Question

2. Find the changes in internal energy (AU) and enthalpy (AH) per kmol, for air going from 10 bar, 277 K and 2.28 m3/kmol (molar volume), to 333 K and 1 atm. Use the following: 4 Cp = 29.3 kJ kmol-? K-1 Cy = 21 kJ kmol-? K-1 PV/T = constant Do not use any additional properties of ideal gases. There is no need to prove the gas is ideal or to use such properties. Hint: split the process up into two processes for which you can calculate AH and AU without assuming ideal gas (such as isothermal, isobaric, isochoric etc.).

Please also explain the difference between delta H and delta U. Thanks!

Answer 1

A solution --* civen: o initial Tempaevaterre (T7) = 2771 initial pressure (P) (P) = lobor initial volume Į (7) = 2.28 mp3 Jiamol molar v atere (T2) = 333K final Temperert fincel pressure (Px) = 1 atm Sincer ce I atm = 1.01325 bar of so P₂ = 1.01325 bar * Since are of the property changes in dependent process that brings them about. base calculation on a Simple telo- step, mechanic nically reversible process in which I kmolo is- cele can g air cooled at constant volume to the final poressure 2. Heated at constant pressure to the finced temperatures

on (pv) Dia grom wulerer as initial condition e 6= intermidicite condition ca fince condition 6 * casero ーター o proces C288 (@ 6 is isochoric procese forocess --18 i coborrie forecase for procese - @- 6 V: consteint so Р T - constant 201110 = Po To since. => Tas T, 2 277 K To = Tax Pb *** Pse Pa = lobor PG=Pc = Pg = 1.01325 bar ogo To = 252 k 1,01395 boet to beto =) To I 98,067K

☺ for final volume ccel colored as- un of Paiva Pex ve = Sincer py = constant Ta Te => c 는 Pava Х Tc Ta Pe > 1/ z Va x Pa Pc T Tai since yinen- Var V, = 2.28 m3/lamol Pa 10bar 1 -७ Pe Pa 1.0132 5 bar To = T = 277K - To 333K Te but in în above equation =) Ve 2.28 6873 x homol 10 bar 1,01375 E best 333K 277 K Ve = 27.051 m3 kmoel

3 case-2 -% -% - for procese la Cor iso choric procere) Vo constant * 6 change calculated as- in internal energy AU, - CAT > AU, = Capo CTG-Te ) + sincer given Cya 21 las Rimolk Ta=277K -> put the value of thei) contain Ts), and To 528.067K. Cr in above equation- =) AU, = 21145 X ( 28.067-279) K kolok =) sugg 2 ou, 5297.593 13 / lemoe AH and OU Related as a AHE OU + Alpv) if Ve long terint AH = AU +V, AP

Song for process.- 6 AH, EAU, TV. AP sincer Au, = -5927.593 kJ/lemoe = -5297.593x1033 y = 2.98 ms Iamue AP = Pb-Pa = (1.01325-10) bors = -8.987bar kmol I bar = 105pas so AP = -8.987 x105 Pas I m3. Pa - J 103 m3. Pa = 1K5 put these value in cibove equation , IRJ=1037 AH, - 5297.593 x103 I + 9.28 m) x (1.01325-105X105 pie kmol amol =) AH, = 5297.593x103. kmol 204.8979X103 m3, pa kmol since- | md. Pe = 13} -) AH, = -5297.593 x 103 2098.999X103 lamel 20-4-7897974039 lemoe =) kmol AH, 7976.572 X1033 kmoe Siniec Ź 1037=1kg =) A4, =-7276.572 KS k moe

(as-(3 ملا ملم for process o (C le p=constant isobutie processo 10 AH for this proces ees calculated aga AH₂ = CP.AT => AH₂ = (p. (Te-To) Sincer given- = 29.3 kJ kmolok Te = = 333K To=98.067K but the value of Te, Tos Co in above equation- A Hg = 29.3 dB x ( 333 – 28.067) K kmolog A Hg 8934. 5369 KJ lamor * © A4 and AU Related as AHE AU + ACPV) = AU = AH- A(PV)

AU - AH - piny p=constante. for forot238-0-0- AU, A Hg - Pexe Ve-V6) Since AH E 8934.5369 14] Knol = 3934.5369910% ] kmol Pc = P = I aim - 1.01325 bort = 1.01325X105 pa Vc = V2 = 27.051 m3 **8210572 V6=N, amol = 2.28 m3 amol ſ put the yalue cap og AH2, Pey ve & Vb in above equation 5) Alla 8934.5369 Kj lamol 1.01325X105 Pax (27.0 51-9,23) m3 kmol -) Δυ, = 8934.5369 X1033 25.0gga m3. Pa x105 lamoe la mol Since 1 m3. Pa = 15$ ñ Aug = 8934.5369X102J 2 509.92 x 10 x 103 u 2. Aug = 6426.6169 x 1033 famol kmol Since 1033=115 HS § Alg = 6424.6169 145 moe

* case-6 -- for Overcell process , Chonge in enternal energy AU = 04, + oua Since- su, E-5227.593 RJ kmol = 6494-6169 Has 402 13 тое put the value of Au, s AUa in above equation- OU = -5227.593 K3 +6424.6169 K3 kmol kmul = Toy = 1197.0239 AJ Answer kmol for overall process, change in enthal by- DHE AH, TAHg AH₂ since AH, = 7.276.572 KI Homoe 8934.5369 RJ mol put in above equation AHF - 7276.572 K3 + 8934.5364 KS Amoe kmal AH = 1657.9649 las Anguiler kmol =)

y case-( *0 internet e energ defined asa * internal enor energy form of energy of the . is the sum of tell the microscopie System they have translationcel one energy. * Since; molecules move, etc. 2 internal energy and the enthal boy do fined ceg enthal by combines the porodes et ġ pressure and volume (Boundory work). the system