Under Pressure

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Under Weight. Dalton's Law of Fractional Weights. The amount Weight?. Q - An inflatable is loaded with unadulterated oxygen. What is the weight of the oxygen in the inflatable? A - Climatic weight. On the off chance that it wasn't, then the inflatable would extend or shrivel.

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Under Pressure Dalton's Law of Partial Pressures

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How Much Pressure? Q - An inflatable is loaded with unadulterated oxygen. What is the weight of the oxygen in the inflatable? A - Atmospheric weight. On the off chance that it wasn't, then the inflatable would extend or recoil. Q - A bigmouth is exploded with breathed out air. What is the weight of oxygen taken care of? A – Around 16 - 21% of air weight (O 2 is 16% of breathed out air, 21% of air) Q - A strong holder is loaded with immaculate oxygen. What is the weight in the holder? A - It could be anything. The holder is strong and in this manner can't shrivel or grow.

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Dalton's law of fractional weights Read pages 460 - 461 In the 18 th century what did numerous researchers accept about the world's environment? The % of which part of air changes the most? Argon makes up around 0.93% of dry air. List the other 3 gasses exhibit in dry air (with their %s). Around what % of become air is made scarce by gasses other than these 4? Give Dalton's law of incomplete weights & the condition. 1 L of N 2 at 50 kPa is blended with 1 L of O 2 at 60 kPa, to frame a 1 L blend of the gasses. What is the subsequent weight? What are the halfway weights? An inflatable contains 75 kPa N 2 , 15 kPa O 2 , 5 kPa CO 2 , and water vapor. On the off chance that barometrical weight is 100 kPa what is the incomplete weight of water vapor?

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Dalton's law of fractional weights That the air was a solitary substance compound. Water N 2 : 78.08%, O 2 : 20.95%, CO 2 : 0.04%. These signify 100% of dry air, so there are no different gasses (truly, 0.002%). The aggregate weight of a blend of nonreacting gasses is equivalent to the entirety of the fractional weights of the individual gasses (P add up to = P 1 + P 2 + P 3 + … ) Total weight = 60 kPa + 50 kPa = 110 kPa 100 - (75 + 15 + 5) = 100 - 95 = 5 kPa

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50 kPa 100 kPa 150 kPa 1 L oxygen 1 L nitrogen 1 L blended gas Summary Dalton found that the aggregate weight of blended gasses is equivalent to the total of their individual weights (gave the gasses don't respond). Take note of: these volumes are the same + = This works as per the KMT in light of the fact that at a similar temperature atoms of various gasses have the same Ek. It doesn't make a difference if the atoms are O 2 or H 2 . Both slam into the holder or different particles with a similar drive.

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Vapor Pressure Defined Vapor weight is the weight applied by a vapor. E.g. the H 2 O(g) in a fixed holder. Inevitably the air over the water is loaded with vapor pushing down. As temperature  , more atoms fill the air, and vapor weight  . However, particles both leave and join the surface, so vapor weight likewise pushes atoms up. To quantify vapor weight we can warm an example of fluid on top of a segment of Hg and see the weight it applies at various °C.

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Vapor weight Temperature Measuring Vapor Pressure 101.3 When the vapor weight is equivalent to the barometrical weight (P atm ), the push out is sufficient to conquer P atm and bubbling happens. Accordingly, water will bubble at a temperature beneath 100 °C if the climatic weight is decreased.

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Collecting gasses over water Many circumstances gasses are gathered over H 2 O Often we need to know the volume of dry gas at STP (valuable for stoichiometry). For this we should make 3 adjustments: The level of water inside and outside the tube must be level (so weight inside is equivalent to the weight outside). The water vapor weight must be subtracted from the aggregate weight (to get the weight of the dry gas). At last, qualities are changed over to STP utilizing the consolidated gas law.

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(P 1 )(V 1 )(T 2 ) (100.4 kPa)(325 mL)(273 K) (P 2 )(T 1 ) (101.325 kPa)(294 K) Sample estimation A gas was gathered more than 21°C H 2 O. After equivalent izing water levels, the volume was 325 mL. Give the volume of dry gas at STP (P atm =102.9 kPa). Step 1: Determine vapor weight (pg. 464) At 21°C vapor weight is 2.49 kPa Step 2: Calculate the weight of dry gas P gas = P atm - P H2O = 102.9 - 2.49 = 100. 4 1 kPa Step 3: List the greater part of the information T 1 = 294 K, V 1 = 325 mL, P 1 = 100. 4 1 kPa Step 4: Convert to STP = 299 mL V 2 =

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Assignment 37.8 mL of O 2 is gathered by the descending relocation of water at 24 °C and an environmental weight of 102.4 kPa. What is the volume of dry oxygen measured at STP? Attempt questions 8 – 10 on page 465. 236 mL of H 2 is gathered over water at 22°C and at a barometrical weight of 99.8 kPa. What is the volume of dry H 2 at STP? In the event that H 2 is gathered over water at 22 °C and a barometrical weight of 100.8 kPa, what is the fractional weight of the H 2 when the water level inside the gas jug is equivalent to the water level outside the jug ?

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P 1 V 1 P 2 V 2 = T 1 T 2 (99.42 kPa)(37.8 mL) (101.3 kPa)(V 2 ) = (297 K) (273 K) (99.42 kPa)(37.8 mL)(273 K) = 34.1 mL (V 2 ) = (297 K)(101.3 kPa) Vapor weight at 24 C = 2.98 kPa Pgas = Patm - Pvapor = 102.4 kPa - 2.98 kPa = 99.42 kPa = P 1) V 1 = 37.8 mL, P 1 = 99.42 kPa, T 1 = 297 K V 2 = ?, P 2 = 101.3 kPa, T 2 = 273 K

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P 1 V 1 P 2 V 2 = T 1 T 2 (97.16 kPa)(236 mL) (101.3 kPa)(V 2 ) = (295 K) (273 K) (97.16 kPa)(236 mL)(273 K) = 209 mL (V 2 ) = (295 K)(101.3 kPa) Vapor weight at 22 C = 2.64 kPa Pgas = Patm - Pvapor = 99.8 kPa - 2.64 kPa = 97.16 kPa = P 1 3) V 1 = 236 mL, P 1 = 97.16 kPa, T 1 = 295 K V 2 = ?, P 2 = 101.3 kPa, T 2 = 273 K

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Answers 4 - Total weight = P H 2 + P H 2 O 100.8 kPa = P H 2 + 2.64 kPa 100.8 kPa - 2.64 kPa = P H 2 = 98.16 kPa For more lessons, visit www.chalkbored.com

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