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Unit 11 – Gases Cont. V. Two More Laws p. 345-347, 366-368 online text book (p. 365-67, 386-388 in blue MC text) Read these pages first!

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Presentation on theme: "Unit 11 – Gases Cont. V. Two More Laws p. 345-347, 366-368 online text book (p. 365-67, 386-388 in blue MC text) Read these pages first!"— Presentation transcript:

1 Unit 11 – Gases Cont. V. Two More Laws p , online text book (p , in blue MC text) Read these pages first!

2 Ptotal = P1 + P2 + ... B. Dalton’s Law
The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases. Ptotal = P1 + P When a H2 gas is collected by water displacement, the gas in the collection bottle is actually a mixture of H2 and water vapor. C. Johannesson

3 B. Dalton’s Law Hydrogen gas is collected over water at 22.5°C. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa. The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H2 and water vapor. GIVEN: PH2 = ? Ptotal = 94.4 kPa PH2O = 2.72 kPa WORK: Ptotal = PH2 + PH2O 94.4 kPa = PH kPa PH2 = 91.7 kPa Look up water-vapor pressure on p.899 for 22.5°C. Sig Figs: Round to least number of decimal places. C. Johannesson

4 B. Dalton’s Law GIVEN: Pgas = ? Ptotal = 742.0 torr PH2O = 42.2 torr
A gas is collected over water at a temp of 35.0°C when the barometric pressure is torr. What is the partial pressure of the dry gas? The total pressure in the collection bottle is equal to barometric pressure and is a mixture of the “gas” and water vapor. GIVEN: Pgas = ? Ptotal = torr PH2O = 42.2 torr WORK: Ptotal = Pgas + PH2O 742.0 torr = PH torr Pgas = torr Look up water-vapor pressure on p.899 for 35.0°C. Sig Figs: Round to least number of decimal places. C. Johannesson

5 C. Graham’s Law Diffusion
Spreading of gas molecules throughout a container until evenly distributed. Effusion Passing of gas molecules through a tiny opening in a container C. Johannesson

6 KE = ½mv2 C. Graham’s Law Speed of diffusion/effusion
Kinetic energy is determined by the temperature of the gas. At the same temp & KE, heavier molecules move more slowly. Larger m  smaller v KE = ½mv2 C. Johannesson

7 C. Graham’s Law Graham’s Law
Rate of diffusion of a gas is inversely related to the square root of its molar mass. The equation shows the ratio of Gas A’s speed to Gas B’s speed. C. Johannesson

8 C. Graham’s Law Determine the relative rate of diffusion for krypton and bromine. The first gas is “Gas A” and the second gas is “Gas B”. Relative rate mean find the ratio “vA/vB”. Kr diffuses times faster than Br2. C. Johannesson

9 Put the gas with the unknown speed as “Gas A”.
C. Graham’s Law A molecule of oxygen gas has an average speed of 12.3 m/s at a given temp and pressure. What is the average speed of hydrogen molecules at the same conditions? Put the gas with the unknown speed as “Gas A”. C. Johannesson

10 C. Graham’s Law An unknown gas diffuses 4.0 times faster than O2. Find its molar mass. The first gas is “Gas A” and the second gas is “Gas B”. The ratio “vA/vB” is 4.0. Square both sides to get rid of the square root sign. C. Johannesson

11 PRACTICE! Work the following problems in your book. Check your work using the answers provided in the margin. p. 324 SAMPLE PROBLEM 10-6 PRACTICE 1 & 2 p. 355 SAMPLE PROBLEM 11-10 PRACTICE 1, 2, & 3 C. Johannesson


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