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Some Gas Law Review Boyle’s Law: PV = k (constant T, m) Units of pressure: 1 Pa  1N/m 2 = (10 5 dyn)/(10 2 cm) 2 = 10 dyn/cm 2 1 torr = 1 mm Hg (Because.

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Presentation on theme: "Some Gas Law Review Boyle’s Law: PV = k (constant T, m) Units of pressure: 1 Pa  1N/m 2 = (10 5 dyn)/(10 2 cm) 2 = 10 dyn/cm 2 1 torr = 1 mm Hg (Because."— Presentation transcript:

1 Some Gas Law Review Boyle’s Law: PV = k (constant T, m) Units of pressure: 1 Pa  1N/m 2 = (10 5 dyn)/(10 2 cm) 2 = 10 dyn/cm 2 1 torr = 1 mm Hg (Because P = mg/A =  Vg/A =  Ahg/A =  gh) 1 atm  760 torr = 101,325 Pa Charles’ Law: V/T = k (constant P, m) Combined Gas Law: Ideal Gas Law: PV = nRT R = 82.06 (cm 3 atm)/(mole K) = 8.314 J/(mole K) = 1.987 cal/(mole K)

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3 Parts per thousand! 273 K

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5 19 8 5. µ nmnm S M

6 The Barometric Formula Imagine a layer of gas of area, A, thickness dz and mass dm. The gas below pushes up with F up = PA. The downward force on the layer is dm g plus the pressure from above, (P + dP)A. Since layer is in equilibrium, F up =F down, PA = (P+dP)A + g dm, and dP = -(g/A)dm From the ideal gas law, PV = P(A dz) = (dm/M)RT, dm = (PMA/RT) dz and dP/P = -(Mg/RT) dz. Integrating (ignoring variation in T, g) gives: P = P 0 e -Mgz/RT From Ira N. Levine, “Physical Chemistry”


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