P21_008 - T versus S ). This leads to (300 K)(15 J / K) =...

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8. (a) It is possible to motivate, starting from Eq. 21-3, the notion that heat may be found from the integral (or “area under the curve”) of a curve in a TS diagram, such as this one. Either from calculus, or from geometry (area of a trapezoid), it is straightforward to Fnd the result for a “straight-line” path in the TS diagram: Q straight = µ T i + T f 2 S which could, in fact, be directly motivated from Eq. 21-3 (but it is important to bear in mind that this is rigorously true only for a process which forms a straight line in a graph that plots
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Unformatted text preview: T versus S ). This leads to (300 K)(15 J / K) = 4500 J for the energy absorbed as heat by the gas. (b) Using Table 20-3 and Eq. 20-45, we Fnd E int = n 3 2 R T = (2 . 0 mol) 8 . 31 J mol K (200 K 400 K) = 5 . 10 3 J . (c) By the Frst law of thermodynamics, W = Q E int = 4 . 5 kJ ( 5 . 0 kJ) = 9 . 5 kJ ....
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