University College - Chemistry Chem 112A Lecture Notes - Lecture 17: Partial Pressure, Kinetic Energy, Ideal Gas
13 Dec 2016
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Equation of the day: if a + b c + d ( h1), and c + d e + f ( h2) This can be overall written as a + b e + f ( h3 = h1 + h2) Final: 1 l, 1. 336 atm, 398k, 0. 0409 mol. Q = ncv t = (0. 0409mol)(3/2r)(100k) = 51. 0 j. Since, qv = u = 51. 0 j, w = 0. Here h has no meaning at constant volume really, but we can still calculate it. H = u + (pv) = u + v p = 51. 0j + (1l * 0. 336 atm * 101. 3 j/l atm) = 85. 0 j: constant pressure heating (really constant pext) Initial: 1 l, 1 atm, 298k, 0. 0409 mol. Final: 1. 336 l, 1 atm, 398k, 0. 0409 mol. Q = ncp t = (0. 0409 mol)(5/2r)(100k) = 85. 0 j. W = -pext v = (1atm*0. 336l*101. 3 j/mol k) = -34. 0 j.
