Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Successfully reported this slideshow.

No Downloads

Total views

507

On SlideShare

0

From Embeds

0

Number of Embeds

2

Shares

0

Downloads

2

Comments

0

Likes

1

No embeds

No notes for slide

- 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. 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. 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. 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 http://www.iupac.org/publications/pac/2001/pdf/7308x1349.pdf. 4

No public clipboards found for this slide

Be the first to comment