Download presentation

Presentation is loading. Please wait.

Published byJosie Todhunter Modified over 3 years ago

1
Gases balloonfestivals.com treehuggerusa.com treehugger.com

2
Pressure The force per unit area of a surface. Units: N/cm 2 N, Newton: SI unit of force As area of contact changes, force changes 500 N = 1.7 N 500 N = 83.3 N 300 cm 2 cm 2 6.0 cm 2 cm 2

3
Christina, Will you go to the prom with me? Steve

4
Barometer Used to measure the pressure of gases. barometerplanet.com

5
Units of Pressure for Gases Millimeters of mercury (mm Hg) Torricelli or 1 torr = 1mm Hg 760 mm Hg = 1 atmosphere at sea level when temp is 0 o C Pascal = pressure exerted by a force of one newton acting on an area of one square meter. Pa = N/m 2 1.013 x 10 5 kPa = 1 atmosphere= 10.1N/cm 2

6
Pressure conversion sample: Express 0.725 atm in a) mm Hg and b) kilopascals (kPa) A)0.725 atm x 760 mm Hg = 551 mm Hg 1 atm B) 0.725 atm x 101.325 kPa = 73.5 kPa 1 atm

7
Standard Conditions (STP) Standard Temperature is 0 o C or 273K Standard Pressure is 1 atm or 760 mm of Hg

8
You must have done by next meeting: A list of the units of pressure (see page 364) and A list of the gas laws: Daltons Law of Partial Pressures: P T = P 1 +P 2 +… Boyles: P 1 V 1 = P 2 V 2 T constant Charless: V 1 = V 2 P constant T 1 T 2 Gay-Lussacs: P 1 = P 2 V constant T 1 T 2 Combined Gas: P 1 V 1 = P 2 V 2 T 1 T 2 Ideal Gas: PV = nRT

9
Daltons Law of Partial Pressures The total pressure of a gas mixture is the sum of the partial pressures of the component gases. Gas collected through water picks up water vapor, so allow inside and outside water levels in a gas collection device to stabilize and: P atm = P gas + P H20

10
Dalton sample problem: Oxygen gas is collected by water displacement. The barometric pressure and the temperature during the experiment are 731.0 torr and 25.0 o C. What was the partial pressure of the oxygen collected? P T = P atm = 731.0 torr P H20 = 23.8 torr (see vapor pressure of water at 25.0 o C from table in handout or book – Table A-8)

11
P T = P atm = 731.0 torr P H20 = 23.8 torr (see vapor pressure of water at 25.0 o C from table in handout or book – Table A-8) P atm = P O 2 + P H20 So P O2 = P atm – P H20 P O2 = 731.0 torr – 23.8 torr = 707.2 torr

12
Boyles Law – at constant temperature, volume of a fixed gas varies inversely with the pressure. If 100.0 mL of a gas, originally at 760 torr, is compressed to a pressure of 800 torr, at a constant temperature, what would be its final volume? P 1 V 1 = P 2 V 2 --> V 2 = P 1 V 1 P2P2 V 2 = 100mL(760 torr) = 95.0 mL 800 torr

13
Charles Law – the volume of a fixed mass of gas at constant pressure varies directly with the Kelvin temperature. A sample of neon gas has a volume of 752 mL at 25.0 o C. What will the volume at 100.0 o C be if pressure is constant? V 1 = V 2 --> V 2 = V 1 T 2 T 1 T 2 T 1 V 2 = 752 mL (100.0 o C) = 300.8 mL 25.0 o C

14
Gay-Lussacs Law – the pressure of a fixed mass of gas at constant volume varies directly with the Kelvin temperature. At 122 o C the pressure of a sample of nitrogen gas is 1.07 atm. What will the pressure be at 205 o C, assuming constant volume? P 1 /T 1 = P 2 /T 2 --> P 2 = P 1 T 2 T 1 P 2 = 1.07 atm(205&273) = 1.29 atm 122+273

15
Gay-Lussacs Law of combining volumes At constant temperature and pressure, the volumes of gaseous reactants and products can be expressed as ratios of small whole numbers. H 2 + Cl 2 --> 2HCl 1L1L 2L H:Cl:HCl = 1:1:2 Formulas must be written correctly and chemical equation balanced.

16
Combined Gas Law – expresses the relationship between pressure, volume, and temperature of a fixed amount of a gas. PV = k T P 1 V 1 = P 2 V 2 --> To find V 2 : V 2 = P 1 V 1 T 2 T 1 T 2 P 2 T 1 Problem: The volume of a gas is 27.5mL at 22.0 o C and 0.974 atm. What will be the volume at 15.O o C and 0.993 atm? Temps to K: 22+273 = 295K and 15+273=288K

17
The volume of a gas is 27.5mL at 22.0 o C and 0.974 atm. What will be the volume at 15.O o C and 0.993 atm? V 2 = P 1 V 1 T 2 P 2 T 1 V 2 = 0.974atm(27.5ml)(288K) 0.993atm(295K) V 2 = 26.3 mL Hint: in solving gas law problems, use the combined gas law and quantities that dont change will cancel out. V 1 = 27.5ml T 1 = 295K P 1 = 0.974atm V 2 = ? T 2 = 288K P 2 = 0.993atm

18
Avogadros Law – equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. Ratios apply here also. 2H 2 + O 2 --> 2H 2 O 2molecules1molecule2molecules 2mol1mol2mol 2volumes1volume2volumes Standard molar volume of a gas is the volume occupied by one mole of a gas at STP. Standard molar volume = 22.4 L/mol

19
Steve, Yes, I will go to the prom with you. Christina

20
At STP, what is the volume of 7.08 mol of nitrogen gas? 7.08 mol (22.4L) = 158 L 1 mol A sample of gas occupies 11.9 L at STP. How many moles of the gas are present? 11.9L (1 mol) = 0.531 mol 22.4L

21
Assuming all volume measurements are made at the same temperature and pressure, what volume of hydrogen gas is needed to react completely with 4.55 L of oxygen gas to produce water vapor? Write and balance the equation. Label known and unknown. Do unit analysis.

22
Gas Stoichiometry – dealing with proportional relationships between reactants and products in a chemical reaction. 2CO 2 (g) + O 2 (g) --> 2CO 2 (g) 2molecules1 molecule2 molecules 2 mol1 mol2 mol 2 volumes1 volume2 volumes

23
Assuming all volume measurements are made at the same temperature and pressure, what volume of hydrogen gas is needed to react completely with 4.55 L of oxygen gas to produce water vapor? Write the correct chemical reaction first: 2H 2 + O 2 --> 2H 2 O Indicate known and unknown: Solve:

24
Ideal Gas Law – the mathematical relationship among pressure, volume, temperature, and the number of moles of a gas. PV = nRT --> R = PV R is the ideal gas nT constant. At STP, R = 1 atm(22.414 L) = 0.0821 L atm 1 mol(273.15K) mol K R is the ideal gas constant.

25
What pressure, in atmospheres, is exerted by 0.325 mol of hydrogen gas in a 4.08 L container at 35 o C? PV = nRT --> P = nRT V T = 35 + 273 = 308K P = 0.325 mol 0.0821L atm) (308K) 4.08 L mol K P = 2.01 atm

26
Gases spread out in a container – diffusion Gases can randomly pass through a tiny opening in a container (leak out) - effusion

Similar presentations

Presentation is loading. Please wait....

OK

Gas Laws.

Gas Laws.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Free ppt on human circulatory system Business templates free download ppt on pollution Ppt on heron's formula free download Ppt on c programming notes Ppt on railway budget 2013-14 Ppt on next generation 2-stroke engine oil Ppt on desert animals and plants Ppt on sustainable rural development Ppt on save environment in hindi Ppt on agriculture and its types