Thermodynamics and an Introduction to Thermostatistics answer 2, Exercises of Chemistry

Download Thermodynamics and an Introduction to Thermostatistics answer 2 and more Exercises Chemistry in PDF only on Docsity! Problem Set 2 Solutions 3.20 MIT Professor Gerbrand Ceder Fall 2003 1 P S(U, V ) : dS = dU + dV T T Problem 1.1 1 PVariables here are U and V and intensive variables are and . T T To go to 1 T Uas a natural variables take the Legendre transform by subtracting from S T U J = S − T So, � 1 � P dJ = −U d + dV T T T FNote that J is equal to − Similarly, go from S(U, V ) to B(U, (F is the Helmholtz energy) T P ) by taking P B = S − V T 1 � P � dB = dU − V d T T A Legendre transform with respect to both U and V defines the Planck function (Yes, you too can be famous by defining your own Legendre transform) P U Y = S − V − T T � 1 � � P � dY = −U d − V d T T One can show that these functions have extrema properties just like the Legendre transforms of energy. Massieu functions are maximum under constant value of their natural variables. Cp Problem 1.2 We know that is defined as � ∂H � Cp = ∂T P Enthalpy can be obtained from the Legendre transform of the internal energy as follows, H = U + PV We know that for an ideal gas U is a function of temperature only and the PV = RT (for one mole), thus � ∂H ∂T � P = � ∂U ∂T � P + ∂ ∂T (RT ) � ∂U � is only a function of temperature and this is not a function of pressure. Also, R is not a ∂T P function of pressure. So Cp of an ideal gas must be independent of pressure. (b) One possibility is � ∂S � � ∂M � = ∂H ∂T T ,P H,P Problem 1.6 (a) U = T dS + τdL G = U − T S dG = SdT + τdL − T dS + SdT Which yields the solution: dG = τdL − SdT (b)We also know if G=G(L,T) that � ∂G � � ∂G � dG = dL + dT ∂L ∂T T L Thus, � ∂G � � ∂G � τ = and S = ∂L − ∂T LT Since we know that dG is a perfect differential, then the second derivatives must be equal independent of the order which they are taken. So, � ∂2G � � ∂2G � = ∂T ∂L ∂L∂T � ∂ � ∂G � � � ∂ � ∂G � � = ∂T ∂L ∂L ∂T T L L T � ∂τ � � ∂S � = ∂T − ∂L TL (c) dU = TdS + τdL � ∂U � � ∂S � = T + τ ∂L ∂L T T Using the relation from part (b), � ∂U � � ∂τ � = + τ−T ∂L ∂T T L (d) If U is only a function of temperature, then � ∂U � � ∂τ � T = 0 = τ − T ∂L ∂T L 1 � ∂τ � 1 = τ ∂T TL LEVEL 2 PROBLEMS Problem 2.1 (a) Start by constructing the differential form of S(T,P) [In Zemansky this is called the second T dS equation] dS = � ∂S � dT + � ∂S � dP ∂T P ∂P T TdS = T � ∂S � dT + T � ∂S � dP ∂T P ∂P T But we know T � ∂S ∂T � P = Cp and from a Maxwell relation we know � ∂S ∂P � T = − � ∂V ∂T � P .So, T dS = CpdT − T � ∂V ∂T � P dP = CpdT − T V αdP For an isothermal compression, Q = TdS = −T � V αdP In the case of a solid or liquid, neither V or α is very sensitive to pressure. So, Q = −TV α � dP = −TV αΔP Now we can use the data given to calculate Q (Replacing V with M/ρ) Q = −T MαΔP ρ = − 298K ∗ 0.5kg ∗ 49.5 × 10 −6K−1 ∗ 5.05 8.96 × 103kg/m3 × 108P a Q = −416J � (b)We now need to calculate the work during the compression W = − P dV � � ∂V � W = − ∂P T P dP � V β W = � V βP dP = V β P dP = � P 2 − P 2� 2 f i Mβ 0.5kg ∗ 6.18 × 10−12P a−1 ∗ �5.05 × 108P a�2 W = �Pf 2 � = 2ρ 2 ∗ 8.96 × 103kg/m3 W = 44J (c) We know the first law, ΔU = Q + W = −416J + 44J ΔU = −372J Thus it can be seen that the extra amount of energy in the form of heat comes from the storage of internal energy. (d) Now we can go back to the relationship we obtained for T dS (which we know is 0 since the process is adiabatic and reversible) T dS = 0 = CpdT − T V αdP � dT � αV = dP T Cp � Tf � αV ln = ΔP Ti Cp � αV � � αM � Tf = Ti exp ΔP = Ti exp ΔP Cp Cpρ � 0.5kg ∗ 49.5 × 10−6K−1 ∗ 5000atm ∗ 1.01 × 105P a/atm � Tf = 298 exp 8.96 × 103kg/m3 ∗ 385J/kg · K ∗ 0.5kg Tf = 300K ΔT = 300 − 298 = 2K Problem 2.2 D(S, f) = U − fX = (1 − a)U(S, X) Now using (4) to get this in terms of S and f only �(1−a)a� f D(S, f) = (1 − a) exp (−bS) exp(bS) a Similarly for F(T,X) F (T, X) = U − TS = U(S, X)− bU(s, x)S = (1 − bs)U(S, X) Using (3) and (1) from above � � T �� T F (T, X) = 1 − ln bXa b Now we do a transformation for to both T and f K(T, f) = U − TS − fX Using (3), (1) and (2) T T � T � Ta K(T, f) = b − b ln bXa − b ⎡ ⎛ ⎞⎤ T T K(T, f) = ⎣1 − a − ln b ⎝ b � T a �a ⎠⎦ bf Problem 2.5 (a) Start by using the chain rule on the left side of the equation � ∂S � � ∂S � � ∂T � = ∂P ∂T ∂P V V V We know that � ∂S � = CV and we can expand the second derivative as follows ∂T V T � ∂T � � ∂T � � ∂V � = ∂P − ∂V P ∂P V T Putting those together, CV � ∂T � � ∂V � CV � 1 � = ∗ −V β T ∂V P ∂P − T ∗ V α T We can replace CV by CP using the relationship TV α2 =CP − CV β Yielding � ∂S � β � T V α2 � CP β = CP − = − V α ∂P T α β Tα V (b) We can expand the derivative as follows � ∂T � � ∂T � � ∂S � = ∂P − ∂S P ∂P S T The first term is T and the second term can be transformed using a Maxwell relation � ∂S � T = ∂P − � ∂V � CP . This gives ∂T P � ∂T � T � ∂V � = ∂P CP ∂T PS � ∂T � T αV = ∂P CPS (c) Start with the left side � ∂2P � ∂ �� ∂P � � T = T ∂T 2 ∂T ∂T V V V Using a Maxwell relation, ∂ �� ∂P � � ∂ �� ∂S � � ∂2S T = T = T ∂T ∂T ∂T ∂V ∂TV ∂VTV V T V Now the right side � ∂S � � ∂Cv � � ∂T � � ∂S � ∂ �� ∂S � � Cv = T = + T ∂T V → ∂V ∂V ∂T ∂V ∂T T T V V T The first term in the above is zero since the change in temperature at constant temperature is zero. The second term can be written as ∂ �� ∂S � � ∂2S T = T ∂V ∂T ∂VT ∂TVV T And since we know the order of differentiation doesn’t matter ∂2S ∂2S T = T ∂TV ∂VT ∂VT ∂TV Problem 2.6 (a) Start with the differential form of G dG = −SdT + V dP � ∂G � � ∂T � � ∂P � = + V−S ∂T ∂T ∂T V V V � ∂G � � ∂P � � ∂V � = −S − V ∂T ∂V ∂T V T P � ∂G � � −1 � = (V α)−S − V ∂T V β V � ∂G � V α = − S ∂T βV (b) Start with the differential form of H dH = TdS + V dP � ∂H � � ∂S � � ∂P � = T + V ∂T ∂T ∂T P P P � ∂H � TCP = ∂T TP � ∂H � = CP ∂T P Problem 2.7 (For this problem Maple is our friend) (a) This part involves taking the indicated derivatives of the given function. The two tricks are knowing what β is and how to change � ∂U � T into something with derivatives of P w.r.t T or V. ∂V β = −1 � ∂V � V ∂P T and using the differential form of U and a Maxwell relation we get, � ∂U ∂V � T = T � ∂P ∂T � V − P The solutions are then The thermal expansivity is defined as 1 � ∂V � α = V ∂T P Using a Maxwell relation, � ∂V � � ∂S � = ∂T ∂P P T And we are told in the problem that S is independent of crystallite structure and pressure, so � ∂S � = 0 ∂P T =0 And thus α = 0 at absolute zero. Problem 2.10 (a) The key property is that � ∂S � T < 0. This can be proven from the Maxwell relation. ∂H � ∂S � � ∂M � = ∂H ∂T T H (this can be proven from the differential of dφ = d(U − TS +PV − MH) = −SdT +V dP − MdH ) and since � ∂M � � ∂S � < 0 = < 0 ∂T ⇒ ∂H TH S(T ) as constant H therefore has the following shape. Adiabatic demagnetization consists of two steps: I. Isothermal application of a field II. insulate material and turn off the field isotropic demagnetization. System moves back to state with H = 0 , hence T decreases. → (b) � ∂S � ∂M � ∂T � ∂H � T ∂T � H = =� ∂S cH∂H S − ∂T � H − T = −T � ∂M � ∂T � ∂T � H ∂H cHS We can evaluate � ∂M � as follows ∂T H κV � ∂M � κV M = H = = T ⇒ ∂T H − T 2 H � = Figure 1: So, � ∂T κV H ∂H S cH T LEVEL 3 PROBLEMS Problem 3.1 Given: � θ � S2 � Rθ � V 2U = R − V0 2 (a) We want to make this the absolute U and not the molar quantity, so multiply by N. � Nθ � S2 � NRθ � V 2 � Nθ �� S �2 � NRθ �� V �2 U = NU = = R − V0 2 R N − V0 2 N 1 �� θ � S2 � Rθ � V 2 � U = N R − V0 2 The equations of state give us � ∂U � 2 � θ � T = = S ∂S N RN,V � � ∂U � 2 N � Rθ � V 2 0 VP == ∂V N,S � ∂U � �� θ � � Rθ � � �� θ � � Rθ � � P,S = − 1 2N S2 V 2 S2 V 2µ = − V 2 0 = − R − V 2 0∂N R or µ = −U (b) Express µ as a function of T and P N � R � S = T 2 θ N � V0 2 � V P= − 2 Rθ Substitute in to the expression for µ �� θ � N2 � R �2 T 2 � Rθ � N2 � V0 2 �2 P 2 � µ = − 1 2N − V 2 0R 4 θ 4 Rθ � R � � V0 2 � � P 2 1 1 T 2 µ = − − 4 θ 4 Rθ Problem 3.2 Start with the expression we used in class � ∂x � � ∂x � � ∂x � � ∂y � += ∂f ∂f y ∂y ∂f zz f For this problem, this equation can relate the constant l heat capacity to the constant P heat capacity as follows, � ∂S � � ∂S � � ∂S � � ∂P � += ∂T ∂T ∂P ∂T l P T l To get heat capacity we need to multiply both sides by T, and when we do this we get � ∂S � � ∂P � Cl Cp + T= ∂P ∂T T l The first derivative can be rewritten using the Maxwell relation � ∂S � ∂P simplifies to T = − � ∂V � which ∂T P