Download Equation of State - Thermodynamics and Statistical Mechanics - Solved Exam and more Exams Thermodynamics in PDF only on Docsity! 1 PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS SPRING 2006 NAME: __________ANSWERS___________________ 1. (20%) The Helmholtz Free Energy is defined as F = U – TS, where U is internal energy, T is temperature, and S is entropy. For a system where dW = PdV, show that VT FS −= ∂ ∂ . dF = dU – TdS – SdT, but the First Law states, dU = TdS – PdV. Then dF becomes, dF = TdS – PdV – TdS – SdT = – SdT – PdV Also, F is a function of T and V, so dV V FdT T FdF TV + = ∂ ∂ ∂ ∂ A term by term comparison of the two equations for dF shows that VT FS −= ∂ ∂ 2. (20%) The equation of state of one mole of a hypothetical substance is: v – aT2 + bP = constant where a and b are constants. a) Find an expression for the entropy change of one mole of this substance when the pressure is increased isothermally from P1 to P2. Express your answer in terms of P1, P2, a, b, and T. dP T vTdTcTds P P ∂ ∂ −= , so dP T v T dTcds P P ∂ ∂ −= For this problem dT = 0, and from the equation of state, aT T v P 2= ∂ ∂ . Then, aTdPds 2−= and )(22 12 2 1 PPaTdPaTs P P −−=−=∆ ∫ )(2 21 PPaTs −=∆ b) Determine whether or not cP is a function of pressure for this substance. (Hint: Start with a Tds equation in both parts a) and b).) The ds equation can be written, aTdPdT T cds P 2+= which is equivalent to dP P sdT T sds TP + = ∂ ∂ ∂ ∂ , so T c T s P P = ∂ ∂ , and aT P s T 2= ∂ ∂ . Then, Docsity.com 2 PT P aT TT c P ∂ ∂ = ∂ ∂ 2 . This says, a P c T T P 21 = ∂ ∂ , or 02 ≠= ∂ ∂ aT P c T P . cP is a function of pressure. (03-2) 3. (40%) One mole of an ideal monatomic gas traverses the idealized Diesel cycle shown in the figure. Process 1→2 is adiabatic, and process 2→3 takes place at constant pressure. Process 3→4 is also adiabatic, and process 4→1 takes place at constant volume. (Hint: for one mole of an ideal monatomic gas, the internal energy is RTU 2 3= .) The temperatures of the gas at points 1, 2, 3, and 4 are: T1 = 300 K, T2 = 1600 K, T3 = 1800 K, T4 = 365 K a) (10%) Find V1/V2, the ratio of the volume of the gas at point 1 to the volume at point 2. (This is the compression ratio for the Diesel cycle.) For an adiabatic process, 122 1 11 −− = γγ VTVT , so 1 2 1 2 1 T T V V = −γ , and 5.113/5 1 1 1 1 2 2 1 33.5 K 300 K 1600 = = = −−γ T T V V V1/V2 = ___12.3_____ b) (10%) Find the magnitude of the heat added to the gas in the process 2→3. Express the answer in terms of the gas constant R. Q2 = Cp(T3 – T2) = K) 1600K 1800(2 5 −R Q2 = _(500 K)R____ Docsity.com
