2. My instructor s name is T. Snee (1 pt)

Size: px

Start display at page:

Download "2. My instructor s name is T. Snee (1 pt)"

Rebecca Bradley
5 years ago
Views:

1 Chemistry 342 Exm #1, Feb. 15, 2019 Version 1 MY NAME IS: Extr Credit#1 1. At prissy Hrvrd, E. J. Corey is Nobel Prize (1990 winning chemist whom ll students cll (two letters 2. My instructor s nme is T. Snee (1 pt 3. Short Answer / Explin in Words! Answer 4 out of 6! (4 pts ech. Wht re the roles of the nd b coefficients in the vn der Wls eqution? Answer: The coefficient describes the ttrction for gs molecules with echother- interctions such s dipole-dipole nd dispersion. The b coefficient ccounts for the finite volume of the gs molecules themselves, nd lso ccounts for the molecules colliding with echother. b. Explin wht the minus sign ( does in the eqution for work: P rev V. Answer: The minus sign mkes compression work positive, which is kin to storing energy. Vice vers for n expnsion. Reversible work is when the exterior pressure is chnged in such wy s to mtch the interior: Pext=P. c. If I compress gs, is the work positive or negtive? Will it get cold or hot (ssume its thermlly insulted? If I compress gs under isotherml conditions, how is it tht the temperture is constnt? If there is het exchnge, is it positive or negtive? Answer: Compressing gs mkes it hot, which is due to positive work. Isotherml trnsitions require het trnsction, whereby energy leves the system. This is negtive energy, which negtes the positive work to keep the internl energy constnt. d. Nme one wy in which the exctness of the prtil differentil of stte vrible hs been useful. Answer: Often there is just one eqution for the chnge of the stte vrible. It lso helped us determine the dibtic eqution of stte. e. Why is the het cpcity t constnt pressure higher thn the het cpcity t constnt volume? Answer: Addition of het energy simply increses temperture under constnt volume conditions, but under constnt pressure the volume of the system must increse. This is work, the energy of which is shved off the het resulting in lower temperture. This is the sme s hving higher het cpcity. f. We told you in freshmn chemistry tht enthlpy is the het relesed by rection (i.e. H = + rections re cold nd H = rections re hot. But is tht relly true? Why or why not? Answer: Enthlpy is het only when pressure is constnt. However, usully pressure is constnt, so it ws lie tht is usully true.

2 4. UNITS! Try just 3 of the four questions. (show you work! (6 pts ech. In the eqution: e U nrt = 1 + V, wht re the units of got? Answer: The got hs units of 1/V, since the term hs to hve the sme units s 1 which hs no units t ll. b. Wht re the units of in the Redlich-Kwong eqution of stte: P = RT V m b T V m (V m +b? Answer: The whole term: must hve units of pressure. So we cn sy hs units of pressure, but it must lso hve units of TV m 2 becuse tht fctor is in the denomintor. Putting this ll together: = P K (m3 2 mol 2 Of course, there re nswers such s: tm K (L2 ; we will be sensible bout it. mol 2 c. The Joule-Thomson coefficient is: T P H = μ J T = 1 H P T Answer: We write the eqution using units: 1 H P T = K J Wht re the units of the coefficient? J = K T. It might hve been esier to use: = K P T P P H P d. It turns out tht U = T(S S. Cn you use this grph of temperture vs. entropy to show wht the chnge in internl energy is t room temperture? (You cn indicte the internl energy using the grph or describe it in words, either is ok. Answer: The chnge in internl energy is the integrl, i.e. T(S S, nd it is shown in the shded re. If you just sid integrte it we would ccept tht. Nchos with Sls! Questions 5-11, you need to only nswer 5 out of 7! (8 pts ech 5. The Gllus eqution of stte is: (P + T V m = RT + b. Thus, cn you show tht the b coefficient rises the compression fctor while the coefficient lowers it? FYI the compression fctor is Z = P, where P is the P perfect gs pressure which mkes Z = PV m RT. Answer: First multiply out the first term of the Gllus Eqution: PV m + V m T nd you cn immeditely identify Z: PV m b which is Z = 1 + compression fctor while lowers it. RT + V m RT 2 = RT RT + b RT = RT + b nd then divide by RT V m RT RT 2. As result, b rises the 6. The compression fctor is Z = P/P where P is the rel gs pressure nd P is the perfect gs pressure. Wht is the compression fctor of one mole of xenon gs in 1 L continer t 200 C? You will need to use the vn der Wls eqution. Answer: First clculte the perfect gs pressure: PV=nRT; P=1mol 8.314J/K/mol ( K / 1L = kp. Now the vn der Wls eqution yields:

3 P = nrt V nb n2 1mol J/K/mol K (1 mol br L 2 /mol 2 100kP V 2 = 1br 1 L 1mol L/mol (1 L 2 = kp. Therefore, the compression fctor is the rtio, which is Z= Does perfect gs (PVm = RT or rel gs tht follows the Berthelot eqution of stte is: P = RT V m b hve greter chnge of pressure with chnging temperture t constnt volume/mol (Vm nd why? (ssume the nd b constnts re lwys positive. Answer: The derivtives re: P T V m = R V m R > R V m b for perfect gs nd P T V M = eqution. Comprting the first terms:, nd the 2 nd term: V m rel gs increses significntly with temperture vs. perfect gs. R V m b + TV m 2 T 2 V m 2 for the Berthelot T 2 V m 2 is only positive. Thus, the pressure of 8. If work is w = P ext V,. Wht is w if the exterior pressure is kept constnt? b. If rel gs follows this eqution of stte: P = nrt wht is w? V V2, nd the exterior pressure mtches the interior, then Answer: In pt., you cn just rewrite the nswer from the chet sheet: w = P ext V. For pt. b, you hve to use some definite integrls; hopefully you looked t the chet sheet nd sw this one: w = V f V i ( nrt V + V2 V = nrt ln ( V f V i + ( V f V i V f V i. x f x i 1 x 2 x = x f x i x f x i. As result: 9. If you hve n dibtic piston tht is both t high temperture nd pressure, it will expnd. But when will it stop? When the interior temperture is the sme s the exterior temperture? Or when the interior pressure is the sme s the exterior pressure? Explin your nswer. Hint: Both re not simultneously true. Answer: It will stop when the outside pressure is equl to the inside pressure. 10. Here is digrm for the chnge in enthlpy for n dibtic trnsition. Since the chnge in enthlpy is exct, we cn follow pths 1 & 2. Cn you determine the eqution for H using this digrm? Answer: Pth 1 represents chnge in temperture t constnt pressure, so the relevnt enthlpy chnge is: H 1 = T. The second pth is n isotherml depressuriztion, for which H 2 = 0 J. Thus, H = T. 11. If I hve n eqution for the differentil chnge in thermodynmic cts: ct = wrm pds brking dogs. Wht re the nturl vribles (nd ssocited conjugtes of cts? b. Cn you define function (let s cll it mice tht hs pds nd brking s nturl vribles? c. Cn you show tht this function (mice hs pds nd brking s nturl vribles? Answer:. The nturl vribles re pds nd dogs; conjugtes re wrm nd -brking, respectively. Yes you hve to get the minus sign.

4 b. The mice function is Legendre trnsform; the mice function subtrcts dogs times its conjugte (-brking from cts: mice = cts ( brking dogs = cts + brking dogs. c. You demonstrte the nturl vribles by determining the differentil: mice = cts + (brking dogs = wrm pds brking dogs + brking dogs + dogs brking = wrm pds + dogs brking Wings with Hbnero Suce. Answer both of them. 12. If I put filment inside n dibtic 1 L piston full of N2 gs t 1 tm pressure nd room temperture (25 C, I cn dd 100 J of energy.. Initilly the piston is fixed into position. Cn you clculte how much the temperture rises? How much work is done? Wht is the new pressure? b. Once relesed, the temperture of the piston will slowly drop s the volume reversibly expnds. Cn you clculte the finl temperture? c. Now cn you show tht the reversible, dibtic work is consistent with the 2 nd Lw? FYI: for nitrogen, C V,m = 20.8 J/K/mol nd,m = 29.1 J/K/mol; nd ssume perfect gs behvior. Answer:. First, we need to solve the number of moles of nitrogen: n = PV RT = P 1 L J/K/mol K = mol. The het cpcity is mol 20.8J/K/mol = J/K, nd thus the temperture rise is q C V = K (the finl temperture is then K K = K. The new pressure is P = nrt mol J/K/mol K = kp. Of course there is no work done. 1 L b. The piston will expnd dibticlly nd reversibly until the exterior pressure is 1 tm. Thus we need to use nr the following from the dibtic equtions of stte: P f = ( T f nr. Better yet: ( P f C P = T f ; plugging in vlues yields: P i T i P i T i mol J/K/mol kp ( mol 29.1J/K/mol T = ( f. Solving for the finl temperture yields Tf = 378 K kp K c. We cn now determine the work vi: w = C V T = J/K (378K K = 32.1 J. This is consistent with the 2 nd Lw becuse the work is less in mgnitude tht the het dded. 13. We solved: = C V + nr in clss using multiple step derivtion. However, there re usully mny wys to do derivtions; for exmple, there is n esier wy to show the sme s bove by solving = q strting T P from: U = P V + q Hint: You re going to wnt to substitute U = C V T, nd use the perfect gs lw. (13 pts Answer: First solve for q nd use the substitution: q = C V T + P V Now divide by T: q = C T T V + P V nd keep P constnt: T T q T T = C V P T + P V P T P Clerly T V = 1, nd P = P nrt P T P T P T P = P nr P = nr. Thus, = q T P = C V + nr V =

5 Extr Credit #2: Cheesecke DO NOT DO THIS FIRST 14. Under dibtic reversible conditions, the chnge in enthlpy is H = T. The chnge in enthlpy is lso: H = V P + q = V P since q = 0 J. Consequently, if you equte the two: V P = T cn you solve for the chnge in temperture vs. pressure in n dibtic reversible trnsition? Hint: you will need to use the perfect gs lw. (5 pts Answer: First, you hve to know tht there is fctor of P nd T hiding in V: V P = nrt there you find: Now you cn move the nrt to the right side: nrt P P = T P P. Going from And integrte: P P = T nr T P f P T f = P P i T i nr T T Now you use the identity ln ( P f P i = nr ln (T f T i ln ( P f = ln [( T f nr ] P i T i Lst, the rguments of the nturl log re equl: P f = ( T f nr P i T i

6 This is ll you need for the test. Seriously, it is. Don t mke things too hrd. Gs Constnts Atmospheric Pressures J / K / mol Pscls tm L / K / mol 1.0 tm mmhg L / K / mol 760 mmhg Torr L / K / mol 760 Torr psi L / K / mol psi br L / K / mol br Force = Newtons = kg m s2 J=Joule=Newton distnce =kg m2 s 2 Equtions: ln ( x y = ln (y x ; ln(xy = y ln(x; If ( x y z = b then (y x z = b nd y x = (b 1 z 1 x x = x 1 x = 1 x 2 x x = x2 x = = x x f x i = x 1 x x 2 x f x i x x2 = 2x = x f x i x f x i 1 x = ln(x x E=½m v 2 = p 2 /2m P V = n R T vdw: P = nrt V nb n2 V 2 lso P = RT V m b V2 m Viril: PV m = RT (1 + B + C V m V2 U = 1 k m 2 BT degrees of freedom U m = 1 RT degrees of freedom 2 U = w + q w = P ext V H=U+P V Irreversible: w = P ext V Reversible nd isotherml: U T V = C V H T P = w = nrt ln ( V f V i C V = nr q = n C V,m T or q = n,m T Adibtic Reversible: V i V f = ( T f C V nr P i ; T i P f = ( T i T f nr ; w = n C V,m T Prtil Derivtive Equtions: f = f x y f x + y y x Euler Test: y f x y x = x f y x y Legendre Trnsform: g(x,z = f(x,y z y where f vn der Wls nd b Constnts Gs (L 2 br/mol 2 b (L/mol Argon Krypton Wter (gs Xenon , Preston T. Snee y x = z This work is licensed under Cretive Commons Attribution-NonCommercil 4.0 Interntionl License.

Math 8 Winter 2015 Applications of Integration

Math 8 Winter 2015 Applications of Integration Mth 8 Winter 205 Applictions of Integrtion Here re few importnt pplictions of integrtion. The pplictions you my see on n exm in this course include only the Net Chnge Theorem (which is relly just the Fundmentl