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Gases Objective I will know the properties of gasses and be able to calculate the quantifiable properties of gasses Success Criteria I will be able to.

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Presentation on theme: "Gases Objective I will know the properties of gasses and be able to calculate the quantifiable properties of gasses Success Criteria I will be able to."— Presentation transcript:

1 Gases Objective I will know the properties of gasses and be able to calculate the quantifiable properties of gasses Success Criteria I will be able to calculate each of the four quantifiable properties using the appropriate equation

2 Video Links Liquid Nitrogen and Balloon Demo Link Charles Law Balloon Demo Link Boyles Law Demo Link

3 Collecting Background 1. What are the physical properties of a gas? 2. What are the quantitative (measurable) properties of gasses? 3. What are the units of pressure? How do you convert between the units? 4. Describe the different ways that the measurable properties can be related (the laws)?

4 Characteristics What are the physical properties of a gas? Gases are fluid – In essence, gases can flow Gases have low density Gases are highly compressible Gases completely fill their container

5 Pressure Pressure is defined as the amount of force exerted per unit area of surface – How hard the gas is pushing on the container There are several units of pressure – Atmosphere (atm) – Kilopascal (kPa) – Millimeters of Mercury (mm Hg) – Torr (torr) – Pounds per square inch (psi) – http://www.youtube.com/watch?v=XfFdNNiIAJw http://www.youtube.com/watch?v=XfFdNNiIAJw – http://www.youtube.com/watch?v=O9p8SRmQlPo http://www.youtube.com/watch?v=O9p8SRmQlPo

6 Pressure Converting Units of Pressure 1.00 atm=101.325 kPa=760 torr=760 mm Hg Practice Problems – The vapor pressure of water at 50.0 o C is 12.33kPa. What is the value in atmospheres? (12.33kPa)(1.00atm/101.325kPa)=0.122atm – In thermodynamics, the standard pressure is 100.0kPa. What is this value in mm Hg? (100.0kPa)(760mm Hg/101.325kPa)=750.1mm Hg

7 Quantifiable Properties There are four measurable properties of gases – P is pressure – T is temperature (must be measured in Kelvin) K = o C + 273 – V is the volume – n is the number of moles of the gas

8 STP Standard Temperature and Pressure (STP) Standard Temperature is 0 o C Standard Pressure is 1.00atm ΔG° = ΔH° – T ΔS° The superscript indicates at STP

9 Boyles Law Boyles Law discusses the relationship of pressure and volume This shows that as the pressure increases, the volume decreases. – This is an inverse relationship PV = k 1 P = k 1 /V where k 1 is the pressure-volume constant And if we keep Temperature and the number of particles constant, but change either pressure or volume we then have: P 1 V 1 =P 2 V 2

10 Boyles Law A sample of oxygen gas has a volume of 150.0mL at a pressure of 0.947atm. What will the volume be at a pressure of 1.000atm if the temperature remains constant? – P 1 V 1 =P 2 V 2 – (0.947atm)(150.0mL)=(1.000atm)V 2 – V 2 =142mL – Be sure your answers are in Significant Figures!

11 Boyles Law A balloon has a volume of 456mL at a pressure of 761torr. It is taken under water in a submarine to a depth where the air pressure in the submarine is 3.3atm. What is volume of the balloon if the temperature remains constant? – Be sure all variables are in the same units (761torr)(1.00atm/760torr)=1.00atm – P 1 V 1 =P 2 V 2 – (1.00atm)(456mL)=(3.3atm)V 2 – V 2 =138mL=140mL

12 Charles Law Charles Law discusses the relationship between Temperature and Volume This shows that as the temperature increases, the volume increases – This is a direct relationship V/T=k 2 At a constant pressure and number of particles, if either temperature or volume change we have: V 1 /T 1 =V 2 /T 2

13 Charles Law A sample of neon gas has a volume of 752mL at 25.0 o C. What is the volume at 50.0 o C if the pressure remains constant? – Change the degrees in to Kelvin 25.0 o C + 273 = 298 K; 50.0 o C + 273 = 323 K – V 1 /T 1 =V 2 /T 2 – (752 mL)/(298 K)=V 2 /(323K) – V 2 = 815 mL

14 Gay-Lussacs Law Gay-Lussacs Law discusses the relationship between temperature and pressure This shows that as the temperature increases, the pressure increases. P = k 3 T where k 3 is the temperature-pressure constant P/T = k 3 At a constant volume and number of particles, if either the temperature or pressure changes, we use this equation: P 1 /T 1 =P 2 /T 2

15 Gay-Lussacs Law At 122 o C, the pressure of a sample of nitrogen is 1.07atm. What will be the pressure at 205 o C, assuming constant volume? – First change temperature to Kelvin 122 o C + 273 = 395 K; 205 o C + 273 = 478 K – P 1 /T 1 =P 2 /T 2 – (1.07 atm)/(395 K)=P 2 /(478 K) – P 2 = 1.29 atm

16 Combined Gas Law When the three different laws just discussed are analyzed, what would they be if they are all combined? PV = k 1 V/T=k 2 P/T = k 3

17 Combined Gas Law PV/T = k 4 If the number of moles remains constant, we can solve for any of the other variables P 1 V 1 /T 1 = P 2 V 2 /T 2 Also used is Standard Temperature and Pressure (STP)

18 Combined Gas Law At a pressure of 780.0 torr and 24.2 o C, Xenon has a volume of 0.350L. What will the volume be at STP? First change temperature to Kelvin – 24.2 o C + 273 = 297.2 K Change the pressure to atmospheres – (780.0 torr)(1.00 atm/760 torr) = 1.03 atm – P 1 V 1 /T 1 = P 2 V 2 /T 2 – (1.03atm)(0.350 L)/(297.2K) = (1.00atm)V 2 /(273K) – V 2 = 0.330 L

19 Diffusion and Effusion Diffusion is the movement of particles from high density to low density – Popping a balloon Effusion is the passage of gas molecules under pressure through a tiny opening – Nail in a tire

20 Grahams Law of Diffusion In Grahams Law of Diffusion, we will analyze the kinetic energy of two different gases – Kinetic energy formula is 1/2mv 2 m is the mass (in this case molar mass) v is the velocity of the molecule – Grahams Law states that the kinetic energy of the first gas will equal the kinetic energy of the second gas m 1 v 1 2 = m 2 v 2 2

21 Grahams Law of Diffusion Oxygen molecules have an average speed of 480 m/s at room temperature. At the same temperature, what is the average speed of molecules of sulfur hexafluoride? – Molar mass of Oxygen is 32.00 g/mol (O 2 ) – Molar mass of SF 6 is 146.07 g/mol – m 1 v 1 2 = m 2 v 2 2 – (32.00 g/mol)(480 m/s) 2 = (146.07 g/mol) v 2 – v = 220 m/s

22 Daltons Law of Partial Pressures Daltons law of partial pressures states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of each gas P T = P 1 + P 2 + P 3 + …… A 42.0 mL sample of hydrogen is collected over water at 30.0 o C. What is the total pressure if the pressure of the dry gas 845.0 torr and the vapor pressure of water at 30.0 o C is 18.6 torr? P T = P 1 + P 2 + P 3 + …… P T = 845.0 torr + 18.6 torr P T = 863.6 torr

23 Essential Information 1.00 atm=101.325 kPa=760 torr=760 mm Hg STP: 273 K and 1.00 atm Daltons Law: P T = P 1 + P 2 + P 3 + …… Combined Gas Law: P 1 V 1 /T 1 = P 2 V 2 /T 2 Gay-Lussacs Law: P 1 /T 1 =P 2 /T 2 Charles Law: V 1 /T 1 =V 2 /T 2 Gramhams Law: m 1 v 1 2 = m 2 v 2 2 Boyles Law: P 1 V 1 =P 2 V 2 Ideal Gas Law: PV=nRT where R is the gas constant Essential


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