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Warm-up 1. What formula will you use if you are given volume and pressure? 2. A sample of gas at 47°C and 1.03 atm occupies a volume of 2.20 L. What volume.

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Presentation on theme: "Warm-up 1. What formula will you use if you are given volume and pressure? 2. A sample of gas at 47°C and 1.03 atm occupies a volume of 2.20 L. What volume."— Presentation transcript:

1 Warm-up 1. What formula will you use if you are given volume and pressure? 2. A sample of gas at 47°C and 1.03 atm occupies a volume of 2.20 L. What volume would this gas occupy at 107°C and 0.789 atm?

2 Motion of gas particles Two types: Two types: –Diffusion – gradual mixing or spreading out of a gas –Effusion – molecules confined in a container randomly pass though a small opening in the container

3 Motion of gas particles The rate (speed) of diffusion and effusion depends on the mass or size of the molecules The rate (speed) of diffusion and effusion depends on the mass or size of the molecules Lighter molecules move faster than heavier molecules Lighter molecules move faster than heavier molecules

4 Avogadro’s Law Says that equal volumes of gases at same temp and pressure contain the same number of molecules Says that equal volumes of gases at same temp and pressure contain the same number of molecules Led to the discovery of the Ideal Gas Law Led to the discovery of the Ideal Gas Law

5 Ideal Gas Law Relationship between pressure, volume, temp, and number of molecules Relationship between pressure, volume, temp, and number of molecules PV = nRT PV = nRT n = number of moles n = number of moles R = Ideal Gas Constant = 0.082 R = Ideal Gas Constant = 0.082

6 Units Pressure - atm Pressure - atm Volume - Liters Volume - Liters Temperature - Kelvin Temperature - Kelvin n - moles n - moles

7 Example What is the pressure in atm exerted by a 0.500 mol sample of nitrogen gas in a 10.0 L container at 298 K? What is the pressure in atm exerted by a 0.500 mol sample of nitrogen gas in a 10.0 L container at 298 K? P=?? P=?? V=10.0 L V=10.0 L n=0.500 mol n=0.500 mol R=0.082(constant) R=0.082(constant) T=298 K T=298 K

8 Example PV=nRT PV=nRT P(10.0) = (0.500)(0.0821)(298) P(10.0) = (0.500)(0.0821)(298) P = (0.500)(0.0821)(298) = 1.22 atm P = (0.500)(0.0821)(298) = 1.22 atm (10.0) (10.0)

9 Graham’s Law The rate of effusion of gas is inversely proportional to the square root of its formula mass The rate of effusion of gas is inversely proportional to the square root of its formula mass


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