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Thermodynamics filled in class notes_Part_56

# Thermodynamics filled in class notes_Part_56 - 119 4.4 HEAT...

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Unformatted text preview: 119 4.4. HEAT CAPACITY First get the entropy: du = T ds = T ds − P dv, du + P dv, (4.173) (4.174) T ds = cv dT + P dv, dT P cv + dv, T T dv dT +R , cv T v dT dv cv +R, T v v T + R ln , cv ln T0 v0 T R RT /P ln + ln , T0 cv RT0 /P0 (4.175) ds = = ds = s − s0 = s − s0 cv = ln T T0 ln T T0 1+R/cv = ln T T0 ln T T0 k = (4.178) (4.179) (4.180) 1+(cp −cv )/cv = (4.177) R/cv T P0 T0 P = (4.176) + ln , R/cv P0 P + ln , P0 P + ln + ln P0 P (4.181) (4.182) (cp −cv )/cv , (4.183) k −1 . (4.184) So s = s0 + cv ln T T0 k + cv ln P0 P k −1 . (4.185) Now, for the calorically perfect ideal gas, one has u = uo + cv (T − To ). (4.186) For the enthalpy, one gets h = u + P v, (4.187) = u + RT, = uo + cv (T − To ) + RT, (4.188) (4.189) = uo + cv (T − To ) + RT + RTo − RTo , = uo + RTo +cv (T − To ) + R(T − To ), (4.190) (4.191) =h o = ho + (cv + R)(T − To ). (4.192) =c p So h = ho + cp (T − To ). (4.193) CC BY-NC-ND. 18 November 2011, J. M. Powers. 120 CHAPTER 4. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS For the Helmholtz free energy, one gets a = u − T s. (4.194) Thus, a = uo + cv (T − To ) − T s0 + cv ln k T T0 + cv ln P0 P k −1 . (4.195) For the Gibbs free energy, one gets g = h − T s. (4.196) Thus g = ho + cp (T − To ) − T 4.5 s0 + cv ln k T T0 + cv ln P0 P k −1 . (4.197) Van der Waals gas A van der Waals gas is a common model for a non-ideal gas. It can capture some of the behavior of a gas as it approaches the vapor dome. Its form is P (T, v ) = a RT − 2, v−b v (4.198) where b accounts for the ﬁnite volume of the molecules, and a accounts for intermolecular forces. Example 4.9 Find a general expression for u(T, v ) if P (T, v ) = RT a −. v − b v2 (4.199) ∂u ∂T ∂u ∂v (4.200) Proceed as before: First we have du = dT + v dv, T recalling that ∂u ∂T = cv , v ∂u ∂v CC BY-NC-ND. 18 November 2011, J. M. Powers. =T T ∂P ∂T v − P. (4.201) ...
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