Chapter 2 Homework Solutions

Transcription

1 . Cl - 3 translations + rotations + vibration Chapter Homework Solutions (a) Um(rigid) = (3/)R + (/)R = (5/)R Hm(rigid) = (7/)R Cpm(rigid) = (7/)R = 9. J/mol-K (b) Um(vib) = (3/)R + (/)R + R = (7/)R Hm(vib) = (9/)R Cpm(vib) = (9/)R = 37.4 J/mol-K. C6H6 (N=) - 3 translations + 3 rotations + (3x-6) = 30 vibrations (a) Um(rigid) = (3/)R + (3/)R = 3R Hm(rigid) = 4R Cpm(rigid) = 4R = 33. J/mol-K (b) Um(vib) = (3/)R + (3/)R + 30R = 33R Hm(vib) = 34R Cpm(vib) = 34R = 8.5 J/mol-K.3 CO (N=3) - 3 translations + rotations + (3x3-5) = 4 vibrations (a) Um(rigid) = (3/)R + (/)R = (5/)R Hm(rigid) = (7/)R Cpm(rigid) = (7/)R = 9. J/mol-K (b) Um(vib) = (3/)R + (/)R + 4R = (3/)R Hm(vib) = (5/)R Cpm(vib) = (5/)R = 6.3 J/mol-K.4 =.4 L = 44.8 L = 73 K (constant) n =.00 mol (a) Reversible U = nc = H = nc = 0 m 0 pm w = nr ln( / ) = mol(8.3 J / mol K)(73 K)ln(44.8 /.4) = 570 J q = U - w = +570 J nr ( mol)(8.3 kpa L / mol K)(73 K) (b) Constant p (final pressure): p = = = 50.6 kpa 44.8 L U = H = 0 w = p( ) = 50.6 kpa(44.8 L.4 L) = 30kPa L = 30 J q = U - w = +30 J

2 (c) p = 0 U = H = 0 w= p ( ) = 0 q = U - w = kPa p =.00 atm = 0.3kPa atm = 300 K = 400 K p p 400 = p = p = 0.3kPa = 35.kPa U = ncm = n R = (.0 mol)(.50)(8.3 J / mol K)(400 K 300 K) = 50 J =.5kJ w = 0 (because = constant) q = U-w = U =.5 kj.6 n =. mol = 00 o C = 373 K vaph = 40.7 kj/mol w = -P(liq - gas) +Pgas = nr = (.0 mol)(8.3 J/mol-K)(373 K) = +300 J = +3.0 kj q = H = n condh = n(- vaph) = ( mol)(-40.7 kj/mol) = kj U = q + w = kj kj = kj.7 = 5 o C = 98 K p =.00 atm Reaction: Mg(s) + HCl(aq) MgCl(aq) + H(g) mol Mg mol H We will need n(h) below: n( H ) = 5 g Mg = 0.67 mol H 4.3 g Mg mol Mg { } { } { } w= p = p = p + p prod rct H MgCl Mg HCl H Simplificaion above because volumes of liquids/solids are negligible compared to volumes of gases. { } H H w = p = p = n R = (0.67 mol)(8.3 J / mol K)(98 K) = 530 J.5kJ.8 mol of W(s) requires mol W x 6 mol CO/ mol W = mol of CO(g). = 50 o C = 43 K w = -P - = +P = +P prod rct rct CO [Consider only gas phase reactants or products] w = -P = -n R = - mol(8.3j / mol K)(43K) = 400 J = +4. kj CO CO

3 .9 n = mol = 5 o C = 98 K = 00 o C = 473 K Cpm = a + b a = 0.7 J/mol-K b = 0.37 J/mol-K (a) Constant Pressure [q = H] q = H = nc pm d = n ( a + b ) d = na[ ] + nb ( K ) 473 (98 K) = ( mol)(0.7 J / mol K)(473 K 98 K) + ( mol(0.37 J / mol K ) = 3530 J J = 8490 J 8.5kJ w = p = p + p = nr + nr = nr( ) = ( mol)(8.3 J / mol K)(473 K 98 K) = 450 J.5kJ U = q + w = 8490 J +(-450 J) =7040 J 7.0 kj (b) Constant olume [q = U and w = 0] For a Perfect Gas U and H depend only upon temperature. herefore the values of U and H are the same as in part (a): U = 7040 J 7.0 kj H = 8490 J 8.5 kj Because = constant w = 0 q = U = 7040 J 7.0 kj.0 Cm = Cpm - R = = 8.8 J/mol-K. mol n =.45 g = 0.056mol 44 g = 7 o C = 300 K = 500 cm 3 = 0.50 L = 3.00 L Let's calculate for an adiabatic expansion. We'll need R/Cm = 8.3/8.8 = 0.89 = R C m R 0.89 C m = = (300 K) = 79 K Adiabatic Expansion: q = 0 w = U w = U = ncm = (0.056 mol)(8.8 J / mol K)(79 K 300 K) = 95 J

4 . n = 3.0 mol = 60 K = 85 K Cpm = 9.4 J/mol-K Let's calculate Cm: Cm = Cpm - R = J/mol-K Because p = constant q = H q = H = nc pm = (3.0 mol)(9.4 J / mol K)(85 K 60 K) = 05 J. kj U = ncm = (3.0 mol)(. J / mol K)(85 K 60 K) = 580 J.6 kj. n =.00 mol Cm = 0.8 J/mol-K = 30 K p = 3.5 atm p =.50 atm γ nr (.0 mol)(0.08 L atm / mol K)(30 K) = = = p 3.5atm C C + R C C 0.8 pm m = = = = m m 7.8 L γ γ γ p γ 3.5 p = p = = 7.8 = 3.06 = p.50 /.40 = (3.06) = 9.4L = p (.50)(9.4) = 87 ()(0.08) = K nr Adiabatic: q = 0 w = U = ncm [ ] = ( mol)(0.8 J / mol K)(87 K 30 K) = 480 J 0.5kJ.3 n = 0.50 mol = 50 K vaph o = 6.0 kj/mol p = const (phase transition). herefore q = H q = H = n H o = (0.50 mol)(6.0 kj / mol) = 3.0kJ vap w = p( ) p = nr = (0.50 mol)(8.3 J / mol K)(50 K) gas liq gas = 040 J.0kJ U = q + w = =.0 kj

5 .4 C6H5CH5(l) + (/) O(g) 8 CO(g) + 5 HO(l) o o o o combh = 8 f H ( CO) + 5 f H ( HO) f H ( C6H5CH5) + (/ ) 0 = 8( 393.5) + 5( 85.8) (.5) = kj / mol [ ]. 5 For N(g) a =.35 L -atm/mol b = L/mol n = mol = 98 K =.00 L = 4.8 L Internal Energy ( U) U π = = na U na U = d = = = d n a n a U = ( mol).35 L atm / mol.00 / 4.8 / L mol L mol 0J = 5.8 L atm = 53. J L atm Work (w) na p + ( nb) = nr nr p = nb n a = = nr n a = + w pd d nr d n a d nb nb ln ln nb nb [ ] w = nr b + n a = nr + n a w = ( mol)(8.3 J / mol K)(98 K)ln (.35 / ) mol L atm mol L L 0J = 6 90 J / mol + (5.40 L atm / mol) = 690 J J L atm = J 5.7 kj / mol Note: Observe that the work is less negative than it would be if there were no attractive forces (i.e. if a = 0). his is because some of the work energy released in the expansion must be used to pull the attractive molecules further from each other.

6 Heat (q) U = +53 J/mol w = J/mol q = U - w = +53 -(-5745) = +668 J/mol +6.3 kj/mol P (.0 atm)(.4 L).6 = 73 K nr = ( mol)(0.08 L atm / mol K) = P (.0 atm)(44.8 L) = 546 K nr = ( mol)(0.08 L atm / mol K) = P 3 3 (0.50 atm)(44.8 L) = 3 73 K nr = ( mol)(0.08 L atm / mol K) = Cm = (3 / ) R =.5(8.3 J / mol K) =.5 J / mol K C pm = Cm + R = = 0.8 J / mol K Step (Constant Pressure) 0J w = p( ) = (.0 atm)(44.8l.4 L) =.4 L atm = 60 J L atm U = ncm = ( mol)(.5 J / mol K)(546 K 73 K) =+ 340 J q = H = nc pm = ( mol)(0.8 J / mol K)(546 K 73 K) = J Step 3 (Constant olume) w = 0 q = U = ncm ( 3 ) = ( mol)(.5 J / mol K)(73K 546 K) = 340 J H = nc pm 3 = ( mol)(0.8 J / mol K)(73K 546 K) = 5680 J Step 3 (Isothermal) otals: U = H = 0 w = nr3ln( / 3) = ( mol)(8.3 J / mol K)(73 K)ln(.4 / 44.8) =+ 570 J q = -w = -570 J Utot = U + U 3 + U3 = = 0 J Htot = H + H 3 + H3 = = 0 J wtot = w + w 3 + w3 = = -690 J qtot = q + q 3 + q3 = = +700 J

7 Note: As expected Utot = Htot = 0 around the cycle. U and H are State Functions However qtot 0 and wtot 0 around the cycle. q and w are not State Functions o within roundoff error wtot + qtot 0 because qtot + wtot = Utot = 0.7 n =.0 mol Note: he initial temperature and pressure (required for the calculation) was not given in the problem. We ll use = 98 K p =.0 atm. nr ( mol)(0.08 L atm / mol K)(98 K) = = = p.0atm 4.4 L Cpm = (7/)R = (7/)(8.3) = 9. J/mol-K Cm = Cpm - R = = 0.8 J/mol-K Step a: Constant volume heating to p = p = atm p = = = = (98) = 596 K p w = 0 q = U = ncm ( ) = ( mol)(0.8 J / mol K)(596 K 98 K) =+ 690 J H = nc pm = ( mol)(9. J / mol K)(596 K 98 K) = J Step b: Adiabatic Expansion from = 596 K back to 3 = 98 K q = 0 w = U = ncm ( 3 ) = ( mol)(0.8 J / mol K)(98 K 546 K) = 690 J H = nc pm 3 = ( mol)(9. J / mol K)(98 K 546 K) = 8670 J We ll need 3 for last step. 3 = 3 R C m 3 3 C m/ R 0.8/ = = = = 5.66 = 5.66(4.4) = 38.3 L

8 Step c: Isothermal Compression from 3 = 38.3 L to =.4 L otals: U = H = 0 w = nr ln( / 3) = ( mol)(8.3 J / mol K)ln(4.4 /38.3) = J q = -w = J Utot = = 0 Htot = = 0 wtot = = -890 J 0 qtot = = +890 J 0