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12 grams of 12C has 6.022  1023 atoms 1 u = 1.6605  10-27 kg.

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Presentation on theme: "12 grams of 12C has 6.022  1023 atoms 1 u = 1.6605  10-27 kg."— Presentation transcript:

1 12 grams of 12C has 6.022  1023 atoms 1 u =  kg

2 Avogadro’s Number 6.022  1023

3 The Mole 6.022 x 1023 1 mol =  1023 particles

4 Notation NA is Avogadro’s number N is the number of particles
n = N/NA = number of moles

5 Ideal Gas Law Assumes that atoms / molecules do not interact much with one another Assumes that molecules do not have much rotational KE

6 When things are constant, group them on one side of the equation:
and so on…

7 Kinetic Theory of Gases
Mechanical model for gas behavior

8 Maxwell distribution for oxygen gas at 300 K and 1200 K

9 Pressure is due to the collisions of atoms or molecules with the walls of the container

10

11 Then P = F/A = F/L2, so

12 Kinetic Theory Pressure is due to collisions of atoms with walls of container Atoms have distribution of speeds, but gas has an average kinetic energy <KE> The <KE> is directly proportional to the absolute temperature More massive atoms will move slower to have same <KE> when at same T as another gas

13 Internal Energy of Monatomic Ideal Gas
Ideal gas is assumed to have little rotational KE and atoms do not interact except by direct collisions For ideal monatomic gas, all of the internal energy (U) is due to the KE of the atoms U = N<KE> = N(3/2 kT) so U = 3/2 nRT [or 3/2 NkT] very important result!


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