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Unformatted text preview: crease in pressure from 600.00 atm to 1.00 atm. Partial credit given even for not so accurate results! Show your work! a) - 4083 J b) + 4083 J c) + 4156 J d) - 4156 J e) 3995 J f) - 5547 J g) - 1206 J h) + 1206 J i) + 5547 J j) - 1206 J k) - 727 J l) + 727 J ΔG = ∫ V dP note that the compound of interest must be a liquid or a solid (from the density value). 1) Quick approximate answer… Assume the volume is not affected by temperature or pressure (not a bad assumption given that the thermal expansion coefficient and the compressibility coefficients are small (See table values). We use V = M/d where M is the molar mass and d is the density. 3 3 - 5 3 V = (78.11 g/mol) / (0.873 g/cm ) = 84.47 cm / mol = 8.447 10 m / mol - 5 3 ΔG = ∫ V dP = V ΔP = 8.447 10 m / mol x (1.00 – 600.00) atm x (101325 Pa/atm) ΔG = - 5430.45 J Since we know this is a good approximation, the correct answer must be fairly close to this value (- 5547 J) o 2) Rigorous answer: First calculate the volume at 60.00 C, TO do so we use, α = (1/V) (dV/dT)P which leads o o to: dV/V = α dT and by integration to: V(60 C) = V(20 C) x exp( α x [60- 20]) o - 5 3 - 3 - 5 3 V(60 C) = 8.447 10 m / mol x exp(1.24x10 x[60- 20]) = 9.40 x 10 m / mol Since we need to integrate V dP, we need to know how the volume is changing with pressure. To do so, we use: κT = (- 1/V) (dV/dP)T which leads to: dV/V = - κT dP and by integration to: SI h ) – 1300 SI Make sure y...
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This note was uploaded on 01/26/2014 for the course CHEM 3615 taught by Professor Aresker during the Spring '07 term at Virginia Tech.

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