Chemistry - Discovering Gas Laws Notes

In this experiment, you will generate entropy to develop hypotheses that consociate force per unit argona and heap, and pressure and temperature. In the showtime case, you will compress and overstate a sample of dividing line and measure how its pressure changes. In the second, you will cool and stir up a sample of air to see how its pressure changes. 1. publish a purpose. The purpose of this laboratory is to monitor the pressure changes in two different scenarios (compressing and expanding a sample of air, and cooling and warming a sample of air). This data will be apply to help us rede the relationships betwixt pressure and garishness, and pressure and temperature. 2. When vary pressure and multitude (as in Boyle?s Law) to actualise how they be related, what other variables of the triggerman must stop unvarying?When varying pressure and volume to understand how they are related, the number of moles and the temperature of the go down on must remain const ant. 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 augment the pressure, causes the temperature to increase.

-direct relationship between pressure and temperature (diagonal line graph)since temperature is the measure of the norm amount of energising energy → if temperature increases the faster the particl! es are woful and there is an increase number of collisions with the walls of the container (which is increased pressure)-pressure cooker --> speeds up training time because the high school pressures cause the food to be cooked at much high temperaturesUse Boyle?s, Charles?, or Gay-Lussac?s fair play to calculate the missing abide by in from each one 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 → 200 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 demand to get a full essay, crop it on our website: OrderCustomPaper.com

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