Ch19-p049 - 49(a From Table 19-3 CV = 5 R and C p = 7 R Thus Eq 19-46 yields 2 2 Q = nC p T = 3.00 7 8.31 40.0 = 3.49 103 J 2(b Eq 19-45 leads to

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49. (a) From Table 19-3, 5 2 V CR = and 7 2 p = . Thus, Eq. 19-46 yields () 3 7 3.00 8.31 40.0 3.49 10 J. 2 p Qn C T §· = = × ¨¸ ©¹ (b) Eq. 19-45 leads to 3 int 5 3.00 8.31 40.0 2.49 10 J. 2 V En C T Δ= Δ = = × (c) From either W = Q – Δ E int or W = p Δ T = nR Δ T , we find W = 997 J. (d) Eq. 19-24 is written in more convenient form (for this problem) in Eq. 19-38. Thus, the increase in kinetic energy is 3 trans avg 3 1.49 10 J. 2 KN K n R T Δ = Δ
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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