ch08 - CHAPTER EIGHT 8.1 a. U (T ) = 25.96T + 0.02134T 2 J...

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CHAPTER EIGHT 8.1 a . ± () . . UT T T =+ 2596 0 02134 2 ±J/mol ± ± ± UU T ref 0 0 100 2809 0 oo o C J/mol C C (since U(0 C)=0) == = o ) 100 o C 0 o C = 0 b. We can never know the true internal energy. U is just the change from U to . ± ( ± ± U 100 o C c. QW U E E kp −= + + ∆∆ ∆ ∆∆ EEW 00 ,, QU = (. ) [ ( ) 30 2809 0 8428 8400 mol J / mol] J J d. C U T dU dT T v V = F H G I K J + ±± [. . ] ± 0 04268 J / (mol C) o ± ( . . ) . . ± ( [ . ( ) . ( ) ] UC T d T T d T T T v T T + = + O Q P P F H G G I K J J =⋅ =⋅ − + =⇒ zz 1 2 0 04268 0 04268 2 2596 100 0 0 02134 100 0 8428 8400 0 100 2 0 100 2 mol) J / mol) mol) (J / mol) J J 8.2 a. CCRC T vp v =−⇒= + °− ° 353 0 0291 8 314 .. [ . bg b g b J / (mol C)] [J / (mol K)] 1 K 1 C g ⇒= + ⋅° CT v 27 0 0 0291 [ J/(mol C)] b. ] 100 100 2 100 25 25 25 ˆ 35.3 0.0291 2784 J mol 2 p T HC d T T ∆= = + = c. . U C dT C dT RdT H R T ==−= = = z z z 25 100 25 100 25 100 2784 8 314 100 25 2160 bg b g Jmo l d. ± H is a state property 8.3 a. v [] . . . kJ / (mol C) o ⋅= + × − × −− 0 0252 1547 10 3012 10 59 T 2 atm L atm L / (mol K) K mol mol) kJ / mol kJ ) k J kJ n PV RT Qn U d T U T d T U T Td T ⋅⋅ = = == ⋅ + × = + ×−× = z z z ) ) [ ] ( ) . ± . ( ) . ± . ] . ± ) . . ] . 200 300 0 08206 298 0245 0 245 0 0252 6 02 0 245 0 0252 1547 10 7 91 0 245 0 0252 1547 10 10 7 67 11 25 1000 22 5 25 1000 33 2 25 1000 %. error in = 6.02 -7.67 7.67 Q 1 100% 215% ×= error in = 7.91-7.67 7.67 Q 2 100% 313% 8- 1
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8.3 (cont’d) b. CC R pv =+ CT TT p [] ( . . .) . .. . kJ / (mol C) o ⋅= + × − × + ×− × −− 00252 1547 10 3012 10 0008314 0 0335 1547 10 10 59 2 2 T QH n C d T d T P T T == =⋅ + × × = × z z 1 2 0 245 0 0335 1547 10 10 9 65 2 25 1000 mol [kJ / (mol C)] 10 J Piston moves upward (gas expands). o3 (. ) [. . . ] . c. The difference is the work done on the piston by the gas in the constant pressure process. 8.4 a. C p l di bg b g bg CH 66 313 K [kJ / (mol K)] × = 0 06255 234 10 313 01360 5 . b. C p v di b g o C [kJ / (mol C) 40 0 07406 32 95 10 40 2520 10 40 77 57 10 40 0 08684 58 2 12 3 °= + × × + × . . .] c. C p s b g b g C 313 K 0.009615 kJ / (mol K)] × × = 0 01118 1095 10 313 4 891 10 313 52 2 . [ d. ± . ... . HT TT T v 3 kJ mol × × + × O Q P P = −−− 007406 32 95 10 2 2520 77 57 4 3171 5 2 8 3 12 4 40 300 e. ± . . HTTT Cs × O Q P P = 0 01118 1095 2 4 891 10 3459 5 22 1 313 573 kJ / mol 8.5 H O (v, 100 C, 1 atm) H O (v, 350 C, 100 bar) 2 o 2 o a. ± H =−= 2926 2676 250 kJ kg kJ kg kJ kg b. ± . . . T × + × × =⇒ z 0 03346 0 6886 10 0 7604 10 3593 10 8845 2 1 2 100 350 kJ mol 491.4 kJ kg T d T 3 Difference results from assumption in (b) that ± H is independent of P . The numerical difference is ± H for H O v, 350 C, 1 atm H O v, 350 C, 100 bar °→ ° b g 8.6 b. C p nCH ( l ) o 61 4 kJ / (mol C) 0 2163 . ± ] . Hd T z 0 2163 1190 25 80 kJ / mol The specific enthalpy of liquid n-hexane at 80 o C relative to liquid n-hexane at 25 o C is 11.90 kJ/mol c. p v ) o 4 kJ / (mol C) +× −× . . . . 013744 4085 10 2392 5766 2 T T 1 2 3 ± . . . ] . T T d T × × + × z 013744 4085 10 10 57 66 10 110 7 2 1 2 3 500 0 = The specific enthalpy of hexane vapor at 500 o C relative to hexane vapor at 0 o C is 110.7 kJ/mol. The specific enthalpy of hexane vapor at 0 o C relative to hexane vapor at 500 o C is –110.7 kJ/mol. 8- 2
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8.7 TT T °= °− = °− CF F bg 1 18 32 05556 17 78 . .. CT p cal mol C F F ⋅° = + = + ′ ° b g 6890 0 001436 05556 17 78 6864 0 0007978 .
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This note was uploaded on 06/07/2008 for the course CHE 2171 taught by Professor Wetzel during the Spring '08 term at LSU.

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ch08 - CHAPTER EIGHT 8.1 a. U (T ) = 25.96T + 0.02134T 2 J...

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