quiz_05_solution - EML 5104 Classical Thermodynamics,...

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Unformatted text preview: EML 5104 Classical Thermodynamics, Spring 2010 Quiz 5, 12 Feb 2010 Name (Print): ___________________________________ UF ­ID: ___________________________________ Please sign the honor pledge: "On my honor, I have neither given nor received unauthorized aid in doing this assignment." _________________________________ Signature Only mark one of the multiple choice answers. Use additional paper if necessary. Q1: Below, the Van der Waals equation of state is given. i) Explain the physical significance of parameter b. ii) Give the metric units of b. iii) Explain the physical significance of the second term on the right hand side. iv) Give the metric units of a. (3 P) p= b: accounts for volume taken up by molecules, (m3/kg). a: accounts for forces between molecules, (J m3/kg3), or (N m4/kg3), or (m5/s2/kg2) Q2: Experiments show that at the critical point (1.5 P) : a) RT v −b − a v 2 ∂T ∂p = 0; v ∂ 2T ∂p 2 2 = 0 v b) ∂p ∂p = 0; = 0 ∂v T ∂v 2 T ∂v ∂2 v = 0; ∂p T ∂p 2 All of the above c) d) = 0 T Q3: Give the definition of the compressibility factor Z (not the virial EOS!) (2.5 P) Z = pv/RT Q4: The ideal gas model is a good approximation if (1.5 P) a) Z is close to 0.375 b) Z is close to infinity c) Z is close to 1 d) Z is close to 0 Q5: A reduced property is the property normalized by the same property at (1.5 P) a) the critical point b) the triple point c) ambient conditions (1 atm, 25 °C) d) none of the above Turn page! EML 5104 Classical Thermodynamics, Spring 2010 Q6: For a simple system in equilibrium the 1st Law for a differential change is given by dU = δ Qrev − δWrev . In addition use the definition of entropy and assume that the reversible work is equal to the pressure work (= p dV) to derive the 1st T dS ­equation (2.5 P). dU = δ Qrev − δWrev ; δWrev = pdV ; δ Qrev = TdS dU = TdS − pdV du = Tds − pdv Q7: Use the result from Q6 and the definition of enthalpy (h=u+pv) to derive one additional T dS equation (2.5 P). du = Tds − pdv h = u + pv dh = du + pdv + vdp dh = Tds − pdv + pdv + vdp = Tds + vdp Q8: The T dS equation for a change in enthalpy is dh = T ds + v dp. Use the fundamental TD function h = h(s, p) to derive two basic relations. (one for T, one v) (2.5 P) dh = Tds + vdp dh = ∂h ∂h ds + dp ∂s p ∂p s comparing: T= v= ∂h ∂s ∂h ∂p p s Q9: Use the results from Q8 and the fact that the order of differentiation may be exchanged, to derive one of the Maxwell relations. (2.5 P) T= v= ∂h ∂s ∂h ∂p p s ⎛ ⎞ ⎛ ⎞ ∂ ⎜ ∂h ⎟ ∂ ⎜ ∂h ⎟ =⎜ ∂p ⎜ ∂s p ⎟ ∂s ∂p s ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ T ⎠s ⎝ v ⎠p ∂T ∂v = ∂p s ∂s p Points: _______/ 20 P Turn page! ...
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