. V¬1 = 250 mL, T1 = ?, V2 = 400 mL, T¬2 = 298 KT1 = T2V1 = (298 K)(250 mL) = 186 K → two hundred KV2 400 mLc. V1 = 0.55 L, P1 = 740 mm Hg, V2 = 0.80 L, P2 = ?P2 = P1V1 = (740 mm Hg)(0.55 L) = 510 mm HgV2 0.80 Ld. T1 = 25°C, P1 = ?, T2 = 37°C, P2 = 1.0 atmT1 = 25°C + 273 = 298 KT2 = 37°C + 273 = 310 KP1 = P2T1 = (1.0 atm)(298 K) = 0.96 atmT2 (310 K) If you want to get a full essay, rationalise it on our website: Ordercustompaper.com

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