ch20-p013

ch20-p013 - 13 The connection between molar heat capacity...

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process 1 2 (d) The work is given by Eq. 19-14: W = nRT 1 ln ( V 2 / V 1 ) = RT 1 ln3 =1.10 RT 1 . Thus, W / nRT 1 = ln3 = 1.10. (e) The internal energy change is Δ E int = 0 since this is an ideal gas process without a temperature change (see Eq. 19-45). Thus, the energy absorbed as heat is given by the first law of thermodynamics: Q = Δ E int + W 1.10 RT 1 , or Q / nRT 1 = ln3 = 1.10. (f) Δ E int = 0 or Δ E int / nRT 1 =0 (g) The entropy change is Δ S = Q / T 1 = 1.10 R , or Δ S / R = 1.10. process 2 3 (h) The work is zero since there is no volume change. Therefore, W / nRT 1 = 0 13. The connection between molar heat capacity and the degrees of freedom of a diatomic gas is given by setting f = 5 in Eq. 19-51. Thus, 5/ 2 , 7/ 2 Vp CR == , and 7/5 γ = . In addition to various equations from Chapter 19, we also make use of Eq. 20-4 of this chapter. We note that we are asked to use the ideal gas constant as R and not plug in its numerical value. We also recall that isothermal means constant-temperature, so T 2 = T 1 for the 1 2 process. The statement (at the end of the problem) regarding “per mole”

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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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ch20-p013 - 13 The connection between molar heat capacity...

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