Question

Please answer the following question completely and correctly. Please show all work and write neatly.

3. A sample of 1.00 mol ideal gas molecules with Cp, m = 7/2 R is initially at p = 1.00 bar and V = 22.44 L and then put thought the following cycle in reversible processes: (a) constant-pressure expansion to twice its initial volume, (b) constant-volume cooling to its initial temperature, (c) isothermal-compression back to 1.00 bar. Calculate q, w, AU, AH, AS for each process and for the whole cycle.

Answer 1

13. Ans: Amount of ideal gas molecule 21.00mole (pm 27/1R Co CP-R> 7/2-R SER. initial, pressure (p) 1100 barro = volume (v) = 22.442 The ideal gas was put the following Cycle in reversible process : (a) Constant prescure expansiad to twice its. Initial volume lie, from a v 22.44L to = 2x22.444 at presswa P, 21:00 ban 44.88L 1 (T) Lt). (AB) pt to (6) Constant volume cont a ct Cooling to its initial temperature . 1. To to BTK. (②) Isothermal Compression back to 1.00 ban. • The following given Cycle is chrown in p-v dingsome. Now Food process (a): A B Constant pressure at Point Ai expansiad. Pa 100 ban = 0.98692 atm. hatma 76 cm 12 22.44L no Imole R20.082 Latmmoli = 1:01325 bar Forsideal gas system, Py = nRT, T2 PM 0.98692 atm x 22.44L Amol x 0.082 Latm mot 41 2 270.079 K 2760 monitor 2 1 tom MR

prouss ] at point Bip-pol bar 0.98692 atm. V202 > 2x22:44L > 44.88L Forsideal gas system, PV₂ = nRT2 -- T2: Piva 0.98692 atmx 44.88€ RR 0.082 Latmmol-Krix imol, 2 540.158K . Wook (W): - Part dr - Posel A+ dp) dv (since reversible - Pint dv (neglecting dpdv small foom) - (-V) --0-986924m (20, -1) 44 > ncv(T.-T) 2-0-98692 *V, Latm 2-0.98692* 22.44 Lalm - Inof x 56R (540.188-270.079) K = - 22.146 Latm 2x8343stk* fx 270.079 -22.6468 x 101-3231 - 2243.9437 5678.310 7 561315923 from 1st law of thermodynamics, Au o qtw 49,240-W 5618.592 - (-2243-943) J 4 7897-563 +7857-535)

AH = ncp (227) Im x 7R(540-158 - 270.079) 7x8.314 * 2701079 J: 7859.028) Entropy of an ideal gas Can be expressed as a functiad of T As a replin (T2) - hRIn(12) and p. Flere To 270.0796 T2 2 540.158 K P = P₂ à 100 bar nolmole -481x7/qR16(2) - nring) भरि - Z Rla2 = 7/X8.34 12 JK 220.169JK Process (6 : B C Constant volume cooling. W s - Pext dv Iz to Tik 2 - Pext (V2_V) (foro finite change) 20 (since volume Constant) Au nov (T-To) - (mol x5 R ( 56 270.079 - -- 5x8314 x 270.0793 - 5613.5923 540 158) *

AH ncp(1, -72) > Trox 70R ( 270.079 - 540-158)K --Zx 834 x 270.079 : -7859.0289 from 1st Law of thermodynamics, 44 qtW Since w20 (as volume is constant) "Alaq - 5613 5927 • Entropy of an ideal gas is expressed as a function of and was follows froom Tz to TK As a neulo (I) thRin (12) Cooling at Here V = V2 Constant volume 2 . As a nculn (h 1 mol x 5f R ln ( 270.079 1 540.158 8 x 8.314 x ln (2) JK - 14.407 Jk1

Process © Isothermal-Compression lech e Fors a reversible isothermal process of an idea gao, Vo final volume at RTI (V point À 2 22.44L 2-Imo/x8.34 x 270.079 12(6) 2 2 2x22 44L) at Imotile K 1122 point c. > + 1556.418197 T2 270.0796 All new (T2-T) 20 (since isothermal process at Titemp.) TH> ncp (T2 - 7,) 0 (at isothermal change trom It law of thermosky namics at Ti temperatures All 29th & Wanq (ad Auso) - 1556.418193 .AS. AEoln (12) Tarn (3) 4S on Tv In (72) + R R ln (V2) aw Temperature is constant at T . T2zT 2. As = nRln (N) o final volume 28.314 13 (6) 222.41 Lata 2~5.7628 Jkl a initial volume atb - 44.88L

for the total Cycle; N = -22131943 J WITROJ Wina + 1556-418191 W total a hyt wy + Will = - 687.524817 • 91 +7857.5353 Cu -1556.41819J 123 qtotal = +687.524813 AU, 561315927 AUD -2613.5927 Autotal = 0 I • GHI +7859 .0281 44a 201 AHIT = -7859.0289 . Atta 207 AUM OJ •ASI +20169 Jkt48 459 -5-76283K! 451 = -14.407JKY 44 Stotal a Asn + Osa tasa 20 JK