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Unformatted text preview: Chemistry 4511 Problem Set 1 Solutions 1. Suppose you have a cylinder as described at right. The gas contained inside has an additional force that is a function of the height where the gas beneath height z repels the gas above it. This changes the expression for dF as shown. a. Derive an expression for P as a function of the height in the cylinder, z, and temperature, T. b. At what height(s) does P = P ? (g = 9.8 m s2 and γ = 30 s2 ) c. If the system is closed and P = 0.5 atm, sketch a plot of the pressure as a function of z from 0 to 1m. (T = 25 o C and M = 0.039 kg/mol) a. we can simplify dF such that dF = ρ gAdz – ργ Azdz = ρ A(g γ z)dz then, following the barometric distribution derivation, which leads to ( ) dz z g RT M P dP ⋅ − − = γ and integrating with appropriate bounds ( ) ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ − − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ⋅ − − = ∫ ∫ 2 2 1 ln 1 z gz RT M P P dz z g RT M dP P z P P γ γ Solving for P at height z at a given T, ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ − ⋅ − = 2 2 1 z z g RT M e P P γ b. P = P when the exponential is equal to one. This occurs for all roots of z. 2 1 2 = ⋅ − z gz γ or z = 0, 2 1 = ⋅ − z g γ which gives m s s m g z 653 ....
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This note was uploaded on 10/11/2011 for the course CHEM 4511 taught by Professor Staff during the Spring '08 term at Colorado.
 Spring '08
 staff
 Chemistry, Physical chemistry, pH

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