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CHEM 444 HW#7                                      Answer key by Jiahao Chen                                         due N...
CHEM 444                                                                                                                  ...
CHEM 444                                                                                          HW#7 Solutions   Note. E...
CHEM 444                                                                                                    HW#7 Solutions...
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  1. 1. CHEM 444 HW#7 Answer key by Jiahao Chen due November 10, 2004 Grading policy: One point is given for each numbered equation shown in the solution, or for written statements and/or equations to the same effect. The total score for this problem set is 25 points plus a maximum of 5 bonus points.Note. If you see PAQ on your script, it means “please answer the question.” ¯ ¯ ¯ 1. (22-8) Determine CP − CV for a gas that obeys the equation of state P V − b = RT . [5 points] Solution. Divide the result of Problem 3 (22-11) throughout by n to convert it into the molar equation ¯ 2 ¯ ¯ ∂V ∂P CP − CV = −T ¯ (1) ∂T P ∂V T The required derivatives of the equation of state are ¯ ∂V R = (2) ∂T P P ∂P RT ¯ = − 2 (3) ∂V T ¯ V −b P2 = − (4) RT 2 ¯ ¯ R P2 ∴ CP − CV = −T − =R (5) P RT Note. You can use Equation (22.23), the way the solutions manual does. However I don’t want you to get the idea that thermodynamics is just about looking up the right equation from the textbook. Here we derived a result in this problem set so I’m perfectly fine with just quoting that result. 2. (part of 22-10) Use Equation 22.22 to show that ∂CV ∂2P =T (6) ∂V T ∂T 2 V ∂CV Show that ∂V T = 0 for an ideal gas and a van der Waals gas. [8 points] Solution. Equation 22.22 is ∂U ∂P = −P + T (7) ∂V T ∂T V 1
  2. 2. CHEM 444 HW#7 Solutions which when differentiated with respect to T at constant V gives ∂ ∂U ∂P ∂P ∂2P ∂2P =− + +T =T (8) ∂T ∂V T V ∂T V ∂T V ∂T 2 V ∂T 2 V but recall that since U is an analytic function, the mixed derivatives are equal, and so by defini- tion ∂ ∂U ∂2U ∂2U ∂ ∂U = = = (9) ∂T ∂V T V ∂T ∂V ∂V ∂T ∂V ∂T V T ∂CV ≡ (10) ∂V T which completes the proof. The equation of state for an ideal gas is P V = nRT (11) and so ∂2P ∂ nR ∂CV = =0⇒ =0 (12) ∂T 2 V ∂T V V ∂V T and for a van der Waals gas, RT a P = ¯ + ¯2 (13) V −b V ∂2P ∂ R ∂CV ⇒ = ¯ =0⇒ =0 (14) ∂T 2 V ∂T V −b ∂V T V Note. Quoting the origin of Maxwell’s relation is OK as long as you define new symbols that you introduce. It is really really important that you state what are y1 , y2 , x1 and x2 , even if you just use them once. 3. (22-11) Consider V to be a function of T and P and write out dV . Now divide through by dT at constant V and substitute the expression for ∂P V that you obtain into Equation 22.23 to get [4 ∂T points] 2 ∂V ∂P CP − CV = −T ∂T P ∂V T Solution. As required, ∂V ∂V V = V (T, P ) ⇒ dV = dT + dP (15) ∂T P ∂P T So differentiating with respect to T at constant V gives ∂V ∂V ∂P 0 = + (16) ∂T P ∂P T ∂T V ∂P ∂V ∂V ∂V ∂P ⇒ = − ÷ =− (17) ∂T V ∂T P ∂P T ∂T P ∂V T Substituting into Equation 22-23: ∂P ∂V CP − CV = T (18) ∂T V ∂T P 2 ∂V ∂P = −T ∂T P ∂V T 2
  3. 3. CHEM 444 HW#7 Solutions Note. Equation 17 follows immediately from the permutation rule for partial derivatives: ∂V ∂T ∂P = −1 (19) ∂T P ∂P V ∂V T Note. Please do not mix intensive and extensive quantities! Either put overbars over all relevant variables, or not at all. What I mean is that ¯ ∂U CV = = CV ∂T V Note. When the question says to “divide through by dT at constant V ” it doesn’t mean literal division. What it means is to change things like dP to become ∂P V . Again, please review multivariable ∂T calculus. 4. (22-14) . Show that the enthalpy is a function of only the temperature for a gas that obeys the ¯ equation of state P V − bT = RT . [4 + 5 points] Solution. Note that since H = H (S, P ) it is sufficient to prove that H for this gas does not vary with pressure [1 point]. To show this, we need Equation (22.34) which we can derive as follows [total of 5 bonus points]: begin with the differential form dH:¯ ¯ ¯ ¯ dH = T dS + V dP (20) and take the partial derivative with respect to P at constant T : ¯ ∂H ¯ ∂S =T ¯ +V (21) ∂P T ∂P T ¯ ∂S We can eliminate ∂P using the Maxwell’s relation from the thermodynamic potential which T has natural variables P and T , i.e. the Gibbs free energy. [1 point] From the differential form ¯ ¯ ¯ dµ = dG = −SdT + V dP (22) the Maxwell’s relation thus obtained is ¯ ∂S ¯ ∂V − = (23) ∂P T ∂T P which when substituted into Equation 21 then gives Equation (22.34) (divided throughout by n); applying this result to the equation of state gives ¯ ∂H ∂V¯ ¯ = V −T (24) ∂P T ∂T P ¯ R = V −T +b (25) P = 0 (26) where the equation of state was again applied to obtain the final result. 3
  4. 4. CHEM 444 HW#7 Solutions 5. (22-18) Show that ∂V dH = V − T dP + CP dT (27) ∂T P What does this tell you of the natural variables of H? [4 points] Solution. Considering H = H (P, T ) gives ∂H ∂H dH = dP + dT (28) ∂P T ∂T P Using Equation 22-34 (none other than Equation 24 above without the overbars) ∂H ∂V =V −T (29) ∂P T ∂T P And that the latter derivative is the heat capacity at constant pressure ∂H ∂U H = U + PV ⇒ = = CP (30) ∂T P ∂T P gives the final result ∂V dH = V − T dP + CP dT ∂T P The result shows that the change of H with respect to P is not independent of the temperature and therefore P and T together cannot be considered natural variables of H. [1 point] Supplement on natural variables. The textbook gives an incredibly hand-wavy definition of nat- ural variables. (If you wanted to be cynical, you could say that neither McQuarrie nor Simon really understands the concept either!) Here I try to explain what exactly is meant by “the coef- ficients... are simply thermodynamic functions.” An IUPAC Technical Report [1] defines natural variables as “[a set of] independent variables [used] to describe the extensive state.” What this means is that if x and y are natural variables of f , for example, then each derivative does not depend on the other variable, i.e. ∂f ∂f ∂f ∂f = (y) and = (x) ∂x y ∂x y ∂y x ∂y x In this problem, we can see clearly from Equation (22.34) that ∂H T is a function of T and ∂P hence the pair (P, T ) does not form a set of natural variables for H. Note that it is meaningless to describe any one variable as being ’natural’. I hope this makes sense to everyone. Note. It is dangerous to say that the derivates are complex. Yes, complex means ’not simple’, but it also means ’not real’!References[1] Alberty, R. A. et. al., Use of Legendre Transformations in Chemical Thermodynamics (IU- PAC Technical Report). Pure Appl. Chem., 73 (8), 2001, 1349-1380. Available online at 4