H C p = T Enthalpy can be obtained from the Legendre transform of the internal energy as follows, U + T = P T (RT )

Transcription

1 roblem Set 2 Solutions 3.20 MI rofessor Gerbrand Ceder Fall 2003 roblem S(U, ) : ds du + d 1 ariables here are U and and intensive variables are and. o go to 1 as a natural variables take the Legendre transform by subtracting U J S U from S So, 1 dj U d + d Note that J is equal to F (F is the Helmholtz energy) Similarly, go from S(U, ) to B(U, ) by taking B S 1 db du d A Legendre transform with respect to both U and defines the lanck function (Yes, you too can be famous by defining your own Legendre transform) U Y S 1 dy U d d One can show that these functions have etrema properties just like the Legendre transforms of energy. Massieu functions are maimum under constant value of their natural variables. roblem 1.2 We know that C p is defined as

2 H C p Enthalpy can be obtained from the Legendre transform of the internal energy as follows, H U + We know that for an ideal gas U is a function of temperature only and the R (for one mole), thus H U + (R ) U is only a function of temperature and this is not a function of pressure. Also, R is not a function of pressure. So C p of an ideal gas must be independent of pressure.

3 roblem 1.3 F F (, ) Compare this with F F df d + d F U S herefore, df Sd d F Using the given formula for F, solve for by taking the derivative w.r.t at constant. F a R f + m b Since f( ) is only a function of, this term drops out and the solution is: F R a m b 2 roblem 1.4 (a) We can write the differential form of the entropy as a function of and 2 m S S ds d + d Multiplying by to get ds (which is equal to dq) S S ds d + d Since we are told the process is adiabatic, ds q 0, so, Or, S S d m d

4 S d α S Cp d C p (b) Using the above relationship, we can get a value for the final temperature, d α d C p f α ln Δ i C p α f i ep C p Δ For this case we can assume we have 1kg of material which means the volumes is m 3. Remember that α 3α L K m 999atm a/atm f 300 ep 2.3J/g K 1000g f K he only trick here is to be careful with your units and the definition of α L and α. roblem 1.5 (a) Start with the internal energy du ds d + HdM Using a Legendre transformation get G(,, H) G U S + HM dg Sd + d MdH And since we know that the order of differentiation does not matter 2 G 2 G H H G G H H H, H,, M H,, H,

5 (b) One possibility is S H, M H, roblem 1.6 (a) U ds + τdl G U S Which yields the solution: dg Sd + τdl ds + Sd (b)we also know if GG(L,) that hus, dg τdl Sd G G dg dl + d L G G τ and S L 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, (c) 2 G 2 G L L G G L L τ S L L L L du ds + τdl L L U L S + τ L

6 Using the relation from part (b), U L (d) If U is only a function of temperature, then LEEL 2 ROBLEMS roblem 2.1 τ + τ U τ 0 τ L L 1 τ 1 τ (a) Start by constructing the differential form of S(,) [In Zemansky this is called the second ds equation] S S ds d + d S S ds d + d But we know S C p and from a Mawell relation we know S.So, ds C p d For an isothermal compression, L L d C p d αd Q ds αd In the case of a solid or liquid, neither or α is very sensitive to pressure. So, Q α d αδ Now we can use the data given to calculate Q (Replacing with M/ρ) Q MαΔ ρ 298K 0.5kg K kg/m a Q 416J

7 (b)we now need to calculate the work during the compression W W d W d β β d β d f i W f Mβ 0.5kg a a 2 2 2ρ kg/m 3 W 44J (c) We know the first law, ΔU Q + W 416J + 44J ΔU 372J hus it can be seen that the etra 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 ds (which we know is 0 since the process is adiabatic and reversible) ds 0 C p d αd d α d C p f α ln Δ i α αm f i ep Δ i ep Δ C p C p ρ 0.5kg K atm a/atm f 298 ep kg/m 3 385J/kg K 0.5kg C p f 300K Δ K roblem 2.2

8 In terms of and, du can be written U U du d + d We can evaluate the two partials U and U using U(S, ) which we know So du ds d U S Which can be simplified since we know C p S and α Similarly U C p α U S which can be simplified using the Mawell relation S utting this all together, U α + β (β α ) du (C α) d + (β α ) d and β roblem 2.3 Given the definition of dh : dh ds + d, we want first to determine how the enthalpy varies with pressure, at constant temperature: H S + By using the Mawell s Relation S (from the epression for the Gibbs Free Energy) and the definition 1 α

9 , we have: H (1 α ) What we are really interested in is to find how H evap H vap H liq changes with temperature: ΔHevap [H vap H liq ] H vap H liq ΔHevap vap liq vap α vap liq α liq vap (1 α vap ) In general, at relatively low pressures and densities, all gases behave ideally. For an ideal gas, α 1 nr 1 herefore, ΔHevap 0 roblem 2.4 Given U (S, X) X a ep(bs) First find the equations of state U X a ep(bs)b bu (S, X) S X U f ax a 1 au (S, X) ep(bs) X X From these we can derive the following which will be useful later S 1 S ln (1) b bx a b f X (2) a U (S, X) b (3) 1 a f X ep ( bs) a (4) Starting with the first function D(S, f ) we perform a Legendre transform as follows

10 D(S, f) U fx (1 a)u(s, X) Now using (4) to get this in terms of S and f only Similarly for F(,X) (1 a)a f D(S, f) (1 a) ep ( bs) ep(bs) a F (, X) U S U(S, X) bu(s, )S (1 bs)u(s, X) Using (3) and (1) from above F (, X) 1 ln bx a b Now we do a transformation for to both and f K(, f) U S fx Using (3), (1) and (2) a K(, f) ln b b bx a b K(, f) 1 a ln a b a b bf roblem 2.5 (a) Start by using the chain rule on the left side of the equation S S We know that S C and we can epand the second derivative as follows utting those together, C We can replace C by C using the relationship C 1 β α

11 Yielding α 2 C C β S β α 2 C β C α α β α (b) We can epand the derivative as follows S S S he first term is and the second term can be transformed using a Mawell relation S C. his gives (c) Start with the left side Using a Mawell relation, S C α S C 2 2 S 2 S Now the right side S Cv S S C v + he first term in the above is zero since the change in temperature at constant temperature is zero. he second term can be written as S 2 S And since we know the order of differentiation doesn t matter 2 S 2 S

12 roblem 2.6 (a) Start with the differential form of G (b) Start with the differential form of H dg Sd + d G S + G S G 1 S ( α) β G α S β dh ds + d H S + H C H C roblem 2.7 (For this problem Maple is our friend) (a) his part involves taking the indicated derivatives of the given function. he two tricks are knowing what β is and how to change U into something with derivatives of w.r.t or. β 1 and using the differential form of U and a Mawell relation we get, he solutions are then U

13 (b)remember R b 1 1 ( b) 2 2 β R 3 2a 2 + 4a b 2ab 2 U R a b 2 1 α R α Epand H in differential form to try and get H into a form you can solve for dh ds + d H S + H + H R + b Now for C C C C α 2 β R 3 2 C C 2 ( b) 2 roblem 2.8 U 1 k + Nf k 1 N k du ds d For a reversible, adiabatic process,

14 (du) S (d ) S Along a reversible adiabatic process: k g(s) so U Again for a reversible, adiabatic process (du) S Integrate from 0 at constant S Moving things around, g(s) k S g(s) k U d g(s) S d k U(, S) U( 0, S) g(s) 1 (k 1) 1 k 1 0 k 1 g(s) 1 g(s) U(, S) + U( 0, S) + k k 1 (k 1) 0 k 1 s) 1 g(s) U(, S) g( + U( 0, S) + k 1 k 1 (k 1) 0 k 1 he second term above is only a function of S, so we can call it some function w(s) k 1 U(, S) + w(s) +w(s) k 1 k 1 (k 1) w(s) is an etensive function, therefore it can be written as N (the number of moles in the system) times a molar quantity f(s) which is independent of the system size k f N k U(, S) +Nf (k 1) N k roblem 2.9

15 he thermal epansivity is defined as Using a Mawell relation, 1 α S And we are told in the problem that S is independent of crystallite structure and pressure, so And thus α 0 at absolute zero. roblem 2.10 S 0 (a) he key property is that S < 0. his can be proven from the Mawell relation. H S M H (this can be proven from the differential of dφ d(u S + MH) Sd + d MdH ) and since M H 0 H S < 0 < 0 H S( ) 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 decreases. (b) We can evaluate M H H S as follows H S H S H S M c H M H c H κ M κ M H H 2 H H

16 Figure 1: So, κ H H S c H LEEL 3 ROBLEMS roblem 3.1 Given: θ Rθ U S 2 2 R 0 (a) We want to make this the absolute U and not the molar quantity, so multiply by N. 2 2 Nθ NRθ Nθ S NRθ U NU S R 0 R N 0 N 1 θ Rθ U S 2 2 N R 2 0 he equations of state give us 2 U 2 θ S S N R N,

17 U N,S 2 Rθ N U θ Rθ θ Rθ µ 1 S 2 2 S 2 2 N N 2 R 2 R 2,S or (b) Epress µ as a function of and Substitute in to the epression for µ roblem 3.2 µ 1 N 2 µ U N R S 2 θ N Rθ θ N 2 R 2 Rθ N R 4 θ Rθ R µ θ 4 Rθ Start with the epression we used in class y + f z f y y f f z For this problem, this equation can relate the constant l heat capacity to the constant heat capacity as follows, S S S + l l o get heat capacity we need to multiply both sides by, and when we do this we get S C l C p + l he first derivative can be rewritten using the Mawell relation S simplifies to which

18 C l And since we know that C p α C l C p α Now we need to epand the second derivative l l l Since the material is isotropic, l 1 α which now leave us with 3 1 C + 2 α 2 l C p 3 l One more partial to go...if we look at the relationship between stress and strain, σ Eε and thus σ ε where E is the Young s modulus for an isotropic material. o get this in a form we can use, we need to multiply by l to switch from ε to l and then invert. his then will give us 1 2 α 2 E C l C p + 3 l l l E roblem 3.3 he total system is at constant volume and temperature hence its Helmholtz free energy is minimal with respect to the internal degrees of freedom. Figure 2:

19 df tot df α + df β α β df tot S α d α p α d α + µ A dn α A S β d β p β d β + µ dn β A A since d α d β 0 and d α d β and dn α dn β A A we can write µ A µ A df tot p α p β d α + α β α β If d α and dn α A were independent, equilibrium would require that p α p β and µ A µ A. However d α and dn α A are NO independent! df tot d α Kdn α A α β µ A µ A p α p β d α K he bracketed term must equal zero for equilibrium. dn α A α β µ A Kp α µ A Kp β roblem 3.4 he heat capacity c is related to the temperature derivative of the entropy under the condition. ( being the conditions defined in the problem statement) Starting with S(, ) we write S c S S ds d + d We can take the temperature derivative of this holding constant S S (ds) d + d S c p S + (Note: he above can also be obtained using y + f f y f z y f z )

20 We know through a Mawell relation that S α v. So now we need to find. We start from the given information that a a constant, thus d ad. his gives us 1 a Continuing from there, we write out the differential form of (, ) But we know So is given by Replacing we get or utting this all together... So we get finally that d d + d α v and β. We now have d α v d + β d α v β β α v a α v a a + β S c p 1 + ( α v ) a S c p 1 αv a + ( α v ) a a + β S 1 αv a c p + ( α v ) a a + β (α v ) 2 c c p a+β