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Vapor Pressure for gasses collected over water (an application of Daltons Law)

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Presentation on theme: "Vapor Pressure for gasses collected over water (an application of Daltons Law)"— Presentation transcript:

1 Vapor Pressure for gasses collected over water (an application of Daltons Law)

2 Vapor Pressure Water has a vapor pressure. This is caused by the evaporation of water into a gas. This water vapor (gas) above a container of water contributes to the air pressure in the container.

3 Vapor Pressure Often gasses are collected in a container above water because the gas can displace the water easily, but cannot escape.

4 Vapor Pressure One can read the volume of water displaced, but some of the gas that has been collected is actually water vapor and must be subtracted from the volume of the gas of interest.

5 Vapor Pressure Water has a substantially lower vapor pressure at lower temperatures, so one needs to take temperature into account. We do this with a vapor pressure chart.

6 Vapor Pressure For example: If a gas is collected over water at a temperature of 23 C and a pressure of 642 mm Hg, what is the pressure of the gas?

7 Vapor Pressure At 23 C the vapor pressure of water is 21.1 mm Hg. Subtract from the pressure: 642 mmHg – 21.1 mmHg = mmHg

8 Vapor Pressure A sample of gas weighing 5.61 grams was collected over water at 42 C at a pressure of kPa, and was found to have a volume of 688 ml. Calculate the molecular weight of the gas. The gas was found to be an element. Identify it.

9 Vapor Pressure We note that the gas was collected over water, so we need to correct for the vapor pressure of water. However, our table is in mmHg and the data is in kPa, so we need to convert.

10 Vapor Pressure kPa 760 mmHg = 1278 mmHg kPa kPa

11 Vapor Pressure Now we can subtract the vapor pressure of water at 42 C. At 42 C, the vapor pressure of water is 61.5 mmHg Subtracting gives: 1278 – 61.5 = 1217 mmHg

12 Vapor Pressure Now we look at what we know: There is information about pressure, volume, temperature and grams. It wants to know molar mass, which involves moles and grams.

13 Vapor Pressure So we need PV = nRT

14 Vapor Pressure PV = nRT P = 1216 mmHg V = 688 ml =.688 L n = x R = T = 315 K

15 Vapor Pressure Plug in and solve (1217)(.688) = x (62.36)(315) 837 = 19643x x =.0426 moles

16 Vapor Pressure Molar mass is grams/moles 5.61 /.0426 = g/mole If this is an element, it should be Xe


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