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HOMEWORK ASSIGNMENT 1 Problems Chapter 1 1.3 Plot V along the y axis and T along the x axis. Then the processes indicated are shown on the figure shown in class. 1.6 Use the data to calculate the temperature according to T 0 = 273.16 PT P T P PT P 100 200 300 349.37 350.33 351.28 400 352.38 As PT P goes to zero the temperature approaches a finite value according to T 0 = T + APT P The coefficients in this linear equation can be determined by the data for the two smallest values of PT P 349.37 = T + 100A, 350.33 = T + 200A which gives T = 2(349.37) − 350.33 = 348. 41, A= 350.33 − 349.37 = 0.009 6 100 This gives an extrapolated value for the temperature of 348. 41. A least squares linear fit to all of the data gives T = 348.35, A = 0.0097 The data and the two linear fits are shown on the curve. The resulting temperatures differ by 1. 722 1 × 10−2 %. 1.9 a) The given linear relationship is X = aT + b Evaluating this at the ice and steam points gives Xi = aTi + b = b, which gives a= Xs = aTs + b = aTs + Xi Xs − Xi , 100 1 b = Xi so the given linear relationship gives T = b) X −b X − Xi = 100 a Xs − Xi The relationship is now given to be T = a ln X + b Repeat the same analysis of part a) Ti = 0 = a ln Xi + b, to get a= 100 , ln (Xs /Xi ) and T = 100 1.11 a) b) c) 100 = a ln Xs + b b = −100 ln Xi ln (Xs /Xi ) ln (Xs /Xi ) ln (Xs /Xi ) T (0 C) = T (K) − 273.15 = 77.35 − 273.15 = −195.8(0 C) T (0 F ) = 95 T (0 C) + 32 = 95 (−195.8) + 32 = −320. 44(0 F ) T (R) = T (0 F ) + 459.67 = −320. 44 + 459.67 = 139. 23 2