ch20-p074 - 74. (a) From Eq. 20-1, we infer Q = T dS, which...

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Q net = 1 2 (2.00)(50) = 50 J . (d) Since we are dealing with an ideal gas (so that ǻ E int = 0 in an isothermal process), then W 1 ĺ 2 = Q 1 ĺ 2 = 700 J . (e) Using Eq. 19-14 for the isothermal work, we have W 1 ĺ 2 = nRT ln V 2 V 1 . where T = 350 K. Thus, if V 1 = 0.200 m 3 , then we obtain V 2 = V 1 exp ( W / nRT ) = (0.200) e 0.12 = 0.226 m 3 . (f) Process 2 ĺ 3 is adiabatic; Eq. 19-56 applies with Ȗ = 5/3 (since only translational degrees of freedom are relevant, here). T 2 V 2 Ȗ -1 = T 3 V 3 Ȗ -1 This yields V 3 = 0.284 m 3 . (g) As remarked in part (d), ǻ E int = 0 for process 1 ĺ 2. (h) We find the change in internal energy from Eq. 19-45 (with C V = 3 2 R ): ǻ E int = nC V ( T 3 T 2 ) = –1.25 × 10 3 J . (i) Clearly, the net change of internal energy for the entire cycle is zero. This feature of a
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