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C) P(Y XjY 025) SOLUTION These problems are most easily solved by drawing pictures and computing the areas of the triangular regions that you see For part (a) the answer is 1 12(2)(3 4)(3 4) = 7 16 (we did this as an example in class) For part (b) the answer is 1 1 2 (1)(4 5) 1 2 (1)(3 4) = 9 40 For part (c), P(Y X;Y 025) = P(Y X) P(Y X(g) 186 kJ Answer We know from the ideal gas law that for one mole of gas, V 1 = RT P 1 and V 2 = RT P 2;1 =y 11 V 1 y 12 V 2 (C1) I 2 =y 21 V 1 y 22 V 2 (C2) Here, the four parameters y 11, y 12, y 21, and y 22 are admittances, and their values completely characterize the linear twoport network Depending on which two of the four port variables are used to represent the network excitation, a different set of equations (and a correspondingly 2 40 'ã "¯Œ^ ƒƒ"ƒY ƒp[ƒ} ƒVƒ‡[ƒg